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Q: Measure of set equal to zero when $n$ goes to infinity Let $(X,A,\mu)$ a measure space and $f:X\to\mathbb R$ measurable. For each $n\in\mathbb N$ let $E_n=\{x\in X:\vert f(x)\vert\gt\frac{1}{n}\}$. Prove that each $E_n$ is measurable and if $\lim\mu(E_n)=0$ then $f=0$ a.e.
I suppose there is not much dificulty proving $E_n$ measurable since if we split $\vert f(x)\vert\gt\frac{1}{n}$ then we get $f^{-1}(-\infty,-1/n)$ and $f^{-1}(1/n,\infty)$ for all $n$. As $f$ is measurable, both preimages are measurable.
I am not sure how to procced on the second part
if $\lim\mu(E_n)=0$ then $f=0$ a.e.
I can see that when $n\to\infty$ then $f=0$
Taking the measure $\lim\mu(E_n)=\mu\{x\in X:f(x)=0\}=0$ ? this clearly does not fit the definition of a.e
A: Since $E_1 \subset E_2 \subset E_3 \subset \cdots$ you have $\mu(E_1) \le \mu(E_2) \le \mu(E_3) \le \cdots$
The fact that $\lim_{n \to \infty} \mu(E_n) = 0$ means that $\mu(E_n) = 0$ for all $n$.
Since $\cup_n E_n = \{x : |f(x)| > 0\}$ you get
$$\mu(\{x : f(x) \not= 0\}) = \mu(\{x : |f(x)| > 0\}) \le \sum_n \mu(E_n) = 0.$$
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Sahnivți (în ) este o comună în raionul Izeaslav, regiunea Hmelnîțkîi, Ucraina, formată din satele Kropîvna, Sahnivți (reședința) și Vlașanivka.
Demografie
Conform recensământului din 2001, majoritatea populației comunei Sahnivți era vorbitoare de ucraineană (%), existând în minoritate și vorbitori de alte limbi.
Note
Sahnivtți
Sahnivtți, Izeaslav
Sahnivtți, Izeaslav, Hmelniîtțkiîi
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Альфред Грандідьє () (1873–1957) — французький натураліст і дослідник.
Біографія
Разом із своїм братом Ернестом Грандідьє відвідав Південну Америку в 1858 і 1859 роках, зокрема, в Андах, Перу, Чилі, Болівії, Аргентини та Бразилії. Під час цієї подорожі вони зібралися значну колекцію зразків, які були проаналізовані в 1860 році Ернестом. Два брата розлучилися після цього. Ернест відправився в Китай і зібрав величезну кількість зразків, які зараз в музеях Лувр і Ґіме. Альфред вирушив до Індії, досягши її в 1863 році. Потім Грандідьє відправився в Занзібар, залишаючись деякий час і здобувши важливу колекцію та опублікувавши звіт про свої знахідки. Потім він відвідав острів Реюньйон і в 1865 році здійснив свій перший візит на Мадагаскар. Він присвятив вивченню острова наступні візити в 1866 і 1868 роках. Він, нарешті, повернувся на постійне проживання в Франції в 1870 році. Під час своїх досліджень він перетнув острів тричі, здолавши 3000 кілометрів в середині острову й 2500 вздовж узбережжя. Він зробив зауваження, які знадобились у виробництві карти острова, які використовувались у майбутніх експедиціях.
Описані таксони
Mungotictis decemlineata, Coua coquereli, Coua cursor, Dyscophus antongilii, Erymnochelys madagascariensis та ін.
Бібліографія
1867 Description de quatre espèces nouvelles de Lepidopteres decouvertes sur la cote sud-oust de Madagascar. Revue et Magasin de Zoologie Pure et Appliquee 19:272-275.
1887 Histoire physique, naturelle et politique de Madagascar,Vol. 18. Imprimerie Nationale, Paris.
1887 Histoire physique, naturelle et politique de Madagascar,Vol. 19. Imprimerie Nationale, Paris.
1885-1887 Histoire naturelle des lepidopteres. Histoire, Physique Naturelle et Politique de Madagascar'' 18 [1887]:i-v, 1-364; 19 [1885].
Джерела
International Union for Conservation of Nature and Natural Resources
Integrated Taxonomic Information System
Французькі зоологи
Автори зоологічних таксонів
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Our bodies work hard for us every day. From sitting in front of the computer screens to standing in queues for hours, we draw out too much service from our bodies. Even after undergoing so much stress and poor nutrition, it still continues working for us. Yet how many times have you paid attention to your muscle tension and the wear and tear of your body muscles? Don't you think it deserves to relieve pain and reduce stress? Well! It does. It is essential to include stress-relieving treatments like massage therapies to flush out toxins and improve blood circulation. Having backache, sore neck, stiff body movements are a clear indication…
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"redpajama_set_name": "RedPajamaCommonCrawl"
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Home / World News / 10 labourers shot dead near China Belt and Road project in Pakistan
10 labourers shot dead near China Belt and Road project in Pakistan
Security officials said militants trying to disrupt construction on the economic corridor have killed 44 workers since 2014, all of whom were Pakistani.
world Updated: May 15, 2017 19:36 IST
The attack took place while the labourers were working on a road project near the Gwadar port.(REUTERS/ Representational Photo)
Ten labourers were gunned down in southwestern Pakistan on Saturday while working on link roads to connect outlying towns to the country's $57-billion Chinese Belt and Road initiative, security officials said.
The attack on the labourers took place some 20 km from the emerging port city of Gwadar in Baluchistan province that forms the southern hub of the China-Pakistan Economic Corridor (CPEC).
"All the labourers were shot at close range," said senior levies official Muhammad Zareef, adding that the shooters were travelling on a motorcycle. The levies are a paramilitary force that oversees security in Baluchistan where police jurisdiction is limited to major urban centres.
Gwadar's deep-water port is the exit point for a planned route from China's far-western Xinjiang region to the Arabian Sea.
Nadeem Javaid, who advises Prime Minister Nawaz Sharif's government and works closely on the CPEC programme, told Reuters earlier in the week that the Gwadar-Xinjiang corridor should be operational from June next year.
He said Pakistan expects up to 4 percent of global trade to pass through it by 2020.
Baluchistan, however, has long faced security concerns. Separatist militants in the province have waged a campaign against the central government for decades, demanding a greater share of the gas-rich region's resources.
Security officials have said previously that militants trying to disrupt construction on the "economic corridor" have killed 44 workers since 2014, all of whom were Pakistani.
Pakistan's military created an army division in 2015, believed to number more than 10,000 troops, specifically to protect CPEC projects and Chinese workers.
The men killed and wounded on Saturday had been working for the provincial government at two separate construction sites on three kilometres apart along the same road. Two labourers wounded in the shootings were taken to hospital where one of them died from his injuries, Zareef said.
The roads the labourers were working on are not specific CPEC-funded projects, but they are part of a network of connecting roads that are part of the corridor.
No group has admitted responsibility for the shootings but past attacks in the region have been carried out by separatists who view construction projects as a means to take over their land.
The shootings come a day after a suicide bomber targetting a Pakistani senator killed 26 people and injured 40, Baluchistan Home Minister Sarfraz Bugti said.
Friday's attack was claimed by Islamic State via its Amaq news agency.
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https://www.hindustantimes.com/world-news/10-labourers-shot-dead-near-china-belt-and-road-project-in-pakistan/story-rA1REuMoV2jaaa3KBSDMXP.html
https://www.hindustantimes.com/world-news/china-reports-third-death-from-sars-like-virus-nearly-140-new-cases/story-RWYg3L4K19mZCGboQkdBrJ.html
https://www.hindustantimes.com/world-news/sad-prince-harry-says-he-did-not-want-to-end-royal-role/story-Td5TNlBsr1RtQkG6sGMIgO.html
https://www.hindustantimes.com/world-news/meghan-markle-s-father-accuses-daughter-of-cheapening-uk-s-royal-family/story-HBuHZClFNX0hvPgTvlLlRM.html
https://www.hindustantimes.com/world-news/spacex-launches-destroys-rocket-in-astronaut-escape-test/story-s8jDFPmtjG9xuh3UI58H0K.html
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Q: jakarta.persistence.PersistenceException: No Persistence provider for EntityManager named error when trying to createEntityManagerFactory I'm trying to setup a backend with an eclipse Dynamic Web Project, and this backend will access a MariaDB using Hibernate ORM. I'm having an issue when creating the Entity Manager Factory. Here is the structure of my project :
project structure
Here is the list of the librairies I've included :
Librairies 1
Librairies 2
Here is the content of my web.xml :
<?xml version="1.0" encoding="UTF-8"?>
<web-app xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="https://jakarta.ee/xml/ns/jakartaee" xmlns:web="http://xmlns.jcp.org/xml/ns/javaee" xsi:schemaLocation="https://jakarta.ee/xml/ns/jakartaee https://jakarta.ee/xml/ns/jakartaee/web-app_5_0.xsd http://xmlns.jcp.org/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" id="WebApp_ID" version="5.0">
<display-name>trakteur-backend2</display-name>
<servlet>
<servlet-name>Jersey Web app</servlet-name>
<servlet-class>org.glassfish.jersey.servlet.ServletContainer
</servlet-class>
<init-param>
<param-name>jersey.config.server.provider.packages</param-name>
<param-value>service</param-value>
</init-param>
<init-param>
<param-name>jersey.config.server.provider.scanning.recursive</param-name>
<param-value>true</param-value>
</init-param>
<load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
<servlet-name>Jersey Web app</servlet-name>
<url-pattern>/*</url-pattern>
</servlet-mapping>
</web-app>
Here is the content of my persistence.xml :
<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.2"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence
http://xmlns.jcp.org/xml/ns/persistence/persistence_2_2.xsd">
<persistence-unit name="persistUnit">
<provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
<class> entities.TrakteurFilm </class>
<class> entities.TrakteurUser </class>
<class> entities.TrakteurComment </class>
<properties>
<property name="javax.persistence.jdbc.driver"
value="com.mysql.jdbc.Driver" />
<property name="javax.persistence.jdbc.url"
value="jdbc:mysql://localhost:3306/trakteur?serverTimezone=UTC" />
<property name="javax.persistence.jdbc.user"
value="user" />
<property name="javax.persistence.jdbc.password"
value="password" />
<property name="hibernate.id.new_generator_mappings"
value="false" />
</properties>
</persistence-unit>
</persistence>
My DAO class is just the following for now :
package jpaDAO;
import java.util.List;
import entities.*;
import jakarta.persistence.*;
public class DAO{
private EntityManagerFactory emf;
public DAO()
{
emf = Persistence.createEntityManagerFactory("persistUnit");
}
public EntityManager getEntityManager()
{
return emf.createEntityManager();
}
}
The error I'm having when executing the project on my Tomcat 10 server is the following :
jakarta.persistence.PersistenceException: No Persistence provider for EntityManager named persistUnit
jakarta.persistence.Persistence.createEntityManagerFactory(Persistence.java:86)
jakarta.persistence.Persistence.createEntityManagerFactory(Persistence.java:55)
jpaDAO.DAO.<init>(DAO.java:13)
service.ServiceTrakteur.<clinit>(ServiceTrakteur.java:14)
java.base/jdk.internal.reflect.NativeConstructorAccessorImpl.newInstance0(Native Method)
java.base/jdk.internal.reflect.NativeConstructorAccessorImpl.newInstance(NativeConstructorAccessorImpl.java:77)
java.base/jdk.internal.reflect.DelegatingConstructorAccessorImpl.newInstance(DelegatingConstructorAccessorImpl.java:45)
java.base/java.lang.reflect.Constructor.newInstanceWithCaller(Constructor.java:499)
java.base/java.lang.reflect.Constructor.newInstance(Constructor.java:480)
org.glassfish.hk2.utilities.reflection.ReflectionHelper.makeMe(ReflectionHelper.java:1356)
org.jvnet.hk2.internal.ClazzCreator.createMe(ClazzCreator.java:248)
org.jvnet.hk2.internal.ClazzCreator.create(ClazzCreator.java:342)
org.jvnet.hk2.internal.SystemDescriptor.create(SystemDescriptor.java:466)
org.glassfish.jersey.inject.hk2.RequestContext.findOrCreate(RequestContext.java:59)
org.jvnet.hk2.internal.Utilities.createService(Utilities.java:2103)
org.jvnet.hk2.internal.ServiceLocatorImpl.internalGetService(ServiceLocatorImpl.java:769)
org.jvnet.hk2.internal.ServiceLocatorImpl.internalGetService(ServiceLocatorImpl.java:732)
org.jvnet.hk2.internal.ServiceLocatorImpl.getService(ServiceLocatorImpl.java:702)
org.glassfish.jersey.inject.hk2.AbstractHk2InjectionManager.getInstance(AbstractHk2InjectionManager.java:160)
org.glassfish.jersey.inject.hk2.ImmediateHk2InjectionManager.getInstance(ImmediateHk2InjectionManager.java:30)
org.glassfish.jersey.internal.inject.Injections.getOrCreate(Injections.java:105)
org.glassfish.jersey.server.model.MethodHandler$ClassBasedMethodHandler.getInstance(MethodHandler.java:260)
org.glassfish.jersey.server.internal.routing.PushMethodHandlerRouter.apply(PushMethodHandlerRouter.java:51)
org.glassfish.jersey.server.internal.routing.RoutingStage._apply(RoutingStage.java:86)
org.glassfish.jersey.server.internal.routing.RoutingStage._apply(RoutingStage.java:89)
org.glassfish.jersey.server.internal.routing.RoutingStage._apply(RoutingStage.java:89)
org.glassfish.jersey.server.internal.routing.RoutingStage._apply(RoutingStage.java:89)
org.glassfish.jersey.server.internal.routing.RoutingStage._apply(RoutingStage.java:89)
org.glassfish.jersey.server.internal.routing.RoutingStage.apply(RoutingStage.java:69)
org.glassfish.jersey.server.internal.routing.RoutingStage.apply(RoutingStage.java:38)
org.glassfish.jersey.process.internal.Stages.process(Stages.java:173)
org.glassfish.jersey.server.ServerRuntime$1.run(ServerRuntime.java:247)
org.glassfish.jersey.internal.Errors$1.call(Errors.java:248)
org.glassfish.jersey.internal.Errors$1.call(Errors.java:244)
org.glassfish.jersey.internal.Errors.process(Errors.java:292)
org.glassfish.jersey.internal.Errors.process(Errors.java:274)
org.glassfish.jersey.internal.Errors.process(Errors.java:244)
org.glassfish.jersey.process.internal.RequestScope.runInScope(RequestScope.java:265)
org.glassfish.jersey.server.ServerRuntime.process(ServerRuntime.java:234)
org.glassfish.jersey.server.ApplicationHandler.handle(ApplicationHandler.java:684)
org.glassfish.jersey.servlet.WebComponent.serviceImpl(WebComponent.java:394)
org.glassfish.jersey.servlet.WebComponent.service(WebComponent.java:346)
org.glassfish.jersey.servlet.ServletContainer.service(ServletContainer.java:358)
org.glassfish.jersey.servlet.ServletContainer.service(ServletContainer.java:311)
org.glassfish.jersey.servlet.ServletContainer.service(ServletContainer.java:205)
org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:53)
I am getting this error no matter what methods are in my DAO or no matter what REST request I do from my service package.
What I am sure of is that it comes from the line emf = Persistence.createEntityManagerFactory("persistUnit"); in my DAO, because if I remove this line, all my methods (that don't need to use the DB obviously) work. With this line, none of them work.
I can't find where I failed with my setup. I searched a lot on this issue but I don't find cases that correspond to mine. Apparently, it appears that the persistence.xml or the persistence unit can't be found, but it's here and without error. My object classes (TrakteurComment, TrakteurUser and TrakteurFilm) don't have any error.
Do you have an idea of what it could be ?
Thanks in advance for any help or any answer.
Have a great day
A: It could be because in your persistence.xml you're specifying version 2.2 (which is an old javax version) and Jakarta is 3.x. I'm assuming this is Hibernate 6. See here: https://itnext.io/whats-new-in-jakarta-persistence-3-1-by-examples-81b292e8b3a4
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{"url":"https:\/\/tex.stackexchange.com\/questions\/390369\/cjk-bibliography-problem-biblatex-chicago-author-date-style?noredirect=1","text":"# CJK Bibliography Problem, Biblatex-Chicago author-date style\n\nThe solution given to this question on how to cite items in East Asian languages using biblatex-chicago's notes option is excellent. However, it doesn't work when instead of notes (= references in footnotes, either abbreviated or full length), one uses author-date (parenthetical citations). The problem is the way in which biblatex-chicago disambiguates between authors with the same surname, first name, and CJK script form of the name. What biblatex-chicago does is to disambiguate by adding the initial of the first name in front of the surname, with the first character from the CJK script form of the name following the name. Ideally, what would be needed is disambiguation by, in case only the surname is identical, printing {Surname} {First name} {Date}, or, in case both surname and first name are the same, but not CJK script form of the name, by printing {Surname} {First name} {CJK script form of name} {Date}.\n\nMWE, with the solution from the aforementioned question included:\n\n\\documentclass[utf8,12pt,a4paper,oneside]{article}\n\\usepackage{setspace}\n\\usepackage{xltxtra}\n\\usepackage[fallback]{xeCJK}\n\\usepackage{fontspec}\n\\defaultfontfeatures{Ligatures=TeX}\n\\newcommand{\\mainfont}[0]{Linux Libertine O}\n\\setmainfont[Ligatures={Required,Contextual,Common,TeX},Contextuals={Alternate},Numbers={OldStyle}]{\\mainfont}\n\\newcommand{\\mainfontCJK}[0]{HanaMinA}\\setCJKmainfont[Scale=0.9]{\\mainfontCJK}\n\\setCJKfallbackfamilyfont{\\CJKrmdefault}[Scale=0.9]\n{NanumMyeongjo}\n\\XeTeXlinebreaklocale \"zh\"\n\\XeTeXlinebreakskip = 0pt plus 1pt\n\\usepackage[bibencoding=utf8,idemtracker=constrict,loccittracker=constrict,compresspages,longcrossref=false,booklongxref=true,authordate,cmsdate=both,strict,isbn=false,backend=biber,hyperref=false,mincrossrefs=2]{biblatex-chicago}\n%Enabling CJK names (http:\/\/tex.stackexchange.com\/questions\/320693\/cjk-bibliography-problem-biblatex-chicago\/320738#320738 ):\n\\newcommand{\\ifnameaffix}[1]{%\n\\ifboolexpr{ test {\\ifrmnum{#1}} or test {\\xifinlist{#1}{\\nameaffixlist}} }}\n% Based on definitions from biblatex.def\n\\newbibmacro*{name:cjk}[3]{%\n\\usebibmacro{name:delim}{#2#3#1}%\n\\usebibmacro{name:hook}{#2#3#1}%\n\\mkbibnamefamily{#1}%\n\\ifdefvoid{#2}{}{\\bibnamedelimd\\mkbibnamegiven{#2}}%\n\\ifdefvoid{#3}{}{\\bibnamedelimd\\mkbibnamesuffix{#3}}}\n\\DeclareNameFormat{given-family}{%\n\\ifboolexpr{ test {\\ifdefvoid{\\namepartsuffix}} or test {\\ifnameaffix{\\namepartsuffix}} }\n{\\ifgiveninits\n{\\usebibmacro{name:given-family}\n{\\namepartfamily}\n{\\namepartgiveni}\n{\\namepartprefix}\n{\\namepartsuffix}}\n{\\usebibmacro{name:given-family}\n{\\namepartfamily}\n{\\namepartgiven}\n{\\namepartprefix}\n{\\namepartsuffix}}}\n{\\usebibmacro{name:cjk}{\\namepartfamily}{\\namepartgiven}{\\namepartsuffix}}%\n\\usebibmacro{name:andothers}}\n\\DeclareNameFormat{family-given}{%\n\\ifboolexpr{ test {\\ifdefvoid{\\namepartsuffix}} or test {\\ifnameaffix{\\namepartsuffix}} }\n{\\ifgiveninits\n{\\usebibmacro{name:family-given}\n{\\namepartfamily}\n{\\namepartgiveni}\n{\\namepartprefix}\n{\\namepartsuffix}}\n{\\usebibmacro{name:family-given}\n{\\namepartfamily}\n{\\namepartgiven}\n{\\namepartprefix}\n{\\namepartsuffix}}}\n{\\usebibmacro{name:cjk}{\\namepartfamily}{\\namepartgiven}{\\namepartsuffix}}%\n\\usebibmacro{name:andothers}}\n\\DeclareNameFormat{family-given\/given-family}{%\n\\ifboolexpr{ test {\\ifdefvoid{\\namepartsuffix}} or test {\\ifnameaffix{\\namepartsuffix}} }\n{\\ifnumequal{\\value{listcount}}{1}\n{\\ifgiveninits\n{\\usebibmacro{name:family-given}\n{\\namepartfamily}\n{\\namepartgiveni}\n{\\namepartprefix}\n{\\namepartsuffix}}\n{\\usebibmacro{name:family-given}\n{\\namepartfamily}\n{\\namepartgiven}\n{\\namepartprefix}\n{\\namepartsuffix}}\n\\ifboolexpe{%\ntest {\\ifdefvoid\\namepartgiven}\nand\ntest {\\ifdefvoid\\namepartprefix}}\n{}\n{\\usebibmacro{name:revsdelim}}}\n{\\ifgiveninits\n{\\usebibmacro{name:given-family}\n{\\namepartfamily}\n{\\namepartgiveni}\n{\\namepartprefix}\n{\\namepartsuffix}}\n{\\usebibmacro{name:given-family}\n{\\namepartfamily}\n{\\namepartgiven}\n{\\namepartprefix}\n{\\namepartsuffix}}}}\n{\\usebibmacro{name:cjk}{\\namepartfamily}{\\namepartgiven}{\\namepartsuffix}}%\n\\usebibmacro{name:andothers}}\n% Based on definitions from biblatex-chicago cbx\n\\DeclareFieldFormat[book]{title}{%\n\\mkbibemph{#1}%\n\\DeclareFieldFormat{booktitle}{%\n\\mkbibemph{#1}%\n\\DeclareFieldFormat{maintitle}{%\n\\mkbibemph{#1}%\n\\DeclareFieldFormat[article]{title}{%\n\\iffieldundef{title}{}{\\mkbibquote{#1}}%\n%Addition for functionality with inbook and bookinbook etc:\n\\DeclareFieldFormat[inbook]{title}{%\n\\iffieldundef{title}{}{\\mkbibquote{#1}}%\n\\DeclareFieldFormat[bookinbook]{title}{%\n\\iffieldundef{title}{}{\\mkbibquote{#1}}%\n\\DeclareFieldFormat[unpublished]{title}{%\n\\iffieldundef{title}{}{\\mkbibquote{#1}}%\n\\DeclareFieldFormat[thesis]{title}{%\n\\iffieldundef{title}{}{\\mkbibquote{#1}}%\n%This makes translation brackets into parentheses:\n\\DeclareFieldFormat{usere}{\\mkbibparens{#1}}\n\\begin{filecontents}{bibliography.bib}\n@book{Book1,\ntitle = {Shuming},\nauthor = {Zhang, \u7ae0\u5b78\u6210, Xuecheng},\nlocation = {Beijing},\npublisher = {Zhonghua Shuju},\ndate = {2017},\n}\n@book{Book2,\ntitle = {Shuming er},\nauthor = 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That makes coding things a bit more intuitive. \u2013\u00a0moewe Sep 8 '17 at 10:08\n\u2022 Unfortunately, there is currently no system to tell which of multiple name parts makes a name unique. So if we consider the given name and the CJK part we only know that adding one of the two will make things unique, not which. \u2013\u00a0moewe Sep 8 '17 at 10:45\n\u2022 @moewe Thank you! I approved the answer in tex.stackexchange.com\/questions\/233053\/\u2026 \u2013\u00a0M\u00e5rten Sep 8 '17 at 14:18\n\u2022 See also github.com\/plk\/biblatex\/issues\/623. I think we need something like that to be able to do what you want here. \u2013\u00a0moewe Sep 8 '17 at 15:31\n\u2022 @moewe I see. Such a function would be very useful for people working with East Asian materials. Thanks for taking the time to look into it! \u2013\u00a0M\u00e5rten Sep 11 '17 at 7:56\n\nbiblatex 3.8\/Biber 2.8 allow you to fully customize the unqiuename function. 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family=Zhao, cjk=\u8d75\u6770, nametemplates=cjk},\ntitle = {Qingdai Manwen de wenzi tese ji yinyun, yinbian tedian},\nusere = {Scriptual specificities of Manchu writing in the Qing period\nand characteristics of phonology and sound change},\nshorttitle = {Manwen de wenzi tese},\njournal = {Manzu yanjiu},\nvolume = {102},\nnumber = {1},\ndate = {2011},\npages = {7--12}}\n@book{li,\nauthor={given=Wuwei, family=Li, cjk=\u674e\u65e0\u672a, nametemplates=cjk and Brown, Junior, Bob and Doe, III, John},\npublisher = {Shangwu yinshuguan},\ntitle = {Riben Hanyu yinyun xue shi},\nusere = {History of the study of Chinese phonology in Japan},\ndate = {2011}}\n\n@book{Book1,\ntitle = {Shuming},\nauthor = {family=Zhang, cjk=\u7ae0\u5b78\u6210, given=Xuecheng, nametemplates=cjk},\nlocation = {Beijing},\npublisher = {Zhonghua Shuju},\ndate = {2017},\n}\n@book{Book2,\ntitle = {Shuming er},\nauthor = {family=Zhang, cjk=\u5f35\u5b78\u6210, given=Xuecheng, nametemplates=cjk},\nlocation = 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\section{Introduction}
This paper is the first one in a series of two, concerning about
the following reaction-diffusion-advection competition system in a periodic habitat
\begin{equation}\label{u1u2}
\begin{cases}
\frac{\partial u_1}{\partial t}=L_1u_1+u_1\left(b_1(x)-a_{11}(x)u_1-a_{12}(x)u_2\right),\\
\frac{\partial u_2}{\partial t}=L_2u_2+u_2\left(b_2(x)-a_{21}(x)u_1-a_{22}(x)u_2\right),
\end{cases}
t>0,~x\in\Bbb R,
\end{equation}
where $L_i:=d_i(x)\frac{\partial^2}{\partial x^2}-a_i(x)\frac{\partial}{\partial x}$,
$u_1$ and $u_2$ denote population densities of two competition species in an $L$-periodic habitat for some $L>0$.
$d_i(\cdot),\ a_i(\cdot)$ and $b_i(\cdot)$ are $L$-periodic functions in $C^\nu(\Bbb R)$ with some $\nu\in(0,1)$,
denoting the diffusion, advection and growth rates of $u_i$, respectively.
$d_i(\cdot)\geq d_0>0$, that is, $L_i$ is uniformly elliptic.
$a_{ij}(x)>0$ are $L$-periodic functions representing inter- and intraspecific competition coefficients, $i,j=1,2$.
Reaction-diffusion models of the type \eqref{u1u2} are widely used to capture the spatial dynamics of two competing species.
Of particular interest is to understand the propagation phenomena as well as spreading properties that the system exhibits.
Over the past decades, quite a few literatures have been devoted to the homogeneous cases,
mainly concerning about persistence, extinction, biological invasions, the minimal wave speeds, traveling waves and entire solutions,
see, e.g., \cite{hosono1989,kan-on1995,Lewis2002,liang2007,morita2009,lv2010,huang2011,guo2011,zhang2012} and references therein.
It is also well known that the periodic environment of space and/or time is one of the very useful approximations
to understand the influence of the environmental heterogeneity on the propagation phenomena arising
from ecological and biological processes. Whereas the homogeneous model has attracted many works in the mathematical literature,
propagation phenomena such as steady state problems, spreading speeds and traveling waves for
spatially and/or temporally heterogeneous systems like \eqref{u1u2} were studied more recently.
More specifically, in spatially heterogeneous environment, Dockery et al. \cite{Dockery1998} studied the effect of
dispersal rates on the survival of two competing species and showed that the slower diffuser always prevails.
Lou \cite{lou2006} proved that the two weakly competing species with some appropriate dispersal rates can no longer exist
in a spatially heterogenous environment.
Lam and Ni \cite{lam2012} considered the interactions between diffusion and heterogeneity of the environment of two species competition diffusion system with spatial heterogeneous growth rates in a bounded domain.
He and Ni further studied in a series of three papers \cite{he2016,he20161,he2017}
the combined effects of dispersal and spatial variations on the outcome of the competition, and
the joint effects of diffusion and spatial concentration on the global dynamics of the competition-diffusion
system with equal amount of total resources.
While Lutscher et al. \cite{Lutscher2007} added the advection term into such a
competition model and discussed spatial patterns and coexistence mechanisms for stream populations.
For a monostable semiflow in one-dimensional periodic environment,
Liang and Zhao \cite{liang2010} proved the existence of the spreading speed and its coincides with the minimal wave speed,
and Fang and Zhao \cite{fang2011} established the existence of bistable traveling fronts by interpreting the bistability
from a viewpoint of monotone dynamical systems to find a link with its monostable subsystems.
One can also see \cite{kong2015,bao2016} for spreading speeds and traveling waves for systems with nonlocal dispersal.
Recently, Yu and Zhao \cite{Yu2017} considered the propagation phenomena of reaction-diffusion-advection system \eqref{u1u2}.
They established the existence of the rightward spreading speed and its coincidence with
the minimal wave speed for spatially periodic rightward traveling waves.
Girardin \cite{Girardin2016} proved the existence of spatially periodic traveling waves for a bistable competition-diffusion system with periodic intraspecific competition coefficients.
In time heterogeneous environment, one can see \cite{zhao2011,bao2013,zhang2013,zhao2014,ma2016} for time periodic traveling waves, \cite{du2016,du20171} for other types of entire solutions, and \cite{liang2006} for spreading speeds and traveling waves of time-periodic semiflows in one dimensional environment.
For time-space periodic environment, one can see a more recent paper \cite{fang2017} for the existence of traveling waves and
spreading speeds of time-space periodic monotone semiflows.
The current paper is the first one in a series of two, mainly devoted to the study of existence,
stability and uniqueness of pulsating fronts (see Definition \ref{def}) for the Lotka-Volterra competition system with advection in a periodic habitat. In the forthcoming paper \cite{du2017},
we are concerned with some other types of entire solutions by considering the interaction of these pulsating fronts.
Let us now make the precise assumptions. Assume that $d(\cdot),\ a(\cdot)$ and $b(\cdot)$ are $L$-periodic functions in $C^{\nu}(\Bbb R)$ and $d(\cdot)>0$ in $\Bbb{R}$. Then the periodic eigenvalue problem
\begin{equation*}
\begin{cases}
\lambda\phi=d(x)\phi^{\prime\prime}-a(x)\phi^{\prime}+b(x)\phi,\quad x\in\Bbb R,\\
\phi(x+L)=\phi(x),\quad x\in\Bbb R
\end{cases}
\end{equation*}
admits a principal eigenvalue $\lambda_0(d,a,b)$ associated with a positive $L$-periodic eigenfunction (see, e.g.,\cite{smith1995}).
Furthermore, $\lambda_0(d,a,b)$ is monotone increasing with respect to $b$ in the sense that,
if $b_1\geq b_2$ and $b_1\not\equiv b_2$ then $\lambda_0(d,a,b_1)>\lambda_0(d,a,b_2)$ (see, e.g., \cite[Lemma 15.5]{hess1991}).
Assume further that $c(\cdot)\geq,\not\equiv0$ is an $L$-periodic function in $C^{\nu}(\Bbb R)$.
Consider the following initial value problem
\begin{equation}\label{e}
\begin{cases}
\frac{\partial u}{\partial t}=d(x)\frac{\partial^2 u}{\partial {x^2}}-a(x)\frac{\partial u}{\partial x}+u(b(x)-c(x)u),\quad t>0,\ x\in\Bbb R,\\
u(0,x)=\phi(x),\quad x\in\Bbb R,
\end{cases}
\end{equation}
where $\phi$ is continuous and positive $L$-periodic in $\Bbb{R}$.
It follows from \cite[Theorem 2.3.4]{zhao2003} that if $\lambda_0(d,a,b)\leq0$,
then $u\equiv0$ is globally asymptotic stable with respect to periodic perturbations, and
if $\lambda_0(d,a,b)>0$, \eqref{e} admits a unique positive $L$-periodic solution $u^*(x)$ which is
globally asymptotic stable with respect to periodic perturbations for any nontrivial initial values.
Denote by $\Bbb P=PC(\Bbb R,\Bbb R^2)$ the set of all continuous and $L$-periodic functions from
$\Bbb R$ to $\Bbb R^2$ equipped with a norm
$\lVert\phi\rVert_{\Bbb P}=\mathop{\max}\limits_{x\in\Bbb R}\lvert\phi(x)\rvert$, then
$\Bbb P_+:=\{\phi\in \Bbb P: \phi(x)\geq0,\forall x\in\Bbb R\}$ is a closed cone of $Y$
inducing a partial ordering on $\Bbb P,$ and $(\Bbb P,\lVert\cdot\rVert_Y)$ is a Banach lattice.
Throughout the paper, we always assume that the trivial solution $(0,0)$
is linearly unstable with respect to the perturbation in $\Bbb P_+$, that is,
\begin{equation}\label{00}
\lambda_0(d_i,a_i,b_i)>0,\quad i=1,2.
\end{equation}
One should note that, by the monotonicity of $\lambda_0(d_i,a_i,b_i)$ with respect to $b_i$,
\eqref{00} holds true in particular if $b_i(\cdot)\geq,\not\equiv0$ for $i=1,2$.
By virtue of \eqref{00} and the above argument,
we know that there exists two positive $L$-periodic functions $u_1^*(x)$ and $u_2^*(x)$ satisfying respectively
$$0=L_1u_1+u_1(b_1(x)-a_{11}(x)u_1)\text{~~and~~}0=L_2u_2+u_2(b_2(x)-a_{22}(x)u_2),\ x\in\Bbb R,$$
such that $(u_1^*(x),0)$ and $(0,u_2^*(x))$ are two semitrivial periodic steady states of system \eqref{u1u2} in $\Bbb P_+$.
To impose a bistable structure on \eqref{u1u2}, we further assume
\begin{itemize}
\item[(H1)] $\lambda_0(d_1,a_1,b_1-a_{12}u_2^*)<0\text{~~and~~}\lambda_0(d_2,a_2,b_2-a_{21}u_1^*)<0$.
\end{itemize}
It then follows that $(u_1^*(x),0)$ and $(0,u_2^*(x))$ are two locally linearly stable periodic steady states
with respect to the perturbation in $\Bbb P_+$. Furthermore, system \eqref{u1u2} has at least one unstable coexistence steady state
(see, e.g., \cite[Theorem 2.4]{ferter1994}).
We would like to mention here that, under the assumptions that $\lambda_0(d_1,a_1,b_1-a_{12}u_2^*)>0$
(that is, $(0,u_2^*(x))$ is a linearly unstable steady state) and that the system has no steady state in Int$(\Bbb P_+)$,
it follows that $(u_1^*(x),0)$ is globally asymptotic stable for all initial values $(\phi_1,\phi_2)\in\Bbb P_+$
with $\phi_1\not\equiv0$ (see e.g., \cite[Theorem 2.1]{Yu2017}), namely,
system \eqref{u1u2} admits a \textbf{monostable} structure.
Yu and Zhao \cite{Yu2017} then considered the propagation phenomena of system \eqref{u1u2} with this monostable structure
and established the existence of spatially periodic traveling waves
connecting the two semitrivial periodic solutions.
In the present work, we focus system \eqref{u1u2} on the standard \textbf{bistable} structure.
To this end, we impose the following
\begin{itemize}
\item[(H2)] System \eqref{u1u2} has no stable periodic steady state in Int$(\Bbb P_+)$.
\end{itemize}
It is clear that (H2) together with (H1) confirms that system \eqref{u1u2} admits the standard bistable structure, that is,
there are two locally stable periodic semitrivial steady states $(u_1^*(x),0)$ and $(0,u_2^*(x))$, and
all the periodic intermediate (coexistence) steady states of system \eqref{u1u2} are unstable.
\begin{remark} \rm
It seems to be a very difficult but interesting problem to investigate sufficient conditions for the existence, uniqueness
and stability of periodic coexistence steady states of system \eqref{u1u2},
we leave it as an open problem, since it is not our point at present.
Indeed, we can conjecture as in \cite{Girardin2016} that the nonexistence of a stable periodic coexistence
steady state is not a necessary condition for the existence of a bistable pulsating front for system \eqref{u1u2},
it is only a required step toward the existence of pulsating fronts obtaining by using the abstract theory developed in
\cite{fang2011}.
Particularly, if we consider system \eqref{u1u2} as following
\begin{equation}\label{G}
\begin{cases}
\frac{\partial u_1}{\partial t}=\frac{\partial^2 u_1}{\partial x^2}+u_1\left(b_1(x)-a_{11}(x)u_1-a_{12}u_2\right),\\
\frac{\partial u_2}{\partial t}=d\frac{\partial^2 u_2}{\partial x^2}+u_2\left(b_2(x)-a_{21}u_1-a_{22}(x)u_2\right),
\end{cases}
t>0,~x\in\Bbb R,
\end{equation}
where $d,\ a_{12}$ and $a_{21}$ are positive constants,
$b_i(\cdot)>0$ and $a_{ii}(\cdot)>0$ are $L$-periodic functions in $C^\nu(\Bbb R)$ for some $\nu\in(0,1)$.
It then easily follows that $\lambda_0(1,0,b_1)>0$ and $\lambda_0(d,0,b_2)>0$,
and then \eqref{G} admits two semitrivial steady states (see, e.g., \cite{berestycki20051}).
Furthermore, it follows from \cite[Propositions 2.1 and 2.10]{Girardin2016} that the two semitrivial steady states are
locally linearly stable and any periodic coexistence steady state is unstable provided that $L$ is sufficiently small
and $a_{ij},\ i\neq j$ are sufficiently large, that is, (H1) and (H2) hold for \eqref{G},
which yields that system \eqref{G} admits the standard bistable structure.
\end{remark}
Let $\mathcal C$ be the set of all bounded and continuous functions from $\Bbb R$ to $\Bbb R^2$ and denote
$\mathcal C_+=\{\bm{u}\in\mathcal C: \bm{u}(x)\geq\bm{0}\text{~~for all~~} x\in\Bbb R\}$. Then $\mathcal C_+$ is a closed cone of $\mathcal C$.
We introduce a norm $\lVert\cdot\rVert_{\mathcal C}$ by $\lVert\bm{u}\rVert_{\mathcal C}=\mathop{\max}\limits_{x\in\Bbb{R}}|\bm{u}(x)|$.
For any $\bm{a}_1,\bm{a}_2\in\mathcal C$ with $\bm{a}_1<\bm{a}_2$, denote
$\mathcal C_{[\bm{a_1},\bm{a_2}]}=\{\bm{u}\in\mathcal C: \bm{a}_1\leq\bm{u}\leq\bm{a}_2\}$ and
$\mathcal C_{[\bm0,\bm{\beta}]}$ by $\mathcal C_{\bm{\beta}}$ for any $\bm{\beta}\in\rm{Int}(\mathcal C_+)$.
Hereafter, we use the usual notations for classical partial order on $\mathcal C$.
Namely, for any $\bm{u}=(u_1,u_2), \bm{v}=(v_1,v_2)\in\mathcal C$, writing $\bm{u}=\bm{v}$ if $u_1=v_1$ and $u_2=v_2$,
$\bm{u}\bm {v}=(u_1u_2,v_1v_2)$, $a\bm{u}=(au_1,au_2)$ for any constant $a$,
$\bm{u}\leq\bm{v}$ if $u_i(\cdot)\leq v_i(\cdot)$ in $\Bbb{R}$ for $i=1,2$,
$\bm{u}<\bm{v}$ if $\bm{u}\leq\bm{v}$ and $\bm{u}\neq\bm{v}$,
$\bm{u}\ll\bm{v}$ if $u_i(x)<v_i(x)$ in $\Bbb{R}$ for $i=1,2$,
and other relations are similarly to be understood componentwise.
In particular, denote by $\bm{0}=(0,0), \bm{1}=(1,1)$
and $\bm{\alpha}\in\langle\bm{0},\bm{1}\rangle$ if $\bm{0}\ll\bm{\alpha}\ll\bm{1}$.
By a change of variables
$$\tilde{u}_1(t,x)=\frac{u_1(t,x)}{u_1^*(x)},\quad\tilde{u}_2(t,x)=\frac{u_2^*(x)-u_2(t,x)}{u_2^*(x)},$$
we transform \eqref{u1u2} into the following (dropping the tilde for the convenience of writing)
\begin{equation}\label{u1u2-0-1}
\begin{cases}
\frac{\partial u_1}{\partial t}=d_1(x)\frac{\partial u_1^2}{\partial x^2}-a_1^*(x)\frac{\partial u_1}{\partial x}+f_1(x,u_1,u_2),\\
\frac{\partial u_2}{\partial t}=d_2(x)\frac{\partial u_1^2}{\partial x^2}-a_2^*(x)\frac{\partial u_2}{\partial x}+f_2(x,u_1,u_2),
\end{cases}
t>0,~x\in\Bbb R,
\end{equation}
where $L_i^*:=d_i(x)\frac{\partial^2}{\partial x^2}-a_i^*(x)\frac{\partial}{\partial x}$ is uniformly elliptic,
$a_i^*(x)=a_i(x)-2d_i(x)\frac{(u_i^*)^\prime(x)}{u_i^*(x)}$, $a_{i1}^*(x)=a_{i1}(x)u_1^*(x)$, $a_{i2}^*(x)=a_{i2}(x)u_2^*(x)$ for $i=1,2$, and
\begin{align*}
f_1(x,u_1,u_2)&=u_1\left[a_{11}^*(x)(1-u_1)-a_{12}^*(x)(1-u_2)\right],\\
f_2(x,u_1,u_2)&=(1-u_2)\left[a_{21}^*(x)u_1-a_{22}^*(x)u_2\right].
\end{align*}
It is easily seen that system \eqref{u1u2-0-1} is cooperative in the region $u_1\geq0$ and $0\leq u_2\leq 1$,
with periodic steady states $(0,1)$, $(0,0)$ and $(1,1)$.
On the other hand, if we let $\phi_i(\cdot)>0$ be positive $L$-periodic eigenfunction associated with
$\lambda_0(d_i,a_i,b_i)$, that is,
$$\lambda_0(d_i,a_i,b_i)\phi=d_i(x)\phi_i^{\prime\prime}-a_i(x)\phi_i^{\prime}+b_i(x)\phi_i,~~x\in\Bbb R,$$
then $\psi_i(x):=\frac{\phi_i(x)}{u_i^*(x)}>0$ satisfies
$$\lambda_0(d_i,a_i,b_i)\psi_i=d_i(x)\psi_i^{\prime\prime}-a_i^*(x)\psi_i^{\prime}+a_{ii}^*(x)\psi_i,\quad x\in\Bbb R,$$
and thus $\lambda_0(d_i,a_i^*,a_{ii}^*)=\lambda_0(d_i,a_i,b_i)>0$ for $i=1,2$.
As a matter of the fact, we can similarly verify that
\begin{equation}\label{i-j}
\lambda(d_i,a_i,b_i-2a_{ii}u_i^*)=\lambda(d_i,a_i^*,-a_{ii}^*),
\ \lambda(d_i,a_i^*,a_{ii}^*-a_{ij}^*)=\lambda(d_i,a_i,b_i-a_{ij}u_j^*),\ i,j=1,2, i\neq j.
\end{equation}
Therefore, one have the following equivalent assumption to (H1) and (H2) on system \eqref{u1u2-0-1}.
\begin{itemize}
\item[(A1)] $\lambda_0(d_1,a_1^*,a_{11}^*-a_{12}^*)<0$ and $\lambda_0(d_2,a_2^*,a_{22}^*-a_{21}^*)<0$.
\item[(A2)] System \eqref{u1u2-0-1} has no stable periodic steady state in $\langle\bm{0},\bm{1}\rangle$.
\end{itemize}
That is, system \eqref{u1u2-0-1} has two locally linearly stable steady states $\bm{0}$ and $\bm{1}$,
and any coexistence steady state between $\bm{0}$ and $\bm{1}$ is unstable,
where we denote $(\hat{u}_1,\hat{u}_2)$ by any periodic coexistence steady state of system \eqref{u1u2-0-1} in the sequel.
Noting that the linearized system of \eqref{u1u2-0-1} at $(\hat{u}_1,\hat{u}_2)$ is cooperative and irreducible,
and thus the Krein-Rutman theorem yields that there exists a positive $L$-periodic eigenfunction $(\hat{\phi}_1(x),\hat{\phi}_2(x))$
associated with a principal eigenvalue $\hat{\lambda}>0$ such that
\begin{equation*}
\begin{cases}
\hat{\lambda}\hat{\phi}_1=d_1\hat{\phi}_1^{\prime\prime}-a_1^*\hat{\phi}_1^{\prime}
+[a_{11}^*(x)-a_{12}^*(x)-2a_{11}^*(x)\hat{u}_1+a_{12}^*(x)\hat{u}_2]\hat{\phi}_1+a_{12}^*(x)\hat{u}_1\hat{\phi}_2,\ x\in\Bbb{R},\\
\hat{\lambda}\hat{\phi}_2=d_2\hat{\phi}_2^{\prime\prime}-a_2^*\hat{\phi}_2^{\prime}
+a_{21}^*(x)(1-\hat{u}_2)\hat{\phi}_1-[a_{22}^*(x)+a_{21}^*(x)\hat{u}_1-2a_{22}^*(x)\hat{u}_2]\hat{\phi}_2,\ x\in\Bbb{R},\\
\hat{\phi}_i(x+L)=\hat{\phi}_i(x),\ i=1,2,\ x\in\Bbb{R}.
\end{cases}
\end{equation*}
Then \eqref{u1u2-0-1} admits the standard bistable structure.
In the sequel, we shall deal with system \eqref{u1u2-0-1} under assumptions (A1) and (A2) due to its equivalence to system \eqref{u1u2}.
The definition of pulsating fronts is given as following.
\begin{definition}\label{def}
An entire solution $$\bm{u}(t,x)=\bm{U}(x,x+ct)=(U_1(x,x+ct),U_2(x,x+ct))$$
of system \eqref{u1u2-0-1} is called a leftward pulsating front with wave speed $c$, if
$\bm{U}(\cdot,\cdot+a)\in\mathcal C_{\bm{1}}$ for any $a\in\Bbb R$, and
$\bm{U}(x,\cdot)=\bm{U}(x+L,\cdot)$ for any $x\in\Bbb R$.
Moreover, we say that $\bm{U}(x,z)$ connects $\bm{0}$ to $\bm{1}$ if
$\mathop{\lim}\limits_{z\to-\infty}\lvert \bm{U}(x,z)\rvert=0$ and
$\mathop{\lim}\limits_{z\to+\infty}\lvert \bm{U}(x,z)-\bm{1}\rvert=0$ uniformly for $x\in\Bbb R$.
Similarly,
an entire solution $$\bm{v}(t,x)=\bm{V}(x,x-ct)=(V_1(x,x-ct),V_2(x,x-ct))$$ of system \eqref{u1u2-0-1}
is called a rightward pulsating front with wave speed $c$, if
$\bm{V}(\cdot,\cdot+a)\in\mathcal C_{\bm{1}}$ for any $a\in\Bbb R$, and
$\bm{V}(x,\cdot)=\bm{V}(x+L,\cdot)$ for any $x\in\Bbb R$.
Moreover, we say that $\bm{V}(x,z)$ connects $\bm{1}$ to $\bm{0}$ if
$\mathop{\lim}\limits_{z\to-\infty}\lvert \bm{V}(x,z)-\bm{1}\rvert=0$ and
$\mathop{\lim}\limits_{z\to+\infty}\lvert \bm{V}(x,z)\rvert=0$ uniformly for $x\in\Bbb R$.
\end{definition}
The rest of the paper is organized as follows.
In Section 2, we establish the existence of pulsating front of \eqref{u1u2-0-1} using the abstract theory
developed by Fang and Zhao \cite{fang2011}.
Section 3 is devoted to the construction of a pair of appropriate sub- and supersolutions.
In Section 4, we obtain the global stability and uniqueness of pulsating fronts
by using the convergence theorem for monotone semiflows (see \cite[Theorem 2.2.4]{zhao2003}),
one can also see a similar argument in \cite{xu2004,jin2008,zhang2012,bao2013,ding2014}.
\section{Existence of pulsating fronts}
In this section, we shall employ the abstract theory developed by Fang and Zhao \cite{fang2011}
to prove the existence of leftward and rightward pulsating fronts connecting $\bm{0}$ and $\bm{1}$.
Let $\bm{\beta}\in\rm{Int}(\mathcal C_+)$ and denote
$$\Pi_{\bm{\beta}}=\{\bm{u}\in\mathcal C_{\bm{\beta}}: \bm{u}(x)=\bm{u}(x+L),\ \forall x\in\Bbb R\},$$
$X=C([0,L],\Bbb{R}^2)$, $X_+=C([0,L],\Bbb{R}_+^2)$
and $X_{\bm{\beta}}=\{\bm{u}\in X: \bm{0}\leq \bm{u}\leq\bm{\beta}\}$.
Define
\begin{align*}
\mathcal{X}&=\{\bm{v}\in BC(\Bbb{R},X): \bm{v}(s)(L)=\bm{v}(s+L)(0),\ \forall s\in\Bbb{R}\},\\
\mathcal{X}_+&=\{\bm{v}\in\mathcal{X}: \bm{v}(s)\in X_+,\ \forall s\in\Bbb{R}\},\\
\mathcal{X}_{\bm{\beta}}&=\{\bm{v}\in BC(\Bbb{R},X_{\bm{\beta}}): \bm{v}(s)(L)=\bm{v}(s+L)(0),\ \forall s\in\Bbb{R}\},\\
\mathcal{K}_{\bm{\beta}}&=\{\bm{v}\in BC(L\Bbb{Z},X_{\bm{\beta}}): \bm{v}(iL)(L)=\bm{v}((i+1)L)(0),\ \forall i\in\Bbb{Z}\},
\end{align*}
where $BC(\Bbb{R},X)$ denotes the set of all continuous and bounded functions from $\Bbb{R}$ to $X$.
Let $T(t):={\rm{diag}}(T_1(t),T_2(t))$, where $T_1(t)$ and $T_2(t)$ are linear semigroups generated by
$$\frac{\partial u_1}{\partial t}=L_1^*u_1+u_1(a_{11}^*(x)-a_{12}^*(x))\text{~~and~~}
\frac{\partial u_2}{\partial t}=L_2^*u_2-a_{22}^*(x)u_2,$$
respectively.
Then $T_1(t)$ and $T_2(t)$ are compact with respect to the compact open topology for each $t>0$.
For any $\bm{u}=(u_1,u_2)\in\mathcal C_{\bm{1}}$, define $F:\mathcal C_{\bm{1}}\to\mathcal C$ by
$$F(\bm{u})=\left({\begin{array}{*{20}{c}}
-a_{11}^*(x)u_1^2+a_{12}^*(x)u_1u_2\\
a_{21}^*(x)u_1-a_{21}^*(x)u_1u_2+a_{22}^*(x)u_2^2\\
\end{array}}\right).$$
Now we can rewrite \eqref{u1u2-0-1} into the following integral form
\begin{equation}\label{Int}
\begin{cases}
\bm{u}(t)=T(t)\bm{u}(0)+\int_0^t{T(t-s)F(\bm{u}(s))ds},\ \ t>0,\\
\bm{u}(0)=\phi\in\mathcal C_{\bm{1}}.
\end{cases}
\end{equation}
Define a family of operators $\{Q_t\}_{t\geq0}$ on $\mathcal C_{\bm{1}}$ by $Q_t(\bm{\phi}):=\bm{u}(t,\cdot;\bm{\phi})$,
where $\bm{u}(t,\cdot;\bm{\phi})$ is the solution of \eqref{u1u2-0-1} with $\bm{u}(0,\cdot;\bm{\phi})=\bm{\phi}\in\mathcal C_{\bm{1}}$.
It can easily seen that for any $t>0$,
$Q_t$ is a monotone semiflow on $\mathcal C_{\bm{1}}$.
Noting also that $\{Q_t\}_{t\geq0}$ restricted on $\Pi_{\bm{1}}$ is strongly monotone in the sense that
if $\bm{\phi}>\bm{\psi}$, then $Q_t(\bm{\phi})\gg Q_t(\bm{\psi})$ for any $t>0$.
Since $T(t)$ is compact with respect to the compact open topology,
$\{Q_t\}:\mathcal C_{\bm{1}}\to\mathcal C_{\bm{1}}$ is continuous and compact with respect to the topology of the locally uniform
convergence. Therefore, assumptions (A2)-(A4) in \cite{fang2011} hold for each $Q_t$ with $t>0$.
Now consider the following two periodic eigenvalue problems
\begin{equation}\label{0-eigen}
\begin{cases}
\mu\phi_{01}=d_1\phi_{01}^{\prime\prime}-a_1^*\phi_{01}^{\prime}+(a_{11}^*(x)-a_{12}^*(x))\phi_{01},\quad x\in\Bbb R, \\
\mu\phi_{02}=d_2\phi_{02}^{\prime\prime}-a_2^*\phi_{02}^{\prime}+a_{21}^*(x)\phi_{01}-a_{22}^*(x)\phi_{02},\quad x\in\Bbb R, \\
\phi_{0i}(x)=\phi_{0i}(x+L),\ i=1,2,\quad x\in\Bbb R,
\end{cases}
\end{equation}
and
\begin{equation}\label{1-eigen}
\begin{cases}
\mu\phi_{11}=d_1\phi_{11}^{\prime\prime}-a_1^*\phi_{11}^{\prime}-a_{11}^*(x)\phi_{11}+a_{12}^*(x)\phi_{12},\quad x\in\Bbb R, \\
\mu\phi_{12}=d_2\phi_{12}^{\prime\prime}-a_2^*\phi_{12}^{\prime}+(a_{22}^*(x)-a_{21}^*(x))\phi_{12},\quad x\in\Bbb R, \\
\phi_{1i}(x)=\phi_{1i}(x+L),\ i=1,2,\quad x\in\Bbb R.
\end{cases}
\end{equation}
We introduce the following assumption.
\begin{itemize}
\item[(B1)]
$\lambda_0(d_2,a_2^*,-a_{22}^*)<\lambda_0(d_1,a_1^*,a_{11}^*-a_{12}^*)$,\quad
$\lambda_0(d_1,a_1^*,-a_{11}^*)<\lambda_0(d_2,a_2^*,a_{22}^*-a_{21}^*)$.
\end{itemize}
Assumption (B1) might be a technique assumption which ensures that
the periodic eigenvalue problem related to the linearized system of \eqref{u1u2-0-1} at $\bm{0}$ and $\bm{1}$
exactly admits a positive eigenvalue with the corresponding eigenfunction positive, respectively.
In fact, we cannot use the Krein-Rutman theorem to confirm the existence of the positive eigenfunction since
the linearized systems of \eqref{u1u2-0-1} at $\bm{0}$ and $\bm{1}$ are not irreducible, one can see a similar assumption (A3) in \cite{bao2013}.
\begin{remark}\rm
In view of \eqref{i-j}, if $d_1\equiv d_2$, $a_1\equiv a_2$, assumption (B1) holds in particular if
$$(a_{21}(x)-2a_{11}(x))u_1^*(x)\leq,\not\equiv b_2(x)-b_1(x)\leq,\not\equiv(2a_{22}(x)-a_{12}(x))u_2^*(x),\quad \forall x\in\Bbb R.$$
\end{remark}
\begin{lemma}\label{0-1-eigen-eigen}
Assume (A1)-(A2) and (B1). Then the periodic eigenvalue problems \eqref{0-eigen} and \eqref{1-eigen}
admit a eigenvalue $\mu_i^-<0$ associated with a unique positive periodic eigenfunction
$\Phi_i^*(x)=(\phi_{i1}^*(x),\phi_{i2}^*(x))$ for $i=0,1$, respectively.
\end{lemma}
\begin{proof}
For problem \eqref{0-eigen}, let $\phi_{01}^*(x)$ be the unique positive periodic eigenfunction satisfying
$\mathop{\max}\limits_{x\in[0,L]}\phi_{01}^*(x)=1$ associated with
the principal eigenvalue $\lambda_0(d_1,a_1^*,a_{11}^*-a_{12}^*)$,
that is,
$$\lambda_0(d_1,a_1^*,a_{11}^*-a_{12}^*)\phi_{01}^*
=d_1(\phi_{01}^*)^{\prime\prime}-a_1^*(\phi_{01}^*)^{\prime}+(a_{11}^*-a_{12}^*)\phi_{01}^*,
\quad\max_{x\in[0,L]}\phi_{01}^*(x)=1.$$
In terms of (B1), there holds
$\lambda_0(d_2,a_2^*,-a_{22}^*)<\lambda_0(d_1,a_1^*,a_{11}^*-a_{12}^*)$.
The same argument as in the proof of \cite[Proposition 4.2]{Yu2017} implies that there exists
a unique positive periodic function $\phi_{02}^*(x)$ such that
$\Phi_0^*(x)=(\phi_{01}^*(x),\phi_{02}^*(x))$ is the unique positive periodic eigenfunction of \eqref{0-eigen}
associated with $\mu_0^-:=\lambda_0(d_1,a_1^*,a_{11}^*-a_{12}^*)<0$.
Similarly, for problem \eqref{1-eigen}, since $\lambda_0(d_1,a_1^*,-a_{11}^*)<\lambda_0(d_2,a_2^*,a_{22}^*-a_{21}^*)$,
there exits a unique positive periodic eigenfunction
$\Phi_1^*(x)=(\phi_{11}^*(x),\phi_{12}^*(x))$ satisfying $\mathop{\max}\limits_{x\in[0,L]}\phi_{12}^*(x)=1$
associated with $\mu_1^-:=\lambda_0(d_2,a_2^*,a_{22}^*-a_{21}^*)<0$.
The proof is complete.
\end{proof}
Next we verify the strongly stability of $\bm{0}$ and $\bm{1}$ from above and below respectively,
and the strongly instability of any periodic coexistence steady state $(\hat{u}_1,\hat{u}_2)$ from both above and below,
for each $Q_t$ with $t>0$, in the sense of the following definition.
\begin{definition}[See \cite{fang2011}] A steady state $\bm{\alpha}\in\Pi_{\bm{\beta}}$ is said to be
strongly stable from below for the map $Q:\Pi_{\bm{\beta}}\to\Pi_{\bm{\beta}}$,
if there exist a positive number $\eta_0$ and a strongly positive element $\bm{e}\in\Pi_{\bm{\beta}}$ such that
$$Q(\bm{\alpha}-\eta\bm{e})\gg\bm{\alpha}-\eta\bm{e},\quad \forall \eta\in(0,\eta_0].$$
The strongly instability from below is defined by reversing the inequality.
Similarly, we can define strongly stability (instability) from above.
\end{definition}
\begin{lemma}\label{0-1}
Assume (A1)-(A2) and (B1). Then for each $Q_t$ with $t>0$, we have
\begin{itemize}
\item[(i)] $\bm{0}$ and $\bm{1}$ are strongly stable periodic steady states from above and below respectively.
\item[(ii)] Any periodic coexistence steady state $\hat{\bm{u}}=(\hat{u}_1,\hat{u}_2)$ is strongly unstable from both above and below.
\end{itemize}
\end{lemma}
\begin{proof}
(i) Noting that $\bm{0}$ is a locally stable steady state of \eqref{u1u2-0-1},
then for any $\varepsilon\in(0,1)$ satisfying
\begin{equation}\label{0-ineq}
\max\left\{\mathop{\max}\limits_{x\in[0,L]}a_{12}^*(x),
\mathop{\max}\limits_{x\in[0,L]}a_{22}^*(x)\right\}\varepsilon<-\frac{\mu_0^-}{2},
\end{equation}
there exists $\rho=\rho(\varepsilon)>0$ sufficiently small such that
for any $\bm{\phi}=(\phi_1,\phi_2)\in B_\rho(\bm{0})\cap\Bbb{R}_+^2$,
$\bm{u}(t,\cdot;\bm{\phi})\in B_\varepsilon(\bm{0})\cap\Bbb{R}_+^2$ for any $t>0$,
where $B_\sigma(\bm{\phi_0}):=\{\bm{\phi}\in\mathcal C:\lVert\bm{\phi}-\bm{\phi_0}\rVert_{\mathcal C}<\sigma\}$ for any $\sigma>0$ and $\bm{\phi_0}\in\mathcal C$, and $\bm{u}(t,\cdot;\bm{\phi})$ is the solution of \eqref{u1u2-0-1} with initial value $\bm{\phi}$.
Let $\bm{e}:=(\phi_{01}^*,\phi_{02}^*)$. Choose some constant $\eta_0>0$ small enough such that
$\eta\bm{e}\in B_\rho(\bm{0})\cap\Bbb{R}_+^2$ for any $\eta\in(0,\eta_0]$.
It then follows from Lemma \ref{0-1-eigen-eigen} that
$$\bm{v}(t,x)=(v_1(t,x),v_2(t,x)):=\eta e^{\frac{\mu_0^-}{2}t}\bm{e}(x),\quad (t,x)\in(0,+\infty)\times\Bbb{R}$$
satisfies
\begin{equation}\label{v1-v2}
\begin{cases}
\frac{\partial v_1}{\partial t}=L_1^*v_1+\left(a_{11}^*(x)-a_{12}^*(x)-\frac{\mu_0^-}{2}\right)v_1,\quad t>0,~x\in\Bbb R,\\
\frac{\partial v_2}{\partial t}=L_2^*v_2+a_{21}^*(x)v_1-\left(a_{22}^*(x)+\frac{\mu_0^-}{2}\right)v_2,\quad t>0,~x\in\Bbb R,\\
(v_1(0,x),v_2(0,x))=\eta(\phi_{01}^*(x),\phi_{02}^*(x)),\quad x\in\Bbb R.
\end{cases}
\end{equation}
On the other hand, since $Q_t(\eta\bm{e})=\bm{u}(t,\cdot;\eta\bm{e})\in B_\varepsilon(\bm{0})\cap\Bbb{R}_+^2$ for any $t>0$,
one can see from \eqref{0-ineq} that $\bm{u}(t,\cdot;\eta\bm{e})=(u_1(t,\cdot;\eta\bm{e}),u_2(t,\cdot;\eta\bm{e}))$ satisfies
\begin{equation}\label{u1-u2}
\begin{cases}
\frac{\partial u_1}{\partial t}=L_1^*u_1+u_1\left[a_{11}^*(x)-a_{12}^*(x)-a_{11}^*(x)u_1+a_{12}^*(x)u_2\right]\\
\qquad\leq L_1^*u_1+\left(a_{11}^*(x)-a_{12}^*(x)-\frac{\mu_0^-}{2}\right)u_1,\quad t>0,~x\in\Bbb R,\\
\frac{\partial u_2}{\partial t}=L_2^*u_2+a_{21}^*(x)u_1-a_{22}^*(x)u_2-u_2\left[a_{21}^*(x)u_1-a_{22}^*(x)u_2\right]\\
\qquad\leq L_2^*u_2+a_{21}^*(x)u_1-\left(a_{22}^*(x)+\frac{\mu_0^-}{2}\right)u_2,\quad t>0,~x\in\Bbb R,\\
(u_1(0,x),u_2(0,x))=\eta(\phi_{01}^*(x),\phi_{02}^*(x)),\quad x\in\Bbb R.
\end{cases}
\end{equation}
In light of \eqref{v1-v2} and \eqref{u1-u2}, one can conclude from the comparison principle for cooperative systems
that $Q_t(\eta\bm{e})=\bm{u}(t,\cdot;\eta\bm{e})\leq\eta e^{\frac{\mu_0^-}{2}t}\bm{e}\ll\eta\bm{e}$ for any $t>0$ and
$\eta\in(0,\eta_0]$. That is, $\bm{0}$ is strongly stable from above for $Q_t$ with any $t>0$.
Similarly, we can show that $\bm{1}$ is strongly stable from below for $Q_t$ with any $t>0$.
(ii) Let $(\hat{\phi}_1,\hat{\phi}_2)>(0,0)$ be the eigenfunction corresponding to $\hat{\lambda}>0$
of the following periodic eigenvalue problem
\begin{equation}\label{coe-eigen}
\begin{cases}
\lambda\phi_1=d_1\phi_1^{\prime\prime}-a_1^*\phi_1^{\prime}
+[a_{11}^*(x)-a_{12}^*(x)-2a_{11}^*(x)\hat{u}_1+a_{12}^*(x)\hat{u}_2]\phi_1+a_{12}^*(x)\hat{u}_1\phi_2,\ x\in\Bbb{R}, \\
\lambda\phi_2=d_2\phi_2^{\prime\prime}-a_2^*\phi_2^{\prime}
+a_{21}^*(x)(1-\hat{u}_2)\phi_1-[a_{22}^*(x)+a_{21}^*(x)\hat{u}_1-2a_{22}^*(x)\hat{u}_2]\phi_2,\ x\in\Bbb{R}, \\
\phi_i(x+L)=\phi_i(x),\ i=1,2,\ x\in\Bbb{R}.
\end{cases}
\end{equation}
Denote $\hat{\bm{e}}:=(\hat{\phi}_1,\hat{\phi}_2)$ and $\eta_1=\eta_1(t):=\frac{\hat{\lambda}e^{-\frac{\hat{\lambda}}{2}t}}
{2\max\{\mathop{\max}\limits_{x\in[0,L]}a_{11}^*\hat{\phi}_1,\mathop{\max}\limits_{x\in[0,L]}a_{21}^*\hat{\phi}_1\}}$ for each $t>0$.
Let
$$\bm{w}(t,x)=(w_1(t,x),w_2(t,x)):=\hat{\bm{u}}(x)+\eta e^{\frac{\hat{\lambda}}{2}t}\hat{\bm{e}}(x),\quad (t,x)\in(0,+\infty)\times\Bbb{R},$$
where $0<\eta\leq\eta_1$ is a constant. Then direct calculation shows that
\begin{align*}
\frac{\partial w_1}{\partial t}&=L_1^*w_1+\frac{\hat{\lambda}}{2}\eta e^{\frac{\hat{\lambda}}{2}t}\hat{\phi}_1+f_1(x,\hat{u}_1,\hat{u}_2)\\
&\quad-\eta e^{\frac{\hat{\lambda}}{2}t}\left(\hat{\lambda}\hat{\phi}_1
-[a_{11}^*(x)-a_{12}^*(x)-2a_{11}^*(x)\hat{u}_1+a_{12}^*(x)\hat{u}_2]\hat{\phi}_1-a_{12}^*(x)\hat{u}_1\hat{\phi}_2\right)\\
&=L_1^*w_1+f_1(x,w_1,w_2)+\eta e^{\frac{\hat{\lambda}}{2}t}\hat{\phi}_1\left(-\frac{\hat{\lambda}}{2}+\eta e^{\frac{\hat{\lambda}}{2}t}(a_{11}^*\hat{\phi}_1-a_{12}^*\hat{\phi}_2)\right)\\
&\leq L_1^*w_1+f_1(x,w_1,w_2),\quad t>0,~x\in\Bbb R,
\end{align*}
and
\begin{align*}
\frac{\partial w_2}{\partial t}&=L_2^*w_2+\frac{\hat{\lambda}}{2}\eta e^{\frac{\hat{\lambda}}{2}t}\hat{\phi}_2+f_2(x,\hat{u}_1,\hat{u}_2)\\
&\quad-\eta e^{\frac{\hat{\lambda}}{2}t}\left(\hat{\lambda}\hat{\phi}_2-a_{21}^*(x)(1-\hat{u}_2)\hat{\phi}_1
+[a_{22}^*(x)+a_{21}^*(x)\hat{u}_1-2a_{22}^*(x)\hat{u}_2]\hat{\phi}_2\right)\\
&=L_2^*w_2+f_2(x,w_1,w_2)+\eta e^{\frac{\hat{\lambda}}{2}t}\hat{\phi}_2\left(-\frac{\hat{\lambda}}{2}+\eta e^{\frac{\hat{\lambda}}{2}t}(a_{21}^*\hat{\phi}_1-a_{22}^*\hat{\phi}_2)\right)\\
&\leq L_2^*w_2+f_2(x,w_1,w_2),\quad t>0,~x\in\Bbb R.
\end{align*}
That is, $(w_1,w_2)$ is a subsolution of \eqref{u1u2-0-1} in $(t,x)\in(0,+\infty)\times\Bbb R$.
The comparison principle then implies that
$Q_t(\hat{\bm{u}}+\eta\hat{\bm{e}})\geq\hat{\bm{u}}+\eta e^{\frac{\hat{\lambda}}{2}t}\hat{\bm{e}}\gg\hat{\bm{u}}+\eta\hat{\bm{e}}$
for any $t>0$ and $\eta\in(0,\eta_1]$, that is, $\hat{\bm{u}}$ is strongly unstable from above for each $Q_t$ with $t>0$.
Similarly, we can prove that $\hat{\bm{u}}$ is strongly unstable from below for each $Q_t$ with $t>0$. The proof is complete.
\end{proof}
Let $E$ be the set of all fixed points of $\{Q_t\}_{t\geq0}$ restricted on $\mathcal C_{\bm{1}}$,
denote by $\hat{E}$ the set of all periodic coexistence steady states,
i.e., $\hat{E}=\{\bm{\alpha}_1\in E: \bm{0}\ll\bm{\alpha}_1\ll\bm{1}\}$, then
$E=\hat{E}\cup\{\bm{0},\bm{\alpha}_2,\bm{1}\}$, where $\bm{\alpha}_2:=(0,1)$.
For any $\bm{\alpha}\in E\setminus\{\bm{0},\bm{1}\}$, in terms of Lemma \ref{0-1},
the bistable system $\{Q_t\}_{t\geq0}$ performs a monostable dynamics on
$\mathcal C_{[\bm{0},\bm{\alpha}]}$ and $\mathcal C_{[\bm{\alpha},\bm{1}]}$, respectively.
It remains to show the so-called counter-propagation assumption in \cite{fang2011}.
Introduce a family of operators $\{\hat{Q}_t\}_{t\geq0}$ on $\mathcal{X}_{\bm{1}}$ as
$$\hat{Q}_t[\bm{v}](s)(\theta):=Q_t[\bm{v}_s](\theta),
\ \ \forall \bm{v}\in\mathcal{X}_{\bm{1}},\ s\in\Bbb{R},\ \theta\in[0,L],\ t\geq0,$$
where $\bm{v}_s\in\mathcal C$ is defined by
$$\bm{v}_s(x)=\bm{v}(s+n_x)(\theta_x),\ \ \forall x=n_x+\theta_x\in\Bbb{R},\ n_x=L\left[\frac{x}{L}\right],\ \theta_x\in[0,L).$$
Then $\{\hat{Q}_t\}_{t\geq0}$ is a monotone semiflow on $\mathcal{X}_{\bm{1}}$.
Since $\{Q_t\}_{t\geq0}$ and $\{\hat{Q}_t\}_{t\geq0}$ are topologically conjecture,
the bistable system $\{\hat{Q}_t\}_{t\geq0}$ performs a monostable dynamics on
$\mathcal C_{[\bm{0},\bm{\alpha}]}$ and $\mathcal C_{[\bm{\alpha},\bm{1}]}$ respectively for any $\bm{\alpha}\in E\setminus\{\bm{0},\bm{1}\}$.
Thanks to Lemma \ref{0-1}, there exist some constants $\eta_i>0$ and
vectors $\bm{e}_i\in\rm{Int}(\mathcal{X}_+)$ for $i=1,2,$ such that
$\hat{Q}_1[\eta\bm{e}_0]\ll\eta\bm{e}_0$ for any $\eta\in(0,\eta_0]$, and
$\hat{Q}_1[\bm{1}-\eta\bm{e}_1]\gg\bm{1}-\eta\bm{e}_1$ for any $\eta\in(0,\eta_1]$.
Define
$$\theta^+:=\sup\{\theta\in[0,1]: \theta\bm{\alpha}\in\mathcal{X}_{[\bm{0},\eta_0\bm{e}_0]}\},\ \
\bm{\alpha}^+=\theta^+\bm{\alpha},$$
$$\theta^-:=\sup\{\theta\in[0,1]: \theta\bm{\alpha}+(1-\theta)\bm{1}\in\mathcal{X}_{[\bm{1}-\eta_1\bm{e}_1,\bm{1}]}\},\ \
\bm{\alpha}^-=\theta^-\bm{\alpha}+(1-\theta^-)\bm{1}.$$
As in \cite{fang2011}, we introduce
\begin{align*}
c_+^*(\bm{0},\bm{\alpha}):
&=\sup\left\{ c\in\Bbb{R}: \mathop{\lim}\limits_{n\to\infty,\ x\leq cn}\hat{Q}_1^n[\bm{\phi_{\bm{\alpha}}^+}](x)=\bm{0}\right\},\\
c_-^*(\bm{\alpha},\bm{1}):
&=\sup\left\{c\in\Bbb{R}: \mathop{\lim}\limits_{n\to\infty,\ x\geq-cn}\hat{Q}_1^n[\bm{\phi_{\bm{\alpha}}^-}](x)=\bm{1}\right\},
\end{align*}
where the initial functions $\bm{\phi_{\bm{\alpha}}^\pm}\in\mathcal{X}_{\bm{1}}$ satisfy respectively
$$\bm{\phi}_{\bm{\alpha}}^+(x)=\bm{\alpha},\ \forall x\geq1,\text{~~and~~}\bm{\phi}_{\bm{\alpha}}^+(x)=\bm{v_{\bm{\alpha}}^+},
\ \forall x\leq0,$$
$$\bm{\phi}_{\bm{\alpha}}^-(x)=\bm{\alpha},\ \forall x\leq-1,\text{~~and~~}\bm{\phi}_{\bm{\alpha}}^-(x)=\bm{v_{\bm{\alpha}}^-},
\ \forall x\geq0.$$
In the following, we are going to show $c_+^*(\bm{0},\bm{\alpha})+c_-^*(\bm{\alpha},\bm{1})>0$
for any $\bm{\alpha}\in E\setminus\{\bm{0},\bm{1}\}$, which nearly assures the propagation of a bistable pulsating front.
For any $\bm{\alpha}_1=(\hat{u}_1,\hat{u}_2)\in\hat{E}$,
to better understand the rightward propagation dynamics of $\{Q_t\}_{t\geq0}$ restricted on $\mathcal C_{[\bm{0},\bm{\alpha}_1]}$,
let $\tilde{v}_i(t,x)=\hat{u}_i(x)-u_i(t,x)$, $i=1,2$,
then this dynamics is equivalent to that of the following system restricted on $\mathcal C_{\bm{\alpha}_1}$:
\begin{equation}\label{u1u2-linearize+}
\begin{cases}
\frac{\partial \tilde{v}_1}{\partial t}=L_1^*\tilde{v}_1
+f_1(x,\hat{u}_1(x),\hat{u}_2(x))-f_1(x,\hat{u}_1(x)-\tilde{v}_1(t,x),\hat{u}_2(x)-\tilde{v}_2(t,x)),\\
\frac{\partial \tilde{v}_2}{\partial t}=L_2^*\tilde{v}_2
+f_2(x,\hat{u}_1(x),\hat{u}_2(x))-f_2(x,\hat{u}_1(x)-\tilde{v}_1(t,x),\hat{u}_2(x)-\tilde{v}_2(t,x)),
\end{cases}
\end{equation}
where $\bm{0}$ is unstable and $\bm{\alpha}_1\gg\bm{0}$ is stable.
Define a family of operators $\{\tilde{Q}_t\}_{t\geq0}$ on $\mathcal C_{\bm{\alpha}_1}$ by
$\tilde{Q}_t(\bm{\phi}):=\bm{\tilde{v}}(t,\cdot;\bm{\phi})$,
where $\bm{\tilde{v}}(t,\cdot;\bm{\phi})$ is the solution of \eqref{u1u2-linearize+} with
$\bm{\tilde{v}}(0,\cdot;\bm{\phi})=\bm{\phi}\in\mathcal C_{\bm{\alpha}_1}$.
Then $\tilde{Q}_t(\bm{0})=\bm{0}$, $\tilde{Q}_t(\bm{\alpha}_1)=\bm{\alpha}_1$ for all $t\geq0$, and
$\tilde{Q}_t$ satisfies all hypotheses (E1)-(E5) in \cite{liang2010} for any $t>0$.
It then follows from \cite[Theorem 5.1]{liang2010} that
$\tilde{Q}_1:\mathcal C_{\bm{\alpha}_1}\to\mathcal C_{\bm{\alpha}_1}$ admits
the rightward spreading speed $c_+^{**}(\bm{0},\bm{\alpha}_1)$, and then
$c_+^{*}(\bm{0},\bm{\alpha}_1)\geq c_+^{**}(\bm{0},\bm{\alpha}_1)$.
Similarly, to better understand the leftward propagation dynamics of $\{Q_t\}_{t\geq0}$ restricted on $\mathcal C_{[\bm{\alpha}_1,\bm{1}]}$,
let $\tilde{u}_i(t,x)=u_i(t,x)-\hat{u}_i(x)$, $i=1,2$,
then this dynamics is equivalent to that of the following system restricted on $\mathcal C_{\bm{\beta}}$:
\begin{equation}\label{u1u2-linearize-}
\begin{cases}
\frac{\partial \tilde{u}_1}{\partial t}=L_1^*\tilde{u}_1
+f_1(x,\tilde{u}_1(t,x)+\hat{u}_1(x),\tilde{u}_2(t,x)+\hat{u}_2(x))-f_1(x,\hat{u}_1(x),\hat{u}_2(x)),\\
\frac{\partial \tilde{u}_2}{\partial t}=L_2^*\tilde{u}_2
+f_2(x,\tilde{u}_1(t,x)+\hat{u}_1(x),\tilde{u}_2(t,x)+\hat{u}_2(x))-f_2(x,\hat{u}_1(x),\hat{u}_2(x)),
\end{cases}
\end{equation}
where $\bm{0}$ is unstable and $\bm{\beta}:=\bm{1}-\bm{\alpha}_1\gg\bm{0}$ is stable.
Define a family of operators $\{\tilde{P}_t\}_{t\geq0}$ on $\mathcal C_{\bm{\beta}}$ by
$\tilde{P}_t(\bm{\phi}):=\bm{\tilde{u}}(t,\cdot;\bm{\phi})$,
where $\bm{\tilde{u}}(t,\cdot;\bm{\phi})$ is the solution of \eqref{u1u2-linearize-} with
$\bm{\tilde{u}}(0,\cdot;\bm{\phi})=\bm{\phi}\in\mathcal C_{\bm{\beta}}$.
Noting that $\tilde{P}_t(\bm{0})=\bm{0}$, $\tilde{P}_t(\bm{\beta})=\bm{\beta}$ for all $t\geq0$, and
$\tilde{P}_t$ satisfies all hypotheses (E1)-(E5) in \cite{liang2010} for any $t>0$.
Then $\tilde{P}_1:\mathcal C_{\bm{\beta}}\to\mathcal C_{\bm{\beta}}$ admits
the leftward spreading speed $c_-^{**}(\bm{0},\bm{\beta})$, and then
$c_-^*(\bm{\alpha}_1,\bm{1})\geq c_-^{**}(\bm{0},\bm{\beta})$.
Next we use the linear operators approach (see \cite{liang2007,weinberger2003})
to estimate these two lower spreading speeds $c_+^{**}(\bm{0},\bm{\alpha}_1)$ and $c_-^{**}(\bm{0},\bm{\beta})$.
\begin{lemma}\label{coun-1}
Assume (A1)-(A2) and (B1). Then $c_+^{**}(\bm{0},\bm{\alpha}_1)+c_-^{**}(\bm{0},\bm{\beta})>0$.
\end{lemma}
\begin{proof}
Consider the linearized system of \eqref{u1u2-linearize+} at $\bm{0}$:
\begin{equation}\label{lin-1}
\begin{cases}
\frac{\partial v_1}{\partial t}=L_1^*v_1
+\frac{\partial f_1}{\partial u_1}(x,\hat{u}_1(x),\hat{u}_2(x))v_1
+\frac{\partial f_1}{\partial u_2}(x,\hat{u}_1(x),\hat{u}_2(x))v_2,\ \ t>0,\ x\in\Bbb{R},\\
\frac{\partial v_2}{\partial t}=L_2^*v_2
+\frac{\partial f_2}{\partial u_1}(x,\hat{u}_1(x),\hat{u}_2(x))v_1
+\frac{\partial f_2}{\partial u_2}(x,\hat{u}_1(x),\hat{u}_2(x))v_2,\ \ t>0,\ x\in\Bbb{R},\\
\bm{v}(0,\cdot)=\bm{\phi}\in\mathcal C.
\end{cases}
\end{equation}
Let $\{M(t)\}_{t\geq0}:\mathcal C_{\bm{\alpha}_1}\to\mathcal C_{\bm{\alpha}_1}$ be the solution map of \eqref{lin-1},
that is, $M(t)(\bm{\phi}):=\bm{v}(t,\cdot;\bm{\phi})$,
where $\bm{v}(t,\cdot;\bm{\phi})$ is the solution of \eqref{lin-1} with $\bm{v}(0,\cdot;\bm{\phi})=\bm{\phi}$.
For any given $\mu\in\Bbb{R}$, letting $\bm{v}(t,x)=e^{-\mu x}\bm{w}(t,x)$ in \eqref{lin-1}, then
$\bm{w}(t,x)=(w_1(t,x),w_2(t,x))$ satisfies the following $\mu$-parameterized linear system
\begin{equation}\label{lin-1-v}
\begin{cases}
\frac{\partial w_1}{\partial t}=L_1^*w_1-2\mu d_1\frac{\partial w_1}{\partial x}
+\left(d_1\mu^2+a_1^*\mu+\frac{\partial f_1}{\partial u_1}(x,\hat{u}_1,\hat{u}_2)\right)w_1
+\frac{\partial f_1}{\partial u_2}(x,\hat{u}_1,\hat{u}_2)w_2,\\
\frac{\partial w_2}{\partial t}=L_2^*w_2-2\mu d_2\frac{\partial w_2}{\partial x}
+\frac{\partial f_2}{\partial u_1}(x,\hat{u}_1,\hat{u}_2)w_1
+\left(d_2\mu^2+a_2^*\mu+\frac{\partial f_2}{\partial u_2}(x,\hat{u}_1,\hat{u}_2)\right)w_2,\\
\bm{w}(0,x)=\bm{\phi}(x)e^{\mu x}.
\end{cases}
\end{equation}
Let $\{M_{\mu}(t)\}_{t\geq0}$ be the solution map of system \eqref{lin-1-v},
that is, $M_{\mu}(t)(\bm{\phi}):=\bm{w}(t,\cdot;\bm{\phi})$,
where $\bm{w}(t,\cdot;\bm{\phi})$ is the unique solution of \eqref{lin-1-v} with $\bm{w}(0,\cdot;\bm{\phi})=\bm{\phi}$.
Noting that system \eqref{lin-1-v} is cooperative and irreducible,
it follows that $\{M_{\mu}(t)\}_{t\geq0}$ is strongly order preserving and
$$M_{\mu}(t)[\bm{\phi}](x)=e^{\mu x}M(t)[e^{-\mu x}\bm{\phi}](x),\ \forall t\geq0,\ x\in\Bbb{R},\ \bm{\phi}\in\mathcal C.$$
Substituting $\bm{w}(t,x)=e^{\lambda t}\bm{\phi}(x)$ into \eqref{lin-1-v},
we obtain the following periodic eigenvalue problem
\begin{equation}\label{ei}
\begin{cases}
\lambda\phi_1=d_1\phi_1^{\prime\prime}-(2d_1\mu+a_1^*)\phi_1^{\prime}
+\left(d_1\mu^2+a_1^*\mu+\frac{\partial f_1}{\partial u_1}(x,\hat{u}_1,\hat{u}_2)\right)\phi_1
+\frac{\partial f_1}{\partial u_2}(x,\hat{u}_1,\hat{u}_2)\phi_2,\\
\lambda\phi_2=d_2\phi_2^{\prime\prime}-(2d_2\mu+a_2^*)\phi_2^{\prime}
+\frac{\partial f_2}{\partial u_1}(x,\hat{u}_1,\hat{u}_2)\phi_1
+\left(d_2\mu^2+a_2^*\mu+\frac{\partial f_2}{\partial u_2}(x,\hat{u}_1,\hat{u}_2)\right)\phi_2,\\
\phi_i(x+L)=\phi_i(x),\ i=1,2.
\end{cases}
\end{equation}
Let $\lambda^+=\lambda^+(\mu)$ be the principal eigenvalue associated with a positive $L$-periodic eigenfunction
for the linear periodic cooperative and irreducible system \eqref{ei}, then a similar argument as in the proof of
\cite[Lemma 3.7]{liang2007} shows that $\lambda^+(\mu)$ is a convex function on $\mu\in\Bbb{R}$.
Furthermore, $\lambda^+(0)=\hat{\lambda}>0$, which together with \eqref{ei} yields that
$\mathop{\lim}\limits_{\mu\to0^+}\frac{\lambda^+(\mu)}{\mu}=\mathop{\lim}\limits_{\mu\to+\infty}\frac{\lambda^+(\mu)}{\mu}=+\infty$,
and hence $\frac{\lambda^+(\mu)}{\mu}$ attains its infimum at some $\mu\in(0,+\infty)$.
Similar observations hold for $\frac{\lambda^+(-\mu)}{\mu}$.
For any given $\varepsilon\in(0,1)$, let $\lambda^+_\varepsilon(\mu)$ be the principal eigenvalue of the following
linear periodic cooperative and irreducible system
\begin{equation*}
\begin{cases}
\lambda\phi_1=d_1\phi_1^{\prime\prime}-(2d_1\mu+a_1^*)\phi_1^{\prime}
+\left(d_1\mu^2+a_1^*\mu+\frac{\partial f_1}{\partial u_1}(x,\hat{u}_1,\hat{u}_2)-\varepsilon\right)\phi_1
+\frac{\partial f_1}{\partial u_2}(x,\hat{u}_1,\hat{u}_2)\phi_2,\\
\lambda\phi_2=d_2\phi_2^{\prime\prime}-(2d_2\mu+a_2^*)\phi_2^{\prime}
+\frac{\partial f_2}{\partial u_1}(x,\hat{u}_1,\hat{u}_2)\phi_1
+\left(d_2\mu^2+a_2^*\mu+\frac{\partial f_2}{\partial u_2}(x,\hat{u}_1,\hat{u}_2)-\varepsilon\right)\phi_2,\\
\phi_i(x+L)=\phi_i(x),\ i=1,2,
\end{cases}
\end{equation*}
and let $\{M_\varepsilon(t)\}_{t\geq0}$ be the solution map of the linear periodic system
\begin{equation*}
\begin{cases}
\frac{\partial \tilde{v}_1}{\partial t}=L_1^*\tilde{v}_1
+\left(\frac{\partial f_1}{\partial u_1}(x,\hat{u}_1(x),\hat{u}_2(x))-\varepsilon\right)\tilde{v}_1
+\frac{\partial f_1}{\partial u_2}(x,\hat{u}_1(x),\hat{u}_2(x))\tilde{v}_2,\ \ t>0,\ x\in\Bbb{R},\\
\frac{\partial \tilde{v}_2}{\partial t}=L_2^*\tilde{v}_2
+\frac{\partial f_2}{\partial u_1}(x,\hat{u}_1(x),\hat{u}_2(x))\tilde{v}_1
+\left(\frac{\partial f_2}{\partial u_2}(x,\hat{u}_1(x),\hat{u}_2(x))-\varepsilon\right)\tilde{v}_2,\ \ t>0,\ x\in\Bbb{R},\\
\bm{\tilde{v}}(0,\cdot)=\bm{\phi}\in\mathcal C.
\end{cases}
\end{equation*}
Let a vector $\bm{\sigma}\in\Bbb{R}_+^2$ be such that
for any $(\tilde{v}_1(t,x),\tilde{v}_2(t,x))\in[\bm{0},\bm{\sigma}]$ and $x\in[0,L]$, there is
$$|a_{12}^*(x)\tilde{v}_2(t,x)-a_{11}^*(x)\tilde{v}_1(t,x)|\leq\varepsilon
\text{~~and~~}|a_{22}^*(x)\tilde{v}_2(t,x)-a_{21}^*(x)\tilde{v}_1(t,x)|\leq\varepsilon.$$
Then we have
\begin{align*}
&f_1(x,\hat{u}_1(x),\hat{u}_2(x))-f_1(x,\hat{u}_1(x)-\tilde{v}_1(t,x),\hat{u}_2(x)-\tilde{v}_2(t,x))\\
&\quad-\frac{\partial f_1}{\partial u_1}(x,\hat{u}_1(x),\hat{u}_2(x))\tilde{v}_1(t,x)-
\frac{\partial f_1}{\partial u_2}(x,\hat{u}_1(x),\hat{u}_2(x))\tilde{v}_2(t,x)\\
&=a_{11}^*(x)\tilde{v}_1^2(t,x)-a_{12}^*(x)\tilde{v}_1(t,x)\tilde{v}_2(t,x)\\
&\geq-\varepsilon\tilde{v}_1(t,x),
\end{align*}
and
\begin{align*}
&f_2(x,\hat{u}_1(x),\hat{u}_2(x))-f_2(x,\hat{u}_1(x)-\tilde{v}_1(t,x),\hat{u}_2(x)-\tilde{v}_2(t,x))\\
&\quad-\frac{\partial f_2}{\partial u_1}(x,\hat{u}_1(x),\hat{u}_2(x))\tilde{v}_1(t,x)-
\frac{\partial f_2}{\partial u_2}(x,\hat{u}_1(x),\hat{u}_2(x))\tilde{v}_2(t,x)\\
&=a_{21}^*(x)\tilde{v}_1(t,x)\tilde{v}_2(t,x)-a_{22}^*(x)\tilde{v}_2^2(t,x)\\
&\geq-\varepsilon\tilde{v}_2(t,x).
\end{align*}
Since $\bm{\tilde{v}}(t,x;\bm{0})=\bm{0}$,
there exists a positive vector $\bm{\eta}\in\Bbb{R}_+^2$ with $\bm{\eta}\leq\bm{\alpha}_1$ such that
for any $\bm{\phi}\in\mathcal C_{\bm{\eta}}$,
$\bm{\tilde{v}}(t,x;\bm{\phi})\in[\bm{0},\bm{\sigma}]$ for any $x\in\Bbb{R}$ and $t\in[0,1]$,
where $\bm{\tilde{v}}(t,x;\bm{\phi})$ is the unique solution of \eqref{u1u2-linearize+}.
By the comparison principle, there holds
$$\tilde{Q}_t(\bm{\phi})\geq M_\varepsilon(t)(\bm{\phi}),\ \forall\bm{\phi}\in\mathcal C_{\bm{\eta}},\ t\in[0,1].$$
In particular, $\tilde{Q}_1(\bm{\phi})\geq M_\varepsilon(1)(\bm{\phi})$ for any $\bm{\phi}\in\mathcal C_{\bm{\eta}}$.
It follows from \cite[Theorem 2.4]{weinberger2003} that
$$c_+^{**}(\bm{0},\bm{\alpha}_1)\geq\mathop{\inf}\limits_{\mu>0}\frac{\lambda^+_\varepsilon(\mu)}{\mu},\ \ \forall\varepsilon\in(0,1).$$
Letting $\varepsilon\to0$ in the above inequality, we have
$$c_+^{**}(\bm{0},\bm{\alpha}_1)\geq\mathop{\inf}\limits_{\mu>0}\frac{\lambda^+(\mu)}{\mu}.$$
By the change of variable $\bm{\tilde{w}}(t,x)=\bm{\tilde{u}}(t,-x)$ in system \eqref{u1u2-linearize-},
it follows that $c_-^{**}(\bm{0},\bm{\beta})$ is the rightward spreading speed of the resulting system for $\bm{\tilde{w}}$.
The similar argument as above implies that
$$c_-^{**}(\bm{0},\bm{\beta})\geq\mathop{\inf}\limits_{\mu>0}\frac{\lambda^+(-\mu)}{\mu}.$$
Let $\mu_1>0$ and $\mu_2>0$ be such that
$$c_+^{**}(\bm{0},\bm{\alpha}_1)\geq\mathop{\inf}\limits_{\mu>0}\frac{\lambda^+(\mu)}{\mu}=\frac{\lambda^+(\mu_1)}{\mu_1},\quad
c_-^{**}(\bm{0},\bm{\beta})\geq\mathop{\inf}\limits_{\mu>0}\frac{\lambda^+(-\mu)}{\mu}=\frac{\lambda^+(-\mu_2)}{\mu_2}.$$
Define $\theta:=\frac{\mu_1}{\mu_1+\mu_2}\in(0,1)$, then $\theta\mu_1+(1-\theta)(-\mu_2)=0$.
The convexity of $\lambda^+(\mu)$ then shows that
\begin{align*}
c_+^{**}(\bm{0},\bm{\alpha}_1)+c_-^{**}(\bm{0},\bm{\beta})&\geq\frac{\lambda^+(\mu_1)}{\mu_1}+\frac{\lambda^+(-\mu_2)}{\mu_2}
=\frac{\mu_1+\mu_2}{\mu_1\mu_2}[\theta\lambda^+(\mu_1)+(1-\theta)\lambda^+(-\mu_2)]\\
&\geq\frac{\mu_1+\mu_2}{\mu_1\mu_2}[\lambda^+(\theta\mu_1+(1-\theta)(-\mu_2))]\\
&=\frac{\mu_1+\mu_2}{\mu_1\mu_2}\lambda^+(0)=\frac{\mu_1+\mu_2}{\mu_1\mu_2}\hat{\lambda}\\
&>0.
\end{align*}
The proof is complete.
\end{proof}
Now we consider the leftward propagation dynamics of $\{Q_t\}_{t\geq0}$ restricted on $\mathcal C_{[\bm{\alpha}_2,\bm{1}]}$.
Let $\mathcal C_1=\{u\in\mathcal C: 0\leq u\leq 1\}$, where $\mathcal C$ denotes
the set of all bounded and continuous functions from $\Bbb R$ to $\Bbb R$.
Since $\bm{\alpha}_2=(0,1)$, we can easily see that this dynamics is equivalent
to that of the following periodic scalar equation restricted on $\mathcal C_1$:
\begin{equation}\label{sca-2}
u_t=L_1^*u+a_{11}^*(x)u(1-u),
\end{equation}
where $0$ is unstable and $1$ is stable steady states of \eqref{sca-2}.
Since $\lambda_0(d_1,a_1^*,a_{11}^*)>0$, we see that
\eqref{sca-2} admits a leftward spreading speed (also the minimal rightward wave speed, see, e.g., \cite{berestycki2002})
$$c_{1-}^*=\mathop{\inf}\limits_{\mu>0}\frac{\lambda_1(-\mu)}{\mu},$$
where $\lambda_1(\mu)$ is the principal eigenvalue of the following periodic eigenvalue problem
\begin{equation}\label{-}
\begin{cases}
\lambda\phi=d_1\phi^{\prime\prime}-(2d_1\mu+a_1^*)\phi^{\prime}+(d_1\mu^2+a_1^*\mu+a_{11}^*)\phi,\ x\in\Bbb{R},\\
\phi(x+L)=\phi(x),\ x\in\Bbb{R},
\end{cases}
\end{equation}
with $\lambda_1(0)=\lambda_0(d_1,a_1^*,a_{11}^*)>0$. Then we have $c_-^*(\bm{\alpha}_2,\bm{1})\geq c_{1-}^*$.
Similarly, the rightward propagation dynamics of $\{Q_t\}_{t\geq0}$ restricted on $\mathcal C_{[\bm{0},\bm{\alpha}_2]}$
is equivalent to that of the following periodic scalar equation restricted on $\mathcal C_1$:
\begin{equation*}
v_t=L_2^*v-a_{22}^*(x)v(1-v),
\end{equation*}
where $\mathcal C_1=\{u\in\mathcal C: 0\leq u\leq 1\}$, and $\mathcal C$ denotes
the set of all bounded and continuous functions from $\Bbb R$ to $\Bbb R$.
Let $w(t,x)=1-v(t,x)$, then $w$ satisfies the periodic scalar equation
\begin{equation}\label{sca-1}
w_t=L_2^*w+a_{22}^*(x)w(1-w),
\end{equation}
where $0$ is unstable and $1$ is stable steady states of \eqref{sca-1}.
In view of $\lambda_0(d_2,a_2^*,a_{22}^*)>0$,
a similar argument as in the proof of \cite[Theorem 1.1]{berestycki20051}
shows that \eqref{sca-1} admits a rightward spreading speed (also the minimal rightward wave speed)
$$c_{2+}^*=\mathop{\inf}\limits_{\mu>0}\frac{\lambda_2(\mu)}{\mu},$$
where $\lambda_2(\mu)$ is the principal eigenvalue of the following periodic eigenvalue problem
\begin{equation}\label{+}
\begin{cases}
\lambda\phi=d_2\phi^{\prime\prime}-(2d_2\mu+a_2^*)\phi^{\prime}+(d_2\mu^2+a_2^*\mu+a_{22}^*)\phi,\ x\in\Bbb{R},\\
\phi(x+L)=\phi(x),\ x\in\Bbb{R},
\end{cases}
\end{equation}
with $\lambda_2(0)=\lambda_0(d_2,a_2^*,a_{22}^*)>0$. Then we have $c_+^{*}(\bm{0},\bm{\alpha}_2)\geq c_{2+}^*$.
We finally need the following assumption to complete the verification of (A6) in \cite{fang2011}.
\begin{itemize}
\item[(B2)] $c_{1-}^*+c_{2+}^*>0$, where $c_{1-}^*$ and $c_{2+}^*$ are the leftward and rightward spreading speeds of
\eqref{sca-2} and \eqref{sca-1}, respectively.
\end{itemize}
\begin{remark}\rm
In the case that
\begin{itemize}
\item[(i)] either $L_i^*u=(d_i(x)u_x)_x$ with $d_i\in C^{1+\nu}(\Bbb{R})$,
\item[(ii)] or $d_i(\cdot),\ a_{ii}^*(\cdot)$ are even, and $a_i^*(\cdot)$ are odd for $i=1,2$,
\end{itemize}
(A1) together with (A2) guarantees (B2). Indeed, in the either of the two cases (i) and (ii),
we know from \cite[Lemma 5.1]{Yu2017} that $\lambda_i(\mu),\ i=1,2$ are even and convex functions
of $\mu$ in $\Bbb{R}$, which together with the fact
$\lambda_1(0)=\lambda_0(d_1,a_1^*,a_{11}^*)>0$ and $\lambda_2(0)=\lambda_0(d_2,a_2^*,a_{22}^*)>0$ yields that
$\lambda_i(\mu)>\lambda_i(0)>0$ for any $\mu>0$ and $i=1,2$. Therefore, we can easily see
$c_{1-}^*=\mathop{\inf}\limits_{\mu>0}\frac{\lambda_1(-\mu)}{\mu}>0$ and
$c_{2+}^*=\mathop{\inf}\limits_{\mu>0}\frac{\lambda_2(\mu)}{\mu}>0$, which directly leads to (B2).
One should also note from \eqref{-}, \eqref{+} and the representations of $c_{1-}^*$ and $c_{2+}^*$ that,
(B2) could not hold generally if the advection terms $a_1^*$ and $a_2^*$ being very large opposite constants.
\end{remark}
Combining Lemma \ref{coun-1} and the above argument, we easily obtain the following result.
\begin{proposition}\label{count}
Assume (A1)-(A2) and (B1)-(B2). Then for any $\bm{\alpha}\in E\setminus\{\bm{0},\bm{1}\}$, there holds
$c_+^*(\bm{0},\bm{\alpha})+c_-^*(\bm{\alpha},\bm{1})>0$.
\end{proposition}
Now we are ready to state our main results of this section.
\begin{theorem}\label{main}
Assume that (A1)-(A2) and (B1)-(B2). Then there exists some $c\in\Bbb{R}$ such that system \eqref{u1u2-0-1} admits a pulsating front
$\bm{U}(x,x+ct)=(U_1(x,x+ct),U_2(x,x+ct))$ satisfying $\mathop{\lim}\limits_{z\to-\infty}\bm{U}(x,z)=\bm{0}$ and
$\mathop{\lim}\limits_{z\to+\infty}\bm{U}(x,z)=\bm{1}$ uniformly for $x\in\Bbb{R}$.
Furthermore, $\bm{U}(x,z)$ is increasing in $z$ for any $x\in\Bbb{R}$ if $c\neq 0$.
\end{theorem}
\begin{proof}
In view of Proposition \ref{count}, it follows from \cite[Theorem 4.1]{fang2011} that system \eqref{u1u2-0-1} admits a pulsating front
$\bm{U}(x,x+ct)=(U_1(x,x+ct),U_2(x,x+ct))$ connecting $\bm{0}$ to $\bm{1}$, that is,
\begin{equation}\label{limit}
\mathop{\lim}\limits_{z\to-\infty}\bm{U}(x,z)=\bm{0},\quad \mathop{\lim}\limits_{z\to+\infty}\bm{U}(x,z)=\bm{1},
\text{~~uniformly for~~} x\in\Bbb{R}.
\end{equation}
Besides, $\bm{U}(x,z)$ is nondecreasing in $z$.
Next we prove that $\bm{U}(\cdot,z)$ is strictly increasing.
Let $$(u_1(t,x),u_2(t,x)):=(U_1(x,x+ct),U_2(x,x+ct)).$$
For any fixed $\tau>0$, define
$$(r_1(t,x),r_2(t,x))=\left(u_1\left(t+\frac{\tau}{c},x\right)-u_1(t,x),u_2\left(t+\frac{\tau}{c},x\right)-u_2(t,x)\right).$$
Then $(r_1(t,x),r_2(t,x))$ is nonnegative in $(t,x)\in\Bbb{R}\times\Bbb{R}$ satisfying
\begin{equation*}
\begin{cases}
\frac{\partial r_1}{\partial t}
=L_1^*r_1+f_1\left(x,u_1\left(t+\frac{\tau}{c},x\right),u_2\left(t+\frac{\tau}{c},x\right)\right)-f_1(x,u_1(t,x),u_2(t,x)),\\
\frac{\partial r_2}{\partial t}
=L_2^*r_2+f_2\left(x,u_1\left(t+\frac{\tau}{c},x\right),u_2\left(t+\frac{\tau}{c},x\right)\right)-f_2(x,u_1(t,x),u_2(t,x)),
\end{cases}
(t,x)\in\Bbb{R}\times\Bbb{R}.
\end{equation*}
Assume now that there exists some $(\bar{t},\bar{x})\in\Bbb{R}\times\Bbb{R}$
such that either $r_1(\bar{t},\bar{x})=0$ or $r_2(\bar{t},\bar{x})=0$. Since $f_{1,u_2}, f_{2,u_1}\geq 0$, one have
$$
\frac{\partial r_1}{\partial t}\geq L_1^*r_1+r_1\left[\int_0^1{f_{1,u_1}\left(x,su_1\left(t+\frac{\tau}{c},x\right)
+(1-s)u_1(t,x),su_2\left(t+\frac{\tau}{c},x\right)+(1-s)u_2(t,x)\right)}ds\right]
$$
and
$$
\frac{\partial r_2}{\partial t}\geq L_2^*r_2+r_2\left[\int_0^1{f_{2,u_2}\left(x,su_1\left(t+\frac{\tau}{c},x\right)
+(1-s)u_1(t,x),su_2\left(t+\frac{\tau}{c},x\right)+(1-s)u_2(t,x)\right)}ds\right].
$$
It then follows from the (strong) maximum principle that either $r_1\equiv0$ or $r_2\equiv0$ in $(-\infty,\bar{t}]\times\Bbb{R}$,
and then in $\Bbb{R}\times\Bbb{R}$ by uniqueness of the Cauchy problem associated to \eqref{u1u2-0-1}, which together with the fact
$f_{1,u_2}\geq 0$ and $f_{2,u_1}\geq 0$ yields that $(r_1,r_2)\equiv(0,0)$ in $\Bbb{R}\times\Bbb{R}$. In particular,
$$\left(u_1\left(\frac{k\tau}{c},x\right),u_2\left(\frac{k\tau}{c},x\right)\right)=(u_1(0,x),u_2(0,x)),\ \forall x\in\Bbb{R},\ \forall k\in\Bbb{Z}.$$
On the other hand, since for each $x\in\Bbb{R}$,
$$\mathop{\lim}\limits_{k\to-\infty}\left(u_1\left(\frac{k\tau}{c},x\right),u_2\left(\frac{k\tau}{c},x\right)\right)=(0,0),\quad
\mathop{\lim}\limits_{k\to+\infty}\left(u_1\left(\frac{k\tau}{c},x\right),u_2\left(\frac{k\tau}{c},x\right)\right)=(1,1),$$
due to \eqref{limit} and $\tau>0$. One has reached a contradiction. Then $(r_1(t,x),r_2(t,x))>(0,0)$ in $(t,x)\in\Bbb{R}\times\Bbb{R}$, and
therefore $(U_1(x,z+\tau),U_2(x,z+\tau))>(U_1(x,z),U_2(x,z))$ for any $(x,z)\in\Bbb{R}\times\Bbb{R}$ and $\tau>0$. The proof is complete.
\end{proof}
\begin{remark}\rm
The pulsating fronts established in Theorem \ref{main} are leftward propagating. Under assumption that $c_{1+}^*+c_{2-}^*>0$,
where $c_{1+}^*$ and $c_{2-}^*$ are the rightward and leftward spreading speeds of \eqref{sca-2} and \eqref{sca-1}, respectively,
we can similarly obtain a rightward pulsating front $\bm{V}(x,x-ct)=\bm{V}(x,z)$ connecting $\bm{1}$ to $\bm{0}$,
which turned out to be decreasing in $z$ for any $x\in\Bbb{R}$.
\end{remark}
\section{Sub- and supersolutions}
In this section, we first give a comparison lemma to an auxiliary system,
which admits the comparison principle in a larger interval and coincides with system \eqref{u1u2-0-1}
in $[\bm{0},\bm{1}]$, since the sub- and supersolutions constructed later
may be unbounded from below by $\bm{0}$ and above by $\bm{1}$.
Then we construct a pair of sub-super solutions preparing for the next section.
Consider the following auxiliary system
\begin{equation}\label{F1F2}
\begin{cases}
\frac{\partial u_1}{\partial t}=d_1(x)\frac{\partial u_1^2}{\partial x^2}-a_1^*(x)\frac{\partial u_1}{\partial x}+F_1(x,u_1,u_2),\\
\frac{\partial u_2}{\partial t}=d_2(x)\frac{\partial u_2^2}{\partial x^2}-a_2^*(x)\frac{\partial u_2}{\partial x}+F_2(x,u_1,u_2),
\end{cases}
t\in\Bbb{R}^+,\ x\in\Bbb{R},
\end{equation}
where for any $(u_1,u_2)\in[\bm{-1},\bm{2}]$,
\begin{align*}
F_1(x,u_1,u_2)&:=f_1(x,u_1,u_2)+D_1\{u_1\}^-u_2,\\
F_2(x,u_1,u_2)&:=f_2(x,u_1,u_2)+D_2\{1-u_2\}^-(u_1-1),
\end{align*}
$\{w\}^-:=\max\left\{-w,0\right\}$, $D_1=\mathop{\max}\limits_{x\in[0,L]}a_{12}^*(x)$ and $D_2=\mathop{\max}\limits_{x\in[0,L]}a_{21}^*(x)$.
We first give the definition of sub- and supersolutions of \eqref{F1F2} as following.
\begin{definition}
A pair of continuous functions $\bm{w}(t,x)=(u_1(t,x),u_2(t,x))$ is said to be a supersolution (resp. subsolution)
of \eqref{F1F2} in $(t,x)\in\Bbb R^+\times\Bbb R$,
if $\bm{w}\in C^{1,2}((0,\infty)\times\Bbb{R},\Bbb{R}^2)$ and
\begin{equation*}
\begin{cases}
u_{1,t}-d_1(x)\frac{\partial u_1^2}{\partial x^2}+a_1^*(x)\frac{\partial u_1}{\partial x}-F_1(x,u_1,u_2)\geq 0 ~(resp. \leq0),~~(t,x)\in\Bbb R^+\times\Bbb R,\\
u_{2,t}-d_2(x)\frac{\partial u_2^2}{\partial x^2}+a_2^*(x)\frac{\partial u_2}{\partial x}-F_2(x,u_1,u_2)\geq 0 ~(resp. \leq0),~~(t,x)\in\Bbb R^+\times\Bbb R.
\end{cases}
\end{equation*}
\end{definition}
Similar to \cite[Theorem 2.2 and Corollary 2.3]{wang2012} (see also \cite[Theorem 3.2]{bao2013}),
we have the following comparison lemma.
\begin{lemma}\label{sub-super method}
\rm(i)
Suppose that $\bm{w}^+(t,x)$ and $\bm{w}^-(t,x)$ are super- and subsolutions of \eqref{F1F2} in
$(t,x)\in\Bbb R^+\times\Bbb R$, respectively, $\bm{-1}\leq\bm{w}^\pm(t,x)\leq\bm{2}$.
If $\bm{w}^-(0,x)\leq\bm{w}^+(0,x)$ for any $x\in\Bbb R$,
then $\bm{w}^-(t,x)\leq\bm{w}^+(t,x)$ for any $(t,x)\in\Bbb R^+\times\Bbb R$.
\rm(ii)
For any $\bm{w}_0(\cdot)\in C(\Bbb{R},[\bm{0},\bm{1}])$, if
$\bm{w}^-(0,\cdot)\leq\bm{w}_0(\cdot)\leq\bm{w}^+(0,\cdot)$ for any $x\in\Bbb R$,
then
$\bm{w}^-(t,x)\leq\bm{w}(t,x;\bm{w}_0)\leq\bm{w}^+(t,x)$ and
$\bm{0}\leq\bm{w}(t,x;\bm{w}_0)\leq\bm{1}$ for any $(t,x)\in\Bbb R^+\times\Bbb R$,
where
$\bm{w}(t,x;\bm{w}_0)$ is the unique classical solution of \eqref{F1F2} with $\bm{w}(0,x;\bm{w}_0)=\bm{w}_0$.
\rm(iii)
For any $\bm{w}_0(\cdot)\in C(\Bbb{R},[\bm{0},\bm{1}])$, the unique classical solution
$\bm{w}(t,x;\bm{w}_0)$ of \eqref{F1F2} with $\bm{w}(0,x;\bm{w}_0)=\bm{w}_0$ is also a
classical solution of \eqref{u1u2-0-1}.
\end{lemma}
Now we are ready to construct a pair of appropriate sub- and supersolutions of \eqref{F1F2} using the pulsating front
$\bm{U}(x,x+ct)=(U_1(x,x+ct),U_2(x,x+ct))$.
Let $\chi$ be a smooth function satisfying $\chi(\xi)=0$ for any $\xi\leq-2$,
$\chi(\xi)=1$ for any $\xi\geq2$, $0\leq\chi^\prime\leq1$ and $|\chi^{\prime\prime}|\leq1$.
Define a positive $L$-periodic vector $\bm{p}(x,\xi)=(p_1(x,\xi),p_2(x,\xi))$ with
\begin{align*}
p_1(x,\xi)&=(1-\chi(\xi))\phi_{01}^*(x)+\chi(\xi)\phi_{11}^*(x),~(x,\xi)\in\Bbb R^2,\\
p_2(x,\xi)&=(1-\chi(\xi))\phi_{02}^*(x)+\chi(\xi)\phi_{12}^*(x),~(x,\xi)\in\Bbb R^2,
\end{align*}
where $(\phi_{i1}^*(x),\phi_{i2}^*(x))~(i=0,1)$ are defined in Lemma \ref{0-1-eigen-eigen}.
Furthermore, denote
$$\rho^*=\max_{i,j=0,1}\left\{\mathop{\max}\limits_{x\in[0,L]}\phi^*_{ij}(x)\right\},
\quad\rho_*=\min_{i,j=0,1}\left\{\mathop{\min}\limits_{x\in[0,L]}\phi^*_{ij}(x)\right\}.$$
\begin{lemma}\label{TWS-sub-super}
There exist positive constants $\beta_0,~\delta_0$ and $\sigma_0$ such that for any $z^\pm\in\Bbb R$ and $\delta\in(0,\delta_0]$,
the functions $\bm{u}^\pm(t,x)=(u_1^\pm(t,x),u_2^\pm(t,x))$ defined by
\begin{equation*}
u_i^\pm(t,x)=U_i(x,x+ct+z^\pm\pm\sigma_0\delta(1-e^{-\beta_0 t}))
\pm\delta p_i(x,x+ct+z^\pm\pm\sigma_0\delta(1-e^{-\beta_0 t}))e^{-\beta_0 t},~i=1,2
\end{equation*}
are a pair of super- and subsolutions of \eqref{F1F2} for any $(t,x)\in(0,+\infty)\times\Bbb R$.
\end{lemma}
\begin{proof}
The proof is analogous to that of \cite[Lemma 3.4]{bao2013}, for the completeness of the present paper, we give the details.
Let $\xi(t,x)=x+ct+z^++\sigma_0\delta(1-e^{-\beta_0 t})$. Then
\begin{align*}
&(u_1^+(t,x),u_2^+(t,x))=\left(U_1(x,\xi)+\delta p_1(x,\xi)e^{-\beta_0 t},~U_2(x,\xi)+\delta p_2(x,\xi)e^{-\beta_0 t}\right)>(0,0),\\
&F_1(x,u_1^+,u_2^+)=f_1(x,u_1^+,u_2^+),\quad F_2(x,u_1^+,u_2^+)=f_2(x,u_1^+,u_2^+)+D_2\{1-u_2^+\}^-(u_1^+-1).
\end{align*}
We only prove that $u^+_{2,t}-L_2^*u_2^+-F_2(x,u_1^+,u_2^+)\geq0$, since the others can be proved similarly.
Direct calculation shows that
\begin{align*}
&u^+_{2,t}-d_2u^+_{2,xx}+a_2^*u^+_{2,x}-F_2(x,u_1^+,u_2^+)\\
&=\delta e^{-\beta_0 t}\left[(c+\beta_0\sigma_0\delta e^{-\beta_0 t})p_{2,\xi}-\beta_0p_2
-d_2(p_{2,xx}+2p_{2,x\xi}+p_{2,\xi\xi})+a_2^*(p_{2,x}+p_{2,\xi})\right]\\
&\quad+\beta_0\sigma_0\delta U_{2,z}e^{-\beta_0 t}+f_2(x,U_1,U_2)-f_2(x,u_1^+,u_2^+)\\
&\quad-D_2\{1-u^+_2\}^-(u^+_1-1)\\
&=\delta e^{-\beta_0 t}\left[(c+\beta_0\sigma_0\delta e^{-\beta_0 t})p_{2,\xi}-\beta_0p_2
-d_2(p_{2,xx}+2p_{2,x\xi}+p_{2,\xi\xi})+a_2^*(p_{2,x}+p_{2,\xi})\right]\\
&\quad+\beta_0\sigma_0\delta U_{2,z}e^{-\beta_0 t}-
\left[\int_0^1{f_{2,u_1}\left(x,\widetilde{U}_1(x,\xi;\theta),\widetilde{U}_2(x,\xi;\theta)\right)d\theta}\right]
\delta p_1(x,\xi)e^{-\beta_0 t}\\
&\quad-\left[\int_0^1{f_{2,u_2}\left(x,\widetilde{U}_1(x,\xi;\theta),\widetilde{U}_2(x,\xi;\theta)\right)d\theta}\right]
\delta p_2(x,\xi)e^{-\beta_0 t}\\
&\quad-D_2\{1-u^+_2\}^-(u^+_1-1),
\end{align*}
where $\widetilde{U}_i(x,\xi;\theta):=U_i(x,\xi)+\theta\delta p_i(x,\xi)e^{-\beta_0t}$, $\theta\in(0,1)$, $~i=1,2$.
Denote
\begin{align*}
\Gamma_0(x,\xi;\theta)&=\mathop{\sum}\limits_{i,j=1,2}\left|f_{i,u_j}(x,\widetilde{U}_1,\widetilde{U}_2)-f_{i,u_j}(x,0,0)\right|,\\
\Gamma_1(x,\xi;\theta)&=\mathop{\sum}\limits_{i,j=1,2}\left|f_{i,u_j}(x,\widetilde{U}_1,\widetilde{U}_2)-f_{i,u_j}(x,1,1)\right|.
\end{align*}
Noting that
$$\mathop{\lim}\limits_{\xi\to-\infty}(U_1(x,\xi),U_2(x,\xi))=(0,0),\
\mathop{\lim}\limits_{\xi\to+\infty}(U_1(x,\xi),U_2(x,\xi))=(1,1)\text{~~uniformly for~~}x\in\Bbb R.$$
Then there exist some $\hat{\xi}>2$ large enough and $\delta_1\in(0,1)$ small enough such that
for any $\theta\in(0,1)$ and $\delta\in(0,\delta_1)$, it follows
\begin{align*}
\mathop{\sup}\limits_{(x,\xi)\in\Bbb R\times(-\infty,-\hat{\xi}]}
\left|\Gamma_0(x,\xi;\theta)\right|&\leq\frac{|\mu_0^-|
\min\left\{\mathop{\min}\limits_{x\in[0,L]}\phi^*_{01},\mathop{\min}\limits_{x\in[0,L]}\phi^*_{02}\right\}}
{2\mathop{\max}\limits_{x\in[0,L]}(\phi^*_{01}+\phi^*_{02})},\\
\mathop{\sup}\limits_{(x,\xi)\in\Bbb R\times[\hat{\xi},+\infty)}
\left|\Gamma_1(x,\xi;\theta)\right|&\leq\frac{|\mu_1^-|
\min\left\{\mathop{\min}\limits_{x\in[0,L]}\phi^*_{11},\mathop{\min}\limits_{x\in[0,L]}\phi^*_{12}\right\}}
{2\mathop{\max}\limits_{x\in[0,L]}(\phi^*_{11}+\phi^*_{12})}.
\end{align*}
Denote
\begin{align*}
d&=\max\left\{\mathop{\max}\limits_{x\in[0,L]}d_1(x), \mathop{\max}\limits_{x\in[0,L]}d_2(x)\right\},\quad
a^*=\max\left\{\mathop{\max}\limits_{x\in[0,L]}|a_1^*(x)|, \mathop{\max}\limits_{x\in[0,L]}|a_2^*(x)|\right\},\\
D&=\max\{D_1, D_2\},\quad
C_1=\max\left\{|p_1|_{C^{2,2}([0,L]\times\Bbb R)}, |p_2|_{C^{2,2}([0,L]\times\Bbb R)}\right\},\\
C_2&=\mathop{\sup}\limits_{(u_1,u_2)\in[\bm{-1},\bm{2}],~x\in\Bbb R}
\left\{\sum_{i,j=1,2}|f_{i,u_j}(x,u_1,u_2)|\right\},\
C_3=\min_{i=1,2}\left\{\mathop{\min}\limits_{x\in[0.L],~\xi\in[-\hat{\xi},\hat{\xi}]}
\frac{\partial U_i}{\partial z}\right\}.
\end{align*}
For any fixed $\beta_0\in\left(0,\min\left\{\frac{|\mu_0^-|}{4},\frac{|\mu_1^-|}{4}\right\}\right)$, let
$$\delta_0=\min\left\{\delta_1, \frac{|\mu_0^-|}{4D\rho^*}, \frac{|\mu_1^-|}{4D\rho^*}, \frac{C_3}{2C_1}\right\},\quad
\sigma_0\geq\frac{2C_1[c+\beta_0+4d+2a^*+\delta_0C_1D+C_2]}{\beta_0C_3}.$$
Consider the following three cases.
\begin{description}
\item[Case~1.] $\xi(t,x)\geq\hat{\xi}$.
Then $p_1(x,\xi)=\phi_{11}^*(x)$ and $p_2(x,\xi)=\phi_{12}^*(x)$, and
\begin{align*}
&\mathcal{F}_2(x,u_1^+,u_2^+)\\
&\geq\delta e^{-\beta_0 t}\left[-\beta_0\phi_{12}^*
-d_2(\phi_{12}^*)^{\prime\prime}+a_2^*(\phi_{12}^*)^{\prime}\right]\\
&\quad-\left[\int_0^1{f_{2,u_1}\left(x,\widetilde{U}_1(x,\xi;\theta),\widetilde{U}_2(x,\xi;\theta)\right)d\theta}\right]
\delta e^{-\beta_0 t}\cdot\phi_{11}^*\\
&\quad-\left[\int_0^1{f_{2,u_2}\left(x,\widetilde{U}_1(x,\xi;\theta),\widetilde{U}_2(x,\xi;\theta)\right)d\theta}\right]
\delta e^{-\beta_0 t}\cdot\phi_{12}^*\\
&\quad-D_2\{1-u^+_2\}^-(u^+_1-1)\\
&=\delta e^{-\beta_0 t}\left\{\int_0^1\left(f_{2,u_1}(x,1,1)-
f_{2,u_1}(x,\widetilde{U}_1(x,\xi;\theta),\widetilde{U}_2(x,\xi;\theta))\right)d\theta\cdot\phi_{11}^*\right .\\
&\quad+\int_0^1\left(f_{2,u_2}(x,1,1)
\left .-f_{2,u_2}(x,\widetilde{U}_1(x,\xi;\theta),\widetilde{U}_2(x,\xi;\theta))\right)d\theta\cdot\phi_{12}^*
-(\mu_1^-+\beta_0)\phi_{12}^*\right\}\\
&\quad-D_2\{1-u^+_2\}^-(u^+_1-1)\\
&\geq\delta e^{-\beta_0 t}\phi_{12}^*
\left[-\frac{|\mu_1^-|}{2}-\mu_1^--\beta_0-D_2\delta\phi_{11}^*\right]\\
&\geq0.
\end{align*}
\item[Case~2.] $|\xi(t,x)|\leq\hat{\xi}$.
Then for any $\delta\in(0,\delta_0]$,
\begin{align*}
&\mathcal{F}_2(x,u_1^+,u_2^+)\\
&\geq-\delta e^{-\beta_0 t}\left[(c+\beta_0\sigma_0\delta)C_1+\beta_0C_1
+4dC_1+2a^*C_1\right]\\
&\quad+\beta_0\sigma_0\delta C_3e^{-\beta_0 t}
-C_2\delta C_1e^{-\beta_0 t}
-D\delta C_1e^{-\beta_0 t}\delta C_1e^{-\beta_0 t}\\
&\geq\delta e^{-\beta_0 t}
\left[\beta_0\sigma_0( C_3-\delta C_1)-C_1(c+\beta_0+4d+2a^*+C_2+D\delta_0C_1)\right]\\
&\geq0.
\end{align*}
\item[Case~3.] $\xi(t,x)\leq-\hat{\xi}$.
Then $p_1(x,\xi)=\phi_{01}^*(x)$ and $p_2(x,\xi)=\phi_{02}^*(x)$. Similar to the proof of
$\bm{Case~1}$, one can prove that $\mathcal{F}_2(x,u_1^+,u_2^+)\geq0$.
\end{description}
Consequently, $\mathcal{F}_2(x,u_1^+,u_2^+)\geq0$ for any $(t,x)\in(0,+\infty)\times\Bbb R$.
With analogous arguments as above, we complete the proof.
\end{proof}
\section{Global stability and uniqueness}
This section is devoted to the study of global stability and uniqueness of the pulsating
front $\bm{U}(x,x+ct)=(U_1(x,x+ct),U_2(x,x+ct))$ with nonzero wave speed.
Without loss of generality, we assume that the wave speed $c>0$, the case $c<0$ can be discussed similarly.
By using the convergence theorem for monotone semiflows (see \cite[Theorem 2.2.4]{zhao2003}),
we show that the solution of the Cauchy problem associated to \eqref{u1u2-0-1} with some proper initial values
converges to a translation of the pulsating front as $t\to+\infty$.
The general strategy of the proof is to trap the solution of a Cauchy problem related to \eqref{u1u2-0-1} between suitable sub- and supersolutions
established in Section 3 which are close to some shifts of the periodic traveling wave $\bm{U}(x,x+ct)=(U_1(x,x+ct),U_2(x,x+ct))$.
The uniqueness result can then be viewed as a consequence of the global stability.
Consider the moving coordinates
$$(t,z)=(t,x+ct).$$
By rescaling system \eqref{u1u2-0-1} into the moving coordinates $(t,z)$, that is, let
$$\bm{v}(t,z)=(v_1(t,z),v_2(t,z))=(v_1(t,x+ct),v_2(t,x+ct)),$$
we transform \eqref{u1u2-0-1} into the following system
\begin{equation}\label{TWS-t-z}
\begin{cases}
\frac{\partial v_1(t,z)}{\partial t}=d_1(z-ct)\frac{\partial^2 v_1}{\partial z^2}
-(a_1^*(z-ct)+c)\frac{\partial v_1}{\partial z}+f_1(z-ct,v_1,v_2),\\
\frac{\partial v_2(t,z)}{\partial t}=d_2(z-ct)\frac{\partial^2 v_2}{\partial z^2}
-(a_2^*(z-ct)+c)\frac{\partial v_2}{\partial z}+f_2(z-ct,v_1,v_2),
\end{cases}
t>0,~z\in\Bbb R.
\end{equation}
Then \eqref{TWS-t-z} is a time periodic system with period $T=\frac{L}{c}>0$.
Define $Q_t(\bm{v}_0):=\bm{v}(t,\cdot;\bm{v}_0)$ for any $\bm{v}_0\in\mathcal C_{\bm{1}}$ and $t\geq0$, where
$\bm{v}(t,\cdot;\bm{v}_0)$ is the unique solution of \eqref{TWS-t-z} with initial value $\bm{v}_0$.
Let $P :\mathcal C_{\bm{1}}\to \mathcal C_{\bm{1}}$ be the Poincar$\acute{e}$ map associated with the
periodic semiflow $Q_t(\cdot)$, that is, $P(\bm{v}_0)=Q_T(\bm{v}_0)=\bm{v}(T,\cdot;\bm{v}_0)$.
For any $\tau\in\Bbb{R}$, since the pulsating front $\bm{U}(x,z+\tau)=\bm{U}(x,x+ct+\tau)$ is a solution of system \eqref{u1u2-0-1},
it follows that $$\bm{V}^\tau(t,z):=\bm{U}(z-ct,z+\tau)$$ is a $T$-periodic solution of \eqref{TWS-t-z} satisfying $\bm{V}^{\tau_1}(0,\cdot)>\bm{V}^{\tau_2}(0,\cdot)$ for all $\tau_1>\tau_2$ by the strictly monotonicity of $\bm{U}(\cdot,z)$ with respect to $z$.
Furthermore, it can be seen that $\bm{0}$, $\bm{1}$ and $\bm{V}^\tau(0,\cdot)$ are fixed points of $P$ in $\mathcal C_{\bm{1}}$, with
the set $\{\bm{V}^\tau(0,\cdot)|\tau\in\Bbb{R}\}$ totally ordered in $\mathcal C_{\bm{1}}$.
Next we shall apply the following convergence lemma to the Poincar$\acute{e}$ map $P$ and
its fixed points $\{\bm{V}^\tau(0,\cdot)|\tau\in\Bbb{R}\}$.
\begin{lemma}[\rm{\cite[Theorem 2.2.4]{zhao2003}}]\label{convergence}
Let $M$ be a closed convex subset of an order Banach space $\mathcal{X}$, and $\Phi :M\to M$ be a monotone semiflow.
Assume that there exists a monotone homeomorphism $\zeta$ from $[0,1]$ onto a subset of $M$ such that
\begin{itemize}
\item[(1)] For each $s\in[0,1]$, $\zeta(s)$ is a stable equilibrium for $\Phi :M\to M$.
\item[(2)] Each orbit of $\Phi$ in $[\zeta(0),\zeta(1)]$ is precompact.
\item[(3)] One of the following two properties holds:
\begin{itemize}
\item[(a)] if $\zeta(s_0)<_\mathcal{X}\omega(\phi)$ for some $s_0\in[0,1)$ and $\phi\in\mathcal{X}_{[\zeta(0),\zeta(1)]}$,
then there exists $s_1\in(s_0,1)$ such that $\zeta(s_1)\leq_\mathcal{X}\omega(\phi)$,
here $\omega(\phi)$ is the $\omega$-limit set of $\phi$ in $\Phi$;
\item[(b)] if $\omega(\phi)<_\mathcal{X}\zeta(r_1)$ for some $r_1\in(0,1]$ and $\phi\in\mathcal{X}_{[\zeta(0),\zeta(1)]}$,
then there exists $r_0\in(0,r_1)$ such that $\omega(\phi)\leq_\mathcal{X}\zeta(r_0)$.
\end{itemize}
\end{itemize}
Then for any precompact orbit $\gamma^+(\phi_0)$ of $\Phi$ in $M$
with $\omega(\phi_0)\cap[\zeta(0),\zeta(1)]_{\mathcal{X}}\neq\emptyset$, there exists
$s_*\in[0,1]$ such that $\omega(\phi_0)=\zeta(s_*)$.
\end{lemma}
Next we provide two lemmas for the preparation of using Lemma \ref{convergence}.
\begin{lemma}\label{G1}
Assume that $\bm{v}_0\in\mathcal C_{\bm{1}}$ satisfies
\begin{equation}\label{initial}
\mathop{\lim\sup}\limits_{z\to-\infty}\bm{v}_0(z)\ll\delta_1\rho_*\bm{1},\quad
\mathop{\lim\inf}\limits_{z\to+\infty}\bm{v}_0(z)\gg\bm{1}-\delta_1\rho_*\bm{1},
\end{equation}
where $\delta_1=\min\left\{\delta_0, \frac{1}{\rho_*}\right\}$ with $\delta_0$ defined in Lemma \ref{TWS-sub-super}.
Then for any $\varepsilon>0$, there exist some
$\hat{\tau}=\hat{\tau}(\varepsilon,\bm{v}_0)>0$ and an integer $\hat{k}=\hat{k}(\varepsilon,\bm{v}_0)>0$ such that
$$\bm{V}^{-\hat{\tau}}(0,z)-\varepsilon\bm{1}\leq\bm{v}(\hat{k}T,z;\bm{v}_0)
\leq\bm{V}^{\hat{\tau}}(0,z)+\varepsilon\bm{1},\ \ \forall z\in\Bbb{R}.$$
\end{lemma}
\begin{proof}
Let $\delta^{\prime}\in(0,\delta_1)$ be such that
\begin{equation}\label{ini}
\mathop{\lim\sup}\limits_{z\to-\infty}\bm{v}_0(z)<\delta^{\prime}\rho_*\bm{1},\quad
\mathop{\lim\inf}\limits_{z\to+\infty}\bm{v}_0(z)>\bm{1}-\delta^{\prime}\rho_*\bm{1}.
\end{equation}
Define a pair of $\tilde{\bm{v}}^{\pm}(t,z)=(\tilde{v}_1^{\pm}(t,z),\tilde{v}_2^{\pm}(t,z))$ as
$$\tilde{v}_i^\pm(t,z)=U_i(z-ct,z+z^\pm\pm\sigma_0\delta(1-e^{-\beta_0 t}))
\pm\delta p_i(z-ct,z+z^\pm\pm\sigma_0\delta(1-e^{-\beta_0 t}))e^{-\beta_0 t},\ i=1,2.$$
Then Lemma \ref{TWS-sub-super} implies that $\tilde{\bm{v}}^{\pm}(t,z)$ are sub- and supersolutions of
\begin{equation*}
\begin{cases}
\frac{\partial v_1(t,z)}{\partial t}=d_1(z-ct)\frac{\partial^2 v_1}{\partial z^2}
-(a_1^*(z-ct)+c)\frac{\partial v_1}{\partial z}+F_1(z-ct,v_1,v_2),\\
\frac{\partial v_2(t,z)}{\partial t}=d_2(z-ct)\frac{\partial^2 v_2}{\partial z^2}
-(a_2^*(z-ct)+c)\frac{\partial v_2}{\partial z}+F_2(z-ct,v_1,v_2),
\end{cases}
t>0,~z\in\Bbb R,
\end{equation*}
By \eqref{ini}, there exist $z^+>0$ and $-z^->0$ large enough such that for $\delta\in(\delta^{\prime},\delta_1)$,
$$\bm{U}(z,z+z^-)-\delta\bm{p}(z,z+z^-)\leq\bm{v}_0(z)\leq\bm{U}(z,z+z^+)+\delta\bm{p}(z,z+z^+),\ \ \forall z\in\Bbb{R}.$$
It then follows from the comparison principle that
$$\tilde{\bm{v}}^-(t,z)\leq\bm{v}(t,z;\bm{v}_0)\leq\tilde{\bm{v}}^+(t,z),\ \ \forall t>0,\ z\in\Bbb{R},$$
that is, for any $t>0$, we have
\begin{align*}
&\bm{U}(z-ct,z+z^--\sigma_0\delta(1-e^{-\beta_0 t}))-\delta\bm{p}(z-ct,z+z^--\sigma_0\delta(1-e^{-\beta_0 t}))e^{-\beta_0 t}\\
&\leq\bm{v}(t,z;\bm{v}_0)\\
&\leq\bm{U}(z-ct,z+z^++\sigma_0\delta(1-e^{-\beta_0 t}))+\delta\bm{p}(z-ct,z+z^++\sigma_0\delta(1-e^{-\beta_0 t}))e^{-\beta_0 t}.
\end{align*}
Now for any $\varepsilon>0$, letting $t=\hat{k}T>0$ in the above inequalities,
where $\hat{k}>0$ is a large enough integer satisfying $\delta\rho^*e^{-\beta_0\hat{k}T}\leq\varepsilon$.
It follows that
\begin{align*}
&\bm{U}(z,z+z^--\sigma_0\delta(1-e^{-\beta_0\hat{k}T}))
-\delta\bm{p}(z,z+z^--\sigma_0\delta(1-e^{-\beta_0\hat{k}T}))e^{-\beta_0\hat{k}T}\\
&\leq\bm{v}(\hat{k}T,z;\bm{v}_0)\\
&\leq\bm{U}(z,z+z^++\sigma_0\delta(1-e^{-\beta_0\hat{k}T}))
+\delta\bm{p}(z,z+z^++\sigma_0\delta(1-e^{-\beta_0\hat{k}T}))e^{-\beta_0\hat{k}T}.
\end{align*}
Let $\hat{\tau}:=\max\{z^+,-z^-\}+\sigma_0\delta>0$, we further obtain
$$\bm{U}(z,z-\hat{\tau})-\varepsilon\bm{1}\leq\bm{v}(\hat{k}T,z;\bm{v}_0)\leq\bm{U}(z,z+\hat{\tau})+\varepsilon\bm{1},
\ \ \forall z\in\Bbb{R},$$
that is,
$$\bm{V}^{-\hat{\tau}}(0,z)-\varepsilon\bm{1}\leq\bm{v}(\hat{k}T,z;\bm{v}_0)
\leq\bm{V}^{\hat{\tau}}(0,z)+\varepsilon\bm{1},\ \ \forall z\in\Bbb{R}.$$
The proof is complete.
\end{proof}
\begin{lemma}\label{lyapunov}
For any $\tau\in\Bbb{R}$, the wave profile $\bm{V}^\tau(0,\cdot)$ is lyapunov stable for system \eqref{TWS-t-z},
that is, for any $\epsilon>0$, there exists some constant $\rho>0$ such that for any
$\bm{v}_0\in\mathcal C_{\bm{1}}$ satisfying $\lVert\bm{v}_0(\cdot)-\bm{V}^\tau(0,\cdot)\rVert_{\mathcal C}\leq\rho$,
there is $$\lVert\bm{v}(t,\cdot;\bm{v}_0)-\bm{V}^\tau(t,\cdot)\rVert_{\mathcal C}\leq\epsilon,\ \ \forall t>0.$$
\end{lemma}
\begin{proof}
For any $\epsilon>0$, let
$$\delta\in\left(0,\min\left\{\frac{\epsilon}{\rho^*+\sigma_0 K_0},\delta_0\right\}\right),\quad\rho:=\delta\rho_*,$$
where $\sigma_0$ and $\delta_0$ are defined in Lemma \ref{TWS-sub-super},
and $K_0:=\mathop{\max}\limits_{x\in[0,L]}\lVert\bm{U}_z(x,\cdot)\rVert_{\mathcal C}$.
Then for any $\bm{v}_0\in\mathcal C_{\bm{1}}$ with $\lVert\bm{v}_0(\cdot)-\bm{V}^\tau(0,\cdot)\rVert_{\mathcal C}\leq\rho$,
we have
$$\bm{U}(z,z+\tau)-\delta\bm{p}(z,z+\tau)\leq\bm{v}_0(z)\leq\bm{U}(z,z+\tau)+\delta\bm{p}(z,z+\tau),\ \ \forall z\in\Bbb{R}.$$
The comparison principle then yields that
$$\tilde{\bm{w}}^-(t,z)\leq\bm{v}(t,z;\bm{v}_0)\leq\tilde{\bm{w}}^+(t,z),\ \ \forall t\geq0,\ z\in\Bbb{R},$$
where $\tilde{\bm{w}}^\pm(t,z)$ are defined as in Lemma \ref{TWS-sub-super} with $z^\pm=\tau$.
Therefore,
\begin{align*}
&\mathop{\sup}\limits_{t\geq0}\lVert\bm{v}(t,z;\bm{v}_0)-\bm{U}(z-ct,z+\tau)\rVert_{\mathcal C}\\
&\leq\mathop{\sup}\limits_{t\geq0}\lVert\tilde{\bm{w}}^{\pm}(t,z)-\bm{U}(z-ct,z+\tau)\rVert_{\mathcal C}\\
&\leq\mathop{\sup}\limits_{t\geq0}\lVert\tilde{\bm{w}}^{\pm}(t,z)-\bm{U}(z-ct,z+\tau\pm\sigma_0\delta(1-e^{-\beta_0 t}))\rVert_{\mathcal C}\\
&\quad+\mathop{\sup}\limits_{t\geq0}\lVert\bm{U}(z-ct,z+\tau\pm\sigma_0\delta(1-e^{-\beta_0 t}))-\bm{U}(z-ct,z+\tau)\rVert_{\mathcal C}\\
&\leq\delta\rho^*+\sigma_0\delta K_0\\
&\leq\epsilon,
\end{align*}
that is, $\lVert\bm{v}(t,\cdot;\bm{v}_0)-\bm{V}^\tau(t,\cdot)\rVert_{\mathcal C}\leq\epsilon$ for any $t>0$.
The proof is complete.
\end{proof}
Now we are ready to prove the globally stability of the pulsating front of \eqref{u1u2-0-1}.
Consider the following periodic initial problem associated with system \eqref{u1u2-0-1}
\begin{equation}\label{u1u2-cauchy}
\begin{cases}
\frac{\partial u_1}{\partial t}=L_1^*u_1+u_1\left[a_{11}^*(x)(1-u_1)-a_{12}^*(x)(1-u_2)\right],
\quad t>0,~x\in\Bbb R,\\
\frac{\partial u_2}{\partial t}=L_2^*u_2+(1-u_2)\left[a_{21}^*(x)u_1-a_{22}^*(x)u_2\right],
\quad t>0,~x\in\Bbb R,\\
\bm{u}(0,x)=\bm{u}_0(x),\quad x\in\Bbb R.
\end{cases}
\end{equation}
\begin{theorem}\label{GS}
Let $\bm{U}(x,x+ct)=(U_1(x,x+ct),U_2(x,x+ct))$ be the pulsating front of system \eqref{u1u2-0-1} connecting $\bm{0}$ to $\bm{1}$
with $c\neq0$,
and let $\bm{u}(t,x;\bm{u}_0)$ be the solution of \eqref{u1u2-cauchy} with the initial value $\bm{u}_0\in\mathcal C_{\bm{1}}$.
Then for any $\bm{u}_0$ satisfying \eqref{initial}, there exists $\tilde{\tau}\in\Bbb{R}$ such that
$$\mathop{\lim}\limits_{t\to+\infty}
\lVert\bm{v}(t,\cdot;\bm{u}_0)-\bm{V}^{c\tilde{\tau}}(t,\cdot)\rVert_{\mathcal C}=0,$$
that is,
$$\mathop{\lim}\limits_{t\to+\infty}
\lVert\bm{u}(t,\cdot;\bm{u}_0)-\bm{U}(x,x+ct+c\tilde{\tau})\rVert_{\mathcal C}=0.$$
\end{theorem}
\begin{proof}
Recall that $P :\mathcal C_{\bm{1}}\to \mathcal C_{\bm{1}}$ is the Poincar$\acute{e}$ map associated with the periodic
semiflow $Q_t(\cdot)$ generated by \eqref{TWS-t-z}, that is, $P(\bm{u}_0)=Q_T(\bm{u}_0)=\bm{v}(T,\cdot;\bm{u}_0)$,
where $\bm{v}(t,\cdot;\bm{u}_0)$ is the unique solution of \eqref{TWS-t-z} with initial value $\bm{u}_0$.
Then $P^n(\bm{u}_0)=\bm{v}(nT,\cdot;\bm{u}_0)$ for any $n\geq0$.
Since $\bm{u}_0\in\mathcal C_{\bm{1}}$ satisfies \eqref{initial},
Lemma \ref{G1} then yields that for any $\delta\in(0,\delta_0)$, there exist some
$\hat{\tau}>0$ and an integer $\hat{k}>0$ such that
$$\bm{V}^{-\hat{\tau}}(0,z)-\delta\rho_*\bm{1}\leq\bm{v}(\hat{k}T,z;\bm{u}_0)=P^{\hat{k}}(\bm{u}_0)(z)
\leq\bm{V}^{\hat{\tau}}(0,z)+\delta\rho_*\bm{1},\ \ \forall z\in\Bbb{R}.$$
By the comparison principle, for any $t\geq0$ and $z\in\Bbb{R}$, we have
\begin{equation}\label{T}
\begin{aligned}
&\bm{U}(z-ct,z-\hat{\tau}-\sigma_0\delta(1-e^{-\beta_0 t}))-\delta\bm{p}(z-ct,z-\hat{\tau}-\sigma_0\delta(1-e^{-\beta_0 t}))e^{-\beta_0 t}\\
&\leq\bm{v}(t+\hat{k}T,z;\bm{u}_0)\\
&\leq\bm{U}(z-ct,z+\hat{\tau}+\sigma_0\delta(1-e^{-\beta_0 t}))+\delta\bm{p}(z-ct,z+\hat{\tau}+\sigma_0\delta(1-e^{-\beta_0 t}))e^{-\beta_0 t}.
\end{aligned}
\end{equation}
Noting that the sequence $\{P^n(\bm{u}_0)\}_{n\geq1}=\{\bm{v}(nT,\cdot;\bm{u}_0)\}_{n\geq1}$ is bounded in
$C^1(\Bbb{R},\Bbb{R}^2)$ for any $\bm{u}_0\in\mathcal C_{\bm{1}}$ by the standard parabolic estimates,
and $\mathop{\lim}\limits_{z\to-\infty}\bm{V}^{\tau}(t,z)=\bm{0}$,
$\mathop{\lim}\limits_{z\to+\infty}\bm{V}^{\tau}(t,z)=\bm{1}$ uniformly in $t\in\Bbb{R}$ and for any $\tau\in\Bbb{R}$,
it then follows from \eqref{T} that the forward orbit $\gamma^+(\bm{u}_0):=\{P^n(\bm{u}_0)|n\in\Bbb{N}\}$ is precompact
in $\mathcal C$ and its $\omega$-limit set $\omega(\bm{u}_0)$ is thus nonempty, compact and invariant for $P$.
Denote $\tau_0:=\hat{\tau}+\sigma_0\delta$ and let $t=nT$ in \eqref{T}, we see
$$\omega(\bm{u}_0)\subset H:=\left\{\bm{\phi}\in\mathcal C:\bm{V}^{-\tau_0}(0,\cdot)\leq\bm{\phi}(\cdot)\leq\bm{V}^{\tau_0}(0,\cdot)\right\}.$$
For any $s\in[0,1]$, define $\bm{\zeta}(s)=\bm{V}^{(2s-1)\tau_0}(0,\cdot)$, then
$\bm{\zeta}(s)$ is a monotone homeomorphism from $[0,1]$ onto a subset of $\mathcal C$ and $H=\mathcal C_{[\bm{\zeta}(0),\bm{\zeta}(1)]}$,
and we see from Lemma \ref{lyapunov} that each $\bm{\zeta}(s)$ is a stable fined point of $P$ for $s\in[0,1]$.
Furthermore, $\gamma^+(\bm{\phi})\subset H$ is precompact in $\mathcal C$ for any $\bm{\phi}\in H$.
Now we verify the last condition in Lemma \ref{convergence}.
Suppose that $\bm{\zeta}(s_0)<_{\mathcal C}\omega(\bm{\phi}_0)$ for some $s_0\in[0,1)$ and $\bm{\phi}_0\in H$, that is,
$\bm{V}^{(2s_0-1)\tau_0}(0,\cdot)\leq_{\mathcal C}\bm{\phi}(\cdot)$ and $\bm{V}^{(2s_0-1)\tau_0}(0,\cdot)\neq\bm{\phi}(\cdot)$ for any
$\bm{\phi}\in\omega(\bm{\phi}_0)$. The strong maximum principle then yields that
$\bm{V}^{(2s_0-1)\tau_0}(t,z)<\bm{v}(t,z;\bm{\phi})$ for $(t,z)\in(0,\infty)\times\Bbb{R}$, and particularly,
$$\bm{V}^{(2s_0-1)\tau_0}(0,z)=\bm{V}^{(2s_0-1)\tau_0}(T,z)<P(\bm{\phi})(z)=\bm{\phi}(z),\quad\forall z\in\Bbb{R},\ \forall\bm{\phi}\in\omega(\bm{\phi}_0).$$
Noting that $\bm{V}^{\tau}(0,z)=\bm{U}(z,z+\tau)$ and
$\mathop{\lim}\limits_{z\to\pm\infty}\frac{\partial\bm{U}(x,z)}{\partial z}=\bm{0}$ uniformly for $x\in\Bbb{R}$,
there exists some $z_0>0$ large enough such that
$$0<\bar{\theta}:=\mathop{\sup}\limits_{s,s^{\prime}\in[0,3/2],s\neq s^{\prime},|z|\geq z_0}
\frac{|\bm{V}^{(2s-1)\tau_0}(0,z)-\bm{V}^{(2s^{\prime}-1)\tau_0}(0,z)|}{|s-s^{\prime}|}
\leq\frac{2\rho_*}{5}\mathop{\min}\left\{\frac{\tau_0}{\sigma_0},\delta_0\right\},$$
where $\sigma_0$ and $\delta_0$ are defined in Lemma \ref{TWS-sub-super}.
Since $\omega(\bm{\phi}_0)$ is compact, there exists some $s_1\in(s_0,1)$ such that
$$\bm{V}^{(3s_1-s_0-1)\tau_0}(0,z)\leq\bm{\phi}(z),\quad\forall z\in[-z_0,z_0],\ \bm{\phi}\in\omega(\bm{\phi}_0).$$
For any $\bm{\phi}\in\omega(\bm{\phi}_0)$, let $n_k\to+\infty$ be the sequence such that
$P^{n_k}(\bm{\phi}_0)\to\bm{\phi}$ as $k\to+\infty$. Then there exists an integer $n_{k_0}$ such that
$\lVert P^{n_{k_0}}(\bm{\phi}_0)-\bm{\phi}\rVert_{\mathcal C}\leq\bar{\theta}(s_1-s_0)$. Therefore,
\begin{align*}
&P^{n_{k_0}}(\bm{\phi}_0)(z)-\bm{V}^{(3s_1-s_0-1)\tau_0}(0,z)\\
&=P^{n_{k_0}}(\bm{\phi}_0)(z)-\bm{\phi}(z)+\bm{\phi}(z)-\bm{V}^{(3s_1-s_0-1)\tau_0}(0,z)\\
&\geq-\lVert P^{n_{k_0}}(\bm{\phi}_0)-\bm{\phi}\rVert_{\mathcal C}\bm{1}
-\mathop{\sup}\limits_{|z|\geq z_0}|\bm{V}^{(2s_0-1)\tau_0}(0,z)-\bm{V}^{(3s_1-s_0-1)\tau_0}(0,z)|\bm{1}\\
&\geq-\bar{\theta}(s_1-s_0)\bm{1}-\frac{3\bar{\theta}(s_1-s_0)}{2}\bm{1}=-\frac{5\bar{\theta}(s_1-s_0)}{2}\bm{1}\\
&\geq-\delta_1\bm{p}\left(z,z+(3s_1-s_0-1)\tau_0\right),\quad\forall z\in\Bbb{R},
\end{align*}
where $\delta_1:=\frac{5\bar{\theta}(s_1-s_0)}{2\rho_*}<\delta_0$, the first inequality followed by
$\bm{\phi}(z)-\bm{V}^{(3s_1-s_0-1)\tau_0}(0,z)\geq\bm{0}$ for all $z\in[-z_0,z_0]$ and
$\bm{\phi}(z)>\bm{V}^{(2s_0-1)\tau_0}(0,z)$ for all $|z|\geq z_0$ and $\bm{\phi}\in\omega(\bm{\phi}_0)$.
Lemma \ref{TWS-sub-super} then implies that
\begin{align*}
\bm{v}(t,z;P^{n_{k_0}}(\bm{\phi}_0))&\geq\bm{U}(z-ct,z+(3s_1-s_0-1)\tau_0-\sigma_0\delta_1(1-e^{-\beta_0 t}))\\
&\quad-\delta_1\bm{p}(z-ct,z+(3s_1-s_0-1)\tau_0-\sigma_0\delta_1(1-e^{-\beta_0 t}))e^{-\beta_0 t}
\end{align*}
for any $t>0$ and $z\in\Bbb{R}$.
Letting $t=(n_k-n_{k_0})T$ in the above inequality, we have
$$\bm{v}((n_k-n_{k_0})T,\cdot;P^{n_{k_0}}(\bm{\phi}_0))=P^{n_k-n_{k_0}}(P^{n_{k_0}}(\bm{\phi}_0))(\cdot)
=P^{n_k}(\bm{\phi}_0)(\cdot)\to\bm{\phi}\text{~~as~~}k\to\infty.$$
Hence we see from $\sigma_0\delta_1\leq(s_1-s_0)\tau_0$ that
$$\bm{\phi}(z)\geq\bm{U}(z,z+(3s_1-s_0-1)\tau_0-\sigma_0\delta_1)\geq\bm{U}(z,z+(2s_1-1)\tau_0),\quad\forall z\in\Bbb{R},$$
that is, $\omega(\bm{\phi}_0)\geq\bm{V}^{(2s_1-1)\tau_0}(0,\cdot)$ in $\Bbb{R}$.
By Lemma \ref{convergence}, there exists $s_2\in[0,1]$ such that
$$\mathop{\lim}\limits_{n\to\infty}P^n(\bm{u}_0)=\omega(\bm{u}_0)=\bm{\zeta}(s_2)=\bm{V}^{(2s_2-1)\tau_0}(0,\cdot),$$
which leads to
$$\mathop{\lim}\limits_{t\to\infty}\lVert\bm{v}(t,\cdot;\bm{u}_0)-\bm{V}^{(2s_2-1)\tau_0}(0,\cdot)\rVert_{\mathcal C}=0.$$
Letting $\tilde{\tau}=\frac{(2s_2-1)\tau_0}{c}$, there is
$$\mathop{\lim}\limits_{t\to+\infty}\lVert\bm{v}(t,\cdot;\bm{u}_0)-\bm{V}^{c\tilde{\tau}}(t,\cdot)\rVert_{\mathcal C}=0,$$
that is,
$$\mathop{\lim}\limits_{t\to+\infty}\lVert\bm{u}(t,\cdot;\bm{u}_0)-\bm{U}(x,x+ct+c\tilde{\tau})\rVert_{\mathcal C}=0.$$
The proof is complete.
\end{proof}
\begin{corollary}
Let $\bm{U}(x,x+ct)$ and $\tilde{\bm{U}}(x,x+\tilde{c}t)$ be two pulsating fronts of system \eqref{u1u2-0-1}
connecting $\bm{0}$ to $\bm{1}$ with $c, \tilde{c}\neq0$.
Then there exists $\tilde{z}\in\Bbb{R}$ such that
$$\bm{U}(\cdot,\cdot+\tilde{z})=\tilde{\bm{U}}(\cdot,\cdot)\text{~~and~~}c=\tilde{c}.$$
\end{corollary}
\begin{proof}
Let $\bm{u}(t,x):=\bm{U}(x,x+ct)$ and $\tilde{\bm{u}}(t,x):=\tilde{\bm{U}}(x,x+\tilde{c}t)$.
In view of $\tilde{\bm{u}}(0,\cdot)\in\mathcal C_{\bm{1}}$,
$\mathop{\lim}\limits_{x\to-\infty}\tilde{\bm{u}}(0,x)=\bm{0}$ and
$\mathop{\lim}\limits_{x\to+\infty}\tilde{\bm{u}}(0,x)=\bm{1}$,
it follows that $\tilde{\bm{u}}(0,\cdot)$ satisfies \eqref{initial}.
The above Theorem \ref{GS} then yields that there exists some $\tilde{\tau}\in\Bbb{R}$ such that
$$\mathop{\lim}\limits_{t\to+\infty}\lVert\tilde{\bm{u}}(t,x)-\bm{U}(x,x+ct+c\tilde{\tau})\rVert_{\mathcal C}=0,$$
that is,
$$\mathop{\lim}\limits_{t\to+\infty}\lVert\tilde{\bm{U}}\left(z-ct,z+(\tilde{c}-c)t\right)-\bm{U}(z-ct,z+c\tilde{\tau})\rVert_{\mathcal C}=0.$$
Letting $t=nT$, then
\begin{equation}\label{U}
\mathop{\lim}\limits_{n\to+\infty}\tilde{\bm{U}}\left(z,z+(\tilde{c}-c)nT\right)=\bm{U}(z,z+c\tilde{\tau}),
\quad\text{uniformly in}\ z\in\Bbb{R}.
\end{equation}
If $\tilde{c}\neq c$, then for any fixed $z\in\Bbb{R}$,
$\mathop{\lim}\limits_{n\to+\infty}\tilde{\bm{U}}\left(z,z+(\tilde{c}-c)nT\right)=\bm{0}$ or $\bm{1}$,
which contradicts to \eqref{U}, and hence $\tilde{c}=c$. Then, we see from \eqref{U} that
$\tilde{\bm{U}}(z,z)=\bm{U}(z,z+c\tilde{\tau})$ for all $z\in\Bbb{R}$. That is,
$\bm{u}(\tilde{\tau},\cdot)=\tilde{\bm{u}}(0,\cdot)$ in $\Bbb{R}$, which implies that
$\bm{u}(t+\tilde{\tau},\cdot)=\tilde{\bm{u}}(t,\cdot)$ for any $t\in\Bbb{R}$,
namely, there exists $\tilde{z}:=c\tilde{\tau}$ such that
$\bm{U}(\cdot,\cdot+\tilde{z})=\tilde{\bm{U}}(\cdot,\cdot)$ in $\Bbb{R}^2$.
The proof is complete.
\end{proof}
\section*{Acknowledgments}
The first author was partially supported by FRFCU (lzujbky-2017-it59), the second author was partially supported by NSF of China (11671180, 11731005) and FRFCU (lzujbky-2017-ct01), and the third author partially supported by NSF of China (11301407) and NSF of Shaanxi Province of China (2017JM1003).
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Dr. Bill Gray from CSU correctly predicted the hiatus 20 years ago, and was rewarded by VP Al Gore by having his long-time NOAA funding cut off.
In 1996, CSU's Bill Gray predicted weak cooling for the next 20-30 years.
His forecast was spot on.
Hurricanes spiked shortly after Bill made his forecast.
Bill Gray is a man of unwavering principle, and has self-funded his efforts to get the truth out about global warming for decades. Bill's wife Nancy (who passed away in 2001) was mayor of Fort Collins in the early 1980's, and was largely responsible for setting up the great set of bike trails which I use several times every day when I am in Fort Collins.
Dr. Brown at Duke had an excellent article over at WUWT – how anyone in any science, if they are to get grants, must mention AGW in their work. It really has nothing to do with science, just politics.
… added to that I was poking around in NASA's current satellite missions and LAGEOS-2 is about the only one that doesn't mention "climate" in it's mission. (Probably because a ball covered with reflectors orbiting the planet can't have it's data adjusted… no data.) Every other mission statement is about the catastrophe our planet can only avoid if we launch another $500 million dollar suite of instruments to track frog droppings in lower Borneo in order to estimate how many of them Mankind's destruction of climate stasis has killed.
Every scientist wants to publish in the top journals, which for biomedical science are Nature and Science. In my own field, the last research papers that landed in these journals were 2005 and 1998, respectively.
Stories about how climate change could impact this same field of research are published in these journals regularly, the latest being 2007 and March 2014, respectively.
It's a pretty sad state of affairs when actual science is rejected in favor of this kind of politically motivated speculation.
Go past Go, Say climate warming and collect $200,000.
My visit to the USSR in 1980 helped prepare me for the future.
Thank you for this tribute to a great scientist.
Both Gray and Gore will be remembered in the future, one as an honourable scientist the other as a sleazy money grabbing political slimeball.
Depends on the bias of the historians. You rate the truth of History, as recorded, to highly.
The victors write history and not the losers, therefore history is always slanted.
I wonder if Nobel Prizes can be revoked? There is more than one, actually more than two, that should be revoked.
Gray and Dyson would be outstanding candidates to receive them instead.
Not just Gore, Obama, and Krugman, but just about every third-world pandering "literature" award in the past several years.
I do not think there is any provision for revocation. Nor can it be awarded posthumously.
Compare Gray's forecast with AGW (IPCC, NASA, NOAA, etc.) 25, or more, model forecasts. BIG difference in accuracy!
And some people still insist science is apolitical.
Your view is tempting, however in actual fact the funding for AGW "research" is increasing geometrically. And polls have shown that while most adults reject the "science", they are still willing to support more government interference in energy supply and consumption, which will of course necessitate higher taxes and more bans on this or that behavior.
As has always been the case throughout U.S. history, we applaud conservative rhetoric, but we continue inexorably to vote for Marxist policies and programs.
Eight miles from where I am sitting at home, the 50th annual Nobel Conference is once again taking place at Gustavus Adolphus College right now. The usual obligatory 'Climate Change' remarks have already been delivered even though it is not the topic of the conference.
I have to agree with Richard. It will not be on its dying legs until the money quits flowing, and the alarmists are the gatekeepers of the money as well as the avenues to publish. Precisely what Ike warned about.
Once politically awake, I do not think all skeptic sites will fade. Too many watchful eyes are now open.
They can lie to the children all they want, but children are remarkably resistant to being told what to think, as long as they have the opportunity to hear that other views exist.
We cannot have a technological society in which the methods of science are not taught to students. So, eventually all the indoctrination will prove a waste of time.
And nothing makes a person more angry than finding out they have been duped and played for a fool.
For a recent, large scale and real life example of the above, consider how effective was the force fed indoctrination of generations of people raised under the various communist regimes post WWII.
Test your hypothesis through experimentation.
Analyze your empirical data objectively & draw conclusions.
Share your data and methods so others can replicate your results.
They substitute models for experiments, they don't use empirical data, their conclusions are not objective, and they refuse to share their data and methods.
Dr. Gray is a true scientist.
That's why they hate him.
And that's why he's one of my heros.
1. young people: The high school graduates of 2014, say 17 & 18 years of age, have not experienced global warming (whatever that is) and severe storms in the US (a least) have not been a big issue during their HS years. The storm named Sandy is an exception but the odd set of circumstances involved were apparent to anyone. The movie, An Inconvenient Truth, AIT came in 2006 and Earth since then has not gone with the script.
2. No longer young: A person would have had to have been born in the early 1970s to have experienced and been aware of the GW movement at its height. A 1973 birth date would make a person 15 when James Hansen's Senate testimony was given and about 25 for the Big El Niño of 1998. Those once young folks are now in their early 40s, say 41 or 42, and most have moved on, have a family, and are more interested in their grand kids and/or the retirement savings, and the once new house that needs a new roof.
Unfortunately the Media, Schools, and Big Official Government Agencies whom we are supposed to trust all push Climate Change… caused by Mankind… a friend of mine who's genius level recently said.. "Who am I to believe? NASA or some guy's Blog?? You really think ALL those Scientists are lying??"… that's what we're up against..
The other day I got lectured by a freshman in high school as to why CAGW was a hoax, so there is hope yet.
In a normal world Dr Gray would have become the go to person for accurate info on the climate.
We got this article out from New Scientist: The world is warming faster than we thought.
i.e. it is entirely worthless and not even newsworthy nor representative of the truth. Noting no comments (that I can see) only presumes everyone will be laughing at them or they have no readers.
The left is gambling they will win. More important the scientific journals and the science departments in Universities are gambling they will win.
CAGW has been front page news for decades.
With luck there will be a massive out cry and a demand for funding cuts to the specific scientists and the universities who were part of the hoax.
Well, to really make an impression, they would need to go and claw back all the public money that went to salaries for the hoaxers during the time that they participated in the hoax. Otherwise, it will just start up all over again, with different players.
They're probably hoping that we'll forget that last "little" part….
Oh yeah, don't forget the benefits also!
No. Then it will be back to the global cooling/ice age scare… which probably has much more validity.
Come 2020 and Bill Gray will still be known to only a few as a great scientist.
And by then the same scamaters will be pitching camp around the next big thing – global cooling. Don't worry too much though as the UN-IPCC will have by that time desided that it is still all your fault, your CO2 did it, and you will pay again.
Money can't buy integrity, but money can buy a lot of publicity to make it difficult for people with integrity to be heard.
It is getting more and more difficult to hide the decline. The Internet never forgets, plenty of alarmist predictions to remind them about.
How did AlGore cut it's funding?
Other than that, Al Gore is untouchable. There was an investigation of V.P. Gore and Maurice Strong over the scam Molten Metal Tech. but nothing came of it.
How do we know that Al Gore was responsible for removing grant money from Dr. Bill Gray. Links?
Gore invited Gray to participate in a global warming meeting in 1993. Gray told him he was a skeptic and might not be what he was looking for. His funding got cut off right after that.
I don't have a link to the conversation I had in Bill Gray's office.
ok! Wow! So Gore used is influence to cut his funding… in 1993… and he calculated this in 1996. So I am assuming he did this research on his own time?
It is not unusual for real scientists to do research on their own time. Most academic research was done that way before WWII.
Consensus "science" emerged right after WWII, together with consensus funding decisions.
Dr. Gray has been self-funded since 1993, despite being one of the best respected and most sought-after hurricane forecasters in the world.
It is what usually happens to people with honesty and integrity whether in Academia or in Industry.
pressure patent. Inventor of the Strouhal Strummer windmill .
I agree with his assessment of the UN's central role in promoting deceptive government science.
John, this is not fair! Go read all I wrote in my blog and twitter… also, when Gore science fiction movie was shown in my kids schools, I fought with teacher and school board to stop this brainwashing and I had to show my kids that all this was make believe and they where being used for some other reason than saving the planet….
When I read this post, which I liked… I did not find the source of information proving that Gore did what he did… So I asked.
We are accusing the other side of ad hominum attacks when they are out of arguments… when I see "our" side do it… I ask for more info.
I am a skeptic of everything and trying to keep both sides honest.
Bill Gray has been MY HERO for the last 10 years.
I have known Bill Gray for more than 20 years and speak with him on a monthly basis. He's a brilliant man and still does research. I value his friendship and wisdom, a meteorological treasure. Had the pleasure of meeting Mrs Gray a couple of years before she passed, they were a perfect match.
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\section{Introduction}
Damped Lyman $\alpha$ systems (DLAs), having column density of {H~I} (N$_{\rm
H\;I}$) $\ge\;2\times 10^{20}$ cm$^{-2}$ have been suggested to be the
high-redshift analogs of disks of nearby luminous galaxies. In spite of having
a high amount of neutral hydrogen these absorbers have very little H$_2$
(Ledoux et al. 2003), which raises questions about their being associated with
high star formation activity. A large fraction of DLAs do indeed appear to have
low star formation rates based on deep emission-line imaging searches (e.g.
Kulkarni et al. 2006a and references therein). Metallicity measurements, using
intermediate to high resolution observations, are available for $>$ 100 DLAs.
These indicate that the metallicity evolution of DLAs is weak; most of the
DLAs, even at $z_{abs}\sim0$ are found to be metal poor (Kulkarni et al. 2005).
It is thus possible that the DLAs may not trace the bulk of star formation in
the Universe and therefore may not be among the leading metal carriers.
Compilation of observed metallicities of individual DLA systems shows a trend
of decreasing abundance with increasing H I column density (Boisse et al. 1998;
Akerman et al. 2005; Meiring et al. 2006a). The trend possibly exists to lower
H I column densities covering the sub-DLAs, having N$_{\rm H\;I}$ between
10$^{19}$ and $2\times 10^{20}$ cm$^{-2}$ (P\'eroux et al. 2003). A similar
conclusion was drawn by York et al. (2006, hereafter, Y06) based on the average
abundances inferred from the composite spectra of sub-samples drawn from a
sample of 809, intervening Mg II QSO absorption line systems (rest equivalent
width of $\lambda$2796 $>$ 0.3 {\AA} and 1 $<z_{abs}<$ 1.86) compiled from the
Sloan Digital Sky Survey (SDSS) Data Release 1 (DR1). The sub-sample with
highest estimated (near solar) Zn metallicity, was found to have a low average
inferred N$_{\rm H\;I}$ of $\sim10^{20}$ cm$^{-2}$. Several arguments were
given in Y06 to suggest that the DLAs may not be the analogs of the local
bright galaxies, rather, the sub-DLAs and possibly the non-DLA Lyman limit
systems (NDLLS) with N$_{\rm H\;I}$ between 10$^{17}$ and 10$^{19}$ cm$^{-2}$
may be associated with large galaxies. Here we consider this scenario further,
using the currently available data on individual absorbers and the results of
Y06. We consider various selection effects at work and discuss the implications
of the recently established mass-metallicity relation for the nature of DLAs
and sub-DLAs.
In section 2 we study the dependence of the Zn abundance on the column density
of H I, for the observations of individual systems taken from the literature
and for the average values of these quantities obtained by Y06 for large
samples. We present other relevant observational information and discuss the
obscuration threshold and other selection effects to argue for the reality of
the observed dependence. The implication of this dependence for the nature of
the absorbing galaxies are presented in section 3 and conclusions are presented
in section 4.
\section{Dependence of metallicity on the neutral hydrogen column density}
\subsection{Observational data} The number of metallicity measurements of
individual sub-DLAs has grown considerably since the compilation of P\'eroux et
al. (2003). We have compiled a sample (the literature sample, hereafter TLS) of
all the measurements of Zn abundance of DLAs and sub-DLAs from the literature.
This sample consists of 119 DLAs with $0.1 < z_{abs} < 3.9$ and 30 sub-DLAs
with $0.6 < z_{abs} < 3.2$. The DLA Zn sample is based on our recent HST, MMT,
and VLT data (Khare et al. 2004; Kulkarni et al. 2005; Meiring et al.
2006a,2006b; P\'eroux et al. 2006b) and those from the literature (data
compiled in Kulkarni et al. 2005, and more recent data from Rao et al. 2005;
Akerman et al. 2005; Ledoux et al. 2006). For the sub-DLAs, we compile the data
from Lu et al. (1995, 1996); Pettini et al. (1994, 1999, 2000); Kulkarni et al.
(1999); Ellison \& Lopez (2001); Srianand \& Petitjean (2001);
Dessauges-Zavadsky et al. (2003); Khare et al. (2004); P\'eroux et al. (2006a);
Ledoux et al. (2006); and Meiring et al. (2006a,2006b). The DLA sample contains
68 detections and 51 limits, while the sub-DLA sample contains 13 detections
and 17 limits. Most of these limits are at high redshifts. The full sample used
here is given in Kulkarni et al (2006b).
In Fig. 1, we have plotted [Zn/H], as a function of N$_{\rm H\;I}$ for TLS. The
trend of decreasing abundance with increasing N$_{\rm H\;I}$ is clear. The
Spearman rank correlation test gives the probability of chance correlation to
be 1.6 $\times 10^{-11}$. It can be seen from the figure that the trend is
similar for systems with $z_{abs}<1.5$ and $z_{abs}>1.5$.
We note that we have only used Zn abundances in Fig.1. There have been studies
which use the Si and S abundances in systems where Zn abundances are not
available due to observational limitations. We, however, refrain from using
these as (i) given the intrinsically higher abundance and higher strengths of
the detectable absorption lines of these elements, line saturation effects may
be important, leading to an underestimate of their abundances; (ii) Si may be
depleted on the dust grains (up to -1.4 dex in cold ISM); (iii) the lines of Zn
II are detectable to lower redshifts; and (iv) lines of S are often in the
Lyman $\alpha$ forest.
As noted in the last section, Y06 inferred the average abundances of various
elements in the composite spectra of several sub-samples of SDSS DR1 QSOs with
strong Mg II absorbers. The sub-samples were chosen on the basis of various
properties of the absorption line systems and QSOs. The column densities of
several species, including Zn II, were estimated from inverse
variance-weighted, arithmetic mean, normalized spectra of individual QSOs in
the samples. The geometric mean composite spectra of the same sub-samples were
compared with similar spectra of matching (in emission redshifts and i
magnitudes) samples of QSOs without absorption lines in their spectra, to
determine the absorber rest frame extinction, $E(B-V)$. The column densities
of neutral hydrogen were estimated by assuming the Small Magellanic cloud (SMC)
dust-to-gas ratio, as the SMC extinction law was found to fit the composite
extinction curves well. In Fig. 1 we have plotted the results for various
sub-samples of Y06. Note that we have used all the sub-samples from Table 1 of
Y06 for which the relevant data were available, some of which were not listed
in their Table A4. Details of their sub-samples which are used here are given
in Table 1. The trend of metallicity dependence on N$_{\rm H\;I}$ is similar to
that in TLS, the Spearman rank correlation test for Y06 points gives the
probability of chance correlation to be 0.098. We have included the main sample
(\#1) of Y06 in our analysis. As the rest of the sub-samples are drawn from
this, its inclusion may cause some bias. We have verified that removing this
sample has very little effect on the correlation noted above.
Data for individual DLAs show a trend of increasing depletion of Cr with
respect to Zn, and thus, higher dust-to-metal ratio, for higher metallicity
(Ledoux et al. 2003; Akerman et al. 2005; Meiring et al. 2006a). We have
confirmed that a similar trend is shown by the results of Y06. Other
distributions, e.g. [Cr/Zn] vs. log(N$_{\rm Zn\;II}$) and [Cr/Zn] vs.
log(N$_{\rm H\;I}$) are also similar for Y06 and TLS, showing that the results
of Y06, which are average values for large samples, are consistent with the
observations of individual systems.
We note that the Y06 results are based on the assumption of a constant
dust-to-gas ratio. This ratio may depend on metallicity in view of the
anticorrelation between [Cr/Zn] and [Zn/H] (and in view of Fig.2 below). We
have tried to estimate the effect of this on the relation between [Zn/H] and
N$_{\rm H\;I}$, obtained by Y06, as follows. We assumed that the abundance of
Zn, (X$_{\rm Zn}$) is proportional to N$_{\rm H\;I}^\alpha$. Thus $E(B-V)$
which is proportional to X$_{\rm Zn}~$N$_{\rm H\;I}$ will be proportional to
N$_{\rm H\;I}^{(1.0+\alpha)}$, giving N$_{\rm H\;I}\propto
E(B-V)^{1.0\over{(1.0+\alpha})}$. Note that the Y06 results assume $\alpha=0$.
For different values of $\alpha$, we determined N$_{\rm H\;I}$ from the
$E(B-V)$ values, using the above relation. Using this value of H I column
density we determined [Zn/H] and then the slope of the best fit line between
[Zn/H] and Log(N$_{\rm H\;I}$). Note that by construction, this slope should
equal $\alpha$. We found that for positive values of $\alpha$ the slope
remains negative for $\alpha<0.81$ and beyond that remains $<\alpha/2$ until
$\alpha=3$. For negative values of $\alpha$ the slope is somewhat smaller than
the assumed value of $\alpha$ and comes close to it (-0.94) for $\alpha~=-0.9$.
Thus we see that assumption of a metallicity dependent dust-to-gas ratio, in
fact, makes the decrease of Zn abundance with N$_{\rm H\;I}$ even steeper
(slope $\sim$ -0.9) than that (slope = -0.45) found for the assumption of
constant-dust-to-gas ratio (Fig. 1).
\begin{figure}
\includegraphics[width=3.15in,height=3.50in]{Fig1.eps}
\caption{Metallicity ([Zn/H]) vs. log(N$_{\rm H\;I}$). Solid (Red) circles
represent average values for large samples as obtained by Y06. Open (Black)
circles and (blue) stars are for $z_{abs}<1.5$ and $z_{abs}>1.5$, respectively,
for TLS. One sigma error bars are shown. Also shown are the best fit lines
(long dashed (black) line and short dashed (red) line for TLS and Y06
respectively) obtained by ignoring the limits. The top solid line (blue)
represents the empirical obscuration bias (N$_{\rm Zn\;II} =1.4\times 10^{13}$
cm$^{-2}$).}
\end{figure}
\begin{figure}
\includegraphics[width=3.15in,height=3.50in]{Fig2.eps}
\caption{Absorber rest frame average extinction, A$_V$, assumed to be
2.93$\times E(B-V)$ vs. log(N$_{\rm Zn\; II}$) for the Y06 results. The linear
best fit relation A$_V$ = 0.1$\times10^{13}$ N$_{\rm Zn\; II}$ is shown as
solid (red) line. The best fit line obtained for Milky Way sight lines, A$_V$
= 0.3$\times10^{13}$ N$_{\rm Zn\; II}$ by Vladilo and P\'eroux (2005) is shown,
as dashed (black) line for comparison.}
\end{figure}
From an earlier version of this diagram (Fig. 1), with 37 points, including
limits, Boiss\'e et al. (1998) noted the trend of decreasing abundance with
increasing N$_{\rm H\;I}$ (their Fig. 19; hereafter, the Boiss\'e plot). They
interpreted this as being due to the observational limitations in detecting
weak Zn II lines at the low N$_{\rm H\;I}$ end and an obscuration bias (caused
by dimming of the QSOs, by the dust in the absorbers, below the limit of
magnitude limited surveys) towards the high N$_{\rm H\;I}$ end, causing a
dearth of points in the lower left and upper right corners of the plot
respectively. Boiss\'e et al. (1998) proposed an obscuration threshold at
N$_{\rm Zn\;II}=1.4\times 10^{13}$ cm$^{-2}$ (above which the background QSO
may be rendered invisible by dust obscuration), shown as a solid (blue)
line in our Fig. 1. This was only an empirical limit based on the sample used
by Boiss\'e et al. (1998). It is thus not surprising that a few systems (with
$z_{abs}<1.5$) in Fig. 1 do lie above this threshold. Most of these systems
were observed recently by Khare et al. (2004), Meiring et al. (2006a,2006b)
and P\'eroux et al. (2006a, 2006b) and many of them have strong Mg II and/or Fe
II lines in the SDSS spectra of the QSO's. Recently, Herbert-Fort et al. (2006)
have reported the presence of two metal strong systems (with $z_{abs}>1.6$) in
SDSS DR3, having Zn II column density greater than the obscuration threshold
(SDSS1610+4724, log(N$_{\rm Zn~II}$)=13.4$\pm{0.03}$ and SDSS1709+3258,
log(N$_{\rm Zn~II}$)=13.19$\pm{0.03}$) and have estimated that $\simeq 5$\% of
the SDSS-DR3 DLA population with $z_{abs}\ge2.2$ in QSOs with $r<19.5$ have
similar Zn II column densities.
\begin{table*}
\begin{center}
\caption{Relevant data for Y06 sub-samples}
\begin{tabular}{|r|l|l|r|r|r|r|}
\hline
$E(B-V)$&Sample &Selection Criterion$^a$&Number&Log(N$_{\rm Zn~II}$)$^b$&Log(N$_{\rm H~I}$)$^c$\\
(SMC)&number&&of systems&&\\
\hline\hline
0.002&24& $\Delta(g-i)^d<$0.2&698&12.4&20.0\\
0.003&23& W$_{\rm Mg\;II}^e<$2.0&558&12.2&20.1\\
0.006&9& $z_{abs}<$1.3127&404&$>$12.4&20.4\\
0.007&5& 1.53$\le$W$_{\rm Mg\;II}<$1.91&139&12.6&20.5\\
0.009&11& i$^f<$19.12&398&$>$12.4&20.6\\
0.010&13& $\beta^g<$0.103&405&$<$12.3&20.7\\
0.011&12& i$\ge$19.12&411&$>$12.4&20.7\\
0.011&14& $\beta\ge$0.103&404&12.5&20.7\\
0.012&26& W$_{\rm Mg\;II}\ge$2.5, $\Delta(g-i)<$ 0.2& 97&12.5&20.7\\
0.013&1& Full sample & 809&12.3&20.8 \\
0.014&16& W$_{\rm Fe\;II}^h$/W$_{\rm Mg\;II}\ge$0.577&369&12.4&20.8\\
0.018&6& 1.91$\le$W$_{\rm Mg\;II}<$2.52&132 &12.8&20.9\\
0.019&20& Fe II $\lambda$2374 present&392&12.5&20.9\\
0.031&7& 2.52$\le$W$_{\rm Mg\;II}<$5.0&134&12.9&21.1\\
0.032&8& W$_{\rm Mg\;II}\ge$2.0&251 & 12.6 &21.2\\
0.034&17& W$_{\rm Al\;II}^i$/W$_{\rm Mg\;I}^j<$1.538&85&12.5&21.2\\
0.034&19& Fe II $\lambda$2260 present&58 & 13.0&21.2\\
0.036&21& Zn II-Mg I $\lambda$2026 present&83&13.1&21.2\\
0.058&22& Zn II-Cr II $\lambda$2062 present&31&13.3&21.4\\
0.081&25& $\Delta(g-i)\ge$0.2&111 &$>$12.7&21.6\\
0.085&27& W$_{\rm Mg\;II}\ge$2.5,$\Delta(g-i)\ge$0.2&48&$>$12.4&21.6\\
\hline
\end{tabular}
\hspace*{1.2in} $^a$ For selection from the full sample (sample 1)\hfill\break
\hspace*{1.2in} $^b$ Estimated from equivalent widths in composite spectra after correction for blends\hfill\break
\hspace*{1.2in} $^c$ Estimated from E(B-V) assuming a constant dust-to-gas ratio\hfill\break
\hspace*{1.2in} $^d$ The difference between the actual colours of QSOs and the
median colours of QSOs at that \hfill\break
\hspace*{1.3in} redshift (Richards et al. 2003)\hfill\break
\hspace*{1.2in} $^e$ Rest equivalent width of Mg II $\lambda$2796 in {\AA}\hfill\break
\hspace*{1.2in} $^f$ i magnitude of the QSO\hfill\break
\hspace*{1.2in} $^g$ Relative velocity w.r.t. the QSO\hfill\break
\hspace*{1.2in} $^h$ Rest equivalent width of Fe II $\lambda$2382 in {\AA}\hfill\break
\hspace*{1.2in} $^i$ Rest equivalent width of Al II $\lambda$1670 in {\AA}\hfill\break
\hspace*{1.2in} $^j$ Rest equivalent width of Mg I $\lambda$2852 in {\AA}\hfill\break
\end{center}
\end{table*}
\subsection{Establishing the reality of the observed trend}In this section we
present some relevant observational facts and discuss the known selection
effects to suggest that the Boiss\'e plot may indeed not be affected by
these and that the trend of decreasing abundance with increasing
N$_{\rm H\;I}$ may be real.
We first discuss the observed extinction in QSOs and point out that it is much
smaller than that observed in typical Milky Way sight lines and may not cause
significant dimming of the background QSOs.
Vladilo \& P\'eroux (2005) have shown that the extinction A$_{\rm V}$ is
proportional to N$_{\rm Zn\;II}$ for Milky Way sight lines. Their relation is
plotted in Fig. 2. Also plotted is a similar relation obtained for the
sub-samples of Y06. It can be seen that a correlation does exist between
A$_{\rm V}$ and N$_{\rm Zn\;II}$ for QSO absorbers but that the average
extinction per Zn ion in the these absorbers appears to be smaller than that in
Milky way by a factor of three.
The maximum value of $E(B-V)$ found by Y06 for their sub-sample (\#27) of most
reddened systems was 0.085 while that for the sub-sample (\#23) of systems with
the rest frame equivalent width of Mg II $\lambda2796 <2.0$ {\AA} is as low as
0.003. The observer frame A$_{\rm V}$, for the average redshift of 1.33 for
their samples, assuming a 1/$\lambda$ extinction law, is thus smaller than 0.6
and 0.02, respectively for the two samples. These values suggest that the dust
obscuration by the absorbers observed towards the SDSS QSOs, may not be very
important and may not cause significant decrease in the brightness of the
QSOs. A similar conclusion is drawn by Murphy \& Liske (2004) from the
analysis of spectral energy distributions of QSOs with DLAs at $z_{abs}\sim3$
in the SDSS DR1. Ellison et al. (2005a) found no significant difference between
the B-K colours of radio selected QSOs with and without DLAs, indicating that
dust obscuration is not very important.
Below we discuss other indirect evidence indicating that dust obscuration is
not significant in QSOs.
Wolfe et al. (2005), from the observed depletion pattern of elements, estimated
the fraction of dust obscured QSOs to be $<$ 10\%. Schaye (2001; also see Zwaan
\& Prochaska 2005) has shown that there is an upper limit of $<$ 10$^{22}$
cm$^{-2}$ on the neutral hydrogen column densities of DLAs because of
conversion of {H~I} to H$_2$ and not because of dust obscuration. Herbert-Fort
et al. (2006) have presented several arguments against the obscuration bias at
redshifts $>1.6 $, most importantly, one based on the magnitude distribution
of the parent QSOs of metal strong systems.
Samples 11 and 12 of Y06 had i magnitude smaller/larger than 19.1.
Coincidently, this is the cutoff for the SDSS targeting algorithm. Thus the
bright sample consists of QSOs which should be unreddened to enter into the
SDSS QSO catalog while the faint sample consists of candidates which were
observed (and found to be QSOs) on the basis of their being X-ray sources,
radio sources etc. Thus one would expect the faint sample to be more reddened.
However, both samples have similar values of $E(B-V)$ and similar distribution
of $\Delta(g-i)$ values. Thus there is no evidence for higher extinction in
faint SDSS QSOs from the Y06 study. Prochaska et al. (2005) measured
40$\pm$20\% higher gas density in DLAs towards bright QSOs than towards faint
QSOs in SDSS DR3 having DLAs ($z_{abs}>2$). This is contrary to the obscuration
bias and they suggest that gravitational lensing due to DLAs may be important.
Vanden Berk et al. (1997) had found a significant excess of C IV systems in
bright QSOs and interpreted this to be evidence for gravitational lensing. Some
of these C IV systems could however be intrinsic to the QSOs (Richards 2001).
Menard (2005) showed that lensing due to intervening Mg II systems with rest
equivalent width smaller than 1.5, could brighten QSOs by up to -0.2 magnitudes.
Though the two effects (lensing and dust obscuration) have different origins
and need to be understood further, the results of Prochaska et al. (2005)
indicate that the effects of lensing dominate over obscuration effects that may
be present.
The above arguments seem to indicate that dust obscuration is not
important even in faint QSOs in the SDSS sample. We, however, note that all
these arguments are based on samples of optically selected QSOs and can only
apply if the dust content in QSO absorbers has a continuous distribution. We
note that a few dusty DLAs have been observed (e.g. Junkkarinen et al. 2004;
Motta et al. 2002; Wild et al. 2006) but in none of these cases $E(B-V)$
exceeds 0.42. Such values of $E(B-V)$ may push bright QSOs below the cutoff
(19.1) of SDSS targeting algorithm but these will still be present in the faint
sample of Y06. Some empty fields have been observed in optical observations
towards radio QSOs by Jorgenson et al. (2006) which could potentially be
obscured by dusty absorbers, though the authors have argued against such a
possibility on the basis of the DLA statistics observed in radio selected QSOs.
Some theoretical studies (e.g. Vladilo and P\'eroux 2005) have argued that up
to 30\% to 50\% of the QSOs may be missed in magnitude limited surveys due to
dust obscuration. Smooth-particle-hydrodynamics simulations do need to
introduce dust obscuration to explain the observed (low) DLA metallicity (Cen
et al. 2003). Our arguments above do not rule out the possibility of a bimodal
distribution of dust columns, with a population of dusty absorbers which have
pushed the background QSOs below the observational limit of the optical
surveys. Though this possibility can only be verified with observations of
fainter QSOs, no compelling evidence for such a population is found from the
study of metallicity, column density distribution and mass density of H I in
DLAs in radio selected samples of QSOs (Akerman et al. 2005; Jorgenson et al.
2006). Their study suggests that the optically selected samples give a fair
census of the population of DLA absorbers. It should, however, be noted that
these samples are still small and cover redshifts $>$ 1.86 only.
From all the above arguments, it seems very likely that not
many points in the upper right corner of the Boiss\'e plot are missed due to
dust obscuration.
The missing points in the lower left corner of the Boiss\'e plot have been
attributed to an artifact of the sensitivity of observations (Boiss\'e et al.
1998). A few systems may indeed be present in this region as seen from the few
upper limits there in Fig. 1. However, it may be noted that in the analysis of
Y06, solar metallicity was estimated in the sub-sample (\#24) of 698 systems
with average N$_{\rm H\;I}$ of about 10$^{20}$ cm$^{-2}$. Most (638) of these
systems did not have detectable Zn II lines in their individual SDSS spectra.
We note that even a solar metallicity system with N$_{\rm H\;I}$ of about
10$^{20}$ cm$^{-2}$, may not produce detectable Zn II lines in an individual
SDSS spectrum which typically has S/N $\sim$ 10-15 and a 3$\sigma$ detection
limit in the observer frame of $>$ 0.3 {\AA}. So the absence of Zn II lines in
an individual SDSS spectrum does not mean low Zn abundance. The composite
spectrum of 698 systems has S/N higher than that in a typical SDSS spectrum by
a factor of $\sim$ 25-30, enabling the detection of Zn II lines. In the
sub-samples of Y06 (\#s 21 and 22) comprised of systems with detectable Zn II
and Cr II lines, the metallicity was estimated to be lower than that in the
sample of 698 systems. It thus appears that their metallicity measurements,
being averages over large samples, are not affected by observational
limitations in detecting weak lines. We also note that the average solar
metallicity in the sub-sample \# 24 can not be an artifact of a possible high
metallicity tail of the metallicity distribution in sub-DLAs. The composite
spectrum, used for abundance determination is the arithmetic mean of individual
spectra. Thus, if 90\% of the 698 systems had subsolar abundance with [Zn/H]=-1
and only 10\% had solar abundance then, assuming an H I column density of
10$^{20}$ cm$^{-2}$ for the systems, the average equivalent width of
$\lambda2026$ line of Zn II will be $0.9\times0.007+0.1\times0.07=0.013
\AA$, which will correspond to highly sub-solar abundance of Zn II for
this sub-sample as opposed to the solar abundance derived by Y06. It
may also be noted that the sub-samples \#23 and 24, do have a large
number of systems with low N$_{\rm H~I}$ which in principle, could have
had low [Zn/H] and could then have been present in the lower left corner
of the Boisse plot. Sub-sample \#23 which is the sample of systems with
W$_{\rm Mg~II} < 2.0$ {\AA}, includes all the systems in sub-samples \#s
2, 3 and 4 (with W$_{\rm Mg~II}<1.91$ {\AA}, all together 397 systems)
which have $E(B-V)<0.001$ and thus N$_{\rm H~I} < 5\times 10^{19}$
cm$^{-2}$. Similarly sub-ample \#24 has 698 systems. Even assuming it has
all the systems with W$_{\rm Mg~II}>1.91$ {\AA} (i.e. all the systems
not included in sub-amples \#s 2, 3 and 4) from the original sample of
809 systems, it will still have $\ge 286$ systems with N$_{\rm H~I}
<5\times 10^{19}$ cm$^{-2}$. However, the high values of the average
abundances for these sub-samples suggest that most of these low N$_{\rm
H~I}$ systems do not have low abundance and thus do not occupy the lower
left corner of the Boisse plot. We also note that it is possible that
these two sub-samples do include some DLAs. However, as argued above, the
DLAs do not have high abundances and can only reduce the average
abundances of the sub-samples, thereby, making the case for higher
abundance of sub-DLAs even stronger.
On the conservative side, we point out that even if some points (in TLS) are
indeed missing in the lower left region of Fig. 1 due to sensitivity limit of
observations, the average metallicity at low N$_{\rm~H~I}$ (in TLS) will still
be higher than that at high N$_{\rm~H~I}$ due to the large number of high
metallicity, low N$_{\rm~H~I}$ systems that have been observed (assuming the
obscuration bias to be absent, as argued above). It thus appears that the
sub-DLA metallicity is indeed higher than that of DLAs and is close to the
solar value, at low redshifts.
It thus seems very likely, that the observed trend of
decreasing abundance with increasing N$_{\rm H\;I}$ is not
due to selection effects and is real and that DLAs
are not among the major metal carriers in the universe. We note that
super-solar abundances have been observed in seven sub-DLAs (Pettini et al.
2000; Khare et al. 2004; Prochaska et al. 2006; Meiring et al. 2006b). Near
solar abundance has been estimated in one NDLLS (Jenkins et al. 2005).
Bergeron et al. (1994), estimated abundances ([X/H]), of several Mg II Lyman
Limit systems with low Lyman limit optical depth, with redshifts between 0.1
and 1.1, to be between -0.5 and -0.3 dex. Super-solar abundances have also
been estimated for three systems with redshifts between 0.7 and 0.95 (Charlton
et al. 2003, Ding et al. 2003; Meseiro et al. 2005) and near-solar abundance
has been estimated in one system at redshift of 0.064 (Aracil et al. 2006);
all these systems have N$_{\rm H\;I}<10^{16}$ cm$^{-2}$. We however caution
that the derived abundance in NDLLS and systems with smaller H I column
densities do depend on the details of photoionization calculations.
Having argued above, for the reality of the observed dependence of Zn
metallicity on H I column density, below, we consider its implications. In
particular, we consider consequences of the hypothesis that the sub-DLAs, and
possibly NDLLS, have higher abundances than DLAs and represent a selection of
galaxies that are the major metal carriers in the universe.
\section{Implications}
Recently a mass-metallicity relation has been discovered by several groups.
Tremonti et al. (2004), from the imaging and spectroscopy of ~ 53,000
star-forming galaxies at $z \sim$ 0.1, found a tight ($\pm 0.1$ dex)
correlation between stellar mass and metallicity spanning over 3 orders of
magnitude in stellar mass and a factor of 10 in metallicity. Savaglio et al
(2005), from a sample of 56 galaxies identified a strong correlation between
mass and metallicity at 0.4 $< z <$ 1.0. They predict that the generally metal
poor DLA galaxies have stellar masses of the order of 10$^{8.8}$ M$_\odot$
(with a dispersion of 0.7 dex) from $z$=0.2 to $z$=4. Erb et al. (2006) have
obtained a mass-metallicity relation in a sample of 87 galaxies at $<z>\;\sim$
2.26 which is similar to the relation for local galaxies (Tremonti et al. 2004)
except for an offset of 0.3 dex in metallicities, indicating that galaxies of a
given mass have lower metallicity at high redshift. They however, note that the
uncertainty in the metallicity offset between the $z\sim2$ and local galaxies
is approximately a factor of 2, about the same as the offset itself. Ledoux et
al. (2006) found a correlation between the metallicity and velocity widths of
lines of low ionization species over two orders of magnitude in metallicity, at
1.7 $< z <$ 4.3. Assuming velocity widths to be a measure of mass, their
mass-metallicity relation is consistent with that found for local galaxies by
Tremonti et al. (2004). We, however, note that Bouche et al. (2006) found an
anticorrelation between halo mass and Mg II $\lambda$ 2796 equivalent widths
which seems to go against the results of Ledoux et al. (2006).
At low redshifts the metallicity of most DLAs is almost an order of magnitude
lower than the solar value, while, as suggested by results of Y06 and those of
Kulkarni et al. (2006b), the mean sub-DLA metallicity seems to be close
to the solar value. {\it The mass-metallicity relation of Tremonti et al.
(2004) would imply stellar masses of about 10$^{11}$ M$_\odot$ and $<10^{9}$
M$_\odot$ respectively for the sub-DLAs and DLAs.} These numbers will be
smaller by a factor of 2 if the mass-metallicity relation of Erb et al. (2006)
is used. If the stellar metallicity is lower/higher than the gas phase
metallicity (as determined by the abundances in DLAs and sub-DLAs) the masses
of the DLA and sub-DLA galaxies may be correspondingly lower/higher, however,
the ratio of the two masses will not be affected.
We thus propose that the sub-DLAs and DLAs in general represent different types
of galaxies. By and large, the sub-DLAs are produced by massive galaxies, with
higher metallicity, while the DLAs are produced by less massive galaxies with
lower metallicity. A few sub-DLAs may indeed arise from lines of sight which
encounter low H I columns through DLA galaxies and therefore, have lower
metallicity. These will give rise to points in the lower left region of Fig. 1.
Similarly, a few DLAs may arise in lines of sight through sub-DLA galaxies and
may have higher metallicities though the probability of this happening may be
small. In our Galaxy, for instance, clouds with log(N$_{\rm H~I}$)$>$20.3 are
very small and represent only a tiny part of the cross section of the entire
Galaxy.
It has been suggested that the low metallicity found in DLAs is caused by
metallicity gradients. Differences between emission line and absorption line
metallicities have been observed in a few cases (Chen et al. 2005; Ellison et
al. 2005b) but not in others (Schulte-Ladbeck et al. 2005; Bowen et al.
2005). The metallicity gradients in nearby spirals are fairly weak (Bresolin
et al. 2004) and can not explain the low metallicity in DLAs with small impact
parameters (e.g. Kulkarni et al. 2005). It has been suggested in several
studies (e.g. Kauffmann 1996; Das et al. 2001; Zwaan et al. 2005) that DLAs
may result for small values of impact parameters while larger impact parameters
through the same absorbers may give rise to NDLLS and Lyman $\alpha$ forest
systems. In this case, abundance gradients can not be invoked to explain lower
abundances in DLAs as compared to sub-DLAs/NDLLS. This scenario has also not
been verified observationally.
Kulkarni et al. (2006b) have shown that the N$_{\rm H~I}$ weighted mean
metallicity for sub-DLAs is a factor of 6 higher than that for DLAs at low
redshifts. They estimate that at these redshifts, the contribution of the ISM
in sub-DLAs to the cosmic metal budget may be several times that of ISM in
DLAs. Prochaska et al. (2006), based on their observations of two metal strong
sub-DLAs (Super LLS in their paper) and results from their sub-DLA survey,
estimated that the ISM in sub-DLAs may contribute at least 15\% to the metal
budget of the universe at $z \simeq2$.
{\it If the galaxies that give rise to sub-DLAs are indeed more massive than the
galaxies that give rise to DLAs, as suggested above, then an individual sub-DLA
galaxy will contribute more to the stellar mass of the universe than an
individual DLA galaxy.} Also, in CDM cosmology, as the massive galaxies form by
mergers of smaller galaxies which triggers star formation leading to a higher
rate of metal production, they are not only expected to show a higher rate of
metallicity evolution (as is found by Kulkarni et al. 2006b) but are also
expected to contribute more mass at lower redshifts. {\it It is thus possible
that the contribution of sub-DLA producing galaxies to the cosmic metal budget
at lower redshifts may indeed be considerably higher than 15\% (as estimated by
Prochaska et al. (2005) at $z \simeq2$) and may help alleviate the missing
metals problem.}
Boissier et al. (2003) compared the observed properties of DLAs with the
predictions of simple models of present day disk galaxies and showed that low
surface brightness galaxies contribute as much as spirals to the number and H I
mass of DLAs. Zwaan et al. (2005) have shown that properties of DLAs are
consistent with their forming in galaxies of various morphological types, with
87\% of DLA cross-section being contributed by sub-L$_*$ galaxies.
Semi-analytic models indicate sub-L$_*$ galaxies to be major contributors to
DLA cross-section (Okoshi \& Nagashima 2005). Thus there seems to be some
evidence for a significant fraction of DLAs to be associated with sub-L$_*$
galaxies.
Deep imaging has shown low redshift DLAs to be often associated with dwarf
galaxies (Rao et al. 2003). Le Brun et al. (1997) suggested that the DLA
galaxies (which were selected on the basis of a damped Lyman $\alpha$ line or 21
cm absorption or a very high Mg II/Fe II ratio) strongly differ from the Mg II
selected galaxies, the latter being mostly luminous galaxies with evidence of
recent star formation activity. Indeed emission-line imaging searches suggest
that a large fraction of DLAs appear to have low star formation rates (Kulkarni
et al. 2006a and references therein). Deep imaging of three sub-DLA galaxies at
$z\;<$ 0.7 shows them to be associated with L$>0.6$ L$_*$, disk/spiral galaxies
(Zwaan et al. 2005). Note that the lines of sight through our Galaxy are,
statistically speaking, mostly metal rich sub-DLAs with N$_{\rm H\;I}
<2\times10^{20}$ cm$^{-2}$, perpendicular to the plane. All these support our
hypothesis about the nature of DLA and sub-DLA galaxies. Chen and
Lanzetta (2003), however, found the DLA galaxies to have a variety of
morphologies. Djorgovski et al. (1996) and Moller et al. (2002, 2004) have
observed a few high-redshift ($z\ge$2) absorbing galaxies in Lyman $\alpha$
emission and found their properties to be similar to disk/Lyman break galaxies.
Chun et al. (2006), using adaptive optics imaging for 4 DLAs and 2 sub-DLAs at
$z<$ 0.5, found most of the candidate absorber galaxies or their companions to
have low-luminosity ( $<$ 0.1 L$_*$).
Thus, it seems that the results of some of the DLA and sub-DLA imaging
studies are consistent with the hypothesis we make here. More systematic
imaging surveys are needed to confirm the ideas presented in this paper. Deep
imaging in K-band for absorbers with high abundances at $z_{abs}<1.5$ should be
able to confirm the presence of massive (red) galaxies, while imaging searches
in narrow, optical emission lines should help in confirming that DLAs are
mostly dwarf galaxies.
\section{Conclusions}
We have studied the dependence of metallicity on N$_{\rm H\;I}$ in QSO
absorbers with N$_{\rm H\;I}>10^{19}$ cm$^{-2}$ and have discussed various
selection effects that may be giving rise to the observed trend. We
have argued that the selection effects are not important and that the observed
trend is real. Our conclusions based on these arguments, and subject to
confirmation by future observations, are as follows:
\begin{enumerate}
\item{} The amount of dust in QSO absorbers is small and is not
responsible for missing many QSOs in magnitude limited surveys. We can not,
however, rule out the possibility of a bimodal distribution of dust columns
such that there may exist a population of dusty absorbers which push the
background QSOs below the observational threshold of current optical
spectroscopic studies and is completely invisible.
\item{} The metallicity in QSO absorbers with N$_{\rm H\;I}>10^{19}$ cm$^{-2}$,
decreases with increase in H I column density of these absorbers. The trend
possibly continues to lower H I column densities. The sub-DLAs thus have
higher metal abundances as compared to the DLAs at redshifts between 0 and 2.
\item{} The observed mass-metallicity relation suggests that most DLAs
are associated with low mass ($<10^9$ M$_\odot$) galaxies while most
sub-DLAs are associated with massive spiral/elliptical galaxies. It is
possible that the non-DLA LLS may also be metal rich and may be associated with
massive galaxies.
\item{} The sub-DLA galaxies will contribute a larger fraction of total
mass (stellar and ISM) and therefore metals, to the cosmic budget, specially at
low redshifts, as compared to the DLAs. The Sub-DLAs and possibly, non-DLA LLS
galaxies, may contain a much larger fraction of the metals at $z< 1$ than has
been appreciated.
\item{} The few imaging studies, of galaxies responsible for quasar absorption
line systems done so far, are ambiguous on the morphology of DLA and sub-DLA
galaxies. More systematic, deep imaging in r-band, Ks-band and in narrow
emission lines is essential to confirming the inferences of this paper.
\end{enumerate}
\begin{acknowledgements}
PK acknowledges support from the Department of Science and Technology,
Government of India (SP/S2/HEP-07/03). VPK and JDM acknowledge support from the
U.S. National Science Foundation grant AST-0206197.
\end{acknowledgements}
|
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{"url":"https:\/\/www.zigya.com\/study\/book?class=12&board=nbse&subject=Chemistry&book=Chemistry+I&chapter=The+Solid+State&q_type=&q_topic=Electrical+Properties&q_category=&question_id=CHEN12043618","text":"\u00ef\u00bb\u00bf Pure silicon is an insulator. Silicon doped with phosphorus is a semi-conductor. Silicon doped with gallium is also a semiconductor. What is the difference between the two doped silicon semi-conductor? from Chemistry The Solid State Class 12 Nagaland Board\n\n### Book Store\n\nCurrently only available for.\nCBSE Gujarat Board Haryana Board\n\n### Previous Year Papers\n\nDownload the PDF Question Papers Free for off line practice and view the Solutions online.\nCurrently only available for.\nClass 10 Class 12\nPure silicon is an insulator. Silicon doped with phosphorus is a semi-conductor. Silicon doped with gallium is also a semiconductor. What is the difference between the two doped silicon semi-conductor?\n\nPhosphorus has one excess valence electron (compared with Si) after forming the four covalent bonds normally with silicon. This excess electron gives rise to electronic conduction. That is why silicon becomes semi-conductor on doping with phosphorus. It is called n-type semi-conductor.\n\nGallium has only three valence electrons. It creates an electron deficient bond or a hole when it is doped with silicon. Such holes can move in the crystal giving rise to electrical conductivity. Thus silicon doped with gallium is also semi-conductor due to movement of holes. It is called-p-type semiconductor.\n\n1060 Views\n\nWhy is glass considred a super-cooled liquid?\n\nGlass is\u00a0amorphous solids and \u00a0have a tendency to flow, though very slowly. Therefore, these are called pseudo solids or super cooled liquids .\n1368 Views\n\nWhy do solids have a definite volume?\n\nA solid have a definite volume because the inter molecular distance \u00a0between its molecules are fixed.\n1439 Views\n\nWhy are solids rigid?\n\nSolids are rigid because the internuclear distance between the \u00a0molecules of a solid are very less and closely packed and their positions are fixed due to the strong forces of attraction between them.\n2488 Views\n\nRefractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show clevage property?\n\nRefractive index of a solid which have the same value along all directions\u00a0are isotropic in nature. It would not show cleavage property.\n1055 Views\n\nClassify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.\n\nAmorphous solids: Polyurethane, teflon, cellophane, polyvinyal chloride, fibre glass.\nCrystalline solids: naphthlene, benzoic acid, potassium nitrate, copper.\n1237 Views","date":"2018-08-15 05:24:01","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5405327081680298, \"perplexity\": 5652.019316382176}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-34\/segments\/1534221209884.38\/warc\/CC-MAIN-20180815043905-20180815063905-00614.warc.gz\"}"}
| null | null |
Q: If Else Expression I am writing if else expression in SSRS like that
=IIF((Parameters!piStartDate.Value = "" and Parameters!piEndDate.Value = ""), "UnExtracted ACH Records","ACH Extracted Record")
Its working for first condition and giving error for second condition. Can any one point out mistake i am making.
A: You are using same condition two time having same parameter 'piStartDate'
Try below
=IIF((Parameters!piStartDate.Value = "") and (Parameters!piEndDate.Value = ""), "UnExtracted ACH Records","ACH Extracted Record")
It will work fine.
A: Many problems.
1st you say you want to check two parameters, but you are only checking the same parameter twice. Both IIF() functions are checking piStartDate.
2nd, in the conditional part of an IIF, you need to write out "AND" instead of "&". So your second example should be:
=IIF(Parameters!piStartDate.Value = "" AND Parameters!piStartDate.Value = "", "UnExtracted ACH Records","ACH Extracted Record")
3rd, you're not checking for NULL, you're checking for empty strings, but maybe that's just what you really meant.
A: =IIF(IsNothing(Parameters!piStartDate.Value) AND IsNothing(Parameters!piEndDate.Value), "UnExtracted ACH Records","ACH Extracted Record")
Isn't the same ask for null that "". In your Report go to property from these Parameter and check the option allow blank value and now your first conditional should work.
A: I wrote the expression like that and it worked for me. Thanks every one for your help.
=IIF(Parameters!piStartDate.Value & Parameters!piEndDate.Value = "", "UnExtracted ACH Records","ACH Extracted Record")
|
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| 2,233
|
Q: Camel unit testing is failing always, I can't explain myself why the expected is 0 or how to assert the expected correctly I have this rest endpoint with Apache Camel to lookup after some values from different sources, first I validate the request then based on path I go to different routes:
restConfiguration()
.component("servlet")
.bindingMode(RestBindingMode.json)
...
rest("/lookup/{path}")
.consumes(MediaType.APPLICATION_JSON_VALUE)
.produces(MediaType.APPLICATION_JSON_VALUE)
.post().type(LookupRequest.class).outType(LookupResponse.class)
.to("direct:validate-request")
from("direct:validate-request")
.routeId(Constants.ROUTE_VALIDATE_REQUEST)
.to("bean-validator://x")
.process(exchange ->
exchange.getIn().setHeader("lookupType", env.getProperty("lookup.path." + exchange.getIn().getHeader("path", String.class)))
)
.to("bean:lookupValidationService?method=validate(${body}, ${header.lookupType})")
.toD("direct:db-lookup-${header.lookupType}");
from("direct:db-lookup-customer")
.routeId(Constants.ROUTE_CUSTOMER_LOOKUP)
.log(LoggingLevel.DEBUG, CamelRouter.CAMEL_BODY)
...
.to("sql:{{lookup.queries.customer}}?placeholder=~&usePlaceholder=true")
.process(new SqlQueryResultProcessor())
...
.endRest();
this is one of my attempts to test the route:
@ActiveProfiles("test")
@CamelSpringBootTest
@SpringBootTest(classes = LookupServiceApplication.class)
@MockEndpointsAndSkip("log:.*")
@UseAdviceWith
class CameltestingApplicationTests {
@Autowired CamelContext camelContext;
@Autowired ProducerTemplate producerTemplate;
@EndpointInject("mock:direct:db-lookup-customer")
MockEndpoint customerMock;
static List<Map<String, Object>> results = new ArrayList<Map<String, Object>>() {{
add(new HashMap<String, Object>() {{
put("customerName", "Test customer");
}});
}};
@BeforeEach
public void before() throws Exception {
camelContext.setTracing(true);
AdviceWith.adviceWith(camelContext, Constants.ROUTE_CUSTOMER_LOOKUP, routeBuilder -> {
routeBuilder.interceptSendToEndpoint("sql:*").skipSendToOriginalEndpoint().process(e -> e.getIn().setBody(results));
});
}
@Test
public void testReceive() throws Exception {
camelContext.start();
// Expect:
customerMock.expectedMessageCount(1);
// And: the process to set the header
customerMock.expectedHeaderReceived("lookupType", "customer.info");
Map<String, Object> headers = new HashMap<>();
headers.put("path", "customer.info");
Map<String, Object> keywords = new HashMap<>();
keywords.put("customerNumber", 1234678);
LookupRequest request = new LookupRequest();
request.setKeywords(keywords);
// When we send the message
producerTemplate.sendBodyAndHeaders("direct:validate-request", request, headers);
// Everything is satisfied.
Assert.assertNotNull(camelContext.hasEndpoint("direct:db-lookup-customer"));
customerMock.assertIsSatisfied();
}
}
this is the log:
2021-07-12 12:10:13.604 INFO 660 --- [ main] o.a.c.i.e.AbstractCamelContext : Apache Camel 3.11.0 (Lookup service) started in 1s246ms (build:160ms init:1s7ms start:79ms)
2021-07-12 12:10:13.623 INFO 660 --- [ main] o.a.c.Tracing : *--> [validate ] [from[direct:validate-request] ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.624 INFO 660 --- [ main] o.a.c.Tracing : [validate ] [bean-validator://x ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.727 INFO 660 --- [ main] o.a.c.Tracing : [validate ] [Processor@0x166ddfb7 ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.728 INFO 660 --- [ main] o.a.c.Tracing : [validate ] [bean:lookupValidationService?meth] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.733 INFO 660 --- [ main] o.a.c.Tracing : [validate ] [direct:db-lookup-${header.lookupT] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.739 INFO 660 --- [ main] o.a.c.Tracing : ---> [customer ] [from[direct:db-lookup-customer] ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.740 INFO 660 --- [ main] o.a.c.Tracing : [customer ] [log ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.740 INFO 660 --- [ main] o.a.c.Tracing : [customer ] [setHeader[customerNumber] ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.891 INFO 660 --- [ main] o.a.c.Tracing : [customer ] [sql:{{lookup.queries.customer}}?p] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.892 INFO 660 --- [ main] o.a.c.Tracing : [customer ] [Processor@0x1feb586d ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupRequest, Body: LookupRequest(keywords={customerNumber=1234678})]
2021-07-12 12:10:13.895 INFO 660 --- [ main] o.a.c.Tracing : [customer ] [Processor@0x34695b23 ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: null, Body: [{customerName=Test customer}]]
2021-07-12 12:10:13.896 INFO 660 --- [ main] o.a.c.Tracing : [customer ] [log ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupResponse, Body: LookupResponse(ok=true, errorMessage=null, results=[{customerName=Test customer}])]
2021-07-12 12:10:13.896 INFO 660 --- [ main] o.a.c.Tracing : <--- [customer ] [from[direct://db-lookup-customer]] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupResponse, Body: LookupResponse(ok=true, errorMessage=null, results=[{customerName=Test customer}])]
2021-07-12 12:10:13.897 INFO 660 --- [ main] o.a.c.Tracing : *<-- [validate ] [from[direct://validate-request] ] Exchange[Id: 32F8D8FB9A7CF84-0000000000000000, BodyType: com.odfl.freight.lookupservice.model.LookupResponse, Body: LookupResponse(ok=true, errorMessage=null, results=[{customerName=Test customer}])]
2021-07-12 12:10:13.900 INFO 660 --- [ main] o.a.c.c.m.MockEndpoint : Asserting: mock://direct:db-lookup-customer is satisfied
...
java.lang.AssertionError: mock://direct:db-lookup-customer Received message count 0, expected at least 1
at org.apache.camel.component.mock.MockEndpoint.fail(MockEndpoint.java:1790)
at org.apache.camel.component.mock.MockEndpoint.assertTrue(MockEndpoint.java:1773)
I'm new with testing Apache Camel routes can somebody please help me to understand where is the problem or how to validate the output as you can see the exchange object body is there and passed. I already tried a few things to mock this endpoint but nothing seems to work and I'm out of ideas.
A: You've mocked only log endpoints with your @MockEndpointsAndSkip("log:.*") and you are injecting customerMock as mock:direct:db-lookup-customer, but if it's not in the route how can it receive a message?
I think all you need is to mock your direct:db-lookup-customer in some way, e.g. within adviceWith or with an annotation
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,077
|
The Baltic Region
2020 Vol. 12 №4
Transformation of global value chains in Russia and the Baltics amid Covid-19: prospects for regionalization and implications for economic policy
Current issue Archive Aims and Scope About the Journal Editorial board Ethical guidleines Peer Review Contacts Author Guide Submit an article online
Simachev Y. V. Fedyunina A. A. Averyanova Y. V.
global value chains Baltic countries Russia Covid-19 pandemic
Although Russia and the Baltics have historically been economic partners, the economic relations between them are tense today. Nearly stagnating bilateral trade contributes little to the development of either side. The Baltics-Russian bilateral trade conducted within global value chains (GVC) and operations of multinational companies is much more resistant to geopolitical and economic shocks than traditional international trade. In particular, the accession of Estonia, Latvia, and Lithuania to the EU and NATO in 2004 and the introduction of reciprocal EU-Russia sanctions in 2014 did not curb GVC activities between Russia and the Baltics. The article discusses factors in the transformation of the Baltics-Russian GVCs amid COVID-19. The research aims to prove regionalisation as a viable prospect for the transformation of global value chains in Russia and the Baltics. In the medium term, regionalisation is possible as (1) part of global trends towards GVC transformation in the industries in which Russia and the Baltics traditionally specialise; (2) as a response to the long-term structural challenges faced by Russia and the Baltics in creating a new generation of internationally competitive firms; (3) as a result of companies tackling the effects of the pandemic against the background of historically stable relationships; (4) as a product of strong social contacts and soft power. GVC regionalisation will be driven by individual companies, regional (local) governments, and Russian-Baltic cross-border cooperation initiatives. Finally, repercussions for Russian and Baltic politics are discussed alongside GVC regionalisation benefits for all the parties involved.
1. Baldwin, R., Freeman, R. 2020, Supply chain contagion waves: Thinking ahead on manufacturing 'contagion and reinfection' from the Covid concussion, VoxEU.org, available at: https://voxeu.org/article/covid-concussion-and-supply-chain-contagion-waves (accessed 8 August 2020).
2. Miroudot, S. 2020, Resilience versus robustness in global value chains: Some policy implications. Covid-19 and trade policy: Why turning inward won't work, VoxEU.org, available at: https://voxeu.org/article/resilience-versus-robustness-global-value-chains (accessed 8 August 2020).
3. Pisch, F. 2020, Just-in-time supply chains after the Covid-19 crisis, VoxEU.org, available at: https://voxeu.org/article/just-time-supply-chains-after-covid-19-crisis (accessed 8 August 2020).
4. Imbs, J., Pauwels, L. 2020, A new measure of openness, VoxEU.org, available at: https://voxeu.org/article/new-measure-openness (accessed 5 August 2020).
5. Banh, H., Wingender, P., Gueye, C. 2020, Global Value Chains and Productivity: Micro Evidence from Estonia, IMF.org, available at: https://www.imf.org/en/Publications/WP/Issues/2020/ 07/03/Global-Value-Chains-and-Productivity-Micro-Evidence-from-Estonia-49376 (accessed 5 August 2020).
6. Hagemejer, J. 2018, Trade and growth in the new member states: The role of global value chains, Emerging Markets Finance and Trade, vol.54, no. 11, p. 2630—2649. doi: https://doi.org/10.1080/1540496X.2017.1369878.
7. Cieslik, E. 2014, Post-communist European countries in global value chains, Ekonomika, vol. 93, no. 3, p. 25—38. doi: https://doi.org/10.15388/Ekon.2014.0.3886.
8. Simachev, Y. V., Daniltsev, A. V., Fedyunina, А. А., Glazatova, М. К., Kuzyk, М. G., Zudin, N. N. 2019, Russia in the changing conditions of world trade: A structural view at the new positioning, Voprosy ekonomiki [Questions of Economics], no. 8, p. 5—29 (In Russ.).
9. Ali-Yrkkö, J., Mattila, J., Seppälä, T. 2017, Estonia in Global Value Chains, ETLA Reports, no. 69, available at: https://pub.etla.fi/ETLA-Raportit-Reports-69.pdf (accessed 8 August 2020).
10. Yashiro, N., De Backer, K., Hutfilter, A., Kools, M., Smidova, Z. 2017, Moving up the global value chain in Latvia, OECD Publishing, Paris, 2017. doi: https://doi.org/10.1787/3a486c5e-en.
11. Klemeshev, A., Fedorov, G., Zverev, Yu. 2011, On the potential and opportunities for cooperation between the Baltics in the field of innovations, Balt. Reg., no. 3, p. 76—84. doi: https://doi.org/10.5922/2079-8555-2011-3-9.
12. Veebel, V., Markus, R. 2008, The bust, the boom and the sanctions in trade relations with Russia, Journal of International Studies, vol. 11, no. 1, p. 9—20. doi: https://doi.org/10.14254/2071-8330.2018/11-1/1
13. Fedorov, G. M. 2018, Russian Federation in the Baltic Region: Political Relations and Economic Development in 1992—2017, Polis. Politicheskie issledovaniya [Policy. Political studies], no. 3, p. 30—41. doi: https://doi.org/10.17976/jpps/2018.03.03 (In Russ.)
14. Djankov, S., Freund, C. 2020, Trade flows in the former Soviet Union, 1987 to 1996, Journal of Comparative Economics, vol. 30, no. 1, p. 76—90. doi: https://doi.org/10.1006/jcec.2001.1752
15. Frankel, J., Stein, E., Wei, S. 1997, Regional Trading Blocs in the World Economic System, Washington DC, Institute for International Economics.
16. Eichengreen, B., Irwin, D. 1996, The Role of History in Bilateral Trade Flows, Cambridge, MA: NBER, doi: https://doi.org/10.3386/w5565.
17. Frankel, J., Stein, E., Wei, S. 1995, Trading Blocs and the Americas: The Natural, the Unnatural, and the Super-natural, Journal of Development Economics, vol. 47, no. 1, p. 61—95. doi: https://doi.org/10.1016/0304-3878(95)00005-4.
18. Freund, C. 2000, Different Paths to Free Trade: The Gains from Regionalism, The Quarterly Journal of Economics, vol. 115, no. 4, p. 1317—1341, available at: www.jstor.org/stable/2586926 (accessed 6 August 2020).
19. Berg E., Ehin P. 2016, Identity and foreign policy: Baltic-Russian relations and European integration, Routledge, 2016.
20. Fedorov, G., Korneevets, V., Tarasov, I., Chasovskiy, V. 2016, Russia among the Countries of the Baltic Region, International Journal of Economics and Financial Issues, vol. 6, no. 4, p. 1502—1506.
21. Bělín, M., Hanousek, J. 2020, Which sanctions matter? Analysis of the EU/Russian sanctions of 2014, Journal of Comparative Economics. doi: https://doi.org/10.1016/j.jce.2020.07.001.
22. Dreger, C., Kholodilin, K., Ulbricht, D., Fidrmuc, J. 2016, Between the hammer and the anvil: The impact of economic sanctions and oil prices on Russia's ruble, Journal of Comparative Economics, vol. 44, no. 2, p. 295—308. doi: https://doi.org/10.1016/j.jce.2015.12.010.
23. Yeliseyeu, A. 2017, Belarusian shrimps anyone? How EU food products make their way to Russia through Belarus, Think Visegrad — V4 Think Tank Platform, available at: https://www.globsec.org/wp-content/uploads/2017/11/Think-Visegrad-Analysis-by-Andrei-Yeliseyeu_Belar... (accessed 5 August 2020).
24. Miroudot, S. 2017, The servicification of global value chains: Evidence and policy implications, UNCTAD Multi-Year Expert Meeting On Trade, Services and Development, available at: https://unctad.org/meetings/en/Presentation/c1mem5_2017_124_S3_Miroudot_2.pdf (accessed 5 August 2020).
25. Broadman, H. 2006, From disintegration to reintegration: Eastern Europe and the former Soviet Union in international trade, Europe and Central Asia reports, Washington, D.C., World Bank Group, available at: http://documents.worldbank.org/curated/en/713901468252863858/From-disintegration-to-reintegration-Ea... (accessed 5 August 2020).
26. Simachev, Y. V., Fedyunina, А. А., Kuzyk, М. G., Daniltsev, А. V., Glazatova, М. К., Averyanova, Y. V. 2020, Rossiya v global'nom proizvodstve [Russia in Global Production]. In: XXI April International Scientific Conference on the Development of Economy and Society, HSE Publishing House, p. 1—147 (In Russ.).
27. Sineviciene, L., Krusinskas, R. 2018, How Dependent Are the Baltic States on Russia? Europe-Asia Studies, vol. 70, no. 8, p. 1264—1280. doi: https://doi.org/10.1080/09668136.2018.1508643.
28. Zvaigzene, A. 2015, EU sanctions against Russia renewed: pros and cons for business and Baltics economy, The Baltic Course, available at: http://www.baltic-course.com/eng/analytics/?doc=114794 (accessed 5 August 2020).
29. Ratso, S. 2015, EU-Russia Trade Relations in Light of Sanctions and Russia´s Import Measures, Diplomaatia, available at: http://www.diplomaatia.ee/en/article/eu-russian-trade-relations-in-light-of-sanctionsand-russias-imp... (accessed 6 August 2020).
30. Morozenkova, О. V. 2019, Prospects for the development of Russian exports of non-primary non-energy products in new markets, Rossiiskii vneshneekonomicheskii vestnik [Russian Foreign Economic Bulletin], no. 9, p. 44—60 (In Russ.).
31. Efremenko, I. N., Karaya, Y. V. 2013, Development of integration processes in the Baltic region countries, Vestnik Rostovskogo gosudarstvennogo ekonomicheskogo universiteta (RINKh) [Bulletin of the Rostov State University of Economics], no. 1—2, p. 126—132 (In Russ.).
32. Khodachek, А. M. 2010, Transgranichnoe sotrudnichestvo makroregiona Severo-Zapad [Cross-border cooperation of the North-West microregion], HSE Saint Petersburg Publishing House (In Russ.).
33. Zhilina, L. N. 2018, Russian Diaspora in the Baltic States and its Role in Russian Foreign Policy, Vestnik diplomaticheskoi akademii MID Rossii. Rossiya i mir [Bulletin of the Diplomatic Academy of the Russian Foreign Ministry. Russia and the world], no. 2, p. 130—116 (In Russ.).
34. Vorotnikov, V. V. Ivanova, N. A. Russian soft power in the Baltic States through the lens of research: traditions, competition, confrontation, Balt. Reg., vol. 11, no. 3, p. 107—124. doi: https://doi.org/10.5922/2079-8555-2019-3-6.
35. Baldwin, R. 2020, The Greater Trade Collapse of 2020: Learnings from the 2008—09 Great Trade Collapse, VoxEU.org, available at: https://voxeu. org/article/greater-trade-collapse-2020 (accessed 5 August 2020).
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
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| 2,147
|
{"url":"https:\/\/sources.debian.org\/src\/r-cran-clubsandwich\/0.5.3-1\/man\/impute_covariance_matrix.Rd\/","text":"## File: impute_covariance_matrix.Rd\n\npackage info (click to toggle)\nr-cran-clubsandwich 0.5.3-1\n 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114 % Generated by roxygen2: do not edit by hand % Please edit documentation in R\/rma-mv.R \\name{impute_covariance_matrix} \\alias{impute_covariance_matrix} \\title{Impute a block-diagonal covariance matrix} \\usage{ impute_covariance_matrix( vi, cluster, r, ti, ar1, smooth_vi = FALSE, subgroup = NULL, return_list = identical(as.factor(cluster), sort(as.factor(cluster))), check_PD = TRUE ) } \\arguments{ \\item{vi}{Vector of variances} \\item{cluster}{Vector indicating which effects belong to the same cluster. Effects with the same value of cluster will be treated as correlated.} \\item{r}{Vector or numeric value of assumed constant correlation(s) between effect size estimates from each study.} \\item{ti}{Vector of time-points describing temporal spacing of effects, for use with auto-regressive correlation structures.} \\item{ar1}{Vector or numeric value of assumed AR(1) auto-correlation(s) between effect size estimates from each study. If specified, then \\code{ti} argument must be specified.} \\item{smooth_vi}{Logical indicating whether to smooth the marginal variances by taking the average \\code{vi} within each cluster. Defaults to \\code{FALSE}.} \\item{subgroup}{Vector of category labels describing sub-groups of effects. If non-null, effects that share the same category label and the same cluster will be treated as correlated, but effects with different category labels will be treated as uncorrelated, even if they come from the same cluster.} \\item{return_list}{Optional logical indicating whether to return a list of matrices (with one entry per block) or the full variance-covariance matrix.} \\item{check_PD}{Optional logical indicating whether to check whether each covariance matrix is positive definite. If \\code{TRUE} (the default), the function will display a warning if any covariance matrix is not positive definite.} } \\value{ If \\code{cluster} is appropriately sorted, then a list of matrices, with one entry per cluster, will be returned by default. If \\code{cluster} is out of order, then the full variance-covariance matrix will be returned by default. The output structure can be controlled with the optional \\code{return_list} argument. } \\description{ \\code{impute_covariance_matrix} calculates a block-diagonal covariance matrix, given the marginal variances, the block structure, and an assumed correlation structure. Can be used to create compound-symmetric structures, AR(1) auto-correlated structures, or combinations thereof. } \\details{ A block-diagonal variance-covariance matrix (possibly represented as a list of matrices) with a specified structure. The structure depends on whether the \\code{r} argument, \\code{ar1} argument, or both arguments are specified. Let \\eqn{v_{ij}}{v-ij} denote the specified variance for effect \\eqn{i}{i} in cluster \\eqn{j}{j} and \\eqn{C_{hij}}{C-hij} be the covariance between effects \\eqn{h}{h} and \\eqn{i}{i} in cluster \\eqn{j}{j}. \\itemize{ \\item{If only \\code{r} is specified,}{ each block of the variance-covariance matrix will have a constant (compound symmetric) correlation, so that \\deqn{C_{hij} = r_j \\sqrt{v_{hj} v_{ij},}}{C-hij = r-j * sqrt(v-hj v-ij),} where \\eqn{r_j}{r-j} is the specified correlation for cluster \\eqn{j}{j}. If only a single value is given in \\code{r}, then it will be used for every cluster.} \\item{If only \\code{ar1} is specified,}{ each block of the variance-covariance matrix will have an AR(1) auto-correlation structure, so that \\deqn{C_{hij} = \\phi_j^{|t_{hj} - t_{ij}|} \\sqrt{v_{hj} v_{ij},}}{C-hij = (ar1-j)^|t-hj - t-ij| * sqrt(v-hj v-ij),} where \\eqn{\\phi_j}{ar1-j} is the specified auto-correlation for cluster \\eqn{j}{j} and \\eqn{t_{hj}}{t-hj} and \\eqn{t_{ij}}{t-ij} are specified time-points corresponding to effects \\eqn{h}{h} and \\eqn{i}{i} in cluster \\eqn{j}{j}. If only a single value is given in \\code{ar1}, then it will be used for every cluster.} \\item{If both \\code{r} and \\code{ar1} are specified,}{ each block of the variance-covariance matrix will have combination of compound symmetric and an AR(1) auto-correlation structures, so that \\deqn{C_{hij} = \\left[r_j + (1 - r_j)\\phi_j^{|t_{hj} - t_{ij}|}\\right] \\sqrt{v_{hj} v_{ij},}}{C-hij = [r-j + (1 - r-j)(ar1-j)^|t-hj - t-ij|] * sqrt(v-hj v-ij),} where \\eqn{r_j}{r-j} is the specified constant correlation for cluster \\eqn{j}{j}, \\eqn{\\phi_j}{ar1-j} is the specified auto-correlation for cluster \\eqn{j}{j} and \\eqn{t_{hj}}{t-hj} and \\eqn{t_{ij}}{t-ij} are specified time-points corresponding to effects \\eqn{h}{h} and \\eqn{i}{i} in cluster \\eqn{j}{j}. If only single values are given in \\code{r} or \\code{ar1}, they will be used for every cluster.} } If \\code{smooth_vi = TRUE}, then all of the variances within cluster \\eqn{j}{j} will be set equal to the average variance of cluster \\eqn{j}{j}, i.e., \\deqn{v'_{ij} = \\frac{1}{n_j} \\sum_{i=1}^{n_j} v_{ij}}{v-ij' = (v-1j + ... + v-nj,j) \/ n-j} for \\eqn{i=1,...,n_j}{i=1,...,n-j} and \\eqn{j=1,...,k}{j=1,...,k}. } \\examples{ library(metafor) # Constant correlation data(SATcoaching) V_list <- impute_covariance_matrix(vi = SATcoaching$V, cluster = SATcoaching$study, r = 0.66) MVFE <- rma.mv(d ~ 0 + test, V = V_list, data = SATcoaching) conf_int(MVFE, vcov = \"CR2\", cluster = SATcoaching\\$study) }","date":"2021-09-17 04:57:04","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6015026569366455, \"perplexity\": 7913.176186152766}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-39\/segments\/1631780054023.35\/warc\/CC-MAIN-20210917024943-20210917054943-00498.warc.gz\"}"}
| null | null |
Course on Justice by Harvard University [12 Weeks, Online]: Enroll Now!
Taught by lauded Harvard professor Michael Sandel, Justice explores critical analysis of classical and contemporary theories of justice, including discussion of present-day applications. Topics include affirmative action, income distribution, same-sex marriage, the role of markets, debates about rights (human rights and property rights), arguments for and against equality, dilemmas of loyalty in public and private life. The course invites learners to subject their own views on these controversies to critical examination.
The principal readings for the course are texts by Aristotle, John Locke, Immanuel Kant, John Stuart Mill, and John Rawls. Other assigned readings include writings by contemporary philosophers, court cases, and articles about political controversies that raise philosophical questions.
What you'll learn?
The fundamentals of political philosophy
An understanding of social justice and criminal justice, and the roles they play in the modern justice system
A deeper sense of the philosophy that underlies modern issues such as affirmative action, same-sex marriage, and equality
The ability to better articulate and evaluate philosophical arguments and ask philosophical questions
Michael J. Sandel Anne T. and Robert M. Bass Professor of Government Harvard University
Course on Justice by Harvard University
JOB POST: Project Manager @ NLU Delhi's Center on the Death Penalty: Salary of Rs. 60,000-70,000; Apply by Feb 22
Ambedkar University, Delhi's MA Programme in Law, Politics and Society: Apply by June 23
Workshop on Right to Information Act 2005 at SLS, Hyderabad [Feb 28]: Submit by Feb 22
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
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| 3,125
|
Moure ist der Name mehrerer Orte und Gemeinden in Portugal:
Moure (Barcelos), Gemeinde im Concelho de Barcelos
Moure (Felgueiras), Ort und ehemalige Gemeinde im Concelho de Felgueiras
Moure (Póvoa de Lanhoso), Ort und ehemalige Gemeinde im Concelho de Póvoa de Lanhoso
Moure (Vila Verde)
Moure ist der Familienname folgender Personen:
Jesús Santiago Moure (1912–2010), brasilianischer Entomologe
Xosé Alvilares Moure (1928–2015), galicischer Schriftsteller, Philosoph und Priester
Luis Moure-Mariño (1915–1999), spanischer Schriftsteller
Manuel Magallanes Moure (1878–1924), chilenischer Schriftsteller und Literaturkritiker
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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| 2,233
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Q: Is server to server email transmission secure? We are running a website with a web to email form. We force https to secure the transmission from the browser to the web server. From the web server we use PHP to generate an email message containing the information received. The destination email server is in another location. I am wondering if the transmission of the email from the web server to the destination email server is secure.
So I have several related questions:
A) Is the transmission of emails from the sender server to the recipient server secured by default? (To be clear, this is NOT a question about the connection from an email client to an SMTP server)
B) if the transmission is not secured by default, how can I check if messages are being transmitted securely?
C) if transmission is currently not secured, how can I request/force a secure connection?
We are using the default PHP mail function (via a Joomla extension)
Thank you!
A: A) initially, a server-to-server smtp connection is always in plain text on port 25.
if both source and target server support the TLS extension then usually the plain connection gets converted into a encrypted connection with the STARTTLS command
B) To check if a mail was transmitted over an encrypted connection, read the "Received"-Headers in the resulting message after it was transmitted.
They look like this:
Received: from X.example.com (X.example.com [y.y.y.y])
by z.example.net (Postfix) with ESMTPS id ......
The important part is the ESMTPS bit. The last S means "SECURED".
If it just says "ESMTP" or "SMTP" instead of "ESMTPS" the transmission was not encrypted.
C) if the target server does not support TLS there is nothing you can do except some sort of end-to-end encryption like PGP (as suggested by Álvaro G. Vicario). Some servers (like postfix) provide configuration options to prevent messages from going out at all if the target can not do TLS.
you can test manually with telnet if a server supports STARTTLS:
telnet gmail-smtp-in.l.google.com 25
Trying 173.194.70.27...
Connected to gmail-smtp-in.l.google.com.
Escape character is '^]'.
220 mx.google.com ESMTP 4si1878861eee.197 - gsmtp
EHLO mail.example.com <--- you have to type that
250-mx.google.com at your service
250-SIZE 35882577
250-8BITMIME
250-STARTTLS <----- GMAIL supports TLS
250 ENHANCEDSTATUSCODES
A: In fact, I think it is a question about connection from an e-mail client to SMTP server after all. When source server connects to destination server to deliver a message it becomes a client. It uses the SMTP protocol just like your desktop e-mail program. There's a very important difference, though:
*
*When you use a regular e-mail client you normally use the same trusted provider that can (and should) offer certain security measures, including authentication and encryption.
*When a mail server connects to a third-party server, it connects anonymously and it needs to establish an unencrypted connection to port 25. They don't have a prior agreement to do things otherwise.
Given that the channel is in clear, server-to-server communication is not secure unless the message itself is encrypted (PGP or whatever). You can think of e-mail messages like snail mail post-cards.
(For this reason, those sites that e-mail your password in clear when you sign up are doing it wrong.)
A: With due respect to the very good answers so far, in Joomla! the SMTP mail function is handled by JMail which extends from the PHPMailer class. When setting up Joomla! you have three different options:
*
*PHP Mail - uses PHP mail() settings via PHPMailer.
*Sendmail - uses sendmail via PHPMailer...
*SMTP - uses PHPMailer class
The PHPMailer class supports both tls and ssl in it's SMTP connection negotiation.
Of course, this is dependent on you setting it up your in Global Configuration->Server. In the pane titled Mail Settings you can turn on SMTP Security (SSL|TLS) and provide your Username and Password. These details are required to authenticate with your SMTP server.
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\section{Introduction}
This paper\footnote{Thanks are due to our colleagues and the anonymous referees for their
useful comments.} considers how to infer a topological representation of an
environment from a trace of {\em place information}. By {\em place
information} we assume that this takes the form of a finite set of
{\em places names}, of unknown location, and a time series specifying
when a mobile agent is at these locations. The places may overlap
spatially (and thus more than one place name may be simultaneously
``active''), but are assumed to be unique (not spatially identical).
One example of such data (which was our original motivation) are the
firing traces of {\em place cells} (PCs) of a rat hippocampus after it has
learned a particular environment
\cite{ODbook}.
After a period of exposure to a particular spatial
environment, cells in the rat
hippocampus are reliably associated with particular regions of space,
{\em place fields} (PFs) such that the PCs ``fire'' (i.e. have a firing
frequency above a certain threshold) when the rat is in the
PF, and only then\footnote{Predictive firings have been reported,
but we ignore that complexity here.}.
Can the spatial layout of
the environment be recovered purely by inspection of the firing
activity of the PCs? Recovering a topological map from
PC data in this way has been called the ``space
reconstructing thought experiment'' (SRE)\cite{archive}. This paper
builds on this unpublished report \cite{archive} where the relevant
neuroscience literature is surveyed in detail. \cite{archive} also suggests the
use of homology theory and mereotopology to analyse the PC data. Here
we explore the second suggestion.
This problem may occur in other domains; one example is the set
of visible SSIDs\footnote{In the case of multiple base stations with
the same SSID, the mac address can be used to distinguish them.
} of wifi base stations\cite{wifiref1,wifiref2}.
We take the position here that places are regions rather than points,
not only because it may be hard to determine when the agent is at the
central ``focus'' of a place, but also because this naturally fits
with the two domains mentioned above. This naturally suggests the use of
a region based representation such as RCC\cite{CohnRenz}.
In both these domains it is unlikely that regions will ever
be tangentially connected (the {\sf EC} or {\sf TPP} relations of the
RCC mereotopological calculus, or similar
relations in other calculi \cite{CohnRenz}). Thus we will
use a purely mereological
calculus, RCC-5\cite{CohnRenz}, illustrated in fig. \ref{rcc5}.
Connectivity between places implies that two regions at least
partially overlap ({\sf PO}).
The RCC relations can all be defined in terms of the connection primitive {\sf C}($x,y$)
\cite{CohnRenz}, ``$x$ is connected to $y$, as can
a predicate {\sf Con}($x$),
which is true when $x$ is a simply
connected region (i.e. is one-piece), and a predicate {\sf P}($x,y$)
which is true when $x$ is a part of $y$ (i.e. either {\sf EQ}($x,y$)
or {\sf PP}($x,y$) holds).
\begin{figure}[tbp]
\begin{center}
\includegraphics{rcc5a.eps}
\end{center}
\caption{The RCC5 relations, from the left: {\sf DC, EC, PO, PP, PPi, EQ}.}
\label{rcc5}
\end{figure}
The underlying question we are trying to answer here, is whether and
how can a topological map be extracted from the sparse time
series data indicated above. In particular we deliberately assume that
there is no information as to the agent's actual metrical movements,
orientation, odometry or heading. This might be because such
information is hard to obtain or compute, but seems to be an
interesting challenge in its own right.
\label{related}
The idea of computing a topological representation of the environment
of a mobile agent is not new, and has been the subject of
active research in robotics.
The work on the
``spatial semantic hierarchy'' is perhaps the most relevant here
\cite{Remolina}. This framework provides a
multilevel approach to computing a topological representation, taking
account of metrical, geometrical information and potentially a wide
variety of sensor readings. Topological information is thus inferred
after several layers of processing, and ``describes the environment as
a collection of places, paths and regions, linked by topological
relations, such as connectivity, order, boundary and
containment''. Places are always points, paths are one dimensional,
and regions two dimensional, e.g. defined by a closed
loop of paths, or by abstraction of a group of places. The process of
inferring the topology of these entities is by abduction using
circumscription: the underlying idea is to abduce a
minimal topological description which explains the underlying sensor
data.
Although relevant to the considerations in this paper, the spatial
semantic hierarchy framework differs in several respects; in
particular it assumes, and makes use of non topological information
and it allows for the possibility of aliasing (i.e. that place names
are not necessarily unique). Finally, we mention the Ratslam approach
to simultaneous location and mapping (SLAM) inspired by the neuroscience work
on the rat hippocampus \cite{ratslam3}. This
uses vision, odometry and neural networks to produce a topological map based on
PCs and is contrasted with the more usual approach to SLAM using a
probabilistic approach based on particles.
Ratslam uses both head direction and
PCs (i.e. both directional and positional information) to produce
its maps. Here we only use PCs in order to
concentrate on the purely topological aspects of the problem.
\section{Computing Connectivity}
\label{computingC}
We assume that we have a finite set of $k$ places, $p_i$,
distinguished by a predicate {\sf Place}($p_i$), whose intended
interpretations are regions of 2D space.
We also assume a finite set of $n$ times, with a primitive
ordering relation $t_1 < t_2$ which specifies when $t_1$ is temporally
before $t_2$ and a predicate {\sf At}($t,p$) which specifies whether
the agent is at place $p$ at time $t$. For the present, we also assume
that time is sufficiently fine grained such that no place transitions
are missed, i.e. not recorded. We will return to this assumption below
in \S\ref{partial}.
If the agent is at two places simultaneously then we can infer that they
are connected:
$\forall x \forall y \exists t [{\sf At}(t,x) \wedge
{\sf At}(t,y)] \rightarrow {\sf C}(x,y) $
\noindent
Note that for the converse to hold, we would have to assume that the
agent had made a complete exploration of the environment, in the sense
that every actually physically connected pairs of places had actually
been visited at consecutive times. For the case of rats running
experimental mazes in laboratories, this is certainly the case after a
relatively short period of time. For large scale geographic
environments this may not hold, though it may be a reasonable ``closed
world'' assumption to make, until contrary evidence comes in,
requiring a {\em non monotonic} revision of the topological map. Of
course map revisions may have to be made in any case as a result of
structural changes in the environment or the number or liveliness of
the place indicators.
The set of pairwise place connections can clearly can be computed in
at most $k^2*n$ time, since it simply requires that each time point is
scanned in turn, checking all pairs of places for whether they are
simultaneously active.
In some domains, determining whether the agent is currently at a
particular place may be more problematic. E.g. in the case of
hippocampal PFs, a certain threshold of firing frequency is
required, or in the wifi domain a minimum signal level might be
required. Such additional constraints would need to be factored into
the computation of {\sf At}. This might be a global threshold, or it
may be a local (spatial) threshold to the particular place, or a
temporal threshold (i.e. different thresholds might be applicable at
different times, perhaps due to changing environmental conditions). We
will not consider this further in this paper, though we note that
calculi for reasoning about regions with indeterminate boundaries,
such as the ``egg-yolk'' calculus\cite{CohnRenz} may be relevant.
From the predicate {\sf C}, it is straightforward to build a
connectivity graph, specifying which regions/places are connected
through the overlap relation, and thus the connection relation
{\sf C}($x,y$) holds between them. From the point of view of having a
representation from which it is possible e.g. to do path
planning, this is all that is
needed. But it would be useful to be able to determine the overall
``topological shape'' of the environment as exposed by the time series
data. Below we show how this might be done; we are not
proposing the particular definitions and concepts below as final, but
rather as illustrative of the kind of analysis possible.
We first consider environments which are essentially linear, such as
the typical mazes run by rodents.
These might be simple linear runs, or with junctions (e.g. in the
shape of a {\sf T, W, H}),or more complex mazes, with loops (e.g. {\sf
O, 8, 9}). In a strictly topological sense, the first set of
examples above are all topologically identical -- they can all be
shrunk to a point. However, by subdividing the shapes into parts
(corresponding to the simple linear stretches), and then specifying
the connectivity between these parts, these shapes can all be
distinguished -- see fig. \ref{simpleshapes}\footnote{Note that this
figure, and other similar ones later in the paper are intended to be
purely illustrative, rather than realistic configurations of actual
PFs recorded from real rats, or from wifi recordings.}, where
three shapes and their connection structure are depicted. Shapes (i)
and (ii) are topologically identical in the sense that they can both
be shrunk to a point, but the connection structure of their places is
different. The connection structure can also of course be given
purely symbolically; e.g. (i) is: {\sf C}(a,b), {\sf C}(b,c), {\sf
C}(b,d), {\sf C}(d,e), and {\sf DC}($\alpha,\beta$), for all other
pairs of regions $\alpha,\beta$. Visualizing this purely symbolic
structure is an interesting problem which we return to below.
\begin{figure}[tbp]
\begin{center}
\includegraphics{shapes3.eps}
\end{center}
\caption{ Three 1D environments formed by overlapping places -- the
darker areas show overlap between regions;
to the right of each shape is the connection structure constructed by the
overlaps between places, which
can be distinguished by virtue of their connectivity
structure.}
\label{simpleshapes}
\end{figure}
In order to formally analyse such linear structures we can make the
following definitions, which group together
collections of regions to form higher level abstractions\footnote{$\exists!^n \alpha
\Phi(\alpha)$ is syntactic sugar for ``there exist exactly $n$
$\alpha$ s.t. $\Phi(\alpha)$''. $\exists^n \alpha
\Phi(\alpha)$ is syntactic sugar for ``there exist at least $n$
$\alpha$ s.t. $\Phi(\alpha)$''.}.
\noindent ${\sf LinearSegment}(x) \equiv_{def}
{\sf Con}(x) \wedge {\sf SumOfPlaces}(x) \wedge$\\
\hspace*{0.2cm}$\forall y [[{\sf Place}(y) \wedge {\sf P}(y,x)] \rightarrow
[{\sf End}(y) \vee {\sf Middle}(y) \vee {\sf Junction}(y)]] \wedge$
\\
\hspace*{0.2cm}$\exists!^2 y [{\sf Place}(y) \wedge {\sf P}(y,x) \wedge
[{\sf End}(y) \vee {\sf Junction}(y)]]
$
\noindent ${\sf End}(x) \equiv_{def}
{\sf Place}(x) \wedge \exists!^1 y [
{\sf Place}(y) \wedge {\sf C}(x,y) \wedge \neg {\sf EQ}(y,x) ]
$
\noindent ${\sf Middle}(x) \equiv_{def}
{\sf Place}(x) \wedge \exists!^2 y [
{\sf Place}(y) \wedge {\sf C}(x,y) \wedge \neg {\sf EQ}(y,x)]
$
\noindent ${\sf Junction}(x) \equiv_{def}
{\sf Place}(x) \wedge \exists^3 y [
{\sf Place}(y) \wedge {\sf C}(x,y) \wedge \neg {\sf EQ}(y,x)]
$
\noindent
$
{\sf SumOfPlaces}(x) \equiv_{def}$\\
\hspace*{0.2cm}$\neg \exists y [{\sf P}(y,x) \wedge \forall z [[
{\sf Place}(z) \wedge{\sf P}(z,x)] \rightarrow {\sf DC}(y,z)]]
$
Thus {\sf End}s are places only connected to one other place, {\sf
Middle}s only to two other places, and {\sf Junction}s are connected
to three or more other places; {\sf LinearSegment}s are composed of
two places being either {\sf End}s or {\sf Junction}s and all other
places in the {\sf LinearSegment} are {\sf Middle}s. The predicate
{\sf SumOfPlaces}($x$) ensures that $x$ is a region every part of
which is part of some place, so that there are no ``extra bits''
of space which are part of $x$ but not part of some place.
These definitions achieve the desired effect in linear environments
providing that no places are long enough to overlap more than one
place in each direction (i.e as in fig. \ref{doublePO}(ii)). E.g.,
consider fig. \ref{doublePO}(i) -- the layout is still
clearly in some sense linear, but as the connection graph shows, it is
not so in any very straightforward way.
\begin{figure}[tbp]
\begin{center}
\includegraphics{doublePO-IP-combined1.eps}
\end{center}
\caption{A linear track composed of (i) doubly overlapping places, \&
(ii) singly overlapping places; beneath are the connection
structures formed by the overlaps. (iii): An environment with two
maximal induced paths: abdfg \& acdfg. }
\label{doublePO}
\label{multipleIPs}
\end{figure}
One approach to this might be
to use the idea of an {\em induced path}
-- i.e. a
sequence of nodes in a graph such that each is connected to two
neighbours in the path, except for the two end nodes, and with the
proviso that there are no ``short cuts'' (i.e. direct links) between
any two nodes in the path using other edges in the graph (which would
be indicated by the presence of a 3-clique)\footnote{In the definition
below, we also rule out trivial induced paths of length one
(i.e. with just two nodes and one edge).}. Thus in the
graph in fig. \ref{doublePO}(i), abde, ace, bce, acd, are all (maximal)
induced paths. However this notion is not entirely satisfactory
since it does not define a unique path for the environment (since there
are two induced paths from a to e). In this case we can note that
there is a unique longest induced path (abcd); of the others, ace has
the same start and end nodes, and the other two each have an end in
common with the longest. None are disjoint. We also note that all the
nodes in the non longest induced paths are within one edge of a node
on the longest induced path (i.e. c is directly linked to a,b,d and
e). It is not guaranteed that there is a unique longest induced path
-- either because there are two entirely disjoint such paths, or
because there are alternatives with common nodes -- e.g. see fig.
\ref{multipleIPs}(iii). We thus propose the following definition for what we will call
{\em quasi linear} segments.
\begin{figure}[tbp]
\begin{center}
\includegraphics{mereo.eps}
\end{center}
\caption{A configuration in which by considering place
trails from than two places simultaneously, a refined map
can be obtained.}
\label{mereofig}
\end{figure}
\noindent${\sf QuasiLin}(x) \equiv_{def}$\\
\hspace*{0.2cm}$\exists w [{\sf P}(w,x) \wedge {\sf InducedPath}(w) \wedge {\sf
SumofPlaces}(x) \wedge$\\
\hspace*{0.4cm}$\forall y [[{\sf P}(y,x-w) \wedge {\sf Place}(y) ] \rightarrow {\sf C}(y,w)]]
$
\noindent$ {\sf InducedPath}(w) \equiv_{def}
$\\
\hspace*{0.2cm}${\sf SumOfPlaces}(w) \wedge {\sf Con}(w) \wedge
\neg \exists v [{\sf 3Clique}(v) \wedge {\sf P}(v,w)] \wedge$\\
\hspace*{0.2cm}$\exists (s,x,y,z) [{\sf Place}(x) \wedge
{\sf Place}(y) \wedge \neg {\sf EQ}(x,y) \wedge {\sf Place}(s) \wedge {\sf P}(s,z) \wedge
$ \\
\hspace*{0.2cm}$ {\sf EQ}(w,x+y+z) \wedge
\forall u [[{\sf Place}(u) \wedge {\sf P}(u,z)] \rightarrow \neg {\sf Con}(w-u)]]
$
\noindent$ {\sf 3Clique}(w) \equiv_{def} \exists (x,y,z) [ $\\ \hspace*{0.2cm}${\sf Place}(x) \wedge
{\sf Place}(y) \wedge
{\sf Place}(z) \wedge {\sf EQ}(w,x+y+z) \wedge \neg {\sf EQ}(x,y) \wedge
$\\
\hspace*{0.4cm}$
\neg {\sf EQ}(x,z) \wedge
\neg {\sf EQ}(y,z) \wedge
{\sf C}(x,y) \wedge
{\sf C}(x,z) \wedge
{\sf C}(z,y)]
$
Returning to fig. \ref{doublePO}(i), ${\sf QuasiLin}$(a+b+c+d+e) is
true since it has at least one induced path in it, and all the other
places in it are directly connected to a place in that induced
path. Exactly what definition of ``linearity'' might be appropriate in
a particular domain
will depend on what is required or suitable for the application.
E.g., ${\sf QuasiLin}(x)$ will be true if $x$ is the union of a
simple path of length $m$ and a clique of size $n$ s.t. exactly two
nodes are in common between the clique and the simple path. It would
be straightforward to eliminate this case, or to allow only cliques of
up to a size $m$ to occur within a quaslinear path.
Another important grouping of places are ``open spaces'' which are
likely to be identified by clusters of many places, though not
necessarily in the form of a clique (though they may well contain
cliques). Having identified cliques, and linear segments, a natural
step would be to replace these by ``super nodes'', and then continue
analysing the environment at this more abstract level. Indeed, there
are already existing approaches to analysing and drawing graphs which
take this approach, e.g. \cite{muller,Friedrich}. E.g., ``open
spaces'' may frequently be represented as cliques of cliques.
One other aspect of connectivity analysis not so far mentioned
explicitly is determining whether there are cycles in the environment
(caused by circular structures in a linear environment, or by
obstacles in an open environment). A graph theoretic approach to this
would be to look for {\em chordless cycles}, i.e. circular induced
paths of length at least four. As a trivial example of this, consider
fig. \ref{simpleshapes}(ii); in practice this approach is likely to require
refinement to properly capture the required notion.
\section{Variants of the mapping task}
\label{variants}
In this section we consider a number of variants of the basic mapping
task and how they might be achieved.
\label{mereo}
\begin{sloppypar}
\noindent {\bf Computing Connectivity Mereologically}:
The approach outlined above shows how we can compute connection
information from knowledge that the agent is simultaneously {\sf At}
two places. This gives rise to binary connectedness information, and
thence (in RCC-5) the knowledge that particular {\em pairs} of places
partially overlap ({\sf PO}). It is not possible from the connection
structure though to infer that any triple (quadruple...) of places
overlap, even though this information might in fact be readily
apparent in the place trail (indicated by an agent being
simultaneously at three (four...) places). We could therefore compute
a more fine grained representation, in which we explicitly represent
those intersections of places which are known to exist, and similarly
those relative complements of places known to exist. E.g. from the
place trail: {\sf At}(t1,a), {\sf At}(t2,a), {\sf At}(t2,b), {\sf
At}(t3,a), {\sf At}(t3,b), {\sf At}(t3,c), {\sf At}(t4,b), {\sf
At}(t4,c), {\sf At}(t5,b), {\sf At}(t5,c), {\sf At}(t5,d), {\sf
At}(t6,d), using pairwise connections, as in \S\ref{computingC},
then a connection structure as in fig. \ref{mereofig}(i) would be
computed. Not knowing about which subregions can actually exist, it
would be reasonable to produce a map such as in fig.
\ref{mereofig}(ii). However, by inspecting the above place trail, we
can infer that the regions a--b--c--d, a+b--c--d, a+b+c--d, b+c-a--b,
b+c+d--a, and d--a--b--c all exist, but there is no evidence to
support the existence of any of the other five Boolean combinations of
the four places, a, b, c, d which exist in fig. \ref{mereofig}(ii).
Thus we can build a simplified map as in fig. \ref{mereofig}(iii),
in which these regions do not exist (indicated by shading).
\end{sloppypar}
\noindent {\bf Partial information}:
\label{partial}
It is believed that PCs form a cover for the environment the
rat has explored, i.e. wherever the rat is, at least one PC
will be firing. However for the SRE, an
external observer is receiving signals from a set of electrodes, and
only a subset of the PCs will actually be recorded and thus only
partial information about the set of places active at any
time. At some times, there may be no active PCs
being recorded, and thus there will be ``temporal gaps''. The question
is, what can we say about the nature of the environment given only
such partial information? In the wifi domain, presumably
full information would always be available; there might still be
temporal gaps, because no base station is in range, but that is
different to not being able to detect a base station which is in
range. One might want to regard the union of all locations where there
is no base station in range as a $k+1$st place. The discussion
below concerns domains where there is only partial information.
First we define the notion of a temporal gap:
\noindent ${\sf Gap}(t_1,t_2) \equiv_{def}
\exists (x_1,x_2) [{\sf At}(t_1,x_1) \wedge
{\sf At}(t_2,x_2) \wedge$\\
\hspace*{0.2cm}$\forall (t_3,x_3) [[ t_1 < t_3 \wedge t_3 < t_2] \rightarrow \neg {\sf
At}(t_3,x_3)]
$
\noindent We can now write a rule which allows us to infer that the agent must
be at at least one place during a gap, and that these places form a
connected region which is itself connected to all the places where the
agent is at at $t_1$ and $t_2$.
\noindent $\forall (t_1,t_2,x_1,x_2) [[{\sf Gap}(t_1,t_2)
\wedge {\sf At}(t_1,x_1) \wedge {\sf At}(t_2,x_2)]
\rightarrow\\
\indent
\exists x_3 [{\sf SumOfPlaces}(x_3) \wedge {\sf C}(x_1,x_3) \wedge
{\sf C}(x_2,x_3) \wedge {\sf Con}(x_3)]]
$
There might be more than one path $x_3$ linking the places at
$t_1$ and $t_2$, however, we cannot infer this without
evidence (such as metric
information about speed and distance travelled, but we do not consider
such possibilities here).
There is (at least) a second way in which the underlying place
information might be partial. We made the explicit assumption (in
\S\ref{computingC}) that time is sufficiently fine grained that
no place transitions are missed. If this is not the case (because the
speed of the agent is fast with respect to the recording granularity),
then gaps may occur even if there are no missing place sensors. In this case
we would need a modified version of the rule above, since it may be
that no new places need to be inferred to fill in the gap, but rather
that at least one of the places at time $t_1$ directly connects with
at least one of the places at time $t_2$.
Some form of non monotonic reasoning is likely to be needed in general
for reasoning in the presence of such kinds of partial knowledge, in
order to perform a domain closure, or to minimize the number of places
assumed to exist (as in the spatial semantic hierarchy discussed in
\S\ref{related}).
\section{Final Comments}
We have discussed the problem of computing topological maps from
knowledge of place trails, a task applicable at least in two
identified domains. There are many ways in which this work could be
extended. E.g. we could consider how to turn the symbolic topological
representations into a graphic visualisation\footnote{Note that we are
interested in a visualisation in which the regions are explicitly
represented as regions; thus although there are standard techniques
for laying out planar graphs (such as the connection graphs above)
with a reasonably balanced vertex distribution and straight line
edges, which could be buffered to produce regions, this makes the
connections into regions rather than the regions themselves.}
automatically. Of course any such depiction will inevitably have
metric qualities, but these must be ignored when interpreting the
visualization\footnote{In fact, in both the domains we are
considering, there are some very approximate metric qualities which
could be inferred, since PFs are known to have a typical size (5cm x
5cm to about 35cm x 35 cm) and wifi base stations similarly have a
maximum range.}. Any qualitative spatial description will always
have many metric realizations. One approach is to
diagrammatic reasoning techniques (e.g. \cite{howse2}) on
visualising Euler diagrams.
We have already conducted some
experimental work with artificial data and are currently collecting
real data and will then evaluate the ideas sketched here, and refine
them as appropriate. We may also consider other variants of the
problem, e.g. scenarios with multiple agents (where the {\sf
At} predicate has a third argument
indicating the agent).
\vspace*{-0.1in}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 1,688
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Q: XBee Wi-Fi module connection directly to UART Sensor I would like to connect a UART sensor directly to an XBee Wi-Fi module using its Rx-Tx, without using another MCU as an intermediate. The XBee would read the data from the sensor and send them via Wi-Fi to the internet (through a router).
Usual Application : Sensor -> MCU -> XBee -> Wi-Fi
My concept: Sensor-> Xbee -> Wi-Fi
From what I have read it is possible to sample analog sensors directly from XBee's analog inputs and send the data via Wi-Fi. It is also possible to sample the states of some GPIO pins of the XBee directly and send them. However, although the XBee has a UART, I have not found an application which uses this UART directly to communicate with a sensor. This UART is only used to communicate with an intermediate microcontroller which reads sensors, and sends commands to the XBee module via this UART. Another problem is that I should be able to transmit a UART command from the XBee module to the sensor (the sensor needs a command to send back a response). My purpose is to send periodically the read command to the sensor, get the response and transmit the response via Wi-Fi.
From my understanding it is not possible to use the UART of an XBee module like that, I would need an intermediate MCU. But I would like to ask, if I am missing something and this is possible. Has someone managed to do it or read something relevant? Is there a way to take advantage of the XBee's UART and program it to do this task?
Xbee: Digi XBee® Wi-Fi - OEM Module with Fully Integrated Support for Digi Remote Manager
Sensor: Dart Sensors WZ-EN CO module - Operation Manual
A: I've not use the wifi model but i don't think there's much difference in how they work.
You just need to send the commands through wifi to your xbee module and it will get passed to the sensor. The sensor should then react to the commands and return the data you requested, which will then be transmitted through wifi.
The baudrate of your sensor need to match with the xbee's baudrate for this to work, and also the logic level.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
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The Kunnattur block is a revenue block in the Kanchipuram district of Tamil Nadu, India. It has a total of 44 panchayat villages.
References
Revenue blocks in Kanchipuram district
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{
"redpajama_set_name": "RedPajamaWikipedia"
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FILMBOX Community Cinema
A rich and diverse world of cinema on your doorstep – at Langley Park Centre for the Performing Arts, Beckenham BR3 3BP
Tag: oscars 2015
See great films on your doorstep SOON @Beckenham #se20 #Bromley
HERE'S WHAT YOU CAN SEE AT FILMBOX SOON IN 2016 – join NOW & come and enjoy our New Years' Members' Social on 14 January – only £2 for a free glass of fizz, nibbles and a film! You can now buy membership easily online! See below…
NON-MEMBERS ARE ALWAYS WELCOME but If you become a member – you and a guest can see each of these films for only £5 each (usual price £8 each), plus a drinks loyalty card and other benefits! All films are screened subject to BBFC film classification. Our box office with details of dates, times & prices is here: https://www.ticketsource.co.uk/filmbox which includes a link to buy membership online! Only £25 if you come and buy at an event, or £28 online (to allow for site commission).
Members have a priority booking window and non-members can book at least 10 days in advance; sometimes tickets are available on the door. Two of you can see each of these films for only £5 each if you become a member – full details here: www.langleyfilmbox.com/members-area
14 January – WHEN HARRY MET SALLY & 27 January – 45 YEARS
Special New Year offer – non-members can see When Harry Met Sally for only £5 each (usually £8). If you become a member, you can see this film for only £2 each and enjoy a free welcome drink and canapés at our members' social before the film.
Email us with any queries: LangleyFilmbox@gmail.com
11 February (members' vote-off between ANNIE HALL & 500 DAYS OF SUMMER) & 24 February – SLOW WEST
10 March – BROOKLYN & 16 March – THE LADY IN THE VAN
We're keeping an eye on the Oscars and BAFTAs nominations for our Summer (April to July 2016) season! We like the look of these so far… but there'll be more to choose from…
See 5 Oscar-winner ONE FLEW OVER THE CUCKOOS NEST + Jack Nicholson 2/7/15 @Beckenham £8/5
Winner of FIVE Oscars in 1976 – One Flew Over the Cuckoo's Nest stars Jack Nicholson in an unforgettable role supported by a superb ensemble cast, including Danny De Vito, Christopher Lloyd, Louise Fletcher and Brad Dourif. Immensely moving and joyously funny by turns, 'Cuckoo's Nest' is a most memorable experience and rarely seen on the big screen – don't miss it!
Book tickets below – £8/£5 – no booking fees. FREE PARKING – NO BOOKING FEES! BOOK NOW – https://www.ticketsource.co.uk/filmbox
"The first movie since Frank Capra's 'It Happened One Night' to sweep the big five Oscars (director, screenplay, actor, actress, film), 'Cuckoo's Nest' remains an unlikely and inspiring triumph. The dream-like novel was considered unfilmable, the co-producer was a TV actor with a famous dad, the Czech director was still largely unknown outside Europe and every 'name' actress in Hollywood had turned down the role of callous Nurse Ratched. And yet, as soon as Jack Nicholson signed on the dotted line, everything fell into place.
He may be a huge, red-headed Irishman in Ken Kesey's book, but the live-wire misfit Randle P. McMurphy, who turns the mental institution upside down, was the role Jack Nicholson was born to play. With or without shock therapy, Nicholson is simply electrifying. But this is far from a one-man show, and indeed, even the non-speaking parts make an indelible visual impression. As direct and simple as it is funny and moving, this is a masterpiece of dramatic naturalism. Verdict: Oscars went to Nicholson, Louise Fletcher as castrating Nurse Ratched, producer Michael Douglas and Milos Forman's unblinking direction. Cuckoo spent 14 years in development limbo but remains the most radical film to emerge from mainstream Hollywood. Too many classic set pieces to mention but keep your ears cocked for that immortal line "Mmmm, Juicy Fruit." Certified brilliance." Colin Kennedy, Empire Magazine
THURSDAY 2 JULY 2015 – ONE FLEW OVER THE CUCKOO'S NEST (Milos Forman, USA, 1975, 133 mins, Cert 18). Bar open 6.40pm, film programme from 7.30pm.
Watch original trailer here: https://www.youtube.com/watch?v=OXrcDonY-B8
FREE PARKING – NO BOOKING FEES! BOOK NOW – https://www.ticketsource.co.uk/filmbox
For full details of FILMBOX Community Cinema – http://www.langleyfilmbox.com
See 5 Oscar-winner ONE FLEW OVER THE CUCKOOS NEST star Jack Nicholson 2/7/15 @Beckenham £8/5
Fly in to see #Birdman Oscar & BAFTA-winner – here this Thur 18/6 @Beckenham
Winner of SIX BAFTAs and FOUR Oscars – Birdman: Or (The Unexpected Virtue of Ignorance) stars Michael Keaton in a barn-storming role with superb ensemble support from Emma Stone, Ed Norton, Lindsay Duncan, Naomi Watts and Zach Galifianakis. A black comedy, Birdman is about a washed-up actor, who once played an iconic superhero, who battles his ego and attempts to recover his family, his career and himself in the days leading up to the opening of his Broadway play. Book tickets below – £8/£5 – no booking fees.
"You'll believe a man can fly. Or you'll believe that believing you can fly and flying are sort of the same thing. Either way, Alejandro González Iñárritu achieves takeoff in a big way with his crazy, freaky-deaky, hellzapoppin' showbiz comedy Birdman (or The Unexpected Virtue of Ignorance). I certainly levitated with enjoyment… At certain moments, watching it felt like inhaling laughing gas mixed with helium…Birdman is a delicious and delirious pleasure." 5 stars – Peter Bradshaw (The Guardian 25/12/14)
THURSDAY 18 JUNE 2015 – BIRDMAN (Alejandro G. Inarritu, USA, 2014, 119 mins, Cert 15)
Bar opens 6.40pm, film programme from 7.30pm.
Watch trailer here: https://www.youtube.com/watch?v=2bqh-UCY6Zg
FREE PARKING – NO BOOKING FEES! BOOK NOW – http://www.ticketsource.co.uk/filmbox
In our Screen 1 – see below (wheelchair accessible at stalls and circle level – please email ahead for our best service). For full details of FILMBOX Community Cinema – http://www.langleyfilmbox.com
See #Birdman Oscar & BAFTA-winner – Thur 18 Jun @Beckenham LangleyFilmbox.com
See #Birdman Oscar & BAFTA-winner – here Thur 18 Jun @Beckenham LangleyFilmbox.com
THIS WED – See electric music drama WHIPLASH 3/6 @Beckenham #Oscars #jazz #BetterOnBigScreen
"…you'll leave the cinema with a spring in your step and a thump in your chest, eager to bang the drum for what deserves to be one of the year's real word-of-mouth hits." Mark Kermode (18/1/15 – The Observer)
Wednesday 3 June 2015 – Winner of Sundance Film Festival & THREE Oscars – WHIPLASH. BETTER ON THE BIG SCREEN! Tickets £8/£5. Bar opens 6.40pm and film programme starts 7.30pm with a brief live display of group and solo drumming in our magnificent Performance Hall. Booking & trailer links below.
Full of tension, energy and dangerous excitement – WHIPLASH (Chazelle, USA, 2014, cert 15) is an edgy drama which features two superb performances by Oscar and Golden Globe winner JK Simmons (Spiderman 1-3) and Miles Teller (Divergent series), as music tutor and student who are locked in a pursuit for excellence – at any cost? The film features "an electrified final act that builds towards one of the most glorious, satisfying finales in the recent cinema… However genius may flourish, you know it when you see it, and Whiplash is it." Robbie Collin (23/2/15 – Telegraph)
WHIPLASH also features a superb original soundtrack of modern jazz composed by Justin Hurwitz & Tim Simonec – including classics by Duke Ellington & Stan Getz.
Trailer: https://www.youtube.com/watch?v=oAhvG3wIE_g
See electric 5star drama WHIPLASH (cert 15) Wed 3 June @Beckenham #Oscars #jazz #BetterOnBigScreen
"There are no two words in the English language more harmful than 'good job'." Terence Fletcher (JK Simmons) in WHIPLASH. The road to greatness can take you to the edge.
Wednesday 3 June 2015 – Sundance Film Festival winner & winner of THREE Oscars.
WHIPLASH. BETTER ON THE BIG SCREEN! Tickets £8/£5. Bar opens 6.40pm and film programme starts 7.30pm with a brief live display of drumming in our magnificent Performance Hall. Booking & trailer links below.
Don't miss electric music drama WHIPLASH 3/6 @Beckenham #Oscars #jazz #BetterOnBigScreen
Wednesday 3 June 2015 – Sundance Film Festival winner – WHIPLASH. BETTER ON THE BIG SCREEN! Tickets £8/£5. Bar opens 6.40pm and film programme starts 7.30pm with a brief live display of group and solo drumming in our magnificent Performance Hall. Booking & trailer links below.
TONIGHT (Wed) get tkts on door THEORY OF EVERYTHING #BestFilm #BestActor @Beckenham
Over 200 booked for tonight's great film! BETTER ON THE BIG SCREEN! Everyone is welcome at FILMBOX – the community cinema on your doorstep. Catch up and make up your own mind about one of the top award-winning films of this year – booking open to all now (link below)! Tickets also available on the door.
WEDNESDAY 20 MAY – THE THEORY OF EVERYTHING (cert 12A) – film programme starts 7.30pm, cafe/bar opens 6.40pm.
Retiring collection on behalf of the Motor Neurone Disease Association http://www.mndassociation.org/
Based on Jane Hawking's book, the film follows her relationship with theoretical-physicist Professor Stephen Hawking. 'The title refers to Hawking's quest for an all-encompassing theory of the physical universe, but the pathos of the film is that in ordinary life, not everything can be made to fit and make sense. Compromises must be made; people must muddle through. It is a gentle, tender story of lovers who found friendship during and after their marriage.' Peter Bradshaw (The Guardian, 1/1/15)
FREE PARKING – NO BOOKING FEES!
BOOK NOW – http://www.ticketsource.co.uk/filmbox
Both films are in our Screen 1 – see below.
Our audience in the larger venue Screen 1 – the Performance Hall
Archive Select Month July 2020 (1) May 2020 (1) April 2020 (1) March 2020 (1) January 2020 (2) November 2019 (3) October 2019 (1) September 2019 (2) August 2019 (1) July 2019 (1) June 2019 (3) May 2019 (2) April 2019 (1) March 2019 (1) January 2019 (1) November 2018 (4) October 2018 (2) September 2018 (4) July 2018 (1) June 2018 (2) May 2018 (3) April 2018 (2) March 2018 (4) February 2018 (3) January 2018 (4) November 2017 (4) October 2017 (8) September 2017 (8) July 2017 (3) June 2017 (15) May 2017 (8) April 2017 (4) March 2017 (7) February 2017 (6) January 2017 (2) December 2016 (4) November 2016 (4) October 2016 (1) September 2016 (2) July 2016 (3) June 2016 (5) May 2016 (3) April 2016 (3) March 2016 (1) February 2016 (1) January 2016 (5) December 2015 (3) November 2015 (3) October 2015 (3) September 2015 (4) July 2015 (2) June 2015 (7) May 2015 (5) April 2015 (9) March 2015 (19) February 2015 (13) January 2015 (16) December 2014 (13) November 2014 (6) October 2014 (3) September 2014 (22) August 2014 (17) July 2014 (8) June 2014 (13) May 2014 (11) April 2014 (7) March 2014 (8) February 2014 (14) January 2014 (22) December 2013 (4) November 2013 (10) October 2013 (8) September 2013 (10) August 2013 (9) July 2013 (5) June 2013 (9) May 2013 (13) April 2013 (11) March 2013 (10) February 2013 (9) January 2013 (11) December 2012 (4) November 2012 (4) October 2012 (3) September 2012 (8)
The Business End
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{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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Q: countif value equals value in another column I want to count the number of values of a column that are equal to a value on the same line in another column. In the following exmaple, I want for each « ans » column the number of values matching the ones on the « ID » column:
ID | ans A | ans B | ans C
------|-------|-------|------
1 | 34 | 1 | 1
23 | 23 | 23 | 32
4 | 4 | 16 | 25
54 | 54 | 54 | 16
32 | 32 | 17 | 12
.../...
TOTAL | 4 | 3 | 1
I tried with COUNTIF but I can only enter a single value as criteria, if I enter a range I always get a 0 as result. Which function should I use?
A: In B8 enter:
=SUMPRODUCT(--($A2:$A6=B2:B6))
and copy across.
The formula in B8 just compares column B to column A (row-by-row) and counts the matches.For more info see:
xlDynamic
EDIT#1:
The formula:
=SUM({1,2,3})
does correctly calculate in:
*
*GoogleSheets
*Excel 365
*the Excel web-app
*Excel 2007 / Win 7
I don't know about other environments.
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{
"redpajama_set_name": "RedPajamaStackExchange"
}
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· I am a United States citizen or lawfully admitted permanent resident, and I am not a federal contractor.
· I am making this contribution using my own personal funds, and not those of another, and not the general treasury funds of a corporation, labor organization, national bank, or entity or person that is a federal contractor, and I will not be reimbursed for this contribution by another person.
· I am making this contribution using my personal credit or debit card, and not with the card of a corporation, business entity, or another person.
· I do not personally hold a federal government contract.
· I am at least 16 years old.
Contributions to Webber for Congress are not tax deductible as charitable contributions for federal income tax purposes. Contributions from corporations, labor organizations, national banks, foreign nationals (i.e., non-green card holders), and federal government contractors are prohibited. Contributions must be made from personal funds and may not be reimbursed by any other person. Federal law requires us to use our best efforts to collect and report the name, mailing address, occupation, and name of employer of individuals whose contributions aggregate in excess of $200 per election cycle. Contributions are subject to federal limits and source prohibitions.
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{
"redpajama_set_name": "RedPajamaC4"
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Pete Chen (* 16. September 1988 in Xizhi; eigentlich Yen Han Chen) ist ein professioneller taiwanischer Pokerspieler. Er ist zweifacher Braceletgewinner der World Series of Poker und gewann 2017 das Main Event der World Poker Tour.
Pokerkarriere
Werdegang
Chen spielt seit 2009 professionell Poker und startete seine Laufbahn mit Cash Games, ehe er zum Turnierpoker wechselte. Auf der Onlinepoker-Plattform PokerStars, bei der er den Nickname psxfrcndhe nutzt, erspielte er sich seit August 2010 Turnierpreisgelder von mehr als 500.000 US-Dollar und gewann 2018 einen Titel bei der World Championship of Online Poker. Darüber hinaus spielt er bei WSOP.com als PeteChen und ist seit Juli 2020 Markenbotschafter der Plattform Natural8, dem asiatischen Ableger von GGPoker, bei dem er als TurboPete in Erscheinung tritt.
Seine ersten Geldplatzierungen bei Live-Pokerturnieren erzielte der Taiwanese ab August 2011 ausschließlich bei Events in Macau, wo er im Oktober 2013 bei der Asia Championship of Poker seine ersten beiden Turniere gewann. Ende Mai 2014 belegte er beim Main Event der Asia Pacific Poker Tour in Macau den mit umgerechnet knapp 75.000 US-Dollar dotierten fünften Platz. Zum Jahresende wurde Chen von PokerStars als Asian Player of the Year ausgezeichnet. Im Juli 2015 war er erstmals bei der World Series of Poker (WSOP) im Rio All-Suite Hotel and Casino in Paradise am Las Vegas Strip erfolgreich und kam bei zwei Turnieren der Variante No Limit Hold'em in die Geldränge. In den folgenden beiden Jahren erzielte der Taiwanese rund um die Welt zahlreiche Geldplatzierungen, jedoch ohne größere Einzelpreisgelder einzufahren. Anfang April 2017 saß er am Finaltisch des Main Events der PokerStars Championship (PSC) in Macau und beendete das Turnier als Sechster, wofür er umgerechnet mehr als 90.000 US-Dollar erhielt. Keine zwei Wochen später entschied der Taiwanese das Main Event der World Poker Tour in Peking für sich und sicherte sich eine Siegprämie von rund 300.000 US-Dollar. Bei der WSOP 2017 belegte er bei einem Event den zweiten Platz und erhielt sein bisher höchstes Preisgeld von über 380.000 US-Dollar. Mitte August 2017 wurde Chen bei einem Side-Event der PSC in Barcelona Vierter und erhielt knapp 140.000 Euro. Bei der WSOP 2018 erzielte er 9 Geldplatzierungen. Ende August sowie Anfang September 2018 gewann der Taiwanese zwei Turniere der European Poker Tour in Barcelona und sicherte sich Preisgelder von knapp 150.000 Euro. Ab Mai 2019 kam er 14-mal bei der WSOP 2019 auf die bezahlten Plätze. Als die Live-Pokerszene aufgrund der COVID-19-Pandemie im Frühjahr 2020 weitestgehend zum Erliegen kam, erzielte Chen bei der erstmals ausgespielten World Series of Poker Online 19 Geldplatzierungen. Bei der WSOP 2021 setzte er sich beim online ausgespielten Ultra Deepstack durch und sicherte sich ein Bracelet sowie den Hauptpreis von über 80.000 US-Dollar. Wenige Tage später gewann er im Wynn Las Vegas das Mystery Bounty und erhielt aufgrund eines Deals mit zwei anderen Spielern eine Auszahlung von knapp 300.000 US-Dollar. Bei der World Series of Poker Online entschied er Anfang September 2022 auf GGPoker das Lucky Sevens Bounty für sich und wurde mit seinem zweiten Bracelet sowie einem Preisgeld von insgesamt über 115.000 US-Dollar prämiert.
Insgesamt hat sich Chen mit Poker bei Live-Turnieren mehr als 4 Millionen US-Dollar erspielt und ist damit nach James Chen der zweiterfolgreichste taiwanische Pokerspieler.
Braceletübersicht
Chen kam bei der WSOP 101-mal ins Geld und gewann zwei Bracelets:
Weblinks
Pete Chen in der Datenbank der World Series of Poker: 2015–2022, 2018–2021, 2020–2022 (englisch)
Einzelnachweise
Pokerspieler (Taiwan)
Braceletgewinner
Gewinner des Main Events der World Poker Tour
Pseudonym
Taiwaner
Geboren 1988
Mann
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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Q: Messages.app on mac: search in history i've a problem, i'm able to search for string in messages.app. It actually tells me all the matches for every conversation, the problem is that is shows only the last match, how can i go through all the results?
A: You can simply use
*
*⌘ + G in order to jump to the next occurrence
and
*
*⌘ + Shift + G in order to jump to the previous one.
A: Have you tried Spotlight Search?
Cmnd + Space
Make sure Messages is enabled for Spotlight Search in your Mac System Preferences.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 794
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Friday night Yuki's mom went to a concert with a friend, so it was up to me to decide where we went for dinner with her dad. In my quest to eat at every Iron Chef's restaurants I was able to convince him that we should head out to one of Chen Kenichi's joints. It wasn't a hard sell. So, the three of us hopped on the train to Roppongi where Chen has one of his four places.
There were a handful of pre fix options, but none of them really had what we wanted, so we ordered a bunch of dishes in typical Chinese family-style dining.
First up was a trio of cold appetizers. On the top were Scallops cooked with chili peppers. Not too spicy, just a nice, slow, gentle burn on the back of the throat. Bottom right was shredded chicken with a sweet miso sauce. Bottom left jellyfish in a light soy. All three were very complimentary of each other and made for a great start to the meal.
Next up was Shark Fin Soup. Not the most politically correct dish, but hey, a little shark fin never hurt anything. A lot does, but a little doesn't. The broth was a thick soy flavor and it had thin slices of pork in it alongside the shark fin.
Then came Abalone, one of the sea's finest of all creatures! Served with shiitake and bamboo shoots it was truly delicious. Judging by Uichiro's (Yuki's dad) reaction when he first bit into it, I'd say it was his favorite part of the meal, next to the beer.
After that we had mixed seafood served on rice cakes. The restaurant manager poured the hot seafood on the rice cakes making it sizzle, much like the classic sizzling rice soup commonly found in American Chinatown restaurants. I need make a correction, I think Uichiro liked this dish more than the abalone. Hard to argue, the shrimp, scallops, and squid were cooked to perfection with all of the natural sweetness brought out.
Next was probably my favorite dish of the meal, beef with mushrooms and lilies in a thick ginger soy sauce. The beef was so tender it almost melted in my mouth. The mushrooms and lilies were nice counterpoints to the salty soy. The only thing missing was white rice to balance a little more of the salt. This was definitely more of what Americans are used to than Japanese. Give me this dish and a cold beer and I'm a happy man!
Here we go, the dish that brought me to Chen's restaurant and the one that will make my brothers very envious, his famous mapotofu! There were two choices on the menu, Kenichi's and Kenmei's. Kenmei was Kenichi's father, the one who brought true Szechuan cooking to Japan. It was a tough choice, but we opted for Kenichi's since it was his restaurant and not his father's. It wasn't quite as hot as I expected, but it was definitely a hot and spicy dish! Packed with Szechuan peppercorns it gives an initial citrusy spice followed by a mouth-numbing burn. Yuki and I added some extra peppercorns to get the full experience, while Uichiro only ate a few pieces of tofu. I think it's a little spicy for him. It was a little oily as it was douced in chili oil, but that's what makes it so delicious. The funny thing is that in the middle of the night Tokyo experienced what it thinks was a small earthquake. It wasn't an earthquake at all though, it was the effects of my trying to digest Chen's mapotofu! I'm still not quite sure exactly what that dish did to my intestines, but it made a city of 16 million rumble a little. And somehow my chest got a little harrier.
We followed the mapotofu with a mild pork and egg noodle dish. It had shiitake, green onions, chinese cabbage, sprouts, and bamboo shoots. It took a few bites to get the burn out of our mouths, but once it was gone this dish's wonderful flavor stood out. It's just too bad I couldn't finish it all because we ordered one too many dishes I think. Oh well, what can you do?
We thought we were finished when the manager brought us some complimentary dessert, Annin Tofu. Annin Tofu is a popular Chinese dessert. It's basically just almond jelly. Very smooth, light, and creamy. It's the perfect way to finish off a meal.
All in all, I have to say that I was a little dissapointed. After watching countless episodes of Chen creating some of the most amazing looking dishes anyone could ever think up, this meal was a very straight forward Chinese meal. Every dish was a classic that you can get at just about any Chinese restaurant. Granted, everything was perfectly balanced, but nothing was off the wall. I was kind of hoping for some Chen originals. This restaurant wasn't the right format for him to create Iron Chef dishes. I have no regrets, but I wasn't blown away like I was at other Iron Chef restaurants.
I have now been to three Iron Chef restaurants (4 if you count Bobby Flay, but I don't consider him an Iron Chef and I never wanted to go to his place). I have done Sakai, Michiba, and now Chen. Next up….Kobe Masahiko.
|
{
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Печенка — река в России, протекает в Гусь-Хрустальном и Судогодском районах Владимирской области. Устье реки находится в 54 км от устья Судогды по левому берегу. Длина реки составляет 13 км.
На истоке реки устроена запруда возле деревни Дубасово в 18 км к югу от города Судогды. Высота запруды — 114,1 м над уровнем моря. Около истока деревня Дубасово и населённый пункт Дом Инвалидов. Ниже река течёт на север по ненаселённому лесу, впадает в Судогду выше деревни Жуковка.
Данные водного реестра
По данным государственного водного реестра России относится к Окскому бассейновому округу, водохозяйственный участок реки — Клязьма от города Владимир до города Ковров, без реки Нерль, речной подбассейн реки — бассейны притоков Оки от Мокши до впадения в Волгу. Речной бассейн реки — Ока.
Код объекта в государственном водном реестре — 09010300912110000032844.
Примечания
Литература
Притоки Судогды
Реки Гусь-Хрустального района
Реки Судогодского района
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 1,606
|
Q: The timer does not stop with clearInterval I'm trying to build a simple start and stop timer feature, however when i click again on the function, the clearInterval does not seem to do any effect. In fact the timer does not stop. However i'm able to start it but i cannot turn it off.
const [seconds, setSeconds] = useState(0);
const [timer, setTimer] = useState();
const handleStartToggle = (seconds) => {
// Start new timer only if it's not run yet
if(!timer) {
setTimer(setInterval(() => {
setSeconds((current) => current + 1);
}, 1000));
// Else, it's already running, we stop it
} else {
return () => clearInterval(timer);
}
}
<div className="row project-task-showcase">
<h2 className="client-name"> {activeClient.name}</h2>
<p className="task-name">{activeTask.name}</p>
<img src={play} className="play" onClick={handleStartToggle} />
{seconds}
</div>
A: Problem in return,
You returning as function, so in function it's not working,
Change code to
const handleStartToggle = (seconds) => {
// Start new timer only if it's not run yet
if(!timer) {
setTimer(setInterval(() => {
setSeconds((current) => current + 1);
}, 1000));
// Else, it's already running, we stop it
} else {
clearInterval(timer); // <-- Change here
// setTimer(null); // <-- To toggle next time
}
}
A: In your handleStartToggle function you are doing different work for if and else statement. For if you are executing few statement but for else you are returning a arrow function that is not correct. To correct this you have to either return function in both
const handleStartToggle = (seconds) => {
// Start new timer only if it's not run yet
if(!timer) {
return () => {setTimer(setInterval(() => {
setSeconds((current) => current + 1);
}, 1000));
}
// Else, it's already running, we stop it
} else {
return () => clearInterval(timer);
}
}
<div className="row project-task-showcase">
<h2 className="client-name"> {activeClient.name}</h2>
<p className="task-name">{activeTask.name}</p>
<img src={play} className="play" onClick={()=> handleStartToggle()} />
{seconds}
</div>
Or use statement in both
const handleStartToggle = (seconds) => {
// Start new timer only if it's not run yet
if(!timer) {
setTimer(setInterval(() => {
setSeconds((current) => current + 1);
}, 1000));
// Else, it's already running, we stop it
} else {
clearInterval(timer); // <-- Change here
setTimer(null); // <-- To toggle next time
}
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,954
|
{"url":"https:\/\/physics.stackexchange.com\/questions\/430694\/parity-conserving-qft","text":"# Parity conserving QFT\n\nI was reading Peskin's QFT text and in Chapter 6 (Radiative Corrections), there is a line in section 6.2, page 185 which I am quoting below. The passage is about the $$\\Gamma^{\\mu}(p^{'},p)$$ and its dependence on possible objects that appear in the Feynman rules. The dependence should be on $$p^{'},p,\\gamma^{\\mu}$$ and constants like $$m, e, etc$$ according to the text. Additionally, the book reads:\n\n\"The only other object that could appear in any theory is $$\\epsilon^{\\mu\\nu\\rho\\sigma}$$ (or equivalently $$\\gamma^{5}$$); but this is forbidden in any parity-conserving theory.\"\n\nMy question is two-fold:\n\n1. Why is the appearance of $$\\epsilon^{\\mu\\nu\\rho\\sigma}$$ forbidden for parity to be conserved? Is it because the antisymmetricity of $$\\epsilon$$ causes a change in sign under parity transformation?\n\n2. How is Peskin relating $$\\epsilon$$ to $$\\gamma^{5}$$?\n\nStarting from Adam and Eve, the space reflection operator (see for example Bjorken & Drell, Relativistic Quantum Field, Vol. I, p. 24) is $$P = \\gamma^0$$ except for an irrilevant phase; this means that while the bilinear $$j^{\\mu} = \\bar\\psi\\gamma^{\\mu}\\psi$$ is invariant under parity transformation, the bilinear $$j^{\\mu}_A = \\bar\\psi\\gamma^{\\mu}\\gamma^5\\psi$$ is certainly not. Since the phenomenological request of $$P$$-symmetry for the EM current implies the total absence of $$\\gamma^5$$ in the QED interaction term $$L_{int} = -e\\bar\\psi\\gamma^{\\mu}A_{\\mu}\\psi,$$ where $$A_{\\mu}$$ is the EM four-potential, Gell-Mann & Low's formula (ibidem, Vol. II, chap. 17) then guarantees us the absence of $$\\gamma^5$$ also from the form factor $$\\Gamma^{\\mu}$$. It is important to notice that this last object can't depend \"directly\" from a four-index tensor such as $$\\epsilon$$, as Peskin seems to state; the only information it can carry around about parity violation is contained in $$\\gamma^5$$.\nThe completely antisymmetric tensor $$\\epsilon^{\\mu\\nu\\alpha\\beta}$$ appears only in the S-matrix squared element (properly contracted in all is indexes), and emerges related to $$\\gamma^5$$ when you take the trace; as a matter of fact, up to a normalization, $$Tr(\\gamma^{\\mu}\\gamma^{\\nu}\\gamma^{\\alpha}\\gamma^{\\beta}\\gamma^{5}) = 4i\\epsilon^{\\mu\\nu\\alpha\\beta}.$$ This accounts for Peskin's supposed equivalence between the presence of $$\\gamma^5$$ in $$\\Gamma^{\\mu}$$, which implies $$P$$-symmetry violation, and the presence of $$\\epsilon^{\\mu\\nu\\alpha\\beta}$$ in your final result, the S-matrix squared element.","date":"2021-12-08 18:17:24","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 27, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8952133059501648, \"perplexity\": 236.09953934399672}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-49\/segments\/1637964363520.30\/warc\/CC-MAIN-20211208175210-20211208205210-00567.warc.gz\"}"}
| null | null |
Q: Renaming part for file using mv command in shell script i want to rename a file using mv command in shell script
now file is in the format foo-<date>.tar.gz i want to rename it to foo1-<date>.tar.gz.
i tried, cut the foo and rename it and concatenate and all but i want to something very simple follows
mv foo*.tar.gz foo1*.tar.gz
date should be maintained only foo should be changed foo1
is it possible ? if yes how?
Thank you in advance!
A: You can use BASH string manipulation:
f='foo-12APR2014.tar.gz'
nf="foo1-${f#*-}"
Test:
echo "$nf"
foo1-12APR2014.tar.gz
PS: If not using BASH then you can use sed:
nf=`echo "$f"|sed 's/^foo/foo1/'`
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 5,378
|
[](https://badge.fury.io/js/parallel-webpack)
[](https://travis-ci.org/trivago/parallel-webpack) [](https://circleci.com/gh/trivago/parallel-webpack) [](https://coveralls.io/github/trivago/parallel-webpack?branch=coverage)
[](https://packagephobia.now.sh/result?p=parallel-webpack)
# parallel-webpack - Building multi-configs in parallel
`parallel-webpack` allows you to run multiple webpack builds in parallel,
spreading the work across your processors and thus helping to significantly speed
up your build. For us at [trivago](http://www.trivago.com) it has reduced
the build from 16 minutes to just 2 minutes - for 32 variants. That performance
improvement naturally comes at the expense of utilizing all available CPU cores.
## Installation
```sh
npm install parallel-webpack --save-dev
```
You can choose whether to install `parallel-webpack` globally or locally.
At [trivago](http://www.trivago.com), we keep our build tools locally to the project
so that we have full control over its versions.
## Basic example
Given a `webpack.config.js` like this:
```javascript
var path = require('path');
module.exports = [{
entry: './pageA.js',
output: {
path: path.resolve(__dirname, './dist'),
filename: 'pageA.bundle.js'
}
}, {
entry: './pageB.js',
output: {
path: path.resolve(__dirname, './dist'),
filename: 'pageB.bundle.js'
}
}];
```
`parallel-webpack` will run both specified builds in parallel.
## Variants example
Sometimes, just using different configurations like above won't be enough and what
you really want or need is the same configuration with some adjustments.
`parallel-webpack` can help you with generating those `configuration variants` as
well.
```javascript
var createVariants = require('parallel-webpack').createVariants;
// Those options will be mixed into every variant
// and passed to the `createConfig` callback.
var baseOptions = {
preferredDevTool: process.env.DEVTOOL || 'eval'
};
// This object defines the potential option variants
// the key of the object is used as the option name, its value must be an array
// which contains all potential values of your build.
var variants = {
minified: [true, false],
debug: [true, false],
target: ['commonjs2', 'var', 'umd', 'amd']
};
function createConfig(options) {
var plugins = [
new webpack.optimize.DedupePlugin(),
new webpack.optimize.OccurenceOrderPlugin(),
new webpack.DefinePlugin({
DEBUG: JSON.stringify(JSON.parse(options.debug))
})
];
if(options.minified) {
plugins.push(new webpack.optimize.UglifyJsPlugin({
sourceMap: false,
compress: {
warnings: false
}
}));
}
return {
entry: './index.js',
devtool: options.preferredDevTool,
output: {
path: './dist/',
filename: 'MyLib.' +
options.target +
(options.minified ? '.min' : '') +
(options.debug ? '.debug' : '')
+ '.js',
libraryTarget: options.target
},
plugins: plugins
};
}
module.exports = createVariants(baseOptions, variants, createConfig);
```
The above configuration will create 16 variations of the build for you, which
`parallel-webpack` will distribute among your processors for building.
```
[WEBPACK] Building 16 targets in parallel
[WEBPACK] Started building MyLib.umd.js
[WEBPACK] Started building MyLib.umd.min.js
[WEBPACK] Started building MyLib.umd.debug.js
[WEBPACK] Started building MyLib.umd.min.debug.js
[WEBPACK] Started building MyLib.amd.js
[WEBPACK] Started building MyLib.amd.min.js
[WEBPACK] Started building MyLib.amd.debug.js
[WEBPACK] Started building MyLib.amd.min.debug.js
[WEBPACK] Started building MyLib.commonjs2.js
[WEBPACK] Started building MyLib.commonjs2.min.js
[WEBPACK] Started building MyLib.commonjs2.debug.js
[WEBPACK] Started building MyLib.commonjs2.min.debug.js
[WEBPACK] Started building MyLib.var.js
[WEBPACK] Started building MyLib.var.min.js
[WEBPACK] Started building MyLib.var.debug.js
[WEBPACK] Started building MyLib.var.min.debug.js
```
## Running the watcher
One of the features that made webpack so popular is certainly its watcher which
continously rebuilds your application.
When using `parallel-webpack`, you can easily use the same feature as well by
specifying the `--watch` option on the command line:
```
parallel-webpack --watch
```
## Specifying retry limits
As a side-effect of using `parallel-webpack`, an error will no longer lead to
you having to restart webpack. Instead, `parallel-webpack` will keep retrying to
build your application until you've fixed the problem.
While that is highly useful for development it can be a nightmare for
CI builds. Thus, when building with `parallel-webpack` in a CI context, you should
consider to use the `--max-retries` (or `-m` option) to force `parallel-webpack` to give
up on your build after a certain amount of retries:
```
parallel-webpack --max-retries=3
```
## Specifying the configuration file
When you need to use a configuration file that is not `webpack.config.js`, you can
specify its name using the `--config` parameter:
```
parallel-webpack --config=myapp.webpack.config.js
```
## Switch off statistics (improves performance)
While the statistics generated by Webpack are very usually very useful, they also
take time to generate and print and create a lot of visual overload if you don't
actually need them.
Since version *1.3.0*, generating them can be turned off:
```
parallel-webpack --no-stats
```
## Limiting parallelism
Under certain circumstances you might not want `parallel-webpack` to use all of your
available CPUs for building your assets. In those cases, you can specify the `parallel`,
or `p` for short, option to tell `parallel-webpack` how many CPUs it may use.
```
parallel-webpack -p=2
```
## Configurable configuration
Sometimes, you might want to access command line arguments within your `webpack.config.js`
in order to create a more specific configuration.
`parallel-webpack` will forward every parameter specified after `--` to the configuration
as is:
```
parallel-webpack -- --app=trivago
```
Within `webpack.config.js`:
```
console.log(process.argv);
// => [ 'node', 'parallel-webpack', '--app=trivago' ]
```
`parallel-webpack` adds the first two values to `process.argv` to ensure that there
are no differences between various ways of invoking the `webpack.config.js`.
## Node.js API
Just like webpack, you can also use `parallel-webpack` as an API from node.js
(You can specify any other option used in [worker-farm](https://www.npmjs.com/package/worker-farm)):
```javascript
var run = require('parallel-webpack').run,
configPath = require.resolve('./webpack.config.js');
run(configPath, {
watch: false,
maxRetries: 1,
stats: true, // defaults to false
maxConcurrentWorkers: 2 // use 2 workers
});
```
You can pass a notify callback as well.
```javascript
var run = require('parallel-webpack').run,
configPath = require.resolve('./webpack.config.js'),
options = {/*...*/};
function notify() {
// do things
}
run(configPath, options, notify);
```
**NOTE:** In watch mode notify callback provided with Node.js API will run **only once**
when all of the builds are finished.
### createVariants
---
#### createVariants(baseConfig: Object, variants: Object, configCallback: Function): Object[]
Alters the given `baseConfig` with all possible `variants` and maps the result into
a valid webpack configuration using the given `configCallback`.
#### createVariants(variants: Object, configCallback: Function): Object[]
Creates all possible variations as specified in the `variants` object and
maps the result into a valid webpack configuration using the given `configCallback`.
#### createVariants(baseConfig: Object, variants: Object): Object[]
Alters the given `baseConfig` with all possible `variants` and returns it.
#### createVariants(variants: Object): Object[]
Creates all possible variations from the given `variants` and returns them as a flat array.
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 7,429
|
National Security Letters Demand Data Companies Aren't Obligated to Provide
Internal documents suggest the FBI uses the secret orders to pursue sensitive customer data like internet browsing records.
Photo: Meg Roussos/Bloomberg/Getty ImagesPhoto: Meg Roussos/Bloomberg/Getty Images
Jenna McLaughlin Cora Currier
Jenna McLaughlin, Cora Currier
January 31 2017, 12:06 p.m.
The FBI's Secret Rules
President Trump has inherited a vast domestic intelligence agency with extraordinary secret powers. A cache of documents offers a rare window into the FBI's quiet expansion since 9/11.
The FBI routinely uses secret orders known as national security letters to demand information that recipients might not actually have to give up, internal documents indicate. The letters are among the FBI's most potent instruments, because they function like subpoenas without requiring the approval of a judge. Internal guidelines suggest that the bureau has been using them to pursue sensitive electronic data and phone records — despite the fact that such attempts overstep the bureau's legal authority.
The Intercept obtained the FBI's rules for national security letters as they are spelled out in two different guides: a document detailing current guidelines for agents using the letters and an uncensored 2011 version of the FBI's main operating manual, the Domestic Investigations and Operations Guide, or DIOG. Both documents are marked "unclassified" or "for official use only." The first document has not been previously released. The DIOG has been made public only in heavily redacted form.
The FBI issues thousands of NSLs each year. They are controversial in part because they carry the force of law but are created entirely outside the judicial system: To issue one, an FBI official just needs to attest that the information sought is relevant to a national security investigation. The letters have also been criticized because they are shrouded in secrecy. Companies that receive them are for the most part forbidden from notifying their customers or the public. The government has fought to keep even basic rules governing them secret.
The FBI's internal guidelines suggest that the bureau uses the letters to demand sensitive information on email transactions — even though the Justice Department has specifically advised the FBI that it does not have the authority to use the letters this way. The documents also indicate that the FBI can use national security letters to surveil a "community of interest" by obtaining information from a business about a customer and every person that customer has contacted. This is a controversial practice that the bureau once halted amid scrutiny. But the documents reveal that a secretive unit that mines phone records can still initiate such requests.
Last June, Congress narrowly rejected a proposal to allow the FBI to use the letters to demand information like browsing history, email headers (not including subject lines), and, depending on your reading of the bill, possibly even some social media information. An amendment to a criminal justice funding bill making that change fell just two votes short of passage.
Even so, the newer document on NSL policy contains a reference to a "model NSL" the FBI uses to request "email transactional" data from companies and other organizations — despite the fact that the organizations are not obligated to provide such information.
The bureau has long used NSLs to obtain basic subscriber information from telecom companies. The Electronic Communications Privacy Act lists four types of information the bureau is allowed to obtain, including the name of the owner of an account, how long that person has owned it, the person's address, and toll billing records, which show phone numbers called, the date and time of each call, and the length of each call. Several years ago, the FBI began using the letters to ask for email headers and internet browsing records — assuming that such queries were consistent with the bureau's right to procure "basic subscriber information." To that end, the bureau requested a broad category of information it sometimes refers to as "electronic communication transactional records."
In 2008, Department of Justice lawyers clarified that the FBI didn't actually have the legal authority to demand that technology companies hand over records outside of the four types listed. However, as The Intercept previously reported, the FBI disagreed with that conclusion and asked for such material anyway in a 2013 NSL it sent to Yahoo.
Some large companies like Facebook and Yahoo have refused to provide email and browsing data in response to such NSLs, but FBI agents may have expected that other companies, especially small ones, would be too ignorant or weak to fight back.
"The government's position is: We can ask for anything analogous to toll billing records" — such as email and browsing data — "and if the providers are dumb enough to give it to us, that's not our problem," said Chris Soghoian, a technologist formerly with the American Civil Liberties Union.
The FBI guide to NSLs obtained by The Intercept references a set of "model NSLs" for agents to choose from; among the options are "email transactional NSL," along with model letters for more conventional requests: "telephone subscriber NSL" and "telephone toll billing record NSL."
"The existence of a standard form in the FBI's NSL system suggests that this is not one or two agents that are misreading the statute, it's policy," said Soghoian of the "email transactional NSL."
The 2011 DIOG obtained by The Intercept does delineate a few "sorts of records" that couldn't be obtained through an NSL, at least in 2011, including social media friend lists and virtual property owned on platforms like Second Life. But neither guide specifies exactly how it defines toll billing records, which are expressly allowed, or "electronic transactional" data, the umbrella term that often appears in the letters.
Even when not explicitly asking for email or electronic transaction records, the FBI implies that toll billing records might include such data, said Al Gidari, a prominent national security attorney who has worked on NSL cases in the past. The language of the letters is ambiguous and "leaves the impression that the provider better think broadly about what a toll record is as opposed to 'Hey, it's up to you as to what you give us,'" Gidari said.
The Department of Justice's inspector general found widespread misuse of NSLs at the FBI in the early 2000s, and the model letters in this case were, ironically, actually part of an effort to reform the NSL process. The idea was that an automated system for generating and submitting NSLs would prevent agents from issuing improper requests for information. In the case of "email transactional" NSLs, however, automation appears to have systematized the bureau's contentious reading of the law.
Chris Soghoian speaks at TEDSummit 2016 held in Banff, Canada, in June.
Photo: Bret Hartman/TED
The unredacted 2011 DIOG obtained by The Intercept sheds light on another worrying use of NSLs — in this case, to obtain multiple individuals' call records in one fell swoop, in order to suss out what is variously referred to as "community of interest," "calling circle," or "second generation" information.
The fact that the FBI made such requests was first disclosed in 2007, at which time the practice was halted under scrutiny from the inspector general of the Department of Justice. The FBI seems to have eventually started it up again, under rules that have not previously been made public.
The DIOG states that "under limited circumstances, one NSL may simultaneously request toll or transactional information for a 'seed number' [normally the target of the investigation] and toll or transactional information for all telephone numbers that have been in contact with the seed number."
Such requests need the approval of the deputy general counsel of the FBI's National Security Law Branch. The guide states that NSLs seeking "second generation/community of interest information" are "used rarely," and specifies that agents cannot ask for second-generation information "if there is reason to believe the 'seed number' has been in contact with members of the news media." (In June, The Intercept published the FBI's standards for NSLs targeting the media, which require a modicum more oversight.)
The guide doesn't detail what circumstances warrant second-generation requests, but it confirms that they are issued by a secretive FBI data-mining unit. Back in 2007, it was reported that the FBI used the information it gleaned from community of interest requests for a technique it called call-link analysis, visualizing phone data to look for patterns and connections and identify previously unknown suspects.
Soghoian said that the FBI might be especially sensitive about community of interest requests because they have the potential to suck up information on a large number of people who may have only a tangential connection to a national security threat; the letters can target people who are not even the subject of an investigation, but merely deemed "relevant" to one.
The FBI abused this power in the past. In 2007, documents released to the Electronic Frontier Foundation showed orders with boilerplate language asking for "a community of interest for the telephone numbers in the attached list." At the time, an unnamed government official told the New York Times that the data "was limited to people and phone numbers 'once removed' from the actual target" of the NSL — the same standard in the 2011 DIOG. A spokesperson for the FBI also told the Times that community of interest data "was used infrequently."
Yet despite this, a 2010 report from the Justice Department's inspector general found that the FBI "often" asked for such data, both through NSLs and other orders and requests. It identified hundreds of NSLs that included language seeking second-generation records. At the time, at least, AT&T was apparently the only company capable of providing community of interest information; Verizon had told Congress it didn't keep that data.
In some cases, the report found that FBI officials who signed NSLs with community of interest language "were not even aware that they were making such requests." Overall, the report concluded the FBI's community of interest practices were "improper," "inappropriate," and "likely resulted in the FBI obtaining and uploading into a [redacted] database thousands of telephone records" without actually certifying that they were relevant to an investigation.
In a 2014 follow-up report, the inspector general concluded that the FBI's policy on community of interest requests as written in the DIOG was a good start in fixing these problems because it required the general counsel's review — but the description of the policy itself was almost entirely redacted.
The FBI recently posted an updated section on NSLs from the DIOG from 2013, but it is heavily redacted, including the section on community of interest requests, making it impossible to know what updates have been made.
"I can say that the DIOG has been updated to reflect privacy concerns re: community of interest information," said an FBI spokesperson. He did not respond to questions about how often the FBI issues community of interest requests, whether they are still always limited to information once-removed from the original target of the NSL, and whether they could be used to obtain email as well as telephone records.
Reporters work on their laptops during a campaign event for then Sen. Barack Obama on May 9, 2008, in Beaverton, Ore.
Photo: Mark Wilson/Getty Images
Some of the instructions in the manuals imply meaningful restrictions on the bureau's use of NSLs. For example, the DIOG includes instructions for what agents should do if a company "overproduces" in response to such a letter and hands over more information than requested, such as data outside the time frame specified in the request, or data on the wrong phone number.
When that happens, the FBI is not allowed to upload the excess data into any of its internal systems except in cases where the information may be obtained via NSL and the bureau issues a second, "curative" letter.
The newer policy document on NSLs also spells out a clear prohibition on "exigent letters" — informal requests, supposedly issued only in an emergency, through which the FBI demanded information without even the internal approval and record keeping required of an NSL, let alone the approval of a judge. Up until 2006, the FBI used these letters to ask for information from companies without any particular authority, sometimes promising to follow up with a subpoena or NSL, but not always doing so. In 2010, the inspector general found that the bureau had illegally collected more than 2,000 call records between 2003 and 2006, often also asking for community of interest information. The report concluded that the FBI's use of the letters flouted the law and internal policies.
The newer NSL guidelines suggest that since 2007, the FBI has complied with this judgment: The document contains a section headed "NO EXIGENT LETTERS," in upper-case letters.
"The practice of using exigent letters to obtain NSL-type information is prohibited," it continues.
Top photo: An employee fixes part of a web server inside the Facebook Inc. Prineville Data Center in Prineville, Ore., on April 28, 2014.
Jenna McLaughlin[email protected]theintercept.com@JennaMC_Laugh
Domestic Investigations and Operations Guide
The rulebook governing all FBI agents' activities, in unredacted form for the first time. This is the 2011 edition, which remains the baseline document today, although the FBI recently released some updates from 2013.
Annotation Sets
Bureau Hid Doubts About Reliability of Stingray Evidence Behind Redaction Marks
Jenna McLaughlin
See Annotations
CIA and NSA Dossiers Are Available to the FBI in the Absence of Any Crime, Raising Privacy Questions
Cora Currier
FBI Spy Planes Must Abide Rules When Looking Into Homes
On Campus, the FBI Sometimes Operates Outside Restrictions
To Probe the Digital Defenses of Targets, the FBI Turns To a Special Program
Confidential Human Source Policy Guide
Detailed rules for how the FBI handles informants. Classified secret. This unreleased September 2015 document is a major expansion and update of a manual from 2007 on the same topic.
How the FBI Recruits and Handles Its Army of Informants
Trevor Aaronson
Counterterrorism Policy Guide
Excerpts from a guide for agents working on counterterrorism cases, which functions as a supplement to the FBI's main rulebook, the Domestic Investigations and Operations Guide. Classified secret. Not previously released. Dates to April 2015.
Disruptions: How the FBI Handles People Without Bringing Them To Court
Confidential Human Source Assessing Aid
A document bearing the seal of the FBI's Anchorage field office that gives tips for agents cultivating informants. It is classified secret, and dates from 2011.
DIOG Profiling Rules 2016
A 2016 update to the Domestic Investigations and Operations Guide's policy on profiling by race, gender, and other factors.
Guidance on Guardian Assessments 2013
A 2013 unclassified communique from the FBI's counterterrorism division explaining the database checks and other steps to be taken as part of low-level investigations.
National Security Letters Redacted
An unclassified internal FBI document explaining the rules for national security letters, orders that the bureau uses to obtain certain information without a warrant. The document is undated but contains references to another document from November 2015.
Editor-in-Chief: Betsy Reed. Series Editor: Ryan Tate. Associate Editor: Andrea Jones. Reporters: Trevor Aaronson, Cora Currier, Jenna McLaughlin, Alice Speri. Research: Alleen Brown, Talya Cooper, Danielle Mackey, Eseosa Olumhense, Miriam Pensack, John Thomason. Art Direction: Stephane Elbaz, Philipp Hubert, Nick Simmons. Additional Photo Editing: Soohee Cho, Shaun Lucas, Chelsea Matiash. Development: Tom Conroy, Andy Gillette, Carl Licata, Cacie Prins, Raby Yuson.
|
{
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{"url":"https:\/\/www.transtutors.com\/questions\/john-and-martha-holloway-are-marnied-filing-jointly-they-are-35-and-31-years-old-res-2820064.htm","text":"# John and Martha Holloway are marnied filing jointly They are 35 and 31 years old, respectively. T... 1 answer below \u00bb\n\nJohn and Martha Holloway are marnied filing jointly They are 35 and 31 years old, respectively. Their address is 10010 Dove Street Atlanta, GA 30294. Additional information about Mr. and Mrs. Holloway is as follows social security numbers: Dohn: 412-34-567e Date of birth: 3\/4\/1982 W-2 for John shows these amounts: Martha: 412-34-5671 Date of birth: 8\/28\/1986 W-2 for Martha shows these amounts: Wages (box 1) $22,600.0e Federal W\/H (box 2)$ 2,200.0 Social security wages (box 3) $22,ee0.ee Social security W\/H (box 4)$ 1,364.80 Medicare wages (box S) $22,869.88 Medicare W\/H (box 6) 319.80 Nages (box 1) 35, 580.80 Federal \u00e0\u00b8\u0153\/H (box 2) _$ 4.10e.ee Social security wages (box 3)-$35,5e8.80 Social security W\/H (box 4) 2,281.80 Nedicare wages (box 5) -$35, 5ee.00 Medicare W\/H (box 6)-514-75 Form 1099-DIV for Martha shows this amount Box la and box 1b $345 00 from MAR Brokerage Form 1099-INT for Martha shows these amounts Box 1$45000 from ABC Bank, Box 4 \\$35.00. Martha is a Human Resources Manager. Prepare the tax return for Mr. and Mrs. Holloway using the John is a maintenance worker and appropriate form. They want to contribute to the presidential election campaign. Mr. and Mrs. Holloway had qualifying health care coverage at all times during the tax year (List the names of the texpayers in the order in which they appear in the problem. Input al the values as positive numbers.)","date":"2019-11-22 06:04:43","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.28209376335144043, \"perplexity\": 13238.555602640725}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2019-47\/segments\/1573496671239.99\/warc\/CC-MAIN-20191122042047-20191122070047-00499.warc.gz\"}"}
| null | null |
Q: When can a pair of groups be embedded in each other? This is a question I made up, but couldn't solve even after some days' thought. Also if any terminology is unclear or nonstandard, please complain.
Given groups $G$ and $H$, we say that $G$ can be embedded in $H$ if there exists an injective homomorphism $\varphi : G \to H$. (Note that the image $\varphi(G)$ is then isomorphic to $G$.) I am interested in the situation where a pair of groups $G$ and $H$ can be embedded in each other. Of course, this is guaranteed to be the case when $G \cong H$. But is the converse true? More precisely:
Q1. Do there exist non-isomorphic groups $G$ and $H$ such that each of them can be embedded in the other?
I am interested in this because, in my mind, this question is analogous to the Cantor-Bernstein-Schroeder theorem in set theory. Of course, this view could be too naive or useless. Oh well.
The only "progress" I could make is to create another question. Let $\varphi_G:G \to H$ and $\varphi_H:H \to G$ be a pair of embeddings as in the question. Then the homomorphism $\varphi := \varphi_H \circ \varphi_G : G \to G$ is also injective; i.e., it is an embedding. I can show that the image of this map ($K := \varphi(G)$) is a proper subgroup of $G$ unless $G \cong H$. This leads me to another question:
Q2. Does there exists a group $G$ that is isomorphic to a proper subgroup of itself?
If the answer to this is negative, then so is the case for Q1. Though both of these seem "obviously false", I cannot prove them. Nor can I construct a counterexample. Any suggestions?
Some remarks:
*
*Nothing is inherently special about groups here. I suppose one could ask the same question for rings, fields, or other structures; I focused on this specific question for clarity.
*I tried to search through Wikipedia and Google books, but I cannot figure out the answer or where I can find the answer.
*I have no idea as to how easy or difficult these questions are. If they are trivial/easy (say, the level of a standard undergrad homework exercise), then please give me hints rather than a complete solution :-).
A: Let $F$ be a free group of finite rank $r > 1$. Then the commutator subgroup $[F,F]$ of $F$ is a free group of (countably!) infinite rank. Similarly but more easily, a free group of countably infinite rank contains as subgroups free groups of all finite ranks.
From this it follows that for any $r_1, r_2$ with $2 \leq r_1, r_2 \leq \aleph_0$, $r_1 \neq r_2$, the free group of rank $r_1$ and the free group of rank $r_2$ can be embedded in each other.
Comment: It is a lot easier to find examples of groups which are isomorphic to proper subgroups of themselves (or, in fancier terminology, non co-Hopfian groups). For instance an infinite cyclic group has this property, as does any nontrivial free abelian group or any infinite-dimensional vector space over $\mathbb{F}_p$ or $\mathbb{Q}$. (Added after seeing Arturo's answer: or, more generally, an infinite direct sum of copies of any nontrivial group!)
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| 9,489
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The 2013–14 San Miguel Beermen season was the 39th season of the franchise in the Philippine Basketball Association (PBA). The team was known as the Petron Blaze Boosters for the duration of the Philippine Cup.
Key dates
November 3: The 2013 PBA Draft took place in Midtown Atrium, Robinson Place Manila.
January 13: San Miguel Corporation president Ramon S. Ang announced that the team will revert to its iconic name "San Miguel Beermen" starting the 2014 PBA Commissioner's Cup.
Draft picks
Rosters
Philippine Cup
Eliminations
Standings
Game log
Playoffs
Bracket
Commissioner's Cup
Eliminations
Standings
Game log
Playoffs
Bracket
Governors' Cup
Eliminations
Standings
Bracket
Game log
Transactions
Trades
Pre-season
Recruited imports
References
San Miguel Beermen seasons
San Miguel
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| 9,077
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THE SITE AND MATERIALS IN THIS SITE ARE PROVIDED "AS IS" AND WITHOUT WARRANTIES OF ANY KIND EITHER EXPRESS OR IMPLIED IN CONNECTION WITH THIS SITE. TO THE FULLEST EXTENT PERMISSIBLE PURSUANT TO APPLICABLE LAW, FORMICA DISCLAIMS ALL WARRANTIES, EXPRESS OR IMPLIED, INCLUDING, BUT NOT LIMITED TO, IMPLIED WARRANTIES OF MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE, AND NON-INFRINGEMENT. FORMICA DOES NOT WARRANT OR MAKE ANY REPRESENTATIONS REGARDING THE USE OR THE RESULTS OF THE USE OF THE MATERIALS IN THIS SITE IN TERMS OF THEIR CORRECTNESS, ACCURACY, RELIABILITY, COMPLETENESS OR OTHERWISE. APPLICABLE LAW MAY NOT ALLOW THE EXCLUSION OF IMPLIED WARRANTIES, SO THE ABOVE EXCLUSION MAY NOT APPLY TO YOU.
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|
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"redpajama_set_name": "RedPajamaC4"
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| 9,665
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Q: PayPal Adaptive Chained Payments JSON setPaymentOptions PayPal Adaptive Chained Payments transactions lacking the functionality of Standard PayPal transactions which contains Order ID,Item ID,Shipping Address by default. Some suggested if we use setPaymentOptions we can implement "inserting Shipping Address" in Adaptive Transactions but that also has been restricted to embed light box. Is there anyway we could achieve Shipping Address inclusion in the Adaptive transactions by JSON? Our Appln uses JSON data. We feel it is must for enabling the Seller Protection policy
|
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"redpajama_set_name": "RedPajamaStackExchange"
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| 4,568
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Q: Given a string, return true if the number of appearances of "is" anywhere in the string is equal to the number of appearances of "not".? This is a correct answer I found to that question above. Can anyone please explain why they subtracted lengths and divided by 3 and 2 ?
String strNotMissing = str.replaceAll("not", "");
String strIsMissing = str.replaceAll("is", "");
int notCount = (str.length() - strNotMissing.length()) / 3;
int isCount = (str.length() - strIsMissing.length()) / 2;
return isCount == notCount;
A: character count of string "not" is 3 and for string "is" is 2 that's why.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
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| 1,277
|
Q: Split and extract data from a field php mysql I have a problem with this data saved in a MySQL row:
34:3,37:2,64:1,85:4
The first number is an ID that identifies a user, the second number (separated by ":") is the arrival order position.
I need to extract and split the data to create a table with the arrivals order:
Something similar to:
name position
Any suggestion?
A: Simple: explode()
explode(",", $yourdata);
PHP Explode
A: $split = explode(';', $myvars);
$id = $split[0];
$whatever = $split[1];
|
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"redpajama_set_name": "RedPajamaStackExchange"
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| 47
|
using System;
using System.IO;
using BizHawk.Emulation.Common;
namespace BizHawk.Client.Common
{
public class CoreFileProvider : ICoreFileProvider
{
public string SubfileDirectory { get; set; }
public FirmwareManager FirmwareManager { get; set; }
private readonly Action<string> _showWarning;
public CoreFileProvider(Action<string> showWarning)
{
_showWarning = showWarning;
}
public string PathSubfile(string fname)
{
return Path.Combine(SubfileDirectory ?? "", fname);
}
public string DllPath()
{
return Path.Combine(PathManager.GetExeDirectoryAbsolute(), "dll");
}
public string GetSaveRAMPath()
{
return PathManager.SaveRamPath(Global.Game);
}
public string GetRetroSaveRAMDirectory()
{
return PathManager.RetroSaveRAMDirectory(Global.Game);
}
public string GetRetroSystemPath()
{
return PathManager.RetroSystemPath(Global.Game);
}
public string GetGameBasePath()
{
return PathManager.GetGameBasePath(Global.Game);
}
#region EmuLoadHelper api
private void FirmwareWarn(string sysID, string firmwareID, bool required, string msg = null)
{
if (required)
{
var fullmsg = $"Couldn't find required firmware \"{sysID}:{firmwareID}\". This is fatal{(msg != null ? $": {msg}" : ".")}";
throw new MissingFirmwareException(fullmsg);
}
if (msg != null)
{
var fullmsg = $"Couldn't find firmware \"{sysID}:{firmwareID}\". Will attempt to continue: {msg}";
_showWarning(fullmsg);
}
}
public string GetFirmwarePath(string sysId, string firmwareId, bool required, string msg = null)
{
var path = FirmwareManager.Request(sysId, firmwareId);
if (path != null && !File.Exists(path))
{
path = null;
}
if (path == null)
{
FirmwareWarn(sysId, firmwareId, required, msg);
}
return path;
}
private byte[] GetFirmwareWithPath(string sysId, string firmwareId, bool required, string msg, out string path)
{
byte[] ret = null;
var path_ = GetFirmwarePath(sysId, firmwareId, required, msg);
if (path_ != null && File.Exists(path_))
{
try
{
ret = File.ReadAllBytes(path_);
}
catch (IOException)
{
}
}
if (ret == null && path_ != null)
{
FirmwareWarn(sysId, firmwareId, required, msg);
}
path = path_;
return ret;
}
public byte[] GetFirmware(string sysId, string firmwareId, bool required, string msg = null)
{
string unused;
return GetFirmwareWithPath(sysId, firmwareId, required, msg, out unused);
}
public byte[] GetFirmwareWithGameInfo(string sysId, string firmwareId, bool required, out GameInfo gi, string msg = null)
{
string path;
byte[] ret = GetFirmwareWithPath(sysId, firmwareId, required, msg, out path);
if (ret != null && path != null)
{
gi = Database.GetGameInfo(ret, path);
}
else
{
gi = null;
}
return ret;
}
#endregion
// this should go away now
public static void SyncCoreCommInputSignals(CoreComm target)
{
string superhack = null;
if (target.CoreFileProvider is CoreFileProvider)
{
superhack = ((CoreFileProvider)target.CoreFileProvider).SubfileDirectory;
}
var cfp = new CoreFileProvider(target.ShowMessage) { SubfileDirectory = superhack };
target.CoreFileProvider = cfp;
cfp.FirmwareManager = Global.FirmwareManager;
}
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
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| 8,343
|
Torsten Hjalmar Lord (* 2. März 1904 in Nacka; † 4. Februar 1970 in Lidingö) war ein schwedischer Segler.
Erfolge
Torsten Lord, der für den Kungliga Svenska Segelsällskapet segelte, nahm in der 6-Meter-Klasse an drei Olympischen Spielen teil. 1936 segelte er bei den Spielen in Berlin unter Sven Salén auf der May Be und erzielte mit unter anderem zwei Siegen in sieben Wettfahrten 62 Gesamtpunkte. Die May Be, deren übrige Besatzung aus Lennart Ekdahl, Martin Hindorff und Dagmar Salén bestand, schloss die im Olympiahafen Düsternbrook in Kiel stattfindende Regatta somit auf dem dritten Platz hinter Großbritannien und Norwegen ab. Auch bei den Olympischen Spielen 1948 in London gewann Lord Bronze. Mit der Ali-Baba II sicherte er sich gemeinsam mit Skipper Tore Holm sowie Karl-Robert Ameln, Martin Hindorff und Gösta Salén hinter dem US-amerikanischen und dem argentinischen Boot den dritten Rang. 1952 segelte er wiederum unter Sven Salén auf der May Be VII, verpasste als Vierter allerdings dieses Mal knapp einen weiteren Medaillengewinn. Mit 3773 Punkten blieb das schwedische Boot lediglich 171 Punkte hinter den drittplatzierten Finnen auf der Ralia.
Weblinks
Torsten Lord beim Sveriges Olympiska Kommitté (schwedisch)
Regattasegler (Schweden)
Olympiateilnehmer (Schweden)
Teilnehmer der Olympischen Sommerspiele 1936
Teilnehmer der Olympischen Sommerspiele 1948
Teilnehmer der Olympischen Sommerspiele 1952
Schwede
Geboren 1904
Gestorben 1970
Mann
|
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| 3,635
|
Luiz Paulo Horta was a Brazilian journalist. He was born in Rio de Janeiro on August 14, 1943. He died on August 3, 2013, in Rio de Janeiro, at the age of 69.
In 1962 he started a law course at PUC-RJ, before moving on to journalism. He joined the Correio da Manhã in 1963, and the Jornal do Brasil in 1964, where he stayed until 1990. He then moved to O Globo, where he continued to work as a music critic. In 1986, he founded and directed the music section of the Museum of Modern Art in Rio de Janeiro. In 2000 and 2001, he directed a group of biblical studies at the Loyola Center of PUC-RJ.
He was a member of the Brazilian Academy of Music and the Brazilian Academy of Art. He was a member of the Development Council of PUC-RJ and of the Cultural Commission of the Archdiocese of Rio de Janeiro.
In 2000 he received the Padre Ávila Prize for Ethics in Journalism, granted by PUC-RJ. In 2010 he received the Medal of the Inconfidente from the Government of Minas Gerais. He was the seventh occupant of Chair 23, elected on August 21, 2008, at the Brazilian Academy of Letters, in succession to Zélia Gattai, and received on November 28 of the same year by academic Tarcísio Padilha.
References
Brazilian journalists
Male journalists
1943 births
2013 deaths
20th-century Brazilian people
|
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| 330
|
Temple Tales #2: 'A Man and his Monument': the real David Davies?
Posted in Hidden histories / Hanesion cudd, League of Nations / Cynghrair y Cenhedloedd, Temple of Peace / Y Deml Heddwch, Wales for Peace / Cymru dros Heddwch by rosajosie
Picture: bronze bust of David Davies in the marble hall at the Temple of Peace
Maggie Smales, volunteer with Wales for Peace, explores different perspectives on Temple of Peace founder David Davies.
I am currently helping to sort out the archives belonging to the Welsh Centre for International Affairs (WCIA). In a box with materials about the Temple of Peace and Health, I found the transcript of a BBC radio programme which provides an unusually intimate portrait of Lord Davies: the Liberal statesman, philanthropist and the force behind the Temple.
The BBC's Home Service in Wales commemorated the occasion of the 25th anniversary of the Temple of Peace and Health in Cardiff in 1963 with an hour-long radio documentary on the life and work of David Davies, "A man and his monument". John Griffiths– who compiled and narrated the programme- used a mixture of interviews and recordings with people who had known Davies well, from his associate Sir Wynn Wheldon to his chauffeur. The documentary built a picture of the man and the two causes to which he devoted his life: social health and international peace.
David Davies was born into a wealthy family and given the same name as his grandfather, David Davies of Llandinam. The Welsh industrialist and Liberal politician had first founded the Ocean Coal Company in the Rhondda, before building Barry Docks and the connecting railways to export his coal. Like his grandfather, David Davies was a Calvinistic Methodist by upbringing, a teetotaller and firm about his Sunday observance.
Davies inherited his grandfather's industrial wealth and landed estate when he was just eighteen and had the confidence which great wealth brings. He was essentially a very wealthy country gentleman, devoted to field sports. His neighbours remembered him as "a big man in a pink coat"… "hunting his own hounds in all weathers in all hours" and "driving open cars at terrific speed – he had 2 Talbot Darracqs…". According to Sir Wynn Wheldon "he was almost unsurpassed in his ability to go on hour by hour…for more than a day…if it would be a fox, he'd find it and would follow that fox until it was 36 hours had expired and he'd have nothing to eat in the meantime."
Davies' private secretary recalled that he was never "really interested in business at all… and left as much work as he possibly could to other people". His real passion was his various philanthropic causes. Here, as his one of his employees noted "He worked himself hard and he expected everyone else to do the same"; "…he was an inconsiderate employer in that he didn't spare himself and he certainly didn't spare other people". "He had no concept of family life – he worked at home and his time over the weekend would be divided between his study, where he would be closeted with secretaries… and the great outdoors."
In 1906, at the age of 26, Davies became MP for Montgomeryshire. Apparently, it was touch and go which party he would join, his grandfather having broken with the Gladstonian Liberals over Home Rule for Ireland. But in the end, as Sir Wynn Wheldon put it "the Liberals got hold of him first and they found him for a good many purposes a very valuable Liberal." In his early days in Parliament Lloyd George apparently dismissed Davies as "the Golden Calf", just a rich son of a rich father, but he soon began to demonstrate the shallowness of this judgement.
Davies was prominent in the 1910 public meeting in Shrewsbury to decide how Wales could best commemorate the recently deceased monarch. He was a driving force behind the decision to set up the King Edward VII Welsh National Memorial Association (WNMA) for the eradication of tuberculosis, a particular scourge in Wales at the time. He himself gave £150,000 towards the eventual £300,000 raised and his sisters Margaret and Gwendoline also made substantial donations. He continued to be heavily involved with this health campaign after the war. In 1920 he endowed a Chair for the study of tuberculosis in the Medical School at Cardiff University. Following the 1921 Public Health (Tuberculosis) Act, which compelled all local authorities to make some provision for TB, he, "moved heaven and earth", according to his private secretary, to ensure that the WNMA would cooperate with the national scheme and become, in effect, a joint board for the whole of Wales.
At the outbreak of the First World War, Davies was given the command of the 14th Battalion of the Royal Welsh Fusiliers. He is recorded as having driven his men and officers to the limits of their endurance when they were training in Llandudno to make them war ready – perhaps he had an inkling of what was waiting for them in France. Once in the field he also, "made good certain deficiencies of supply out of his own pocket, providing his men with fishing waders (to wear in the trenches) …and his unit with field telephones". It is recorded that he did not get on well with the Higher Command and was perhaps "too original for them".
Towards the end of 1916, Davies-who was still a serving MP- was brought back from the front to become Lloyd George's Parliamentary Private Secretary. He lasted for less than a year in the post as he was far too opinionated for Lloyd George.
Davies' experience in the trenches made him a passionate advocate for peace-making. In 1917, before the end of the First World War, he founded the League of Free Nations Association with HG Wells and Gilbert Murray. In 1918 this merged with another group, the League of Nations Society, to form the League of Nations Union, with former Foreign Secretary Sir Edward Grey as President. Davies was later to endow the organisation in Wales with enough money to meet the cost of its administrative staff. In 1919, he also funded the world's first Chair of International Relations at Aberystwyth University College.
Davies had many other interests. He realised the importance of better housing, especially in the fight against TB, and encouraged T Alwyn Lloyd to develop model housing estates in places such as Machynlleth and Wrexham, and after the First World War, Rhiwbina. He was interested in agriculture, bailing out the debts of the Royal Welsh Agricultural Show on more than one occasion, and loaning out prize bulls and stallions to local farmers in his county to improve the quality of livestock locally. He gave away parcels of his own land to form playing fields in Machynlleth and Caersws and encouraged neighbours in mid-Wales to do the same; he helped fund village institutes including the pavilion in Newtown and also sent one of his staff to help set up the first Miners' Welfare Scheme in the Rhondda.
But the League of Nations Union was his abiding passion. Davies was never a pacifist – his vision of the League of Nations included the concept of an International Court of Equity to intervene in disputes and an International Police Force to carry out its decisions. Here he parted company with much of the interwar peace movement, including the Peace Pledge Union, and what he saw as "negative disarmament". After resigning as an MP in 1929, he spent much of the 1930s campaigning for an armed League of Nations, including founding an organisation called New Commonwealth in 1932 to promote the idea of an International Police Force and writing several books and political tracts.
In the late 1920s a plan was emerging to build a headquarters for the WNMA in Cathays Park in Cardiff on a site donated by the City Corporation. Davies saw this as another opportunity to promote the cause of peace by including in the proposal offices for the League of Nations Union in Wales and a Temple of Peace "which will be an example for the whole world". This inevitably involved considerable extra expense, especially as Davies envisaged a grand Hall of Peace built with material from all parts of the world. The WNMA agreed to contribute just £12k (the capitalised value of its £500 annual rent) towards the eventual £85k cost, and David Davies provided much of the remaining funds whilst furnishings in the Temple were the gift of Misses Margaret and Gwendoline Davies. Davies' factotum Captain Glen Jones noted that while Davies had a proper familial affection for the two "he always brought in his sisters… he always wanted to screw out of them any contributions" for whatever project was occupying him at the time.
The foundation stone for the Temple of Peace was laid by Viscount Halifax in 1937. Ironically, Halifax went on to be Foreign Secretary and one of the architects of the policy of appeasement towards Hitler which David Davies himself resolutely opposed. By the time the Temple was formally opened in November 1938, Hitler had already marched into the Sudetenland, and despite PM Chamberlain's hope of "peace in our time", war was in the offing.
Sir Wynn Wheldon believed that Davies' premature death from lung cancer in June 1944 was caused at least in part by his feelings of impotence in the face of the second great war of his lifetime. "He had to pay the price physically – he didn't live to be an old man". Davies had become embittered in the 1930s when he could see that war was imminent and governments and politicians refused to heed his warnings. As Wheldon explained, "if you have one thing burning in you all the time, people tend to avoid you. He didn't always find the welcome that the splendour of his own ideas and the amount of work, money and time he had given to it, really deserved. He became in a sense tedious".
Posthumously, of course, Davies' ideas influenced the writing of the United Nations Charter, especially with regards to sanctions and the transition of national armies to an international police. In his own words: "We shall never get prosperity and security until we get peace: we shall never get peace until we get justice and we shall get none of those things until we establish the rule of law by means of a really effective International Authority equipped with an equity tribunal and an international police force."
"A man and his monument" was broadcast on 21 November 1963 by the BBC Home Service in Wales.
« Temple Tales #1: The Cold Case of the Russian Doll
Welsh among the ANZACs: WW1 in Palestine on the Centenary of Beersheeba, 31st Oct 1917 »
One thought on "Temple Tales #2: 'A Man and his Monument': the real David Davies?"
Ann Lowe
Very interesting article a piece for our time also
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
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| 3,662
|
// Copyright 2022 Google LLC
//
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
/**
* MobileDeviceTargeting.java
*
* This file was auto-generated from WSDL
* by the Apache Axis 1.4 Mar 02, 2009 (07:08:06 PST) WSDL2Java emitter.
*/
package com.google.api.ads.admanager.axis.v202205;
public class MobileDeviceTargeting implements java.io.Serializable {
private com.google.api.ads.admanager.axis.v202205.Technology[] targetedMobileDevices;
private com.google.api.ads.admanager.axis.v202205.Technology[] excludedMobileDevices;
public MobileDeviceTargeting() {
}
public MobileDeviceTargeting(
com.google.api.ads.admanager.axis.v202205.Technology[] targetedMobileDevices,
com.google.api.ads.admanager.axis.v202205.Technology[] excludedMobileDevices) {
this.targetedMobileDevices = targetedMobileDevices;
this.excludedMobileDevices = excludedMobileDevices;
}
@Override
public String toString() {
return com.google.common.base.MoreObjects.toStringHelper(this.getClass())
.omitNullValues()
.add("excludedMobileDevices", getExcludedMobileDevices())
.add("targetedMobileDevices", getTargetedMobileDevices())
.toString();
}
/**
* Gets the targetedMobileDevices value for this MobileDeviceTargeting.
*
* @return targetedMobileDevices
*/
public com.google.api.ads.admanager.axis.v202205.Technology[] getTargetedMobileDevices() {
return targetedMobileDevices;
}
/**
* Sets the targetedMobileDevices value for this MobileDeviceTargeting.
*
* @param targetedMobileDevices
*/
public void setTargetedMobileDevices(com.google.api.ads.admanager.axis.v202205.Technology[] targetedMobileDevices) {
this.targetedMobileDevices = targetedMobileDevices;
}
public com.google.api.ads.admanager.axis.v202205.Technology getTargetedMobileDevices(int i) {
return this.targetedMobileDevices[i];
}
public void setTargetedMobileDevices(int i, com.google.api.ads.admanager.axis.v202205.Technology _value) {
this.targetedMobileDevices[i] = _value;
}
/**
* Gets the excludedMobileDevices value for this MobileDeviceTargeting.
*
* @return excludedMobileDevices
*/
public com.google.api.ads.admanager.axis.v202205.Technology[] getExcludedMobileDevices() {
return excludedMobileDevices;
}
/**
* Sets the excludedMobileDevices value for this MobileDeviceTargeting.
*
* @param excludedMobileDevices
*/
public void setExcludedMobileDevices(com.google.api.ads.admanager.axis.v202205.Technology[] excludedMobileDevices) {
this.excludedMobileDevices = excludedMobileDevices;
}
public com.google.api.ads.admanager.axis.v202205.Technology getExcludedMobileDevices(int i) {
return this.excludedMobileDevices[i];
}
public void setExcludedMobileDevices(int i, com.google.api.ads.admanager.axis.v202205.Technology _value) {
this.excludedMobileDevices[i] = _value;
}
private java.lang.Object __equalsCalc = null;
public synchronized boolean equals(java.lang.Object obj) {
if (!(obj instanceof MobileDeviceTargeting)) return false;
MobileDeviceTargeting other = (MobileDeviceTargeting) obj;
if (obj == null) return false;
if (this == obj) return true;
if (__equalsCalc != null) {
return (__equalsCalc == obj);
}
__equalsCalc = obj;
boolean _equals;
_equals = true &&
((this.targetedMobileDevices==null && other.getTargetedMobileDevices()==null) ||
(this.targetedMobileDevices!=null &&
java.util.Arrays.equals(this.targetedMobileDevices, other.getTargetedMobileDevices()))) &&
((this.excludedMobileDevices==null && other.getExcludedMobileDevices()==null) ||
(this.excludedMobileDevices!=null &&
java.util.Arrays.equals(this.excludedMobileDevices, other.getExcludedMobileDevices())));
__equalsCalc = null;
return _equals;
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private boolean __hashCodeCalc = false;
public synchronized int hashCode() {
if (__hashCodeCalc) {
return 0;
}
__hashCodeCalc = true;
int _hashCode = 1;
if (getTargetedMobileDevices() != null) {
for (int i=0;
i<java.lang.reflect.Array.getLength(getTargetedMobileDevices());
i++) {
java.lang.Object obj = java.lang.reflect.Array.get(getTargetedMobileDevices(), i);
if (obj != null &&
!obj.getClass().isArray()) {
_hashCode += obj.hashCode();
}
}
}
if (getExcludedMobileDevices() != null) {
for (int i=0;
i<java.lang.reflect.Array.getLength(getExcludedMobileDevices());
i++) {
java.lang.Object obj = java.lang.reflect.Array.get(getExcludedMobileDevices(), i);
if (obj != null &&
!obj.getClass().isArray()) {
_hashCode += obj.hashCode();
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__hashCodeCalc = false;
return _hashCode;
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// Type metadata
private static org.apache.axis.description.TypeDesc typeDesc =
new org.apache.axis.description.TypeDesc(MobileDeviceTargeting.class, true);
static {
typeDesc.setXmlType(new javax.xml.namespace.QName("https://www.google.com/apis/ads/publisher/v202205", "MobileDeviceTargeting"));
org.apache.axis.description.ElementDesc elemField = new org.apache.axis.description.ElementDesc();
elemField.setFieldName("targetedMobileDevices");
elemField.setXmlName(new javax.xml.namespace.QName("https://www.google.com/apis/ads/publisher/v202205", "targetedMobileDevices"));
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elemField.setNillable(false);
elemField.setMaxOccursUnbounded(true);
typeDesc.addFieldDesc(elemField);
elemField = new org.apache.axis.description.ElementDesc();
elemField.setFieldName("excludedMobileDevices");
elemField.setXmlName(new javax.xml.namespace.QName("https://www.google.com/apis/ads/publisher/v202205", "excludedMobileDevices"));
elemField.setXmlType(new javax.xml.namespace.QName("https://www.google.com/apis/ads/publisher/v202205", "Technology"));
elemField.setMinOccurs(0);
elemField.setNillable(false);
elemField.setMaxOccursUnbounded(true);
typeDesc.addFieldDesc(elemField);
}
/**
* Return type metadata object
*/
public static org.apache.axis.description.TypeDesc getTypeDesc() {
return typeDesc;
}
/**
* Get Custom Serializer
*/
public static org.apache.axis.encoding.Serializer getSerializer(
java.lang.String mechType,
java.lang.Class _javaType,
javax.xml.namespace.QName _xmlType) {
return
new org.apache.axis.encoding.ser.BeanSerializer(
_javaType, _xmlType, typeDesc);
}
/**
* Get Custom Deserializer
*/
public static org.apache.axis.encoding.Deserializer getDeserializer(
java.lang.String mechType,
java.lang.Class _javaType,
javax.xml.namespace.QName _xmlType) {
return
new org.apache.axis.encoding.ser.BeanDeserializer(
_javaType, _xmlType, typeDesc);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,516
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package qbg.jtrack.sound;
import java.io.Closeable;
import java.util.Arrays;
import javax.sound.sampled.AudioFormat;
import javax.sound.sampled.AudioSystem;
import javax.sound.sampled.DataLine;
import javax.sound.sampled.SourceDataLine;
/**
* Wrapper around Java audio
*/
public class JavaAudio implements Closeable, AudioDriver {
/**
* The line the system uses
*/
private SourceDataLine line;
/**
* Array of registered operations
*/
private SoundOperation[] operations = new SoundOperation[0];
/**
* Common buffer
*/
private int[] buffer = new int[16];
/**
* Mix buffer
*/
private int[] mix = new int[16];
/**
* Buffered used by sound I/O
*/
private byte[] target = new byte[2048];
/**
* Ready the audio system
*/
public void open() {
AudioFormat format = new AudioFormat(44100, 16, 1, true, false);
DataLine.Info info = new DataLine.Info(SourceDataLine.class, format);
if (!AudioSystem.isLineSupported(info)) {
throw new RuntimeException("Line type not supported");
}
try {
line = (SourceDataLine) AudioSystem.getLine(info);
line.open(format);
line.start();
} catch (Exception ex) {
throw new RuntimeException(ex);
}
}
public void close() {
line.stop();
line.close();
}
/**
* Remove all registered operations from this system
*/
public void removeAllOperations() {
operations = new SoundOperation[0];
}
/**
* Register an operation with this system
* @param operation
*/
public void addOperation(SoundOperation operation) {
int len = operations.length;
operations = Arrays.copyOf(operations, len+1);
operations[len] = operation;
}
/**
* Generate a number of frames of sound
* @param frames
*/
public void generateFrames(int frames) {
int idx = 0;
for (int frame = 0; frame < frames; frame++) {
for (int op = 0; op < operations.length; op++) {
operations[op].fillBuffer(buffer);
for (int i = 0; i < 16; i++) {
mix[i] += buffer[i];
}
}
for (int i = 0; i < 16; i++) {
int val = mix[i];
val = val < -32767 ? -32767 :
val > 32767 ? 32767 :
val;
mix[i] = 0;
target[idx++] = (byte)val;
target[idx++] = (byte)(val >> 8);
}
if (idx == 2048) {
line.write(target, 0, 2048);
idx = 0;
}
}
line.write(target, 0, idx);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,785
|
About the Author
When Melissa Cynova was fourteen, a kid in her class gave her a deck of tarot cards for unknown reasons. She's been reading ever since. In addition to being a prolific tarot reader, she teaches classes at her kitchen table and at tarot conferences. She lives in St. Louis, Missouri, with her husband, Joe, their two amazing kids, two dogs, four black cats, and a tortoise named Phil. When she's not slinging cards, she's reading, watching superhero movies, and hanging out with her people. You can find her online at www.melissacynova.com.
Llewellyn Publications
Woodbury, Minnesota
Copyright Information
Tarot Elements: Five Readings to Reset Your Life © 2019 by Melissa Cynova.
All rights reserved. No part of this book may be used or reproduced in any matter whatsoever, including Internet usage, without written permission from Llewellyn Publications, except in the form of brief quotations embodied in critical articles and reviews.
As the purchaser of this e-book, you are granted the non-exclusive, non-transferable right to access and read the text of this e-book on screen. The text may not be otherwise reproduced, transmitted, downloaded, or recorded on any other storage device in any form or by any means.
Any unauthorized usage of the text without express written permission of the publisher is a violation of the author's copyright and is illegal and punishable by law.
First e-book edition © 2019
E-book ISBN: 9780738758466
Book design: Donna Burch-Brown / Ted Riley
Cover art: Harry Briggs
Cover design: Kevin R. Brown
Interior art: Llewellyn's Classic Tarot, illustrated by Eugene Smith © 2014. Used with permission from Llewellyn Worldwide Ltd., no further reproduction permitted.
Tarot for Beginners: A Practical Guide to Reading the Cards by Barbara Moore © 2010 Llewellyn Worldwide, Ltd. 2143 Wooddale Drive, Woodbury, MN 55125. All rights reserved, used by permission.
Spread on page 154 © 2012 Beth Maiden. Used with permission.
Llewellyn Publications is an imprint of Llewellyn Worldwide Ltd.
Library of Congress Cataloging-in-Publication Data (Pending)
ISBN: 978-0-7387-5840-4
Llewellyn Publications does not participate in, endorse, or have any authority or responsibility concerning private business arrangements between our authors and the public.
Any Internet references contained in this work are current at publication time, but the publisher cannot guarantee that a specific reference will continue or be maintained. Please refer to the publisher's website for links to current author websites.
Llewellyn Publications
Llewellyn Worldwide Ltd.
2143 Wooddale Drive
Woodbury, MN 55125
www.llewellyn.com
Manufactured in the United States of America
To Sarah Kate Buckles,
who turned her home into a writer's retreat for me,
helped me breathe, made whole meals for me with only
ingredients in the house (how did you do that?),
and helped me birth this book.
You are the best book doula in the entire world,
and it wouldn't be here without you.
I love you more than five, Sk8.
Disclaimer
You weren't born with or handed the tools to deal with life. Sometimes it's a good idea to see a counselor or therapist who has the training and the tools, and they can help you out. No shame and no blame—just chasing joy.
The information in this book is not a substitute for mental, medical, or financial care from a licensed professional. The publisher and the author assume no liability for any consequences to the reader that may result from the reader's use of content contained in this publication and recommend common sense when contemplating the practices described in the work.
Contents
Introduction: Can You Fix My Life?
Chapter 1: How Element Readings Work
Chapter 2: How to Read Tarot Cards
Chapter 3: Earth: A Reading to Shelter You
Chapter 4: Air: A Reading to Help You Breathe
Chapter 5: Fire: A Reading to Light the Way
Chapter 6: Water: A Reading to Help You Float
Chapter 7: Spirit: A Reading to Bring You Connection
Chapter 8: Additional Spreads for Clarity
Chapter 9: Self-Care
Conclusion
Appendix: Tarot Card Meanings
Recommended Reading
Acknowledgements
Introduction
Can You Fix
My Life?
A long-time tarot client asked me for a reading. She'd been going through a really difficult period of time. She'd gotten into a car accident after losing her job. She'd become disabled, and while going through this experience, her marriage ended. She had to find a new place to live that was wheelchair accessible, and then her son went into rehab after overdosing. Oh—and both of her parents died.
All of this happened in six months. Think about that. When she called me, she was in her new home (that she hated), her divorce was pretty much final, and her parents had both been put to rest and their estate finalized. Her son was in rehab, doing well, but damn. She was exhausted—mentally, spiritually, physically, and emotionally. She said she was doing "Okay? I guess." You know the way you say that you're "Okay?" and your voice goes up at the end? It becomes a question that's trying to answer itself. It's the polite thing that you say when you feel like your heart has been run over. You don't want to overburden your friend with the absolute train wreck your life has become, but you also don't want to lie. You go to this half-honest, shadowy place and you test out this answer. "Okay?" really means "I can't breathe. Please help. Unless, you know, you're too busy, because I know you've got a lot going on."
This "Okay?" is generally followed with a solid "I guess" because saying that things are okay is such a complete and utter untruth that your whole self wants to fight it in the parking lot after school. "I guess" turns the presumption that "Okay?" is even in your wheelhouse into a laughable fantasy.
So we talked for a little bit. Turns out she was not, in fact, okay. She wasn't even "Okay? I guess." She didn't have a job to distract herself with. She didn't have a partner to talk to, and her son was unable to help. She had no siblings, and all her friends had briefly vanished. She laughed about this, saying, "Every time I called them, something even more terrible happened. They need a break from me."
I asked her what she wanted to ask about, and she said, "I don't know. Everything? Nothing? Can you just fix me? Can you fix my life and hit a reset button or something with the universe? Because I just need one thing to be okay. And right now? Nothing is okay."
I said that I couldn't fix her life, but that maybe I could give her some tools that she could use to do the heavy work herself. The reading was pretty gentle, and the best/worst card that she got was the Ten of Swords: "Hey, sis. Listen. You're at rock bottom. You're here. It will literally only go uphill from here. I promise. I promise."
We finished the reading and were saying goodbye when she said it again. "It would be great if I could just get, like, a cosmic do-over, you know? New outlook on love, on my stupid house, on my job situation..." I remember being disappointed in myself and my cards for not knowing how to answer that. It would be nice. Hell, I'd love to do that for myself sometimes, but that's not how my readings work.
I kept thinking about her, though.
A few weeks later, I might have hallucinated a tiny bit. No drugs or anything, Dad, I promise! I have a friend named Aidan Wachter who is a talismanic jeweler. He posted a picture of a pentacle that he'd made and said, "It's too small." I couldn't stop looking at this tiny pentacle. I was tossing the image around in my head and I saw it split apart into pieces, then come back together as a whole. I saw the integration of earth, air, fire, water, and spirit. They all come together to make the pentacle. To make us.
| Earth is our home.
---|---
| Air is our mind.
| Fire is our body.
| Water is our heart.
| Spirit is our soul.
They make us. They're the bones of who we are.
But they're buried, right? Our bones are stuck deep, deep beneath self-esteem issues and tragedy and heartbreak and frustration and boredom. You can't see them. You can't feel them. So you react to things in a way that is Not-You. You carry around all the detritus and bullshit that people lay on you. You add to it with self-doubt and fear. You carry it and carry it and then you look into the mirror and you're not sure who you are anymore in your core. My client was suffering from a serious case of "Who am I, and how in the hell did I get here?"
It is true that our broken parts are what catch the light to make us shine, but it is also true that if there are too many broken parts, we can shatter into bits. I think that we spend a lot of time going around the hurt instead of healing. We shove it to the side and pretend that we're fine. It's easy and difficult at the same time. If we compartmentalize our pain, we can tiptoe around it and pretend it never happened. Then, when we least expect it, it frickin' explodes all over our lives. When we put our broken bits back together, it creates a beautiful mosaic that is us. However, if we put our broken bits together with lies, it's not going to stay. The cracks will grow and the pain will find us again.
When I ask you during a reading whose sadness you are carrying around and you say you don't know, there is healing to be done. When you tell me that you know you should get healthy, but you can't make it worth your time, there is a reason. That reason is not that you're stupid, incompetent, or lazy. That might have been what you were told, but it's just not true. We do this thing where we'll invest countless hours in the problems of a friend and completely ignore our own. Why do we do that? Are we not as valuable as our people?
With all of this swimming around in my head, I called my client. She was surprised and a little weirded out to hear from me, because you don't often have your tarot reader call you, but she was interested in my idea.
What if we broke down her life into these elements? What if we did five readings, one for each part of her life? What if we used these readings to look specifically into what her frustrations were? Where are the supports in her life? What resources were out there for her? We talked about it for a bit, and she agreed to be my guinea pig. Brave woman. What I taught my client who lost so much is what I'm going to teach you.
I designed a reading for each element based on its alchemical symbol. I created a kit to use in an exercise designed to help her start the process of making change. I compiled community and online resources for continuing the change. I wanted to get to the heart of the problems. Not "Why can't I find a partner?" but "What can I fix in me emotionally that will lead me to being happy without a partner?" Not "Why can't I lose weight when I want to?" but "How can I see myself as worthy of good health?" When I was ready, we started with earth, talking about her home and how unhappy she was. After we figured that issue out, we moved on to air and took a good look at her mind and how tired she was of being at home. Fire addressed her physical self, and water her emotions. We finished with the spirit reading and helped her find a spiritual practice that resonated.
I did not fix her. This is so important. I cannot fix people that I didn't break. She fixed herself. I gave her the tools she needed—that's my job as a tarot reader. I'm the guide, but all the hard work starts with the client.
This program is a reset button for your life. For your heart, mind, body, soul, spirit, and home. This is not a replacement for counseling or meds if you need them. Tarot readings are excellent for helping you figure out what's next, and sometimes the next step is therapy. This is not a bandage for a gaping wound; this is essentially a New Age self-help book. We're going to talk about each part of your life and figure out what is helping you, what is hurting you, and which tools you have available to help you move forward to your most authentic self.
That's what it's all about, right? Living an honest, uncluttered life in which you can breathe freely, stand tall, and roll with the punches. Part of living, though, is picking up damaging habits, other people's opinions, and misinformation. Tarot Elements will help you sort through the detritus of living and pull away those things that don't serve you. We'll get to the root of the problem.
These readings and this journey can be intimidating. It can be full of tears and longing and wondering if your reset button is stuck or lost or broken. That's where I come in. We're going to do this together. We're going to walk through the scary stuff to the other side, and take a deep breath, and wipe our eyes, and realize that the way forward might still be hard, but it will be good. And we'll be carrying less baggage, so our steps will be lighter.
Each element reading process will follow this formula:
* Description of the element
* Story to illustrate what happens when that element goes bajiggedy, using examples from my awesome yet intense life
* Tarot reading to help you make sense of the element in your life
* Two sample readings to use as examples
* Your homework
* Resources and additional tarot spreads to help you bring it on home
Let's do it.
XO,
Melissa
[contents]
Chapter 1
How Element
Readings Work
I tried to keep the structure of each reading very basic. These spreads are based on the alchemical signs for each element. I found that this simplicity worked beautifully with the particular elements. For water, for example, there are three things/cards that can "get in your way." In my readings, I find that it's seldom just one thing in our history that keeps us from being emotionally healthy. In the air reading, you have one thing to hold on to. One focus. What is your truest, most honest desire. Can you rely on your skills, your degree, your mental acuity? What is that thing that will open the door to the rest of your satisfaction with work and with exercising your mind?
What Are We Asking?
Each reading, except for spirit, has the same three questions in different card configurations. When you aren't sure where to go or what to do, you want to start with the basics:
* What can you hold on to?
* What can pull you forward?
* What gets in your way?
These three questions will help you find your reset button.
What Can You Hold On To?
What are the things in your life—in relation to heart, body, mind, home, and soul—that are currently supporting you or are things that you can begin to lean on. These are people, organizations, or spiritual beliefs that are already in your life that you can rely on. Sometimes it's as simple as faith. Sometimes it's reaching out to people you really don't want to reach out to.
For example, in the fire reading, you get the Three of Pentacles and the Queen of Wands in this position. You can hold on to your crew—the Three of Pentacles talks about teamwork. You can reach out to your support system, or if you haven't got a great one in regard to health, to a gym or a physician. Reach for someone or something that is very near to you that will be easy to hold. The Queen of Wands tells us that you have confidence and motivation. She's the kick-start that you've already got in your life. She shows up to remind you that you can do hard things and tells you to get out of your own way.
What Can Pull You Forward?
Think about the tools that can unstick you from a place of stasis. What are the resources that you have access to that will move you out of this spot? These don't have to cost a lot of money or take a lot of time to find. You can ask friends, colleagues, and family members for assistance. The thing that can unstick you might already in your possession. That thing might also be a change of mind or heart.
In the air reading, cards 4, 5, and 6 ask this question. What are the things that can move you away from that boring job? What do you need to finish that degree? If you pull the Fool, the King of Swords, and the Three of Cups, you would see that you need to take a chance (Fool), that you need to talk to someone who is "in charge" (King of Swords), such as a headhunter or someone one step up from your current boss, and that you can find answers in your group of friends (Three of Cups). You would ask among your friends if they know of any positions or if they know anyone who could have a conversation with you. I've had this happen in my own life. After posting on Facebook that I was being laid off, a friend immediately reached out and said, "I know a guy who's looking for an HR person." Within two weeks I had a new job.
What Gets in Your Way?
These cards represent the internal and external obstacles that you come across. They can be old challenges or new, internal or external. These can be people or situations, emotional or mental states. Sometimes your obstacles will be memories. Honestly, memories are sometimes the hardest things to work through. The past will swoop in out of nowhere like a specter and tell you don't deserve or need good things and blam—you're back to square one.
In the water reading, there are three cards that ask what gets in your way. Let's say you pulled the Devil, the Six of Cups reversed, and the Five of Wands. The Devil tells a story of temptation and vice that you haven't been able to shake. The Six of Cups reversed say that you have been feeling a little petty and pouting a bit. And there is a component in the Five of Wands of getting involved in drama that doesn't belong to you. All these things together can distract you from your goal and keep you looking behind or beside you instead of straight ahead.
The Use of Elemental Dignities
Since these are elemental readings, you can find another layer of interpretation to the spreads. This is optional, but can help you decide which part of the reading to focus on first. The best explanation of elemental dignities I've ever seen is by my beloved Barbara Moore in her 2012 Llewellyn Unbound post titled "Elemental Dignitaries," so I totally lifted it (with permission, of course):
Applying elemental dignities to a reading is simple. When interpreting a card, consider the cards that are next to it. Using the elemental relationships described below, weave the effect into the meaning of the card.
For example, let's say we are looking at the 3 of Cups (a spontaneous, unexpected joy or pleasure). Next to it is the 5 of Wands. Technically, it doesn't matter if it is 5 or any other number. What matters is the suit. In this case, it is Wands, or Fire. Wands/Fire is the opposite of Cups/Water, therefore weakening the card. The spontaneous joy or pleasure will not be quite as joyful or pleasurable as it could be; it is weakened or diminished by the presence of Wands/Fire.
One way to think about elemental dignities is to think of a card as a word and the card next to it as a modifier. The modifier can indicate an intensification of meaning like underlining or adding an exclamation point does in writing. A modifier can also diminish the impact by adding "kind of" or "meh" to the sentence.
* Cards of the same element strengthen each other greatly.
* Wands (fire) and Swords (air) are both considered active and support each other.
* Cups (water) and Pentacles (earth) are both considered passive and support each other.
* Wands (fire) and Cups (water) are opposites and weaken each other.
* Swords (air) and Pentacles (earth) are opposites and weaken each other.
* Wands (fire) and Pentacles (earth) have little effect on each other.
* Swords (air) and Cups (water) have little effect on each other.
When an element is strengthened, it does not always mean that it is a positive situation. It means that the experience is stronger or intensified, whether positive or negative.
Fire and Air are considered active. This means the energy is active, it moves, it creates, it acts; it also indicates swift movement. That is, if this energy is present, it is moving around making things happen and happen quickly.
Water and Earth are considered passive. That is, the energy is passive, it is still, it is reactive, it is shaped; it also suggests slower movement. Passive energy waits for something to happen, hence the idea of slowness, and then reacts.
When either Water and Fire or Air and Earth are present, the result is a weakening of both. They are opposites, fighting against each other, causing conflicting energy that will be evidenced in the situation.
The combinations of Fire and Earth or Air and Water are considered neutral. They have little or no effect on each other.
So, if you had a lot of wands (fire) in a water reading, you would want to take your time and be very deliberate with your choices, as these elements oppose each other. This could mean that you will eventually need to cut ties with a friend or partner, but not just yet. It might not be safe or in your best interest to do so right away. You may have more work to do to untangle the situation before you can walk cleanly away.
If you had lots of swords (air) in your fire reading, it's time to move. These two elements encourage each other to make things happen. This could indicate the need for a medical checkup or that the time is exactly right to start walking while listening to Zombies, Run! This is a walking app that puts you in the story of a zombie apocalypse and helps you stay motivated while you exercise.
Using the elemental dignities is not required for these readings, but it can add some depth to the work that you're doing. If you're not sure which step to take first, or if you want to do all five readings and then are completely overwhelmed and confused about which to address first, this can help.
Where do I start?
If you're like my client from the introduction and your whole life has blown wide open (or has gotten locked into a rut), it can be tricky to figure out which reading to start with. If you are looking for a whole-life reset, there are a few ways you can decide which reading to do first. I do encourage you to truly complete one element before you a start another. This isn't a race, and there isn't a prize for whoever tears through them first. The idea is to take a deliberate, rational view of your life, and then write a map for aligning it with your most authentic self. This ain't nothin', folks. You can't wing it and expect lasting results.
Sometimes it's blatantly obvious which element to start with. Your emotional self is just beaten down—water—or you're desperately seeking a spiritual home—spirit. You feel like your house is so full of stuff that you just can't breathe? Earth.
Sometimes, though, you just don't know. Maybe you're in the same house, job, and relationship that you've had for twenty years. You go to the same church or temple. Your friends are the same, you're about the same physically, and so on. You're looking for something to shake things up a little, right? So where to begin?
Process of Elimination
I've designed the following tarot spread to help you decide which element reading to do first. It's kind of like that basketball thing where the teams play against each other to see who wins (googles frantically): the BRACKET. Thinger. I don't know sports, you guys. Just go with me on this.
First, you need to pick your "teams." I've picked the cards that I think have the most to do with the individual elements. You can pick other cards if you'd like. Remember, this is your process, so you do what feels right.
Earth: Ace of Pentacles
Air: Ace of Swords
Fire: Ace of Wands
Water: Ace of Cups
Spirit: The Fool
We are going to use a five-team single-elimination bracket. (By the way, before I wrote this book, I had no idea that you could string all of those words together to make a cohesive sentence. Learn something new every day.)
This is a series of short readings to see which you should start with. Also, it's really fun to play with the cards like this.
Round 1: Earth vs. Air
Pull one card for each of your element indicator cards. This will tell you which element is the most urgent.
In this example, the Chariot carries with it more intent and motion than the Three of Cups, so you would choose air.
|
---|---
Ace of Pentacles (Earth) | Ace of Swords (Air)
|
Three of Cups | The Chariot
Round 2: Air vs. Spirit
In this matchup—the Devil isn't really interested in moving you away from your vices, and the Eight of Wands is about being decisive, so spirit wins.
|
---|---
Air | Spirit
|
The Devil | Eight of Wands
Round 3: Fire vs. Water
The Knight of Pents is kind of stuck, and the Six of Wands is about celebration. The winner is water!
|
---|---
Fire | Water
|
Knight of Pentacles
Reversed | Six of Wands
Round 4: Spirit vs. Water
The Two of Swords is a card of inaction, and the Queen of Wands is about to jump up, so our winner is water once again.
|
---|---
Spirit | Water
|
Two of Swords | Queen of Wands
And so, the best reading to do first is the water reading.
Determine Your Stressors
Another way to decide which to do first is to look at your stressors. Make a list of the things that are causing unhappiness and unrest in your life and see which problems align with each element.
What Is Stressing You Out? | Which Reading Should You Use?
---|---
Unhappy at work/school | Air
The stress of success | Earth if it's financial Water if it's emotional
Starting a new business | Earth if it's financial Air if it's planning
Injury or illness | Fire
Arguments in personal life | Air if you're bored Water if you're emotionally stressed
Can't sleep | Air for anxiety Fire if it's physiological
Family issues | Water
Legal issues | Air for planning Water if it's an emotional stressor
Should I move? | Earth
Trying to become healthier | Fire for body
Empty nest | Earth for downsizing Water for emotional upset
Unemployment | Pentacles for housing stability Air for job hunting
Relationship issues | Water, water, water
Financial worries | Pentacles
Can't find a partner | Water
Breaking them down like this could seem simplistic or trite, but we're looking to simplify your life here.
Look at your problems—at the root of them—to see what is causing you the most frustration. That's where you begin.
If you want to do more
than one reading
This is actually pretty cool. You can start with the most pressing thing and move on to the next element when you're finished. The important thing about this is to not rush anything. Sometimes you can finish all that you need to in your emotional life in three months; sometimes it will take a year. Remember that the only person you're competing against is yester-you.
I have a client who has been working on her air reading for over a year. She found that she doesn't feel challenged in her position and is going back to school. She already has a master's degree in one field, but that field didn't serve her, so she's moving to another. She's said that she's not going to do another element reading until she finds a new job in her new field. This is great for her. She is single-minded and really focused and wants to close the circle that she started before moving on to another project. That is the correct mindset to have with these readings. It truly is a project and the project is you.
[contents]
Chapter 2
How to Read Tarot Cards
Assuming you haven't done a tarot reading before or are unfamiliar with how to read new spreads, I'm going to give you a quick and dirty resource for how to read tarot. If you're just starting out with tarot, you might want to grab Kitchen Table Tarot, available at all fine booksellers and written by yours truly, to help you with the meanings of the card. I included a short description of each card at the end of the book as a reminder, but it would be helpful to have a reference book on hand.
Step 1
It's helpful to have a question in your mind but not necessary. I find that if I don't have a question, I get kind of a general overview of my life. Those are nice, but sometimes you get to the end of a general reading and say, "Well, yeah. No kidding. Is that it?" If you want something more specific, you need a question or at least an idea. For example, "Am I going to die alone and my cats will gnaw on my face, probably starting with my eyes?" That's fair. I've actually given this reading before... a few times. Let's go with it.
Step 2
Decide which spread you're going to use. The spread doesn't really matter, what matters is that you believe that the cards in those exact spaces mean an exact thing. We'll use a three-card spread for the cat-eating-your-face question. We'll decide that the first card will mean the past, the middle card will represent the present, and the third card will represent the possibly faceless future.
Step 3
After you have your question and your spread, shuffle the cards. There is nothing mystical about this. It gives you something to do with your hands and is relaxing, and somehow (not magical at all, I swear) the cards end up in the perfect position to answer your question with a high degree of accuracy that borders on spooky. If you want to cut the cards after you shuffle, by all means, do the thing.
Step 4
Lay the first three cards down in the three positions we decided on before. If they come up reversed, but you don't read inversions, flip them right-side up. It's really okay.
| |
---|---|---
1 | 2 | 3
Step 5
Next, flip to the appendix and find the handy-dandy list of card descriptions I wrote for you. If you find that your idea of the card differs from mine, let's go with yours, okay? It's your reading, not mine.
| |
---|---|---
1 | 2 | 3
Card 1 (Past): Three of Swords
Ouch. Pain. Like, searing, heart-falling-out pain. The thing about pain this severe is that it's purifying. That's literally the only good thing about this card. Sometimes you have to feel all the pain before it will go away. Remember that holding pain inside can cut you to pieces. Releasing it, even if only to yourself in a journal, is one of the most amazing things you can do. If you don't want those words lying around, write them down and burn them.
What this means is that in the past, you got your heart broken, and the pain of that broken heart has prompted the question at hand. You might think that you're over it, but it's still here, squatting in the corner of your brain like a toad. (Nice visual, eh?)
Card 2 (Present): The Four of Cups
Don't settle. Don't pout either, but make sure that this is what you really want. Fussy and ripe for hissy fits and eye rolling. Apathy is easy to ignore. Don't wanna. Don't want to change, don't want to take my medicine, whatever that medicine may be. You have to look deeper than the apathy. Why don't you want to change? You know these behaviors make you sad, and yet you won't put them down. Consider the why of holding on to the listlessness and figure out how that's paying off for you. There's a payoff for "meh." Whether it's the reward of not failing (you can't fail if you don't try) or the reward of being comfortable, decide if it's worth being stuck.
Since you got your heart broken, you've been pouting a bit. This is okay, really, but it doesn't really serve you after a few weeks. If you truly don't want to die alone but you're still tender from the earlier breakup, you have to do some work to get yourself ready to meet people again. If you need counseling, get it. Meds? Get those, too. Exercise? Awesome. Whatever medicine you need that will help you get your balance back is the right thing. Lying on your couch watching Agents of SHIELD while checking the same three apps over and over and over on your phone is not the right thing.
Card 3 (Future): Two of Cups
You have support in the relationship that's most prominent in your life. Allow this person to help you find the right way. Equal partnership. Respect. Esteem. No one is lacking here. No one is wondering. This love is enough. This relationship—with work, friend, sweetheart, whatever—is enough. And you can feel it.
Oh, hey! There is someone in this card with you! Looks like you're not dying alone after all, and your life won't end like it's a terrifying Polish fairy tale. (Have you ever read those? Jesus.) You have assurance now that you are going to find a person, but if you look to your present activities, you can see that it's not going to happen from the couch. Start the path toward "okay" and then make a list of what you absolutely have to have in a relationship.
For example, the person must be good to their family, know the name of the cleaning staff at their office, like Doctor Who, not make balls of their socks and then throw them in the wash so they never get really clean or really dry, not ask you for money, like petting your hair, and get along with your cat. Then date, and if the person on the other side of the table isn't this person? That's not your person. Keep looking. You'll find them.
Step 6
Now that the reading is over, do not give yourself another reading with the same question asked a different way. You're not fooling anyone, and after a bit, the cards will start messing with you. I tried this with a question once (it was probably this one), and after the third time, I got the Ten of Swords (destruction), the Tower (destruction), and the Lovers reversed (Thelma & Louise–level drama). Point taken, tarot. You didn't have to be a jerk about it.
Step 7
For the element readings, there will be two or three cards that answer the same question. For example, if you're doing the earth reading and the cards for "What gets in your way?" are the Queen of Swords reversed and the Moon, the two things that get in your way are your cranky mother and your anxiety. Once you identify the things, you can work on a plan to release some of that power those two things have on you.
Step 8
The most important part of the reading occurs afterward. You actually have to do the thing. You have to make a plan to get off the couch and seriously should feed your cat just in case. You have to take action to get the results that you want, because wishing and getting are two different things. Fix the idea in your head, then move on to real life.
[contents]
Chapter 3
Earth: A Reading
to Shelter You
You hear a lot about being grounded in the energetic sense but might not have any idea what that means. Essentially, being grounded in your body is being wholly present and in your skin. The element of earth is the essence of being grounded, and your home is a huge part of feeling safe and grounded. Finances are also reflected in this element, compounding that feeling of safety and security. One of the most notable things about the earth element is that if you are secure in the present and in your home, it doesn't matter, really, what happens outside of that space. If your family is going through a tough time but you've got a couch that's just perfect for snuggling, you have a home-field advantage. If you're going through emotional turmoil, pulling up your Doctor Who blanket and reaching over, then knowing exactly where the light switch is by touch can help you settle down. When you feel at peace in your home, things outside of you don't have as much power to shake you up.
Instead of living in the future with anxiety or staring into the past, you are truly grounded when you are right here, right now.
Earth
Home
Financially, when you have a certain amount of money in the bank, you can relax. If the water heater breaks or the car overheats, you won't have that moment of panic and can just take care of it like a grown-up. I've found that in my own life and in the lives of my clients, there is always a number that someone can think of that will make them feel at ease. If you have that number in savings, it releases the anxiety that debt can cause. It lends that same grounded energy to your life.
Safety, security, and peace of mind—these are the goals of the earth reading.
First things first: if you don't like or feel comfortable in your home, it's going to be difficult for you to deal with the rest of your life. A few years ago, I met one of my clients at her new house. We got there at the same time and I followed her in. She tossed her keys toward a table and fell short. Dammit. She couldn't find a pen at first. She couldn't find a notepad and got cranky about it. It had been four months since she gotten the house, and she still hadn't finished unpacking. All her stuff was present, but everything wasn't in its home. It made her homecoming uneven and annoying. Not impossible, but difficult. There is something about coming home and feeling at peace that helps you deal with the stress of the day.
As a society, many of us are far removed from living off the land. Now we live on it. Even though I'm a Pagan, honestly, I'm a terrible one. I have allergies, I hate bugs, and the heat makes my asthma crazy. I swear, I've looked at the full moon more out of my kitchen window than I've seen it outside. My idea of "roughing it" is no restaurant in the hotel. Nightmare. I do have friends who go camping on purpose, so I know that there are others like them. They feel both free in and connected to the outdoors, and they strive to bring that energy into their homes. I think that we're all looking for that feeling of having a nest. A sanctuary where everything makes sense.
I feel connected to the earth in my home. It took a while to get where I wanted to be, but after a few moves and life changes, I've reached a place in my life where I can navigate my kitchen without turning the lights on. I have artwork that I love on the wall and little collections and tchotchkes around the house that make me feel better every time I see them. I feel comfortable walking around in my underwear and eating leftovers at 3 a.m. My kids and my partner feel a sense of ease when they get here. Our house isn't perfect, by any means, but it feels perfect to us. We don't have a lot of money or really nice stuff, but it's ours. I always remember my dad saying when I was a kid, "It might not be fancy, but it's ours, and we're going to take care of it." Regardless if you have 600 square feet or 6,000, building a nest in which you feel safe is paramount to peace of mind.
When I do readings for people who are not at home in their house, I can pick up on that discomfort. They've inherited furniture that's stacked in the basement. They haven't gotten rid of their kids' outgrown clothing (or their own). All over their house, you'll find things that take up square footage without paying rent. This is a problem. I don't know a lot about feng shui or how energy works in homes, but I do know that if your home is filled with things that don't matter, there is less space for those things and people who do.
Another problem turns up if they've gotten a divorce and have a big, empty house that doesn't suit them anymore. Or they got a divorce and realized that their house never really matched them. When I was divorced, it was difficult for me to be alone in my own house. My kids were spending time at their dad's house, and he was gone, too. I remember walking around the empty house sobbing. Even though the house was technically mine, it didn't feel like it. It felt like the house that I used to own with my husband. It held arguments and sadness. It held the heartbreak of realizing that he wasn't my person anymore. That maybe I didn't have a person. It was quiet, lonely, and alien. I hated to be home without my kids.
It took me a while to pull my head up and look around. Do you know what I saw? White walls. No bookshelves upstairs. It was really stark, and I'm a person who dances in bright colors. I saved up and hired a fella to paint every room in my house. My house. The living room was bright orange and soft yellow. Green bathroom, peacock master bath, pink and purple for the Girl Child's room, and green and yellow for the Boy Child's room. I bought a beautiful comforter made of recycled saris for my bedroom and painted the walls light brown and avocado. The basement? Sky blue!
Just this simple thing—adding color—made this my house. Mine. I added bookshelves in nearly every room and filled them up, creating the library I'd always wanted. After I was finished, I felt good. I liked coming home and I loved feeling that I had a nest. A home instead of a house. This wasn't as expensive as new cabinets, flooring, or siding. It was affordable and made such a huge difference to me. It helped me realize that the more my home reflected who I was, the more comfortable I would feel in it. I never quite thought about that before.
I also feel better when my house is clean. Call it compulsive, but I can't think clearly in a dirty house. I just keep thinking about the mess and how it's just sitting there... mocking me. (I literally just got up to put dishes in the dishwasher because writing about it reminded me and bothered me so much.) I'm sure this is a leftover impulse from my childhood, but knowing that the sinks are empty and the floor is swept makes me sleep better.
Another thing that I do as medicine for my home is use candles and incense. And before you give me any trouble about cultural appropriation for using sage or smudge sticks, please recognize two things. One, I am a descendant of Elsie Lake, who was Native American, likely Seneca. Two, so many spiritual practices use smoke to clear energies that it's commonplace outside of the traditions of Native American peoples. Actually, I literally can't think of a spiritual practice that doesn't rely on incense, candles, censers, or even charcoal to realign energies and provide focus. Even in the Catholic church in which I grew up, the priest used smoke in ritual to gain attention at sacred times and candles to carry intent. As long as someone isn't wandering around with a sacred symbol on the back of their phone or calling what they do an "authentic smudging ceremony" when it's actually them walking around with a sage and cedar stick they got off of Amazon, asking all evil spirits to get the hell out of their apartment, I don't care.
Using smoke and candles to clean the energy in your home is a really old and really easy tradition. You can find ethically sourced sage bundles online or grow your own. You can pick up incense almost anywhere. Think of these tools as energy erasers. They can cancel out any lingering energy. I've outlined a way to clean your house energetically later in this chapter.
I often read for people whose lives have changed in many ways. They're empty-nesters, their relationship has changed, they have moved out of their folks' house or are home from college. Their lives have changed, but their homes haven't changed to reflect it. One of my best friends is sending his second kid off to college, and the first thing he and his wife did was sell their big house. He's one of the most centered fellas I know, and he doesn't like unnecessary things in his home. He and his wife decided that they had excess house and found a much smaller place out in the woods where the kids still had rooms, but it wouldn't feel as empty after they went back to school. Brilliant. Now they have a forever home, and they're not even close to retirement yet.
I've had clients who have moved into apartments after living in a house. Who have inherited all of their Great Aunt Ida's china and had no room for it. Who have never let go of a magazine or a plastic butter dish (because you never know) in their entire life. I understand this impulse. My family is blue collar all the way. Farmers and police officers and nurses. I understand not releasing things, because they can build a wall of security. The problem comes up when the stuff has no value—not to you, not to a collector. If it has no value, it becomes burdensome.
This reading can find that you need to move, declutter, or even just add some paint. If you are happy in your home, I'm happy to tell you that there are four other readings for you in this book! And way to go—that's not easy. If you can't feel at peace in your home, it's difficult to feel at peace in any other part of your life. We long for home. We sing songs about it. Our hearts live there and our families connect there. If our home is lacking, our life is, too.
As far as the financial problems that can come up and make everything unsettled, this reading will work as well. Just shift your question and thinking to financial security instead of your home. Also, I've put in some tips at the end of this chapter for financial stability that might help out.
Earth Reading
The tarot reading for this element is going to break down the ways in which you already feel secure in your life, the places that could use a little fine-tuning, and the tools that you have available to make changes. We'll also look at what's standing in your way.
| Cards 1, 2, and 3: What can
you hold on to?
Cards 4, 5, and 6: What can pull
you forward?
Card 7: What gets in your way?
---|---
Earth Reading Example #1: Ryan
Ryan is a buddy of mine who is happily married to his partner of a billion years and is daddy to two toddlers. He has a busy career and a full life and is looking for some assurance in his life. With two toddlers, who can blame him?
Card 1: The Ace of Cups
What can you hold on to?
Your cup runneth over with blessings. Be sure to be grateful. Happiness is afoot. Blessings abound. Accept them, acknowledge them, and say thank you. The universe conspires to shower you with blessings. Let it. If you're not in a place to say thank you, start with "at least." At least you have power. At least you woke up today. If things are solid, kick into the thank yous. Thank you for the little things and the big things, too.
One of the things that is in Ryan's favor is a full heart. He and his husband have a beautiful family and have made a lot of changes in their home to help it expand as the kiddos came along. He regularly has friends and family over to build that feeling of home, and everyone feels comfortable in his home. He and his partner have decorated in a way that lightens their spirits and that makes coming home a joy. You can feel the love in this home.
Card 2: The Hanged Man
What can you hold on to?
Sometimes, you need to unstick your tongue from the roof of your mouth, relax all your muscles, and just be. You will not dissolve if you aren't moving forward for a few minutes. Take a beat. Catch your breath. Take time to assess the entire situation. Even though we are seeking opportunities, opportunity is seeking us, as well. Sometimes our job is to remain still and calm so that when it finds us, we are ready.
Ryan used to feel as if his house had to be clean all the time. He felt that it was a reflection of who he was as a person if the kitchen needed sweeping. You know what, though? He found that he would rather play Legos with his daughter or color with his son than sweep the floor. He has shifted his personal values so that even though the house is clean, it's not spotless. Parenthood can be defined by long days and short years. He knows that if he doesn't live in the present and spend as much time with his littles as possible, he's going to miss something truly amazing. He can clean the house tomorrow.
Card 3: Strength
What can you hold on to?
Strength takes many forms. Figure out what yours is and feed it. You have reserves you might not be aware of. Use your strengths—compassion, kindness, humor, resting bitch face, active bitch face. Use your gifts to remain strong—whatever that means for you. Sometimes it means sitting still until you can stand up under your own power. Sometimes it means faking it till you're making it. Sometimes it means saying no and making that a complete sentence. No.
Ryan loves his house, and although he knows there are a few problems, there is no urgency to make a major change. He is content in his home. He feels supported by his husband and by knowing that they are financially secure. His strength comes from his family support, his financial security, and knowing that he's been through harder times. He's also become more reflective and less reactive as he's gotten older, making his decisions more logical and less emotional.
Card 4: Five of Pentacles
What can pull you forward?
Shelter and sanctuary exist, but you have to look for them. If you keep your head down, you'll miss all the good stuff. Sometimes you get pulled so far into your own darkness that you miss the lights shining near you. Sanctuary is available. You are not alone. Stop walking into the unknown and find your resources. It's going to be okay, but you have to let it be okay.
One of the tools that Ryan needs to move forward is a clear budget to make sure he's pointed in the right direction for home improvement. This can be solved by hanging out with a contractor for an afternoon and getting a handful of estimates. Rather than waiting for something to go wrong, he can upgrade his house as his budget allows. He should ask for help before he needs it.
Card 5: Page of Cups
What can pull you forward?
The pages are messengers, and the Page of Cups brings good news! Romance, sweetness, and emotional availability. It's called a crush for a reason. Whether it's a piece of music or a book or a person, whatever makes your heart leap out of your chest is found here. Now, is it going to be forever? Probably not, but it's going to be a hell of a ride.
Ryan and his family really love their home. You can feel it when you walk in the door. It makes you feel immediately comfortable. This is an energy that he can encourage by removing clutter and making sure that the items in his home are as beloved as the people who live there. The Page of Cups in this position encourages him to treat his home like the sanctuary it is.
Card 6: The Moon
What can pull you forward?
Sometimes you should be afraid. Be sure to get a mental health check and make sure your depression/anxiety isn't lying to you. Walk your path carefully because you are not in charge right now. Listen to your intuition. You can be a warrior when you're strong again. I promise.
Ryan has some anxieties that keep him from really relaxing. Whether it's because he's had financial issues in the past or because he's hyper-protective of his children, there is some anxiety there. The best way to beat back this anxiety is to solve your worries. If he is worried about money, he needs to ask a financial advisor how much he should have in savings and then work his tail off to get that number in the bank. Then relax. Don't just stress—worrying about something without doing anything about the source of your worry just eats your life.
Card 7: The Wheel of Fortune
What gets in your way?
Either you'll ride the top of the wheel or go to the shadow side for a while. Remember that you're in charge of how you behave regardless of where the wheel takes you. There is power in knowing where you stand. There is value in all these positions. The only constant is change, so if you're being smushed into puddin' by life right now, wait till you feel the weight to start to lift and scramble to get back on top.
Ryan has already expanded his home to hold his larger family. His biggest problem is realizing when his family changes in smaller ways and adjusting for that. When the kids have grown out of diapers, get rid of the changing table (even though that's where you stack the laundry). When the kids (or their dads) stop playing with certain toys, get rid of them! He needs to make space for the people they're becoming.
Summary
In summary, Ryan is doing okay with his home but can take little steps to be sure that it stays the warm, loving environment he and his husband have created. By staying ahead of the clutter, doing scheduled maintenance, and actively seeking calm and comfortable vibes in his house, he can count on his children growing up in a nurturing and safe environment.
Earth Reading Example #2: Debbie
My friend Debbie is a small-business owner with her husband. They have an awesome son and a really old, lovely home. She's a mama to not only her kiddo but also to most of her friends and her community. She's got two full-time jobs and reads tarot when she can.
Card 1: The Chariot
What can you hold on to?
Use your drive and devotion to get out of the rut. You've got stuff to do. Remember that you're in charge of you. Move forward. Don't stop. The two steeds in front of the chariot will do what you ask, but you have to ask. Otherwise, they'll just start wandering down the paths of habit and take you where you've already been. It's important to focus on the destination and then do everything you can to pull yourself in that direction.
Debbie's house is the spiritual home of her family and pretty much anyone who walks into it. It motivates folks and helps them feel stronger. You know that house in your childhood neighborhood that you just walked into without knocking, and you got cookies out of the jar and took the trash out without being asked? That's this house. It makes her people and community stronger.
Card 2: Eight of Cups Reversed
What can you hold on to?
You know that thing when you get really angry at work and start planning out exactly what you'll be wearing when you quit and exactly what you'll say and ALL their jaws will drop and you will walk out like Angela Bassett's character walking away from her husband's flaming car in Waiting to Exhale? Yeah, don't let that be a real thing. If you're unhappy, take a second to make sure that you actually want to leave the situation and aren't just really pissed off.
The energy in this house is like a fountain. People come here to be refreshed and bring their good vibes along with them. They come when they're stressed or scared, and they feel stronger for walking through the door. The energy is an extension of the happy family that lives here and the strong magic that Debbie uses to protect her home. When it falters, people drag their garbage through her front door.
Card 3: Seven of Cups Reversed
What can you hold on to?
Congratulations, you finally know what you want. That's a huge step. Now that you want, get to gettin'. It's really important, now that you know which direction to go, that you go now. This card shows up as the green light, and ignoring it means that you'll go back to that mushy, indistinct thing where you can't decide. It was a struggle to get here, and dawdling will only make it into a struggle again.
There are a few people—not many—who come to her home that shouldn't. They're not nice or trustworthy. Debbie needs to strengthen the protection around her house, and make sure that people with ill intent are stopped at the door. Luckily, the energy of the house is such that those characters feel uncomfortable from the moment they walk in the door. Good. If she lets her protections get too lax, this might change, and we don't want that. She knows how to do this.
Card 4: Eight of Wands
What can pull you forward?
Time to make a move. Don't let insecurity get you trapped. Decide. If it's the wrong decision, make another one. You decide your course of action. Which obstacles will you step over and which will stop you? How badly do you want to move forward? Break free of stasis. You have to understand that if you don't decide the course of your life, the universe will decide for you.
There are some things that need to be finished around the house, and rather than set her own timeline, Debbie has hired friends (who could use the money) to do the work. The problem with this is that they don't have a schedule. She's stuck with wanting the bathroom to be finished and is waiting for her friend to get their act together or fit her in. Debbie needs to either renegotiate with the friend or hire another professional to do the work. This waiting around stuff is for the birds.
Card 5: Seven of Wands
What can pull you forward?
Tenacious. Literally, this exchange: "Are you done being sassy?" "No." This is the honey badger card of the tarot. Not backing down. Not asking permission. Not going away or sitting down.
This is another reminder to put up protection magic in her house. It's become a seasonal practice for Debbie to protect her house, but life gets life-y and that isn't happening as often as she'd like. It's like brushing your teeth twice a day. You know that you need to do it, but sometimes you're too tired and your bed is too inviting. Debbie needs to be reminded of her self-imposed responsibilities so she can yell at me for reminding her and then go do what needs to be done.
Card 6: The Hierophant
What can pull you forward?
You may need to find a teacher or counselor to unlock the mysteries. Remember that you are not always the smartest person in the room. Every person has something to teach you. You have to study—people, nature, feelings, whatever—so you can open the world. The world doesn't open to the lazy, and with immersion in a subject, you can begin to own it.
Debbie needs to listen to someone else for a bit. Defer for a bit. She is in charge of so many things so much of the time that handing the reins to someone (even if she hates releasing control) will help. So, so much.
Card 6: Ten of Wands
What gets in your way?
You are doing it wrong. All of it. Even the things you think you're doing right are actually getting overshadowed by the things that you're doing wrong. Put down everything, get some advice from someone you love, then try to pick up just a few things and see how it works out. Baby steps. Decide what your priorities are, say no to everything else, and move forward. And please stop carrying other people's expectations around—they're surprisingly heavy.
There is just so much stuff. Debbie needs to be married to lists. A list for each room that needs renovation or clearing out. Reminders on her calendar for quarterly cleansing/blessing of her home. She needs to have orderly, timelined lists for things that will make her house one of rest and upkeep instead of thirty half-finished tasks that drive her mental.
Summary
Debbie's house is emotionally sound and healthy. She needs to be more practical with her to-do list, building a budget and a schedule instead of allowing things to happen on other people's schedules. She also needs to stick to a schedule with her protection magic, considering it standard sacred maintenance for her home.
Exercise
This is a very simple exercise that's going to help you isolate those parts of your house that feel off. I've found that it's easier to have a visual when you're talking about a broad concept like home. Although home can be an idea, for our purposes, we're going to bring you back to the second grade. Draw a picture of your house.
Draw each room and spend time in them. If you're home when you do the exercise, walk into each room and sit down.
* Do a quick survey of the room.
* How does it make you feel?
* What do you like about the room and its contents?
* What don't you like?
* Is it serving its purpose?
* Are you excited about what's in the room?
On your picture, mark with colors or symbols how each room makes you feel. When I did this for my house, I put smiley faces and frowns in each room. I haaaaated our basement, for example. It has, throughout the life of my home, been a room with an ill-fitting pool table, a sitting area, a spare bedroom, and a storage space for a bunch of crap that we didn't have homes for, and, coming soon, a home gym and library. When I sat in it, it was in the middle of the crap-o-lanche stage. I sat in the middle of the floor and drew a distinctly surly frowny face on my drawing. I liked nothing about this room. I hate clutter, the whole space made me anxious and unhappy, and I avoided it as often as I could.
Now, what to do about it?
First, you have to clean your house. And I mean clean it. You can have someone else clean it if you give them dollars. You can enlist your friends and family and make a night of it. You can pick one room a day and clean it to the best of your abilities. What's important is that the entire house is clean. My favorite website for housekeeping is called unfuckyourhabitat.com. It's magical.
Now pick a room. This is going to be easy and fun, if you do it right.
Step 1
Sort through everything in that room. Decide what's going to happen with it.
1. Keep (only if it makes you happy!)
2. Trash and recycle
3. Redistribute to family members
4. Donate
5. Sell
Step 2
Do the thing you've decided. Give it away. List it on eBay and move it to the garage until it sells. Put the donate and redistribute piles in your car. Put the sell stuff in a place next to your computer so you can post it on eBay. Make it easy to get rid of this stuff.
Let's talk about all the stuff in your house. Why is it there? Why are you holding on to it? How does it serve you? I mentioned earlier that people who've got clutter in their house are letting things take up valuable real estate in their houses without paying rent. Think about it in terms of the payoff for you. If you have a gigantic china cabinet full of china that you never use, and will never use, give it to your cousin who will enjoy it, and put a sewing machine that you love in the place of the thing that was just taking up space. I inherited my grandpa's piano, which was lovely, except that no one in my family played piano. It just sat there, all lonely and unused for a few years, until I found out my cousin's kid was taking piano lessons. We got the piano down to her house, and it's become the center of her home. It makes sense in her house. Didn't make sense in mine.
If you need help with this—letting go of things is difficult, I know—get a good friend to walk through the house with you. Have them ask you the hard questions. Prepare to answer them.
Step 3
Go into your clean, less-cluttered room and look around. Do you like the flooring? Do you like the color on the wall? Listen, we don't all have money lying around to use for redecorating. What we do have, though, is initiative. Maybe you can't repaint or pull up the carpet, but you can get a say-something lamp from Ikea for $10 and a nice new-ish couch on Craigslist for $25. Just make it your own. You don't have to go all HGTV on it.
Step 4
Do a smoke blessing. The best way that I've found to clean the house with smoke is this:
1. Start at the bottom level of the house.
2. Walk counterclockwise with your sage bundle or incense stick and "paint" the walls with the smoke. Reach as high as you can and as low as you're able, and paint each wall, window, and door in the room.
3. Pay special attention to the in-betweens: the door frames, windows, etc. This is where energy gets sticky.
4. As you clean the room, ask whatever negative energies that are around to please leave and ask good, healthy energies to stay. Or you can tell them. It's your house. If an energy or a spot feels especially gross, you can throw some salt in the corners of the room and the in-betweens as well.
5. Light a candle or a stick of incense in each room as its "finished." This helps hold that good energy in place.
6. I always try to end the ritual at the front door, which I then slam. I ask that nothing with ill intent be able to enter our home, and I throw some salt on the front step for good measure.
Work with your particular spiritual tradition if you have one. As I said, I'm a witch and I integrate my Paganism into this exercise to make it a ritual. If you are Jewish, you can end the exercise at your front door and pray the mezuzah. If you are Christian, you can get holy water from church and bless each room with a prayer. Make it your own.
7. The basic idea is that you're asking bad energy to leave your house and welcoming good energy into it. You don't want to just do the first part. That leaves kind of an energetically empty space, and nature abhors a vacuum and all that. After you're finished, carry the trash out.
The complete process is not going to happen overnight unless you are supremely motivated. Make a timeline. Make it accessible. Do the thing at your own pace and remember that you're doing this for yourself.
The last part of this is the homework. It's pretty simple. I don't want you to complete this gigantic self-exploratory project and then lose all the progress in six months. Put these prompts on your calendar or in your journal to be sure you don't go backward.
Homework
* How will you strengthen the protections on your home?
* What active steps will you take to make the suggestions that came up in your reading actually happen? Write them down.
* What will your home look like in three months?
In six months?
In one year?
In five years?
* What will you do to maintain this level of home care?
It's not enough to just do the exercise and the reading. You've actually got to schedule and do something about it. This is your home. This is where your heart lives. Investing time and energy to make sure that it bolsters instead of drains your energy is so important. It helps you stay grounded and safe and pulls your family and friends together. You deserve a quiet place to rest your head.
Resources
How to Clean Your House
This is something that you aren't really taught unless one of your parents took the time to do so. You can kind of figure it out, but there comes a time in your life when you truly have to learn how to clean your home. There are some great websites that will help you strategize house cleaning. It's a constant demand, and you have to come at in a way that works for you. Get organized in your organizing. I find it easiest to start with one room at a time.
1. Pick up everything off the floor and put it where it belongs. If you find dishes, put them in the sink. Shoes in the closet. Clothes in the laundry room.
2. Start laundry—that can go while you're doing everything else.
3. Start at the top of the room—dressers, counters, tabletops—and move down.
4. After each room, take the trash out. Don't leave it sitting around.
5. Clean the floors last, then light a candle in the room. You're awesome.
6. Pick another room and do it again. Take your time. Don't hurt yourself or frustrate yourself out of closing your circle. Don't move to another room until you've finished the first one.
If you're not sure how to clean something, look it up. If you can't clean by yourself, hire someone to do it. If you can't afford that, see if you can barter for the service. Trust me, it's worth it.
Blessing Herbs
You can find ethically sourced sage in so many locations. Some kinds of sage are endangered, so be certain to check your sources. You can also use sweetgrass, lavender, copal, or frankincense. Or incense. If you have allergies to these, you can find sprays that will do the same thing—they can clear the energy just as well as the herbs. These will reset the energy balance in your house, and they'll really help after a cleaning.
Plants that deflect negative energy include basil, frankincense, calendula, and chamomile. I love having fresh flowers in the house, too. If you have a black thumb, air plants are great and also clean the air while they're eating it—or however that works.
Teamwork!
If other people live in your home, cleaning is their job, too. The emotional labor involved in keeping your partner or kids or roommates in line with a clean house is completely exhausting. Make a chore list, assign tasks, and then stop. If it doesn't get done, the kids get consequences and your partner gets a conversation about shared spaces, resources, and workload. Your roommate gets their mess dumped in the middle of their bed. That's it. No negotiation. No cleaning up after grown people.
As a social worker, I worked with the poor and disenfranchised, and I saw living conditions that were unimaginable. A family heating their home by leaving the oven open because the gas got shut off. A hole in the floor of the living room feet away from where grandma was sleeping, because her house was owned by a slumlord who wouldn't fix it. The reason I'm telling you about these problems is because there are always resources out there. Every city has resources for people whose housing has become tenuous. Please reach out and be an advocate for yourself if this applies to you.
If you're like me and this doesn't apply to you, please stop for a moment and say thank you to the universe for letting you live without fear about where you will sleep tonight.
Finances
A lot of readings that I do show that one of the limiting factors people struggle with financially is the idea that they deserve to be poor. I can use myself as an example. I grew up poor, then lower working class, then lower middle class. I remember the distinct feelings that came from living in a place with mice and leaky windows and eating welfare bricks of cheese and food we grew in the garden 60 percent of the time (or not eating). Then we were able to eat out at a restaurant once a month—that was a big deal!
Each of these transitions brought with it a huge amount of feelings. Did I deserve to eat at Bonanza? I knew how much it cost, and I knew that my folks were working their tails off. Did I really need new jeans? There was so much guilt in the little things that my reaction as a teen and in my twenties was either to blow all my money in one weekend and then guilt trip myself for the next three weeks or scrimp and save on "fun stuff" and then overpay on bills out of guilt.
This left me in the same place every month. Broke. Exactly where I was the most comfortable and used to.
So, how did I get over the guilt and the feeling that I deserved to be broke?
A few years ago, my buddy Mark noticed that I always said, "That's why I can't have nice things." And before you think I'm going to go all The Secret on you, I need to tell you that I think that's a crock. I don't think that people get sick because they don't want to be well.
What I do believe is that thoughts have roots, and if you don't believe you deserve nice things, you're not going to behave as if you do, either. Your head will be down, you'll miss opportunities, and you'll accept failure as a matter of course.
The best thing you can do to combat this is to monitor your thoughts and behavior. I quit saying "I can't have nice things" and quit acting like it, too. I got a job that paid better, a nicer car, more money in the bank, and a boyfriend upgrade, too. The better I felt, the better I did, and so on and so on. Do I think that the universe heard me when I said that sentence? No. I do believe that I heard it—and I believed it, too.
So, from my experience and observations in my readings, the first step to financial success and stability is rerouting your thought process and your relationship with money. And when you do see the dollars coming in, say these three things that my friend Limitless Megan taught me:
thank you.
i am grateful.
more, please.
Budgets = Freedom
Do you know that thing where you stop looking at your bank balance midweek because you're afraid of what you'll see? Well, I do. I've spent a lot of years hiding from my financial decisions because I didn't want to see how deep of a hole I was digging. I would splurge on things I wanted without thinking about where the money was coming from. I used credit cards like they were real money (they're not real till you have to pay them back at 24 percent interest). I would have such anxiety about what I was doing to myself financially that I would just not look. If you can't see it, it can't hurt you, right?
Now the first thing I do when I'm up and dressed in the morning is check both of our bank accounts. One is personal for family stuff, and the other is for Melissa Cynova LLC. Every Friday, Joe and I figure out our bills for the week. We keep a running total of how much post-bill-paying money we've got by texting the balance back and forth after we spend money.
$754... I got gas.
$730... I got a book.
$720... and so on. At the end of the week, if we have "extra" after budgeting, we decide whether it goes into savings or toward a bill that we're paying off. Then we do the whole thing again the next week. That's a lot of work, so every morning I check all the deductions and see if anything looks weird. I make sure that the bills getting paid this week are all lined up. I'm not afraid anymore.
The difference between those two approaches is pretty intense, and the time span from getting from one place to another was about six years. It wasn't easy, or fast, and I screwed up a lot and slid backward a few times, but after doing readings for folks in the same position for thirty years, I've found there are a few things that always help make the transition.
1. Realign your relationship with money. You get to be in charge, not money. One trick that sounds really stupid is to write it a letter. Tell it that you appreciate it being there when you need it, and that you're ready to take a stronger role in your relationship.
2. You gotta have a budget. Have to. The first step is to list out all your income and your bills and your debt. Have to. You can't know where you're going unless you know where you are. (There's a lot of pentacle magic going on here.)
3. There is a great book called The Total Money Makeover by Dave Ramsey that has a great plan to getting your finances in shape. It's a bit preachy in places, but it's been a lifesaver to so many of my clients.
4. Walk the walk. It isn't enough to have a budget, you have to stick to it. Expect that you're going to screw up now and again, and then get right back to the budget.
5. Keep setting goals for yourself. Pay off the smallest bill first and cross it off of the list (it feels amazing). Save up for vacations or whatever pampers you. Make your money work for you!
Just remember that if you're not in charge, the money will be, and that bitch doesn't care about your future.
One of the most important things that comes up in money readings is that you have to stay present. So many people either live in anxiety (future) or regret (past) and forget to enjoy where they are here and now.
Let's assume that you've gotten into right relations mentally with your money. You've got a budget and you've made some plans for your future finances. Now what? Now you would really like to get that tattoo sleeve you've been mentally designing for like five years.
Since you're all fiscally responsible now, you can't just throw $600 at your tattoo artist, for example. (Plus, a really good tip for them. Don't be an asshole.) That would derail your budget and land you right back where you were before you worked so hard. Here's what you do:
1. You need to talk to your partner, if you have one, and if you share financial responsibilities. Joe and I touch base with each other about anything over $100. We have a conversation. Sometimes I get what I want and sometimes I don't, but the conversation is important.
2. Let's say your tattoo is going to be $700 (including the tip). You don't want to mess with your budget too much, but you can find some room if you try.
3. Look at two weeks of expenses. Can you cancel a subscription box for a few months? What about
Netflix? Stop eating sushi for lunch for a few weeks? Can you sell some stuff on eBay?
4. Once you figure out how much you can save or earn, mark it on your calendar. You are saving $20 a week by skipping Starbucks? Awesome! It will take you 35 weeks to save up for your tattoo. It will take less if you find more things to get rid of or more things to ignore for a while.
5. The decision that you truly have to make is this: How badly do you want the tattoo? If it's worth it, you'll do what you have to.
The best part of this? The absolute best part is walking out of the tattoo parlor with your new tattoo, no debt, no bills left unpaid, and no interest rates following you around. It's yours. No regrets.
It is really good to be fiscally responsible, but if you forget to have fun, you're more likely to go crazy on Amazon Prime at 2 a.m. after you've taken a Xanax, I hear (cough). Not that I've ever... um. Anyway.
This year I was able to pay for a tarot retreat and my flight without panicking or using a credit card. I can't tell you what a relief that was. Melissa Cynova LLC has only been an official business for five years and is now completely self-supporting. It's got its own savings account and is taking care of all the expenses that come from running a business.
If you think that the best thing about this is that you'll be rolling in money, it's not. I'm not a millionaire, but I'm comfortable. The best thing about this is that I'm not afraid anymore. If something happens, we're able to take care of it without panicking or wondering what we're going to sacrifice so we can handle the emergency.
[contents]
Chapter 4
Air: A Reading to
Help You Breathe
When you breathe in, you pull into yourself the energy around you. When you breathe out, that sense of release and relief fills you. When you do this mindfully, you can pull yourself into a meditative state. We breathe in and out about 23,000 times a day (hopefully). How often do you think about it? How often do you consider what a gift it is to have access to an unlabored breath? How often do you consider what you're pulling into you and what you're pushing out?
Air
Mind
Probably not often, unless the air around you is foul in some manner. A bad smell hitting your nose causes everything else to stop until you can find the source. I remember pulling my apartment to pieces looking for That Smell, until I found a small potato that had rolled to the back of the pantry and had become its own ecosystem. I stopped everything because the smell hitting my face shouted, "Wrong, something's wrong! Fix it!"
I think of the kinds of breaths we take. I have a certain sigh that my partner is most attuned to. I'll breathe out in a certain way, and he'll immediately say, "Honey, what's wrong?" My kids will puff their cheeks out and blow their breath out, and I'll know that they're frustrated. We light incense and candles so the smells around us are soothing or stimulating. We breathe out to make wishes on birthday candles, to calm ourselves, to flirt, to express anger or disappointment. We hold our breath when we're scared and release it in one huge exhalation when the fear passes. Breath is life. Cliché? Cliché. But it can be so much more than that.
Your breath can clear your mind and help you focus. Manipulating your breath can help you get rid of a headache, calm your aching heart, and prepare for a difficult experience. In this chapter, we're going to talk about air and its connection to your mind. How you think. How you process life and ideas. How you rise up to challenges. Your breath will carry you there.
Most of the readings that I do regarding air/mind land in four distinct areas:
1. I am bored.
2. I hate my job
3. I love my job, but I hate the energy in my workplace.
4. I have a passion, but I can't leave my job to pursue it.
Whether we're talking about your job or about your intellectual pursuits outside of work, the same can be said for both: you have to feel alive. You have to feel as if you matter and that your work is valued. This is nonnegotiable, and yet...
I have found that being bored is one of the most catastrophic forces in people's lives. It causes self-destructive behavior, limits prospective growth, and disallows personal growth. We are challenged all through school. We are given goals and tests and we achieve great things. Even if it's outside the school curriculum, we learn how to fail, how to interact with people, how to create our life. After graduation, when we leave that hothouse of growth and encouragement, usually the only person who cares about our growth is us, and we become too busy to tend to it. It becomes so easy to hit the couch after work and let the screen in front of us dictate how we stimulate our mind.
In regard to non-employment-related stimulation, unless you deliberately choose an activity that interests you—knitting, chess, reading, mediation, whatever—you get caught in this lockstep routine that makes your whole life become like furniture. The question "What are you excited about?" should only take a few seconds to answer. I'm excited about tarot cards. I'm also excited about books and writing and the tarot meet-ups that we do and playing board games with my family. I am excited about decorating my house and playing with my pets and coloring. These are things that I have to make time for, because if I don't, they go away. If I don't make time to read, I become unhappy and stale. I don't think I would be a good writer if my brain weren't constantly stirred up by others' words. Terry Pratchett, Patrick Rothfuss, Margaret Atwood, Sara Benincasa, Neil Gaiman, J. G. Ballard, Liz Gilbert—these authors make me think differently. Their work influences my work and makes my brain light up. If I don't do things outside of work that make me happy, my life will become beige, and I can't have that.
The reason I can rattle off all of the things that I'm excited about is because I consciously make space in my life to pursue things that make me light up. I could have a life in which I only work, only spend time with my family, occasionally see friends, and spend a lot of time on the couch. I could. But that's not all that I am. It's a limited, less colorful self, and I just can't handle that idea.
Regarding your nine-to-five, this has to be a place that enhances your life instead of diminishing the quality of it. You spend more time at your workplace than you do with your family. That time should matter. It should challenge you, and you should be able to walk out of the office with your head up and your heart lighter. I've had lots of jobs in my life. Some jobs suited me and some did not.
Once upon a time, a long, long time ago, I had this one job...
Okay, let's face it—I've had lots of jobs. Some were good and some were bad and some were so bad they caused stress-related alopecia (my hair was falling out in clumps). It's advisable to never burn a bridge, every step is an opportunity, and so on. For the purposes of this chapter, I'm creating an amalgam of jobs, pulling out the worst parts of them to emphasize my point and taking creative license as author.
Before we go any further, I want to emphasize that there is a difference between a job and a career. I never thought I was one of those people who were born with Purpose. My father is. My dad wanted to be a police officer since he was four years old. He was a junior police officer, then a police officer, and then a police sergeant, lieutenant, and captain. Then he retired and became a bailiff, because of course he did. He's my dad. He is a cop. He is one of the good ones. He told me once that there are two kinds of police officers: those who protected the little ones on the bus and those who picked on the little ones. He's one of the protectors. It's part of who he is in life and in his work, and you can't really separate them. Also, he's never questioned in his life what he was supposed to do.
Me? I questioned everything. I wanted to be Indiana Jones. I wanted to be an author. I wanted to work in museums. I wanted to be a docent. I didn't know exactly what that was, because I was eight, but I read about it in a book and it sounded cool. I wanted to help people, and I wanted to be a hermit. I got a psychology degree and became a social worker. After twenty or so years of that, I became an HR person. It's like social services with fewer heart-stopping moments of human tragedy. Anyway, I took jobs because they were open. I took jobs because they paid well. I took jobs because I was moving to a new city and it was the first job I found. I never had a job that I trusted would be my forever job. I've never—before I decided that writing and tarot reading were things I wanted to do really, really badly—had a career, in the huge sense. I was a social worker for a really long time, at a series of jobs that I enjoyed but never defined myself by. If one job ended, I would find another job that was similar but never the same. I went from social service worker to office manager of a social service agency to operations manager of a laboratory. Easy transitions to jobs that I liked but that didn't define me. I went from investing my entire self into my social work jobs to clocking out at 5 p.m. and meaning it.
Back to the fairytale-esque narrative.
Once upon a time, I had this job...
I found this amazing job, and I started work ready to change the world and kick some tail. I was full, my friends, of piss and vinegar. I was earnest and made notes and had ideas. A few weeks in, I realized that my boss didn't like me. It was kind of jarring, finding out that the person who signs your checks doesn't like you as a person. It happened in a meeting. We were brainstorming, and I brought up an idea. "That's stupid," she said, in front of two other people, and my jaw dropped. I let it go, but it really hurt my feelings. I remember walking out of the meeting feeling stupid and devalued and pissed off. Things like this kept happening. Not making eye contact with me during meetings. Never asking me my opinion. Acting like I was wasting her time every time I was in the room. It started to undermine my self-
esteem, and I started to make mistakes.
You know, most people don't start a job intending to do a bad job. If you treat people like they're stupid, though, you're going to cause them to make mistakes. The pressure of that failure will compound the anxiety, which will lead to more mistakes, and so on. Pretty soon, you have someone crying in the parking lot and hoping to get into a fender bender on the way to the office because they just can't face it today.
It wasn't long before I hated my job. The thing was, it wasn't the work. I liked the work. I was good at the work—usually. It was the environment in which I worked. People were treated like things there. I was treated like a thing. It was important that the trains arrived on time, but no one cared that the people driving the trains were broken. I started looking for a new job and felt like a failure. Even though there were people who had been there much longer than I had with the same problems. They were afraid to leave and heartbroken about staying. They judged themselves based on the behavior of a few bullies who'd cowed most people into submission. Why do we do this to ourselves? Objectively and with hindsight, I was a very bad fit for this company, which had serious morale and staffing issues. I didn't cause the issues, and I was blaming myself for not fitting into a place that made me miserable. Good lord, what a mind game.
I should have left right after my idea was called stupid. I should have left after I realized that most of my coworkers were miserable, too. I should have left after I noticed that people crying in the office was not noteworthy anymore. I can remember people talking on their office phones in a really low voice, urgently cheering each other on so they could get through the day. Instant messenger was used as a vehicle for encouragement.
Guys. You shouldn't have to be coached to finish your work day. That's not okay.
People would come to work sick because they felt guilty using their benefit days. They would shorten vacations, work through lunches, come in early and leave late. Yet they were only as valuable as their last mistake. One friend, upon asking to leave early due to a funeral was told, "Listen, people die (eyeroll and gigantic sigh). I get it, but you're needed here." Are you fucking kidding me? She was then given the silent treatment by this boss for about a week. Because of a funeral. Because someone she loved died.
In my life and in readings for clients, I see that people stay in jobs that don't serve them because they are afraid. Fear is really strong, and it's difficult to pinpoint it at times. Fear can feel like so many other emotions. Insecurity—maybe I am stupid and incompetent. Exhaustion—I'm too tired to do a good job. Anxiety—what if I'm really screwing things up? There goes my job, then my house, probably my relationship. I'm totally screwing up my entire life.
We look inside before we look outside. We blame ourselves instead of our work culture or coworkers. I worked one time doing data entry for an insurance company. I did. Me. Little Miss Explosive Vocabulary Who Bounces in Her Seat. Me. It was terrible. No one told twenty-year-old me that there were going to be places that I just didn't belong. No disrespect to data entry or to insurance (well, some disrespect to those guys), but I wasted so much time in a place that just didn't fit me. I am meant to write and to be around people who like to laugh and work as hard as they play. I am not shushable. I spent so many years in places that didn't feed me because I didn't know that there was an option.
We are not taught to leave if it's not working out. Why is that? Why did I have to learn to take care of myself in the workplace, and, further, why did it take me so long to figure it out?
I'm going to tell you this. There is an option. You are not like everyone else. You do not have to work in a place that doesn't feel like home. Even if it is just a job, you deserve basic human respect. You deserve support and encouragement and high fives. "Just a job" is something that can pay for your dreams, and money is important, but you don't have to feel like garbage just to earn a paycheck. There is no honor in taking verbal and emotional abuse from someone whose only accomplishment is keeping his head down long enough to earn his pension. I have a lovely friend who was working for about $9 an hour at a job for three years. She put in extra time, had only one day off, no benefits, and was treated like garbage. She found another job, doing something totally different for $12 an hour plus commission and bonuses. This new place invested in her well-being. The company took time to make sure she was okay, provided lunch and breakfast on busy days, and gave her insurance and a benefits package. All I can think is that for three years, which I'm sure were important in other ways, her employer got to have their way with her self-esteem and ability to care for herself. It makes me angry that her bosses were allowed to treat her like garbage and will continue to treat her replacement like garbage. Until all the workers rise up and leave, however, it will not change. It changed for her, though. She decided it was time.
air Reading
| Card 1: What can you hold on to?
Cards 2, 3, and 4: What gets in your way?
Cards 5, 6, and 7: What pulls you forward?
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Air Reading Example #1: Ellen
Ellen has worked for some real jerks. Recently, however, she has finally found a job that not only treats her well but also gives her time to pursue other interests. Whaaat! She's anxious, though, that the other shoe is going to drop, and she'll get hit by a surprise that takes the wind out of her sails.
Card 1: Ace of Swords
What can you hold on to?
Follow your great idea. You're being given inspiration. The aces are the heart of their suit. The Ace of Swords is the brain of the tarot. Allow yourself to be inspired and enlightened. Be open to influences and your muse and be sure to follow this fantastic head start to its (already lucky) conclusion.
Is she pointed in the right direction? Astoundingly, yes! She is. She is working at a job that she really enjoys with a company that values her. She is expected to only work during her allotted hours and spends the rest of the time working and playing with her friends on hobbies that she adores. She's found a weird kind of balance, intellectually at least, that allows her to work two jobs that she loves at the same time. This has never happened for her before.
Card 2: Queen of Swords
What gets in your way?
Be as direct as possible. No emotions come into this decision, just a sense of fairness and of what is the right thing to do. To quote my friend Sara Benincasa, "I gave up nice a long time ago. Nice did me no good as a woman. Niceness is a lie they teach you to keep you sweet and compliant while you're screaming inside. You know what I picked instead? Kindness. I chose to be kind. Kind means I respect your boundaries and you respect mine."
This represents making sure that she has time to do what's important and keeping her focus. She's so easily distracted that things that don't matter bulldoze the things that do. She doesn't need to be on Facebook. She doesn't need to get involved with her friends' family drama. She doesn't need to do these things, but they're easier than sitting down to work, honestly. They're handy distractions that get in the way. Since she got the Queen of Swords, this tells her that for now she is in charge, but she needs to remain aware of her distractions.
Card 3: Four of Pentacles
What gets in your way?
Hold on to those things that belong to you and be generous when you can. Be sure that you have everything you need before you share. Be sure that you come first. This card also indicates that you should keep an eye on the things that belong to you. Watch your money. Watch your possessions. Be safe.
Ellen is afraid of being poor. She has been very, very poor before, and the possibility of it happening again keeps her up at night. So she has to remember that she has a plan. She has a savings account that she protects fiercely. She is not wealthy, but she is secure, and she needs to remind herself of this on the daily so that the money anxiety doesn't eat her. When I have clients who call about money, I tell them to read The Total Money Makeover by Dave Ramsey. Follow his advice. Find a number in your mind to have in the bank that will help you sleep at night and then make that goal happen. Stop letting money drive you. Start driving your finances. It's only scary at first, and then you get to be in charge.
Card 4: Ace of Cups
What gets in your way?
Your cup runneth over with blessings. Be sure to be grateful. Happiness is afoot. Blessings abound. Accept them, acknowledge them, and say thank you. The universe conspires to shower you with blessings. Let it. If you're not in a place to say thank you, start with "at least." At least you have power. At least you woke up today. If things are solid, kick into the thank yous. Thank you for the little things and the big things, too.
Ellen is way, way too nice. She wants to please even those people who treat her like a thing. In Terry Pratchett's Carpe Jugulum, Granny Weatherwax says, "There's no grays, only white that's got grubby. I'm surprised you don't know that. And sin, young man, is when you treat people as things. Including yourself. That's what sin is." By allowing folks to treat her poorly, Ellen is harming herself. Remember, be kind, not nice.
Card 5: Six of Swords
What can pull you forward?
Focus on where you're going and get out of where you are now. Don't look back. You don't live there anymore. Part of going from rough waters into smooth is that you don't always trust that transition. You wait for the other shoe to drop, and in doing so, sometimes you manifest that shoe right on your head. Trust that you're going to be okay. Trust that when okay comes, you might not know what it looks like right away. Trust that if you feel threatened, you can grab one of those swords to protect yourself. We never go backward. I know it's hard, and I know you're tired. Just keep going.
Ellen needs to remember to never move backward. If this day job goes sour, she will find another one. She will not stay in a place that is unhealthy for her. She will never, ever move backward in respect to how she is treated by coworkers. She spends forty hours a week in that place, and it has a profound effect on her well-being and on her family. The most important thing to her should be her, and she will remember that and never again trade financial security for dignity. Ever.
Card 6: The Emperor
What can pull you forward?
Time to dot your i's and cross your t's. Be precise. Consider briefly the emotional impact of the decision, and then decide what is practical and smart. Sometimes you have to do the right and difficult thing. Even though it might break you a little bit. The Emperor wants you to be okay but realizes that sometimes the path to okay is painful (but necessary). He is willing to go through the pain to make things happen for the greater good.
Ellen must remember that she is in charge of herself. She can't blame unhappiness on anyone except for her. This is in regard to things she can control, mind you. Part of achieving the focus she needs intellectually is to pare away those things that are outside her control. She has a lot of privilege in her life. She is white. She is middle class. She has a house and food on the table and a car. She has a supportive, loving partner and enough money to get sushi pretty much whenever she wants. Part of being a privileged person is understanding that her voice is strong, that her influence is felt, and that if she doesn't like something, she has it within her to change it. I reminded her of the advice I got from my friend, Andrew McGregor, who said, "Don't ask questions if you already know the answer." Ellen spends too much time flailing about, looking for help when, honestly, if she were to take the time to be still and quiet, she would already know exactly what to do. She needs to trust herself and her abilities and just do the thing. The most important part of being a privileged person is using that privilege to do whatever she can to help others. She can be a strong voice for our brothers and sisters in the margins. She can look for opportunities to help her friends and family and those in her community rise up. We all do better when we all do better. Ellen has a part to play in that.
Card 7: The Fool Reversed
What can pull you forward?
Just because you can just doesn't mean you should. Just because the ledge is there doesn't mean you shouldn't look over the edge and double check your parachute. The Fool reversed throws caution to the wind. And common sense, a sense of pride, decency, and intellect. Just blundering stupidly into the next mistake.
The Fool is telling her, very clearly, "Stop jumping, dummy. Do the work that's in front of you. Do the work and then move forward. If you keep bouncing around, you're going to knock over things and create more obstacles. Sit down. Breathe. Do your work with a clear mind and an open heart."
Summary
Ellen needs to stop "stopping" just when things go well. She works her butt off, and just when she's about to crest into happiness, she loses her momentum and eases back into "okay." I feel it's important to share with you that when your tarot cards talk smack to you, like they did with Ellen, it's absolutely permissible to tell them to shut up. (You should still listen, but you'll feel better.)
Air Reading Example #2: Pixie
Thank you to my friend Pixie for allowing me to read for her! That's not her name; I'm just fairly certain that she's part pixie. Pixie is working at a job that's just a job and has her eyes focused on a career that is beyond her reach for now.
Card 1: Six of Pentacles
What can you hold on to?
There is an exchange here. Giving and receiving. Be sure that the balance remains. Can you give gracefully? Can you receive with equal grace? Can you allow balance and generosity into your life? Give a damn. We all do better when we all do better. You have to invest in yourself first, then your family, and then your community. It has to flow outward. If it sticks with just you, your riches spoil and stagnate. If they flow out to only your family, they puddle and pool. If they reach your community, however, they flow and flow and flow. A rising tide raises all boats.
Pixie just got a new job, and although it's adjacent to her dream job, it's not her dream job. What it does, though, is pay the bills. She needs to focus on the practical aspects of the job—learning a different part of her field, making connections, and so on—and stop worrying about taking time away from her dream. This is a part of her dream, albeit a more boring part. Also, this part helps her keep the lights on, which ain't nothing.
Card 2: The Tower Reversed
What gets in your way?
This can be a very tense situation that feels like it's going to explode at any second. It can sustain itself for years, though. I once gave a reading to a wealthy, beautiful couple with a beautiful house and beautiful children. They were also both addicts, both cheating, and both ignoring their children (and in debt up to their eyeballs). That level of delusion can only last so long. Holding your hands against the tower to keep it from falling will only leave you with bruised and bloodied hands.
Feeling stuck makes Pixie pout, and Pixie pouting makes for poor performance. She has a job. This is a good thing. If she doesn't embrace the gratitude necessary to keep herself afloat in a job, she's gonna not have a job anymore. It's okay to be disappointed that you're still in a job and not a career. However, your job is paying your bills. Someone is giving you money to do something. This is a blessing. Find a reason to be happy to be there, even if that reason is labeling every hour of work. "This hour pays for my Netflix. This hour pays for my gym membership." Keep it real.
Card 3: Page of Pentacles Reversed
What gets in your way?
You not only dropped the ball, but you dropped it, kicked it, and then looked away before you saw where it landed. No good, man. You've got to get back on track. If you're disillusioned or disheartened, it is absolutely okay to take a small break, but then you need to realign yourself with your goals and try again. No one is going to chase your dreams for you. Move it.
Pixie was unemployed for a few months, and it scared the bejeezus out of her. She needs to know that this is the lowest amount of money that she's ever going to make. It's all uphill from here. No more unemployment, no more deciding which bill to pay. Things might be tight for a while, but she's on her way up. Anxiety about money is distracting her from the here and now.
Card 4: The Moon
What gets in your way?
Sometimes you should be afraid. Be sure to get a mental health check and make sure your depression/anxiety isn't lying to you. Walk your path carefully because you are not in charge right now. Listen to your intuition. You can be a warrior when you're strong again. I promise.
Pixie tends to freeze when she's afraid. This piggybacks on some of the other cards—being stuck, being anxious, and so on. The most important part of this is to know that this is what happens. If she can see this reaction coming, she'll be better able to identify the behavior and work around it.
Card 5: Queen of Cups
What pulls you forward?
The queen of emotion. She's in control of her feelings and surrounded by love. You can love someone but not so much that you lose your mind about it. Can you be near drama without being in drama? Can you shed those emotions that no longer serve you? She loves everyone, but I'm 100 percent sure that she loves herself first. She holds sacred space for herself and takes care of herself without apologies.
Pixie works in the entertainment industry, and it's notoriously difficult to get a solid career path in that industry. You have to start very early and work your way up, or know someone, or chance into a spot that opened that didn't have someone groomed for it. She needs a mentor who has the career that she wants so she can grill them. Does she need classes? Contacts? More experience in X, Y, or Z? She needs to have lunch with someone and pick their brain to see how to get on that track.
Card 6: Six of Swords
What pulls you forward?
Focus on where you're going and get out of where you are now. Don't look back. You don't live there anymore. Part of going from rough waters into smooth is that you don't always trust that transition. You wait for the other shoe to drop, and in doing so, sometimes you manifest that shoe right on your head. Trust that you're going to be okay. Trust that when okay comes, you might not know what it looks like right away. Trust that if you feel threatened, you can grab one of those swords to protect yourself. We never go backward. I know it's hard, and I know you're tired. Just keep going.
Pixie is feeling that going to a different part of her field is holding her back from her passion. What it is doing instead is teaching her more practical information about the thing she loves. She's looking at the industry from an entirely different perspective, which can only help her. She's paving the way to becoming more informed and a better asset to any future employers. Go, Pixie!
Card 7: The Sun
What pulls you forward?
It's all okay. The sun has a way of shining light into all the dark corners of our lives. All the shadows and mistakes and flaws are illuminated. Those broken bits of you catch the light and make you shine. Be confident that the world is waiting to shower you with blessings. Know that things will be okay even if the situation is ambiguous. Feel the support of your people and of spirit, and share this light with everyone who comes into its warmth. Remember that fire burns away the darkness and that you are a strong and gifted fire dancer.
This woman is solid sunshine. In the time I've known her, she's had her heart broken, lost a job, and had family issues. Through it all, she's still dancing. She is tough and resilient and a true optimist. The best thing that Pixie has in her corner is Pixie.
Summary
Basically, Pixie needs to suck it up and work at this bill-paying job until her career can start. She has time, and she would be much happier if she were to focus on the present instead of being so anxious about the future. (Also, she totally got her dream job three months after this reading. Way to go, Pixie! She stuck it out with the Muggle job and worked hard and was able to professionally transition to her new gig.)
Exercise
For the exercise, we want to calm your breathing and your brain so that you can rationally think through your problem. You're going to meditate in whichever way is most comfortable to you, traditional or active meditation. I personally suck at traditional meditation, and I choose active meditation. I deliberately think about problems while cleaning my house or working in my yard. I put myself wholly into a mindless task so that I can be busy while concentrating on big issues. Whether sitting in traditional meditation or working in active, light some incense or a candle and find your quiet center. The purpose of this exercise is to figure out what you should be shifting in your life. Do you need a new job? Do you need to go back to school to pursue something that lights you up? Or to increase your salary? Do you need more hobbies so that your life isn't just work, home, work, home? Do you need to retire? (Like, really retire—not like my dad, who will never retire ever.)
Think about those areas of discontent in your life. Picture all those problems dissolving and blowing away in the wind. All that's left is you and your big sassy brain. I want you to imagine your brain—the wrinkles, the curves, the sparks of electricity. Picture your brain connecting to your spine and every part of your body. Understand that you have unlimited potential, and that you can overcome any obstacle that might be in your way. Understand that this exercise will help you push away all those distractions so that you can see the clearest way to move forward successfully.
Let your mind drift to those things that light you up. What have you always wanted to be when you grow up? Do you want a job or a career? Do you want to change fields or just positions? Think about all the possibilities that are out there for you.
After spending time with only these thoughts, I want you to sit down to do your homework.
Homework
First, write down every thought or idea or even a sliver of an idea that came to you during your mediation. Did you think "I want to pursue an MFA in poetry" out of the blue? Write it down. Did you fixate on current work stressors? Make a list of them. Did you think about your day job and how you really don't mind Monday mornings? That's important, too.
Answer these questions:
* Is your job truly supporting you?
If no, what are two action steps that you can take right now to be sure that your job is working for you, not against you?
* How are you stimulating yourself outside of work?
* Do you feel challenged and fulfilled?
If no, how can you fix this?
The next step involves other people. Grab a friend, a colleague, a mentor, or even an adviser from a local college. Ask them for advice. Ask them about their observations of you and your life. Have dinner with them, just talk about where you are in your career and intellectual life, and ask for their experiences and thoughts. I think that most of us forget that we don't live in a vacuum and that other people can tell if we're happy or challenged.
Everyone's "doing" homework is going to be different here. Some examples that I've come across in readings include the advice to begin looking for a new job that day. Get another opinion on your résumé and send it out immediately. Some folks need to finish or pursue degrees that fell to the wayside. Some people realized that their jobs are actually great fits but that they needed to join a meet-up or a book club to stimulate them outside of work. Just do the work. Please don't spend emotional and physical energy on trying to fix this problem and then say, "Yeah, that would be great." It's your time. This is your life. Don't waste it.
Resources
Meditation
Choose which form of meditation works best for you. You don't have to trance out immediately—this is a skill that takes time to learn, just like everything else. Start with ten minutes in the morning and ten at night. Find an app to help you out if you need to. If it's active meditation, save the dishes for when you're alone and can focus on one mindless task. Schedule this time into your life so you can help yourself stick to the program.
Journaling
Keep track of your moods and your happiness with work and your level of boredom outside of work. The best person to compare yourself against is yourself, and if you can look back at your engagement in your job over the last year, it will help to have an objective measurement of your happiness. I have one client who gives themselves engagement points every day. Did they start a new project? Did they empty their inbox? Did they surf the internet most of the day after they finished their work in two hours? These types of things are indicative of engagement.
Incense
Find a scent that clears your mind. One of my favorites is Amethyst by Shoyeido. When I'm writing or doing readings, the first thing that I do is light incense. It makes me happy, calms me down, and sets the mood for my work. I work from home at my Muggle job and have taken to lighting incense at 8 a.m. every morning, too. It gives me a mental cue that I'm supposed to be focused. Plus, it smells pretty.
The readings I give to people who are bored are some of the most heartbreaking. The bored tend to cheat on their partners, slack off at their job, take desperate risks. The bored will poke at a glass until it falls off the table and shatters, just so they have something to do. When they do this with people instead of things, their lives can spin out of control. I encourage you to do an honest assessment of your life. If you are bored, are you taking it out on someone else? Are you treating people like things? Are you being a jerk at work because you can't stand what you do? If so, thank you for your honesty. Now it's time to fix it.
[contents]
Chapter 5
Fire: A Reading to
Light the Way
Of the elements in this book, fire is the only one that is outside of us, except for cases of spontaneous combustion. We hold earth in our bones, water in our blood, air in our lungs, and spirit in our soul. That fire, though, is tricky. It flickers and dances around us. Is it the spark of life? Is it an external flame that draws us close? I've connected it with body in this book because to me, fire is the thing that makes us dance. Think of all the expressions: We light up. She's got fire in her eyes. Someone is all fired up and ready to go. And oooooh, they are hot.
Fire
Body
Remember that fire can create as well as destroy. It is transformative, turning wood into ash and darkness into light. It is the only element that we create ourselves. It warms our food and our bodies and can cause excitement or terror. From the smallest spark of hope to a raging inferno, fire is best when it's in our control. When we tame it and give it specific boundaries, fire can enhance and brighten our lives. When we lose control is when it either goes out or consumes us.
This lines up with our bodies as well. When we are focused and care for them, we can be strong and lithe. We can sleep soundly and feel good most of the time. We can meet the challenges that face us without worrying about losing our breath or our strength. It's when that control slips that we run into problems.
Let's call the day-to-day a brightly burning fireplace fire. It has a place to grow, has food to feed it, and gives us heat and light without making us afraid. Fire represents our energy source and the expression of that energy. Physically, either you're able to walk for as long as you'd like, bend over comfortably, and rely on your body to get you where you need to go, or you might have a disability and employ the use of a wheelchair or other supportive tech to help you get around—if you can physically do what you want to do during the day, this applies to you.
When we start to run down, the flames can die down to embers. Once you're an ember, it's hard to build back up to a flame. When we think about the body, this would be represented in someone who has been unwell or has been unwilling or unable to move their body in the way they want. The day starts with an assessment of wellness and ability. It usually starts with a to-do list: I need to go shopping, mail some things, clean the house, and do laundry. As you assess your physical, just-woke-up self, things start dropping off that list. I've heard this called the "spoon theory" when applied to emotional well-being. I like to think of my physical reserve of energy as a gas tank.
Keep a journal of your percentages of how you're feeling each day. Wake up and assess: if you're at 20 percent, you don't do the things that will take up all your strength. If the next day you're at 60 percent, take a stab at some of the more difficult tasks. After a while, you'll have a chart of the things you can do during your high-percentage days. This way, you won't wear yourself out on 20-percent days and you won't beat yourself up for those things you just don't have energy for.
Rekindling your physical energy can be difficult, but if you properly assess your abilities, you'll find that your expectations will easily match your abilities after a while. The key is to accept where you are—right now—and move on from there, without blaming yourself or over- or underestimating yourself.
In a spiritual journey, it's easy sometimes to forget that you're dragging a meat suit along with you. You want to do all the things your spirit calls for you to do, but your back hurts and it's easier to hang out on the couch and binge Broad City. Fire is the element that pulls you to action. I've found in the readings that I give that a lot of people have an imbalance in mind, body, and spirit. They're either super stimulated intellectually and have a robust spiritual life but no way to exercise their bodies, or they're firing on all cylinders intellectually and physically but have no spiritual activity.
I didn't grow up exercising. I mean, we rode bikes and ran around the neighborhood, but I had asthma and didn't like high school sports. Or other sports. My idea of exercising was carrying the hardcover copy of Stephen King's It to and from school so I could read it during lunch. It has been a challenge to bring that energy into my life. Having that balance, though, is so helpful.
I'm talking a lot about myself and my own personal experiences in this book, which makes it a little vulnerable. It also makes it really True—like, capital T True—and really honest. This chapter is about having broken parts and then figuring out how to help them (and you) heal. If you are content with the skin you're in, excellent. Skip this section and move on to water.
I became a mom in 2004. I've always been a healthy gal with good breeding hips, and I didn't have many physical ailments. I'm clumsy as hell, so I'd broken bones and every single toe at least twice, but overall, no worries. When I became pregnant, I was so tired. All the research I did said that tired was normal, so I thought nothing of it when I would come home from work, sleep until 8 p.m., wake up, eat, and sleep until 7 a.m. Almost every day in the first three months. Tired was normal, right?
What I didn't know was that I had diabetes insipidus. My body wasn't getting rid of water like it was supposed to. I kept getting bigger, I was drinking water like crazy, and I was about to suffer multisystem organ failure while delivering my daughter.
Oh, and I was dying. They couldn't treat me aggressively or they'd risk hurting my baby, so they were trying to figure out how to get me to carry her to term. She was almost two months early. That's when her heart stopped.
I've never seen people move that quickly before. Within minutes, I was in an operating room being prepped for a C-section. By this time, my lucid moments were few and far between. One minute I was begging for ice chips or water and teasing my doctor about his hair, and then I started blacking out. I'd overheard two people talking about "extracting the baby," which they don't do when the mom is alive, so I knew I was probably going to die. I didn't know if I was going to get to see my daughter first, though, and that really bothered me. I held on as best as I could until I saw her sweet, scrunched up, pissed off little face, and then I went away.
It's a very strange thing to go away like that. I felt like I would either wake up later or wake up somewhere else, and I was really fine with either. I got to see my Monkey. I knew her dad and our families would take care of her, and it was okay.
I did wake up, thank goodness, to the news that in addition to the baby, they had drained seven gallons of fluid from my body, using the medicine they couldn't while she was still in there. The baby was perfectly healthy, if tiny, and I would eventually be healthy, too.
Becoming healthy was a struggle because I was so tired. I wasn't allowed to carry the baby around, so I spent a lot of time on the floor with her. It wasn't bad, but it was different. I couldn't clean my house. I wasn't able to spend time with my friends. I just couldn't. Someone else had to clean my house. God, the humility that hit me was palpable. My then mother-in-law cleaned for me, and I pretended to nap with the baby while crying my eyes out, embarrassed that I couldn't do it myself.
Now, thirteen years later, I'm revisiting this hot-faced shame with a critical eye. I nearly died. A few times, actually, and had just had a painful C-section. OF COURSE I COULDN'T CLEAN THE HOUSE. No kidding. But do you think I believed that then? Nope. I felt like I was lazy and ungrateful and demanding. I was raised that you clean your own messes and you just handle things. Now I can look back and I wonder how I didn't lose my mind in addition to my physical strength.
I got a little better, day by day, and was able to go back to work after two months or so, but it was hard. I was still so tired all the time, and when I came back from work, I wanted to spend time with my new little family. Then there was laundry. There was always laundry. It was so overwhelming.
Then, because who doesn't love a plot twist, I got pregnant when my daughter was ten months old. I had just gotten back to feeling like myself, and my son—this determined little soul—clearly heard me and his dad talking about vasectomies and snuck in right under the wire, in spite of two forms of birth control being in use. My son is not a miracle baby. He's a goddamned ninja.
We decided to keep the baby, and my doctor said, "The odds of contracting diabetes insipidus again are slim to none." That was accurate. I did not get diabetes insipidus again.
I got kidney failure, kidney stones, liver failure, pregnancy-
related tachycardia, and sepsis. I also had a nursing toddler at home with me part time and was trying to clean and upkeep my house, too. While I was dying of sepsis and working forty hours a week. They couldn't lock the kidney infection and sepsis down entirely because the antibiotics needed would have hurt the baby. Sepsis has a 60 percent fatality rate. I had it at least six times. I literally lost count. I'm not sure of the math on this, but it seems like I had at least a 360 percent chance of dying, and I was worried about the laundry. I also couldn't nurse anymore because of the medication I had to take to get rid of the sepsis, and, of course, felt like a failure here as well.
I eventually recovered from this early birth, too. My son is awesome. He is a happy teenager now and worth every percentage of almost dying and then some. I was left, though, with not just another amazing kid, but with kidney stones every three months or so. For ten years. This caused my back muscles to scrunch up so much that I had chronic back pain and my hips to become misaligned. The pulmonary edema and congestive heart failure from my first pregnancy contributed to my developing a pretty nasty case of asthma. Other than that, though, I'm golden.
All this physical trauma and the thing that I keep coming back to is that I was embarrassed I couldn't clean my house. I was embarrassed that laundry wasn't finished or that dishes weren't done. I felt like my disability (I didn't even call it that) was keeping me from doing my Job. My Job, you guys, was to not die. To raise my kids. To heal. I couldn't see that, though, because of whatever nonsense I'd absorbed over the years about what a wife and mother should be doing. My day-to-day responsibilities completely overshadowed how sick I really was.
What I needed to do, and what I need you to do, is meet yourself where you are. Right now. I kept holding myself to the expectations set by the physical performance of yester-me. I could rollerblade in the past (shut up, it was the '90s). I could go up the stairs without losing my breath. I could even run. I could stand at the sink and wash dishes without my incision pulling and my back screaming in agony. I could—in the past—do those things. I couldn't right after the babies. I still can't stand for very long because of my back and, you guessed it, I'm still embarrassed when I have to ask for a chair.
I am, however, less embarrassed than I used to be. I've gotten to a pretty good place with who I am now and how my body moves, but it was not always that way. The weight that I couldn't get rid of was no small part of this. Two babies in seventeen months and about three years of really, really slow physical movements left me overweight. The asthma makes it hard to exercise. Even if I eat a piece of lettuce every day, I'm going to be heavy until my body heals more and I'm able to move it more often. This me, right here? This is perfect. I can now see myself as perfect, just as I am. I will continue to try to become healthier, but I'm sure as hell not going to waste my life holding out for "when I'm thinner, I will..." That day might never come. The day that I'm strong enough, skinny enough, whatever enough might not come at all, and I will have wasted all that potential happiness by beating myself up because it's not here yet.
I tell you this. I might not be society's physical ideal of woman, but I am here. And I am strong. And I am enough.
We need to get you to the place where you feel like enough, too.
There are a lot of body-related issues that I've dealt with in the fire reading. I've talked with people about disability, unhappiness with their weight at either end of the spectrum, body strength, gender transitions, or just a general "meh" unhappiness about the size of their belly. This reading is not about what other people think of your body. This is about you and what you think about your body. If we can get to the place where you feel solidly in your skin—where you feel present and engaged and in control—whatever your challenge is, you will have a sturdy place from which to face it.
I want you to get to a place where you can look in the mirror and say, "Okay. This is me. I am here. The alternative to being here sucks, so let's do this."
Fire Reading
| Card 1: What gets in your way?
Cards 2 and 3: What can you hold on to?
Cards 4, 5, and 6: What pulls you forward?
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Fire Reading Example #1: Amy
Amy is in her forties and has a degenerative chronic illness. She is still mobile and is handling changes in her abilities like a champ. She's figuring out how to exist in her current body. She has a full-time job as well as a family, so the expectations on her are high. She's anxious about falling. Anxious about the illness speeding up and taking her mobility. Pretty much anxious in general.
Card 1: Five of Cups
What gets in your way?
Remember to give your old heartaches respect and grieve them properly. Honor that time, and then move forward without them. It's okay to grieve, really. You can miss what you had and mourn what could have been. You can feel sorry for yourself, but you can't live there. You are seeing the world through lenses of pain and will interpret it incorrectly. If you find yourself here, make only small decisions. Drink some tea. Take a nap. Wait for the pain-shaped hole in you to heal. Let it get to a place where you can breathe without pulling strands of the past into your lungs. Don't stay in the in-betweens for too long.
Amy's main problem is fear that she'll injure herself. She's been going to physical therapy, but the idea of overdoing it keeps her from doing anything outside of the PT at all. Her main challenge is that she knows she needs to move and is pretty sure she's able to, but she's afraid of going backward. This problem started when she fell down the steps and hurt herself pretty badly. She's recovered just fine, but the fear backs up on her. Instead of working out safely, she decides to watch The Craft for the nine thousandth time. (It is a really good movie, you guys.) Regardless of the anxieties, if Amy wants to maintain her mobility, she has got to get stronger.
Card 2: Seven of Wands
What can you hold on to?
Tenacious. Literally, this exchange, "Are you done being sassy?" "No." This is the honey badger card of the tarot. Not backing down. Not asking permission. Not going away or sitting down.
Amy's latent "come at me, bro!" attitude is a major asset. Good luck telling her what to do. If she decides that it's time to do something, then by God, it's going to be done. For some reason, she hasn't yet applied this to her physical self. At all. Ever. As we were doing the reading, it was the first time she'd ever considered it. As we were sitting there, doing this reading, this was literally the first time she'd even considered it. A friend once said that even when she's down, she comes back up grinning—with blood staining her teeth. Why the hell hasn't she taken this tenacity with her to maintain her strength?
Card 3: Nine of Pentacles
What can you hold on to?
Appreciation of what you have and fulfillment of your wishes. A time to give thanks and swim around in your happiness for a while. Part of being comfortable in the skin you're in is moving away from a mindset of scarcity to one of enough. What you have now? Be grateful for it. If you seek more, be grateful for the ability to find it. You have worked hard and you deserve good things. There is no shame or blame in being comfortable. Just remember to help those still on the path.
This is telling Amy that she has the means to hire a trainer to teach her how to work out without hurting herself. Not, like, an everyday trainer (she's not Oprah rich), but she has enough to have someone show her the best way to get strong and not thrust herself backward into the near-bedridden state she was in. She can also ask her physical therapist for exercises she can do that won't aggravate her illness.
Card 4: Knight of Cups
What pulls you forward?
Romantic but fickle. Use your heart but keep your head on your shoulders. This knight is sweet and emotional. He can be shy and introverted but is also a flirt. When you get this card, remember that you can keep things light in love. You don't have to get married after the first date. Have fun and relax.
Amy should continue treating herself with love and respect. She's accepted the shape that she's in—mostly. She's accepting that her mobility will stay about the same as long as she cares for herself. She knows that life is a blessing. And not in the cheesy #blessing #mochalatte sense. It's a true blessing. Every day that she gets to be here, surrounded by love and her family and friends—my God, she is so blessed. She's going to continue to honor who she is where she stands. Right here and now. This person is tenacious and bold and strong. She is enough.
She has always been enough. She will always be enough.
Card 5: Ace of Pentacles Reversed
What pulls you forward?
It's not going to happen. The raise, the new house, the business plan. It's not going to come to fruition right now. It could be that it's not time or that you planned poorly. It could be because someone is actively working against you. Whatever it is, you need to walk away from the project and try again another time. Banging yourself against the gate that's closed against you will only leave you bruised.
She doesn't know everything. She doesn't have the knowledge that she needs to take great care of herself, so she needs to go get it. She also doesn't have to do everything. She can drop the ball sometimes. She needs to be realistic and erase this idea of perfection from her brain. It is absolutely not attainable. She just has to do enough so that she knows she's working toward her goal. She can be slow about it. She can make space for exercise in her life and let that space grow as she's able.
Card 6: Six of Wands
What pulls you forward?
Celebration! You did the thing! Well done. Go have a nice dinner. Be gracious, but still accept praise. Don't slap it away. Just say thank you. Remember to sit still in this moment and actually feel it. Don't speed into the next goal or spend time regretting what didn't work out. Just enjoy the thing and don't let other people cast a shadow on your sunshine, Sunshine. Shine, shine, shine.
And then she triumphs! Legit, this gave her so much hope for the process. The Six of Wands is a card for celebration, and if she can embrace this as her future, she will be able to embrace the possibility of looking and feeling on the outside like she does on the inside. In her head, she's already the image of perfection. Just gotta get in gear in real life to keep this vessel strong. She has a lot to do.
Summary
Amy has this in the bag. She just needs to apply this hard-earned work ethic to herself instead of just her work, family, and home.
Fire Reading Example #2: Renee
Renee is fifty years old and looks thirty. She takes excellent care of herself and is looking for advice regarding keeping her health foremost in her mind (and also how to quit jogging because she hates it). She is recovering from surgery and always has that in the back of her head.
Card 1: The Tower
What gets in your way?
Everything is going to fall down. It needs to. This doesn't mean you have to be buried under it. Sometimes things need to fall so new things can grow. Batten down the hatches, and collar those new emotions that come with change and inspect them. Make sure they're yours and you're not borrowing stress. Sometimes you need to set the past on fire and let it light the way to the future. It's not comfortable, but it's a lot more comfortable than cuddling up with decay. Get in right relations with the things that serve you so you can move forward with alacrity.
Renee is in great shape but had some health problems and surgeries that slowed her down. She has to realize that the person she is now is not the same person who went into those surgeries. The person who came out can't run like she used to without causing more damage than benefits. She has to plan a workout that allows her to stay in shape without the impact that her old training caused. Pilates came to mind. The Tower says that we have to rebuild, so let's do it right.
Card 2: The Magician
What can you hold on to?
Have confidence. You have all the tools you need to get ahead. Make it happen. You put in the time. You are experienced. You are talented. You've got secrets up your sleeves, and in your pockets, and behind your back. Now is the time to act like the expert you are and stop hiding behind false modesty. There is a difference between confidence and arrogance. Find it.
Renee has friends who teach Pilates! She's got the drive and the motivation to change up her exercise regime but hasn't considered it. Remember that the way you have done things in the past isn't necessarily the way you should do them now.
Card 3: Ace of Wands
What can you hold on to?
Move. The universe is giving you a window for action. Take it. It only takes one match to light a fire. Whether that's a controlled burn or a wildfire is up to you. This is the universe's hand on the small of your back, urging you forward. Sometimes it shoves you. Move forward with confidence. Know that you're where you are for a reason. If you grow roots at the starting gate, the finish line will never come closer.
Renee can be lazy. There, I said it. She's used to not having to earn her healthy body. It was just there. Functional and beautiful and exactly what she wanted. As she gets older and has had surgery, she can't bounce back like she used to. Her option is to move, move, move.
Card 4: Ace of Swords
What pulls you forward?
Follow your great idea. You're being given inspiration. The aces are the heart of their suit. The Ace of Swords is the brain of the tarot. Allow yourself to be inspired and enlightened. Be open to influences and your muse and be sure to follow this fantastic head start to its (already lucky) conclusion.
Renee is a smart woman and knows intellectually what she needs to do. She might be worrying a little more than necessary when the tools are right in front of her. She would benefit from writing out a training program that incorporates several types of exercise that won't bore her to tears.
Card 5: Nine of Swords
What pulls you forward?
Anxiety lives here. Remember that stupid thing that you did in 2001? Well, so does your brain, and it's going to mess with you every now and again to remind you of it. The fix for this is remember how much further along you've come since then. If I am in an anxious place, the anxieties that rush in mostly have nothing to do with reality. It's hard to discern what is real and what is just your brain poking at you. Luckily, you can rationalize your way out of this and stab your brain with Q-tips if necessary.
Similar to Amy, Renee is worried about reinjuring herself. She's hesitant to move forward with exercises she's unfamiliar with. This is where her trainer friends come in. Fear is a really powerful obstacle. She needs to allow people who have studied the body to advise her on hers.
Card 6: The Wheel of Fortune
What pulls you forward?
Either you'll ride the top of the wheel or go to the shadow side for a while. Remember that you're in charge of how you behave regardless of where the wheel takes you. There is power in knowing where you stand. There is value in all these positions. The only constant is change, so if you're being smushed into puddin' by life right now, wait till you feel the weight to start to lift and scramble to get back on top.
In order to maintain her good health and her peace of mind, Renee needs to combine several workout plans throughout her days, including meditation and some group activities—maybe a hiking club? Getting in a rut is counterintuitive for her as a person and would slow down her progress.
Summary
Renee has recovered from her surgery but needs to believe that she's okay. She needs to stir up the routine that she's been in since her thirties and adapt it to something more interesting and accessible to the amazing fifty-year-old she's become. There is no reason her health should move backward, and the only obstacles she needs to be aware of are boredom and anxiety. She's got this.
Exercise
Light a candle for ambiance, babies. This is going to be fun.
Lie down on the floor and stretch your body. Each arm, each leg. Sloooowly stretch. Don't hurt yourself; do what you can. I want you to really pay attention to your body. Stretch your feet and each toe. Stretch your hands and fingers. Your neck and back. When you're ready, slooooowly sit up. Sit as close to crisscross-applesauce as you can comfortably. Place your hands on your knees and tilt your chin to the sky. Stretch your spine and neck. Rotate your head gently.
To help pay attention, name the parts of your body as you stretch them. When you're ready, stand up and reach your hands as high in the air as you can. Slowly, with intent. After you're all stretched out, go for a walk (blow out the candle first) for as long as you're comfortable. If you aren't able to walk, go sit outside.
On your walk, I want you to think about your body. I want you to remember every decade you've spent in it. All the surgeries, stretch marks, and scars. Really see how far you've come since childhood. Acknowledge and offer gratitude for being able to walk as far as you can. I can walk for about twenty minutes before Auntie Asthma comes to knock me over, but that's twenty minutes longer than I could last year. I want you to do an assessment of this amazing body you have. Then, when you're ready, come back home and work on your homework. Now that you've met your body again, you can begin a new relationship with it.
Homework
Write a letter to ten-year-old you. How strong you were! You could run and jump and fall spectacularly and get back up again. Write a letter to twenty-year-old you—all the misuses and abuses you put yourself through and still made it. Thirty-year-old you—so tired and still so strong. Forty-year-old you—new lines from laughing with old friends. Fifty-year-old you—aches and pains and beauty that comes from knowing better than pretty much anyone else. Include the surgeries, the scars, the hurts, and accidents. Include those things that you wish hadn't happened.
Write letters to all of the yester-yous. Thank them for getting through the hard times. Compliment them on getting through the scrapes and bangs. Tell them that everything turned out okay. Tell them that you forgive them their trespasses and that everything turned out beautifully in the end.
When you've finished, I want you to burn these letters. One by one, the oldest you to the youngest you. Say thank you. Tell yourself that you're going to take good care of yourself because you deserve it. Believe it when you say it. Release everything you were told about yourself that was not true. Release all the shame and guilt that weighs on you. Let it burn and let the ashes blow away. You don't need any of it.
Accepting who you are is hard. We hold ourselves up against the person we used to be or against other people. Don't do this. Allow yourself to have a starting point. We can't all look or feel as amazing as P!nk, but we can figure out where our baseline is. Compare yourself to yourself. Allow yourself to acknowledge the hardships you've been through and how far you've come. Set realistic, achievable goals, and when you've reached those, set some more. You've done amazing things and come through on the other side. You can do this, too.
After all, you're still here, right? This tells me that you're already a success.
Resources
Move Your Body
Dr. Andrew Weil recommends walking for twenty minutes a day. That's it. If you want or need to challenge yourself further, go for it, but if not, see if you can do twenty minutes a day. Walk in your neighborhood or your local park, or drive to a national park and walk there. Grab a buddy if you need one (even a furry one with paws). Just move your body. Feel your feet on the ground. Allow yourself the time to take care of yourself. Put "walk" on your digital calendar every day at the same two times. When the notification comes through, do not delete it.
Don't Eat Garbage
I am not great at this. I generally eat well, but then I'll wake up hungry, get caught up in taking the kids to school, and forget to eat. Then when I'm on the way home to work, I'll pass McDonald's and remember they have bacon. I pull in, get "food," and feel sick for the rest of the day. Awesome. If I would have prepared myself for the morning by making a shake or making eggs when I got home, I wouldn't have this problem at all. Remember that cutting corners in your diet only makes you unhappy, and that fresh food is the best food.
Stretch
When is the last time you really stretched? I mean arms-wide, bending-over, feeling-your-back-pop stretched? Get up and do it now, just like when you were in gym class. Stretch your arms, hands, and fingers. Flex your feet and your legs. Do this whenever you've been sitting down for an hour or when you have finished a task. Do this stretching as often as you can, because it seriously reinvigorates you and keeps you from sitting on your butt so long.
Have Good Sex
With yourself or with a partner, have healthy, fun, consensual sex. Be sure that it's legal and all, but definitely have good sex. The endorphins alone are great for your complexion. You'll relax, you'll feel more in touch with your body, and you'll feel more confident. It doesn't get much better than that.
[contents]
Chapter 6
Water: A Reading to Help You Float
Water is emotion. It's our heart. The moon affects the tides and we're 60 percent water, so it stands to reason that we're affected by the moon, too. The waxing and waning cycles affect us just as they affect the sea. If we allow ourselves to become dehydrated, everything slows down. We can't think as clearly, and our body doesn't process anything the way it's supposed to. Our need to refresh ourselves is biological, but there is also a pull to surround ourselves in water. Swimming and bathing and showers—the buoyancy and the feeling of being cradled in the water pull to us.
Water
Heart
The prevailing thought when someone calls to get a reading about love is that they're looking for their person. They are, but they're really not. They're looking for themselves. They're searching for something to put in that hole in the middle of them. The hole was created by hurt feelings and judgement. By abuse and broken trust. In looking to become whole, we seek someone—anyone—to fill in that emptiness inside us. So it won't hurt anymore. So we won't feel so lonely all the damned time. We take whatever person shows a fleeting interest in us and try to figure out how to get them to fit in that hole. Sure, they're verbally abusive (they're just tired) or distant (keep their emotions inside) or neglectful (they're just busy). We tiptoe around their broken bits and blame ourselves when we get cut. And most of the readings I do about broken hearts come at the full moon. Oh, she does pull on us. We expand and reach out and look for someone or something to connect to.
This reading is not about finding your person. I feel like I need to stress that point. Like all the element readings, this is about hitting a reset button on your life. Although emotional readings are often romantic, that's not the whole of it, and they can often be a distraction from problems that we've been carrying around with us.
Water connects us to the moon and the waters of the earth. We are born of it, and it sustains us. Water is also the element that aligns with emotions. Sloshy, messy, weeping emotions. Sometimes they dry up and sometimes they flood us, and sometimes other people's emotions surround us and drown us until our fingers get all pruney. Who wouldn't need a reset button for their emotional life? Our water exercise is going to address where you are emotionally and look for ways to maintain balance and avoid carrying around another person's baggage.
There are parts of my past that stick in my brain. They show up when I'm feeling vulnerable or unsure and remind me of alllllll the stupid stuff I've done in my life. Trusted that one person who ended up reading my diary to her friends. Believed a person who swore they would never hurt me. Believed a person who said I was worthless. Each of these things is a time (or times) in my life that I accepted someone else's assessment of who I was and where my abilities lay.
I spent so much of my time just believing people and allowing their assessments to shape who I was. One of my first emotional letdowns came really early: my mother lost custody of me and my sister in 1976. I was eighteen months old and my sis was almost four. I don't really like to go into it, but suffice to say that in the '70s, in central Kansas, a court gave full custody to my father instead of my mother. So, you know, that's noteworthy. My dad, however, is the most amazing person in the world, and I was lucky enough to get a pretty awesome stepmother when I was seven or so.
The loss of my mother, though, was significant. I was too little to be told why she left, so I filled that space myself. She left after I was born, so clearly it was because of me. I didn't have anyone telling me anything different, so I believed the logical, if inaccurate, story that I wrote in my head. To little me, my mother didn't want to see me, be around me, watch me grow. My fault. Because I was so small, all these reasons pushed into my skin until I started wearing them like tattoos. I wasn't enough to keep her around. I was the reason that my sister didn't have a mother. I was the reason that my father didn't have a wife. I filled the vacuum of information with insecurity, and that became a part of who I was. (God, I'm thinking now that I need to tell my kids that kids aren't responsible for adult decisions about anything ever, because I sure as hell wish someone would have told me.)
I did have mother figures in my family, and they made sure I knew I was loved and valued. They held space until my stepmother found our little family, and for that I am eternally grateful. I also co-opted Princess Leia and Wonder Woman to help me in my quest to find my mother. These women taught me to fake a haughty accent if I was feeling small and to be bulletproof.
Even Wonder Woman was not enough to fill that mother-
shaped hole that was inside of me, though, and that was something I struggled with for a long time. That feeling of "it's okay to leave me" and "I am not worth fighting for." Completely constructed from my own little brain but manifesting itself in so many ways throughout my life. I would put up with friends who bullied me, teachers who intimidated me, and partners who treated me like garbage. To strangers, I was fierce and self-protective and sassy. If they loved me even a little, though? I would do whatever I could to be sure that they wouldn't leave me.
I became this hotbed of insecurities that would extend to work, friendships, and my first marriage. It was very narcissistic, in that I thought that everything that happened was because of me. The house isn't clean enough, so he's angry. I wasn't present enough in conversation, so my friend is ghosting me. I probably deserve it. Even when dating, if someone didn't call me back, clearly it was me. I don't think I ever looked up from myself to see that maybe someone else was the problem.
When I was in my twenties, I needed someone to fill that hole. I tried filling it with booze and sex and weed or with whatever person would spend time with me. I convinced myself that I was lacking, not the people I chose as my partners. They didn't stick around because I wasn't smart enough or pretty enough or skinny enough. I just wasn't enough. I poured all of myself into the next person who came along, hoping all this love I was handing out would stick. I continued throwing my heart as hard as I could at people, and when it slid off and smacked onto the ground, I would pick it up, dust it off, and look for the next person I could take aim at.
I'm writing this and wandering around my friend Sarah Kate's house asking, "Why did I do that? Why did I subdue who I was and allow people to use me? Why was I so awful to myself?" I was about to go through some serious existential stuff when she said, "Honey, it's because you were in your twenties. We all did that."
Oh. Thank god for Sarah Kate.
That habit I had of putting other people's happiness ahead of mine changed, for the most part, when I was about thirty-four. I realized that I wasn't happy in my marriage and that I couldn't fix it. I tried, but I couldn't do it. I started looking up, you guys. I saw that I couldn't change someone else for the first time in my life. I realized that I could only change myself. So I did. It was heartbreaking and traumatic and all the things that a divorce is, but at the end, I was looking up. I considered getting a tattoo that read "you can't fix someone you didn't break." I instead got one that said "but then there was a star danced, and under that was I born," from Much Ado About Nothing.
This is such a huge experience that I am currently in awe of that simple self-transformation. Memories of times when I allowed people who loved me to hurt me are flooding me. I took it and I took it because I thought that pain was a part of love. I thought that you struggled through whatever came with love because it was worth it just to not be alone. I was wrong. It's so much better to be alone than lonely with the wrong person.
The divorce was the most significant shift, but I also broke up with some friends along the way. Some had been around for over a decade, but I noticed (head up, remember) that the flow of friendship only went one way. I would bend over backward for them, and when I had a crisis, they were nowhere to be found. It was painful and I wanted that kind of pain to stop. I cleaned house. I kept close to me those friends who would answer the phone at 2 a.m. with "What's up? Do you need me?" and I released those friends who called me so often but never asked how my day was going. Completely one-sided relationships are so exhausting. One of the best side effects of this was that I was able to be a good friend to the ones I held on to. I wasn't constantly helping drama queens put out their fires. I swear to god, some people are made of fire. I was able to concentrate on myself and my friends who weren't roving garbage fires and, in doing so, had enough energy left over at the end of the day.
I still struggle on occasion with feelings that don't serve me. It still takes me a while to realize that one of my friends is too dramatic or rude or inconsistent, and I would benefit from some space between us. I'm not perfect, by any means, but I feel like I can withstand a lot more than I could in the past. I find myself observing drama and then moving away from it as quickly as I can. I feel like the holes that were in me—created by loss—have been filled by confidence, self-love, and common sense. Common sense that seemed to be a last resort back in the broken days is my first choice when I need a weapon. My first observation is no longer "What did I do wrong?" but "Why are you such a jerk? Not my problem."
Water Reading
| Cards, 1, 2, and 3: What gets in your way?
Cards 4 and 5: What pulls you forward?
Card 6: What can you hold on to?
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Water Reading Example #1: Sarah
Sarah is a mom and an artist and is pretty laid back, in spite of having been through some rather serious issues in her life. She has had some rough romances and is taking a break but wonders if she's just better off alone. What if she finds another stinker of a partner? What if her heart gets broken again? On the other hand, she doesn't want to be without affection, so she's feeling really stuck.
Card 1: Six of Swords Reversed
What gets in your way?
It's all well and good to say that you're going to move on, but sometimes you just can't. The past sticks to you like glue and won't give you peace. The answer is not to just deal with it. The answer is to confront it—in counseling, with ritual, or with prayer. Deal with the problem of yester-you. Forgive yourself and let yourself heal.
This feeling that she might be moving backward in her emotional journey is terrifying. Just getting this reading, recognizing that she'd have to sort out her past to be wholly present—that's scary stuff. If you spend a long, long time being afraid of something, losing that fear of it doesn't just automatically make everything better. You have to get used to not being afraid. You have to fill the hole that fear left in you with something that will help you heal. Like crying. Or eating ice cream. Or reading a really good book. You have to acknowledge the hole and make friends with it until it heals. That's hard. It's worth it, though. Can't have unattended emotional holes just lying all over the place, waiting for her to step in them.
Card 2: Strength
What gets in your way?
Strength takes many forms. Figure out what yours is and feed it. You have reserves you might not be aware of. Use your strengths—compassion, kindness, humor, resting bitch face, active bitch face. Use your gifts to remain strong—whatever that means for you. Sometimes it means sitting still until you can stand up under your own power. Sometimes it means faking it till you're making it. Sometimes it means saying no and making that a complete sentence. No.
Sarah is strong. She's one of the strongest people that I know, and yet it's as if she doesn't see it. She needs a reminder on a daily basis. I asked her to write a reminder on her bathroom mirror. I got a tattoo that says "fireproof" to remind myself. I can't not see it or feel it. Sarah and I have been through some serious life experiences, and we're still standing. We're still resilient and joyful and looking for our balance. As long as we seek it, as long as we recognize who we really are, we're going to be just fine. We are fireproof.
Card 3: Two of Swords Reversed
What gets in your way?
Listen, if you don't want to decide, you don't get to complain. There are a lot of reasons to remain in stasis, and you can for a while, but if you allow your decision-making to fall to other people, you get to experience their path. Do you want that? Do you want to walk someone else's road? Even if they love you and mean the best for you, it will still be inauthentic.
Sarah has spent too much time quiet and embarrassed about her past failings. A lot of her healing comes from talking, writing, and making art. As long as she can take the hurt and anger and make something with it. Get it out. It won't stick to her as much and she won't carry her past mistakes with her. Keep being loud. Keep getting readings that put another piece of the puzzle in place. The right way to release herself from the trap that she is who she is and that she is incapable of change is however it makes sense for Sarah.
Card 4: Four of Pentacles
What pulls you forward?
Hold on to those things that belong to you and be generous when you can. Be sure that you have everything you need before you share. Be sure that you come first. This card also indicates that you should keep an eye on the things that belong to you. Watch your money. Watch your possessions. Be safe.
What can strengthen Sarah's emotional self is protecting and being grateful for what she has. She's got a solid relationship with her person. She has a warm and welcoming support system in her friends and family. She has hobbies that light her up and a job that she loves and is great at. All these things make her stronger. All these things are earned. If you'd asked twenty-year-old Sarah what her life would look like in twenty years, the details would have been different but the qualities would have been the same. To be loved, respected, and supported. To be able to live honestly and with integrity. These are things to hold on to and things of which she is fiercely protective. The other stuff that life offers is nice, but at the end of the day, knowing that you're secure and happy beats the other stuff hands down. The four pentacles tell her to see how valuable her life is and to protect it.
Card 5: The Fool Reversed
What pulls you forward?
Just because you can just doesn't mean you should. Just because the ledge is there doesn't mean you shouldn't look over the edge and double check your parachute. The Fool reversed throws caution to the wind. And common sense, a sense of pride, decency, and intellect. Just blundering stupidly into the next mistake.
Sarah is encouraged to stick it out. She will not jump. She will not be skittish. She will stay put and weather the slings and arrows of "Did you take the trash out?" or "Why did you say it like that?" and realize that sometimes she can be highly reactionary. She needs to realize that she's going to do stupid things and then forgive herself. Most mistakes are temporary. She needs to stop carrying them around after their best-by date and stop defining herself by her lowest moments.
Card 6: Ten of Wands Reversed
What can you hold on to?
When you end a relationship, the shadow of that person follows you around for a while. When you get hurt, the shadow of that pain lingers. This card reversed asks you to sever the connection to the things that weigh you down. Have a ceremony. Write it/them a letter and burn it. Do something that tells the universe that you are finished with that person, the booze, that habit. Make a statement and then have a funeral for the thing that no longer serves you.
I hope to god my husband never reads this, because he's an Aries and his ego might need to have a permanent room in our house after this, but here we go. Sarah's advice here is from Joe.
When we first started dating, Joe told me that I needed to learn to put things down. I still haven't learned this lesson. I like to chew on pain until I've sucked all the marrow out of it and then fashion a necklace out of the bones and wear them around, clanking into the night with tales of my idiocy. Sarah is exactly the same as me in that both of us can be complete idiots.
This is a stupid, narcissistic exercise, and we just need to stop it. Sarah doesn't have this particular problem (ha—neither do I! Waits for three emails to arrive...), but she does have some problems that she needs to work on. She needs to make space in her life to address them and then put them down!
God. My husband is always right (eyeroll for years).
Summary
Sarah needs to go to counseling to work out some of this hurt that she's carrying around with her. It's gotten so heavy and has been around for so long that it feels normal to her. She should not seek out a partner until she is sure that she is mostly healed and won't be looking for a partner to "complete" her. No one can do that. She has to complete herself, and the best way to do that is to work with a counselor on forgiving herself for past mistakes, learning how to move forward as a confident single person, and stop basing her future happiness on whether she'll be able to hold hands with someone. She should aim for happiness, and if the hand-holding comes along, that's a bonus.
Water Reading Example #2: Bella
I got to read for a friend named Bella. She travels a lot, for business and pleasure. She is single, young, and accomplished. She's also bored to tears. She seems to date guys that have the same pattern: they're very interested, they have a very good time, and then poof—they disappear. Or just as quickly, she leaves. She can't seem to connect below the surface and it's driving her crazy.
Card 1: Four of Cups
What gets in your way?
Don't settle. Don't pout either, but make sure that this is what you really want. Fussy and ripe for hissy fits and eye rolling. Apathy is easy to ignore. Don't wanna. Don't want to change, don't want to take my medicine, whatever that medicine may be. You have to look deeper than the apathy. Why don't you want to change? You know these behaviors make you sad, and yet you won't put them down. Consider the why of holding on to the listlessness and figure out how that's paying off for you. There's a payoff for "meh." Whether it's the reward of not failing (you can't fail if you don't try) or the reward of being comfortable, decide if it's worth being stuck.
Oh, Bella is so bored. Not just romantically, although that is a large part of it. Mostly because she has to, because of work and social obligations, hang out with a bunch of shallow people with whom she tries to dive deeply. She looks for connection with folks who are impossible to connect with. The advice here is to invest in a few people who have shown their loyalty and only deal shallowly with others. You can engage in relationships with acquaintances on a surface level and only dive into those friendships that are true. You won't bonk your head that way.
Card 2: Seven of Wands
What gets in your way?
Tenacious. Literally, this exchange: "Are you done being sassy?" "No." This is the honey badger card of the tarot. Not backing down. Not asking permission. Not going away or sitting down.
One of the problems that Bella is running into is that she's been let down so many times by other people. When she meets someone new, even if there is a connection, her mind begins the countdown of when this person will screw her over. It's difficult to make new friends, much less have a new relationship, with someone when you're waiting for the other shoe to drop.
Card 3: King of Cups Reversed
What gets in your way?
All the intensity and power of the king's emotion flooding out of control. This is drama, you guys. Temper tantrum, hissy fit, manipulative, Real Housewives of Wherever They Are Now drama. The worst part of this is the power that the king has. He can make things really bad for you. He can trample cities under his feet. It's not a good thing, and unless this card represents a true crisis, you should seriously take cover. Egad, the drama.
Another problem is that Bella continues dating but seems to be finding the same guy in different thousand-dollar suits. Same age, same MO, same ridiculous problems. She is looking for Mr. Right and keeps finding Mr. Right Now. Consequently, I grounded her from dating for six months. Not to be mean but so that she can invest in herself wholly and then come back to dating with a solid idea of who she is and what she wants. She'll be more able to dismiss those who don't belong in her world and engage with those who do, without the suspicion.
Card 4: The Hermit Reversed
What pulls you forward?
It's normal to pull away from folks when you're feeling sad. Absolutely normal. It is not normal to make it a long-term thing. If you find yourself isolating yourself unhappily (some people are just fine with solitude!), you should ask yourself why. What is making you push folks away, and what are you seeking in your self-inflicted exile?
One of her strengths is that she is a mighty extrovert. She needs to keep seeing her friends, continue building relationships, and get out of her house. Just because she's not dating for a while (sorry, Bella!) doesn't mean she needs to turn that part of her personality off. She has been pulling back because of the romance, but she still needs to hang out with her people. Crappy partners don't get to define your social life.
Card 5: Page of Swords
What pulls you forward?
This card has so much potential. It is the idea that wakes you. The lightning bolt whose afterimage lingers like a neon light behind your eyes. The 4 a.m. scribble on the piece of paper by your bed that says "moon cheese" and nothing else. You've got the seed. You just have to give it care and space to grow, and it's going to change the world.
While she's on this dating embargo, she needs to dive into intellectual pursuits that excite her. Go to museums. Start a book club with friends. Find something that isn't work related and doesn't have a thing to do with romance. She's bored. There are some great ways to rid herself of boredom that don't include fellas.
Card 6: King of Wands
What can you hold on to?
Those boundaries you are working on have to remain strong. Your job is to keep Negative Nellies away and protect those things most valuable to you. This king suggests that you stop half-assing everything and start whole-assing one thing at a time. Finish what you started. Close your circle. Recognize that even though you want to be friends with everyone, you need to have a pretty keen eye for recognizing who people really are. When they show you—with their behavior, their promises, and their accountability—who they are, you need to believe them.
Bella finds her balance in travel. She has a safe and warm nest in Europe and friends all over the world. She is a wanderer and needs to continue wandering and pursuing the next beautiful thing. Her strength, essentially, is being herself.
Summary
Bella has been sticking with the formula that a lot of us learn when we begin dating. You become attracted to someone, you exchange information, you flirt online for a while, and then you date each other exclusively until someone gets bored or distracted by another person.
What I'd love for her to do is break this pattern. Take a break from dating, hang out with her friends, and be sure her life is fulfilled. When her break is over, she can start dating more than one person at a time. Keep it very surface, like it's the '50s. Dinner and a movie. Texting every few days or so. Full disclosure that it isn't exclusive. Give it a few months to grow and if you have an idea that it's not going to work out, shake hands and walk away. It's much more civil and doesn't start you down the path of "But we were talking every day and I thought it was getting serious." You decide when it's getting serious. You don't guess or wonder.
Exercise
I learned this from my friend Keva. I was feeling pretty low one day, and Keva told me to write down everything that I was afraid of. I want you to list all your concerns. What's bothering you? What came up in your reading?
Concerns
I'll lose my job.
I'm screwing up my kids
I'm going to get a divorce.
We're going to go broke.
I'll lose the house.
My parents will die.
After you write down all those horrible things, make another list on the other side.
Concerns | Hopes
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I'll lose my job. | I will always be happily employed.
I'm screwing up my kids | My kids are healthy and smart.
I'm going to get a divorce. | My marriage is solid and we are crazy in love.
We're going to go broke. | The universe will always provide us with enough.
I'll lose the house. | We will always have a happy home.
My parents will die. | My parents will live a long, healthy life.
Now, tear that page right down the middle. Or use scissors. Your choice. Go crazy.
Take the left side outside and burn it in a safe place. Watch that page of fear catch fire and burn away.
Take the right side and decorate it. Get out your colors and gel pens and stickers and adorn your hopes and reality. Either post your hopes someplace you'll see them every day, or tuck them in your journal as a bookmark. Make them prominent in your life. Believe in your hopes.
Hopes
I will always be happily employed.
My kids are healthy and smart.
My marriage is solid and we are crazy in love.
The universe will always provide us with enough.
We will always have a happy home.
My parents will live a long, healthy life.
After this exercise, go take a bath. Soak up the all the good energy from the compliments you've paid yourself and the hopes you've invested in. Soak it in and then let those complaints and problems go with a quick shower. (You shower after a bath, right? Cuz it's like people soup?)
Important
If you think that you might be depressed or suffering from anxiety, please call someone to get help. You are not alone, and there is always a way out. Remember that depression and anxiety lie, and sometimes your serotonin reuptake inhibitors need a hand. Practice being kind to yourself. If you catch a nasty thought going through your brain, counter it with something positive and take a deep breath. You're doing the best you can with what you have.
Homework
Make a list of what you're looking for in a partner. If you don't see those things in the person you're dating, stop dating them. Just stop. If you're married and things are not okay—don't keep pretending they're okay. That helps no one. If you are single and okay with being single, just be happy. Honestly. Your homework here for wherever you are is to be sure that where you are is happy.
Resources
If you need someone to talk to, you can always reach out. The people who man these hotlines genuinely care about their jobs and the people they serve. Just remember that whatever it is, it's just for now. It really does get better.
* SAMHSA (Substance Abuse and Mental Health Services Administration) National Helpline: 1-800-662-4357
* GLBT National Hotline: 1-888-843-4564
* GLBT National Youth Talkline: 1-800-246-7743
* Trans Lifeline: 877-565-8860 (US), 877-330-6366 (Canada)
* National Domestic Violence Hotline: 1-800-799-7233
* National Suicide Prevention Lifeline: 1-800-273-8255
Music
I can't tell you how often music has saved me from myself and from my dark and twisties. Ani DiFranco, the Indigo Girls, Brandi Carlile, Johnny Cash, Prince, Al Green, Digital Underground, Blackstar, Neil Diamond, Lauryn Hill. Honestly, the list goes on and on. I have playlists for when I'm having a bad day (Queen, the Avett Brothers), when I need to be comforted (lots of John Denver and Joni Mitchell), and when I'm angry (Talib Kweli and Liz Phair). I have playlists that help me feel my feelings and scream (the Clash) and cry (Tracy Chapman) if I need to. Then I've got playlists to help lift me back up (the Beatles). I've got Spotify on my phone and it lifts me up every day.
Art
Invest in some art supplies. Coloring books, crayons, gel pens. This is the most relaxing pastime. Just color for a while. Or go to an art museum or a gallery. Surround yourself with beauty. When my kids were little, one of my favorite things was coloring on the sidewalk with brightly colored chalk. We would create entire oceans on our driveway and rainbows up the stairs. Just seeing it made me smile. The magic of art is that it is external but affects us internally. Being in the presence of beauty can shift your mood and allow you to feel lighter without a lot of internal struggle.
Flowers
Having freshly cut (or living) flowers in your house will help cheer you immensely. I can't keep plants alive, so I try to get cut flowers every time I go to Trader Joe's. I can usually get a pretty bouquet for about $7. This is a small price to pay for a splash of joy in your day.
Restraint for Empaths
Use it! The next time you're around some crazy friend of yours who is going through drama, please watch the conversation as if you are watching a movie. Have emotional distance. Consider their position rationally and then respond to them calmly. It feels very different from joining them in emotional upheaval, and you walk away less exhausted.
Gratitude Journal
This part sounds stupid, but it works. Once a day, write down three things that you're thankful for. It's more difficult to hold on to those things that bring us joy than it is to hold on to grief. It can be as simple as a soft blanket, watching Deadpool, and hearing a great song on the radio. The point is to keep yourself present and notice the shiny things as well as the garbage that rolls in and out of your day.
Community
I have a lot of clients who are lonely. Either they've lost friends over the years or moved to a new place and haven't really connected with anyone. I take the lead here from my friend Kelly. She moved to Nashville when she was in her thirties with two small kids and her husband. She knew no one. The next thing I knew, she was part of the neighborhood Bunco group (I hear it's a game with booze), had signed up for a bunch of meet-ups for moms with young kids, joined a book club, and joined a meet-up that was designed to help you be a tourist in your own town. It's ten years later, and Kelly has a thriving social life and has only held on to those friends and clubs that continue to serve her. These groups are an outstanding way to find people who share your interests.
[contents]
Chapter 7
Spirit: A Reading to Bring You Connection
When I was writing this book, I put off writing about spirit until the end. We are all so diverse, and we all seek spiritual enlightenment in different ways. I didn't want to presume that all readers have a spiritual path, and if you did, I didn't want to presume to guess at what that was. I'm not trying to tell someone how to believe or if they should seek anything in their spiritual life, so why did this feel so big and important? I think that I've figured it out: regardless of the path you're on, everyone is seeking to feel connected. To our communities, to our earth, to each other. That sense of connection is what makes us the social beings that we are. It is why we have compassion and empathy and sympathy. The connection that we feel in this world is what makes us part of something greater than ourselves. It forces us to look outward instead of inward and focus on how we can make the world better. How can we help? How do we align ourselves in right relations with our people?
Spirit
Soul
You can use this spirit reading to look for direction in a religious or spiritual sense as well as for which direction you can go with your spiritual gifts.
I find that the spirit element relates directly to Spidey-sense. Whatever you'd like to call it—intuition, psychic energy—it is a connection to something outside of us. This kind of spirit connection can be baffling if you're not sure what's going on. I remember seeing spirits when I was a kid and singing this children's hymn called "Be Not Afraid" under my breath until they went away. I would get flashes of intuition about grown-up problems and concerns that would scare the hell out of me. I would dream and the dreams would become reality. In college, I tried to drink it away—didn't work for long. What I needed was someone to tell me that I could ask it to stop until I was ready. I needed someone to tell me that it was okay, that I wasn't crazy, and that I could make this make sense with some time. You can use this reading to find your path regarding your psychic gifts. Are you heading in the right direction? What can you bring into your daily practice to sharpen these gifts? What can you leave behind?
In the traditional spirit reading, you can look into your faith. Where you've been and where you are now. Where you're going and what needs to stay behind. In a nontraditional reading, you can figure out where to find resources to help you strengthen your intuition.
I was raised Catholic. It didn't take, but I did take some of it with me. I love ritual and incense and the kindness that Jesus taught. I found, however, that I had serious misalignment with the Catholic Church as I grew up. Not just in adulthood, but I was the kid asking why people who haven't heard of Jesus were going to go to hell. That seems like a dick move. Why do we have to ask the priest for forgiveness if God forgives us as soon as we sin? Why bother hating the sin if the sinner is in front of us and the sin didn't hurt me?
As I grew older, my problems got bigger. Why should I hate gay people? Why should I pray to take away a woman's right to choose what to do with her own body? Why can't women have a role in the church? Why do I just need to accept this information from the Bible when the Bible contradicts itself? Why should I stop asking so many questions, Sister Rita Mary? I thought this was a school. That one earned me a smack with a textbook. Clearly, I should stop asking questions so I wouldn't get hit anymore. Jesus, indeed.
Now, before the emails start flying, my family has been Catholic for as long as I know. Some of my best friends are Catholic! I have respect for the help that the church has given communities in its time. But it's not for me. It doesn't and hasn't suited me, so I left it.
I did the requisite wandering around and talking smack about the faith that I left, without putting any thought into why I left it and what I would do with this spirit-shaped space left in me. I liked praying. I liked feeling connected to God and feeling like part of a community. Just not like that. I went to a Unitarian Universalist meet-up in college and felt the stirrings of God there, but I was in college and drunk a lot, so I didn't pursue it.
When I was in my late twenties, a very good friend took me on a Pagan retreat in the woods. I didn't understand a lot of what was going on, but I knew that I felt God there. Off to the bookstore with me—you wouldn't believe the stack of Paganism books that I came home with. I did research, tried little spells, and tried on being a witch. Turns out, it fit. I connect to spirit through candle magic, having an altar, following the cycles of the moon and the seasons. I did not connect with those books, though. I don't do Ritual. I do lowercase r ritual. If I'm "cleaning" my house energetically, I'm basically stomping around in my jammie pants, tossing salt in windows and telling negative energy to get the hell out of my house.
My ex-husband is Jewish, so when the kids came along, I wanted to find a place for them to worship. I was puzzled, though, as to how to find a place of worship for the children of a Jewish fella and an ex-Catholic witch. I went to this website called Beliefnet.com, and took the amazingly named "Belief-O-Matic" quiz. Essentially, I answered a bunch of questions about my moral values, spiritual beliefs, and personal feelings, and the Belief-O-Matic spit out the top five organized religions that I aligned with:
1. Liberal Quakerism
2. Pagan
3. Reform Judaism
4. Unitarian Universalism
5. Neopagan
I went to a Quaker Friends meeting and nearly fell asleep. No disrespect meant; I'm just not meant to hold still and be quiet. I already knew the Pagan thing, but I was looking for more structure. I attended a Reform Jewish service and it was beautiful and aligned with me, but I felt like the language (don't speak Hebrew) divide was too wide. I visited a Unitarian Universalist chapel and found my home. Their book had readings from the Koran, the Talmud, the Bible, the Upanishads, Native American elders, and Martin Luther King Jr. They also pray with their feet. My church has a bowl of Black Lives Matter buttons in the entrance of the church. I'm down with this.
I have friends from all walks of faith, and they tell me that they have a consistent worship practice because of connection and community. One of my friends goes hiking every week. Every week. She finds God in solitude. One of my friends volunteers with an animal rescue organization. She sees God in the faces of abused animals and does all she can to help them. Another friend is a devout Christian but is gay and had problems aligning themselves with the church they grew up in. They found the United Church of Christ, who welcome all regardless of who they love.
If you have found yourself looking for a better fit for your spiritual life, even within your current place of worship or faith, recognize that some form of growth and transition is normal. You are not the six-year-old who took Holy Communion or the thirteen-year-old who completed their Bar Mitzvah. You are older and have more layers to you. Revisiting your spiritual connection at several points in your life should be embraced and explored.
spirit reading
You might have noticed that this one is a bit bigger, and therefore a bit different from the patterns drawn by the other readings. This is because while I tried to simplify the others, doing the same to spirit just doesn't work. It's a complicated thing and has a more complicated spread. The mechanics are the same, though. You got this.
| Card 1: What can you hold on to?
Cards 2 and 3: What gets in your way?
Cards 4 and 5: What pulls you forward?
Cards 6 and 7: Who or what in your life can assist you with this journey?
Card 8: What choice can you make right now to get closer to your path?
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Spirit Reading Example #1: Shavonne
Shavonne is a single mom and an artist who went through a very difficult time a few years ago and is just getting back on her feet. She is confident but sometimes worries that things are going to go backward. She's anxious but doesn't really have anything to pin that on except for fretting over "what if?" It might not sound like a big deal, but it keeps her up at night.
Card 1: Six of Pentacles
What can you hold on to?
There is an exchange here. Giving and receiving. Be sure that the balance remains. Can you give gracefully? Can you receive with equal grace? Can you allow balance and generosity into your life? Care. We all do better when we all do better. You have to invest in yourself first, then your family, and then your community. It has to flow outward. If it sticks with just you, your riches spoil and stagnate. If they flow out to only your family, they puddle and pool. If they reach your community, however, they flow and flow and flow. A rising tide raises all boats.
Shavonne is now in a place where she can give back and make a difference. She was in a place a few years ago where she had to receive from her church. She was unemployed and had two kids and no money. She struggled for several months until she was able to get back on her feet again. Her church sent grocery money to her every month, and her minister checked in to make sure she was okay. At one service, someone who'd heard she was struggling slipped twenty dollars into her purse. She doesn't know who did it, but they should know that she used that money to put gas in her car for a job interview. Now that she's doing okay, it's her turn to reach out.
Card 2: The Chariot Reversed
What gets in your way?
If you don't believe in yourself and in your goals, you will drive your chariot in circles. You will spin and wander and sit down suddenly, wondering what the next step is. If you have this vehicle for change, whether it be a life decision or a new job or a new partner—you have to really focus on what's good for you and what isn't. Pay attention. You're spinning.
Shavonne is distracted by life pulling her in many directions. She has a full-time job, two kiddos, and a partner and is working on her home business. She needs to remember that her spiritual connection is important, too. She needs to make space to honor that. Mind, body, and spirit balance is so very important, and the spirit side always seems to slip away.
Card 3: Two of Pentacles
What gets in your way?
You often get this card when you're keeping your balance well amid rough weather. Keep light on your feet and don't lose sight of your goals. Regardless of how crazy the world can be, as long as you maintain your balance, you'll be just fine. Remember, the Buddha said be kind to everyone. You are part of everyone. Don't involve yourself in other people's issues. Just mind your business and reach out if your business involves getting assistance.
Shavonne doesn't have to do everything, but she does need to have a little bit more balance in her world. She can't just work and go home. That's not enough living for her. She likes the outdoors but hasn't found time to go hiking for a year or so. She loves going to the movies but doesn't make space for this. She's also been wanting to check out some volunteer opportunities, but hasn't made space or time for this, either. Shavonne needs to get off her butt and go do something instead of wondering.
Card 4: Three of Pentacles
What pulls you forward?
Teamwork makes the dream work. Cheesy, but true. Collaborate and join up with like-minded people to make things happen. We are stronger together. Put your heads together and make magic happen.
This card is telling Shavonne that it's okay to do the seeking for a spiritual home alone, but the point of the search is to find a connection to a community. We are social animals, and we can't live in a vacuum. Whether Shavonne decides to volunteer or engage in her spiritual community, she needs to be out there.
Card 5: The Hierophant
What pulls you forward?
You may need to find a teacher or counselor to unlock the mysteries. Remember that you are not always the smartest person in the room. Every person has something to teach you. You have to study—people, nature, feelings, whatever—so you can open the world. The world doesn't open to the lazy, and with immersion in a subject, you can begin to own it.
This reminds Shavonne that she has a lot to learn. She needs to break open her mind and her heart and connect with other people. There are so many paths to learn from and teachers to find. She is interested in Reiki and tarot as well as becoming more politically active. These are pursuits that will enrich her life and help her grow her community and her spiritual self.
Card 6: The Devil
Who or what in your life can assist you with this journey?
Listen, you know what your demons are and what tempts you. Identify your personal devil and be aware of it. Don't let it get the upper hand. Just because you can sit on your phone all day doesn't mean you should. If you do, there will be consequences, and they'll suck pretty hard. If you know that a thing/person/cyclical thought process isn't good for you, put it down. You are a precious, shining treasure, and that thing/person/cyclical thought process is dimming your shine. So quit it.
The Devil stuck his head in to remind Shavonne that the religion of her youth does not equal the spirituality of her now. She is smart and open to education. She can create a spiritual practice that fulfills her and drop the guilt and shame that she learned in her past.
Card 7: Four of Pentacles
Who or what in your life can assist you with this journey?
Hold on to those things that belong to you and be generous when you can. Be sure that you have everything you need before you share. Be sure that you come first. This card also indicates that you should keep an eye on the things that belong to you. Watch your money. Watch your possessions. Be safe.
Shavonne eliminated her spiritual life and filled the space with things that could go away tomorrow with no trace left behind: Facebook, video games, and her phone. These things are time-fillers. This is not to say that I'm judging that they exist, but they shouldn't take up the lion's share of your day. The return on investment isn't great. The thing that can help her with this journey is consciously turning away from the mundane and turning toward the spiritual.
Card 8: The Fool
What choice can you make right now to get closer to your true path?
Jump! Take a chance. The universe will catch you. Be bold. Jump even if you're afraid. If you keep moving forward, you're winning. Even if you're only moving by inches. Chaos behind me, and probably chaos ahead. The power of being wholly in the present, though, is what gives us the ability to say "okay then" and move forward anyway. Yes, we've screwed up. Sometimes in the most spectacular fashion. So what? We still have today and tomorrow and all the rest of our lives. Let's decide. Let's stop defining ourselves by our lowest moments and focus on the shiny parts. Let's jump.
If Shavonne starts feeling stuck, she can remember a number of platitudes, such as "it's not the destination, it's the journey." She needs to remember that she is her own spiritual guru and that she can lead the way in her studies and adventure. She doesn't need a guru or a life coach or even a tarot reader (egad). She just needs to choose a direction and start exploring.
Summary
Shavonne needs to find a spiritual home. Reaching out to a community leader and testing out a few places that align with her soul will do the trick. She also needs to be more curious about the world around her.
Spirit Reading Example #2: Aidan
Aidan is a young, single professional. They were involved in a church when they grew up but felt very restricted by it. After they outgrew it, they didn't put anything else in its place and have been thinking about how to fill that space.
Card 1: Temperance
What can you hold on to?
Balance in all things. Mind, body, and spirit. If one is out of whack, the others will soon follow. I made a checklist in my head that I follow every day to find balance. Have I tended to my altar? Did I move my body? Am I writing or reading or having on-fire discussions? These things keep me balanced. Find your things. Also, stop letting your brain feed you a litany of have tos and calm down and do one thing at a time. Simmer down.
Aidan's life is pretty balanced—usually. They managed to keep their work and emotional life so full that they didn't notice that their spiritual life disappeared somewhere during the last decade. This is why they chose the spirit reading.
Card 2: King of Swords
What gets in your way?
You don't always have to be nice. You just have to be honest with yourself and with others. Move forward with focus and take no prisoners. This card reminds you to get your ducks in a row or the ducks will be fired and replaced with better, stronger ducks. Not even ducks, geese. Highly trained, efficient attack geese. Get to work—it will be worth it.
One of the distractions that Aidan runs into is that the only "church" experience they've ever had was one that they didn't like. At all. They weren't invested, the message didn't resonate, and they didn't agree with most of the things that they heard in church.
Card 3: Knight of Pentacles Reversed
What gets in your way?
If you are too "nose to the grindstone," you're going to wear that nose right off your face. You have got to look up and around. You have to stop the sleep/work/eat/watch TV pattern that you've fallen into. Snap out of it and start looking around at your life. Are you happy? Are you bored? Figure that out. Also, stop looking at your goddamned phone so much.
The other distraction that Aidan hit was that the spiritual side wasn't a priority until they decided to make time for it. They're very busy and very efficient, and have been very good at efficiently ignoring their spiritual self for about a decade. It was working just fine for them until they started paying attention.
Card 4: The Tower
What pulls you forward?
Everything is going to fall down. It needs to. This doesn't mean you have to be buried under it. Sometimes things need to fall so new things can grow. Batten down the hatches, and collar those new emotions that come with change and inspect them. Make sure they're yours and you're not borrowing stress. Sometimes you need to set the past on fire and let it light the way to the future. It's not comfortable, but it's a lot more comfortable than cuddling up with decay. Get in right relations with the things that serve you so you can move forward with alacrity.
Aidan is ready for their life to change, and they're actively doing and saying things that will bring change. The act of getting a tarot reading invites change. Conversations with friends. Looking up stuff online to revisit later. The unrest is growing. They're ready to make things happen. This quest for truth and change is one of the strengths that Aidan brings to this journey.
Card 5: The Queen of Pentacles
What pulls you forward?
The Queen of Pentacles welcomes you into her lap for a snuggle. Comforting, loving, and compassionate. She is calm wherever she is because she chooses to be there. She doesn't overreact. She doesn't lose her balance. She decides how she's going to feel, how she's going to act, and who she's going to allow in her life. She has the BEST BOUNDARIES EVER. Feet on the ground, steady and reliable.
Aidan is going to start the hunt for a faith community (like we discussed earlier with the Belief-O-Matic). They are looking at it as an adventure and aren't going to settle for a place that is stuffy or staid or boring or that makes them uptight. They're going to look for a place that feels like home.
Card 6: Page of Pentacles Reversed
Who or what in your life can assist you with this journey?
You not only dropped the ball, but you dropped it, kicked it, and then looked away before you saw where it landed. No good, man. You've got to get back on track. If you're disillusioned or disheartened, it is absolutely okay to take a small break, but then you need to realign yourself with your goals and try again. No one is going to chase your dreams for you. Move it.
Aidan's biggest problem will be that they'll come out of the gate ready to go on this spiritual journey and then falter and then forget. They need to have a buddy system in pace. Aidan and a friend—a spiritual Sherpa, if you will—should go to interview and experience several different places of worship, and they should take their friend to brunch afterward. They need to stay accountable.
Card 7: Three of Cups
Who or what in your life can assist you with this journey?
Find your community! Do what you can to surround yourself with like-minded people and enjoy the companionship. Lean on them, be silly with them. Call your bestie right now and make plans to hang out. It's good for you. Sometimes you have to dance because the real world has gotten so dark that if you don't chase down a spark and make it grow, you'll become part of the darkness. Don't join the darkness.
Another support they'll have is the warm and welcoming folks that they'll meet on the way. They're going to visit a lot of places of worship and are a little afraid of meeting so many new people. Even though not every place will be the one for them, they'll meet amazing people who are truly looking out for their best interests.
Card 8: Six of Wands Reversed
What choice can you make right now to get closer to your path?
To quote Captain America from Spider-Man: Homecoming, "So, you got detention. You screwed up. You know what you did was wrong. The question is, how are you gonna make things right? Maybe you were trying to be cool. But take it from a guy who's been frozen for sixty-five years—the only way to really be cool is to follow the rules." Cap is right: if you drop the ball, you gotta figure out how to pick it up again.
The choice that they can make right now to bring this desire for a spiritual home to reality is to accept that they need more sanctuary in their life. They need something holy. Something sacred. They put all of this away so long ago, and it's calling for them now. Aidan needs to open themselves to the Divine and see what role it will play in their life.
Summary
Aidan is going to become a spiritual explorer. Much like I did, they're going to take the quiz, grab their spiritual Sherpa, and start planning trips to those places that match their moral compass. They're going to go to at least two places of worship a month until they find one that speaks to them—one with ceremony and tolerance. If they don't find one, they're going to find a charity near them that speaks to their moral compass and will start volunteering every week. Your spiritual path isn't going to knock on your front door and invite you to service. You have to explore who you are, what you need, and where you fit. After that, it's a cakewalk. (Which we totally did in Catholic school. It was like musical chairs, but you won a cake. What's an eight-year-old going to do with an entire cake? Besides go into a sugar frenzy.)
Exercise
The hardest part of all this is to quiet yourself. To sit and be still and see yourself as part of a pattern. You are not alone. When you are truly quiet and at peace, you're able to separate all the noise and clutter of the world from yourself and truly listen to your own thoughts, feelings, and spirit. Whether you are a devout church-goer, an atheist, a spiritual-but-not-religious person—wherever you stand—we are going to quiet your spirit so you can hear what the universe is trying to tell you.
You get to play with fire again! Try to find some flying wish paper. If you can't find that, regular paper is fine. Just be safe and don't burn your house down.
I want you to write a letter to your spiritual self. It doesn't have to be long, but it does need to be specific. The point of this letter is to decide what you want to leave behind in your spiritual life and what you want to welcome in. You can leave apathy behind and welcome passion. Leave behind a childhood religion that no longer serves you and welcome a new place of worship that does. After your letter is finished, find a pencil and summarize your lists on the wish paper, one piece for each list.
Release | Bring In
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apathy | community
negative influences | support
judgement | peace of mind
hate | friends
When you're finished, crumple up the wish paper, straighten it out, and then roll it into a chimney shape. Put the paper onto the wish paper burning platform or other fireproof surface and light it. The paper will blow away and take your concerns and hopes with it. (If you used regular paper, just burn it and let it go. Seriously, don't burn the house down.) Please release the things you want to leave first. After they blow away, sit in silence for a few minutes and just breathe. When you feel it's time, do the same with your welcome paper and send your hopes out to the universe.
Homework
* Who do you need to speak to in order to start your new spiritual journey?
* How will you eliminate the distractions we've identified?
* What specifically will your new journey do to make your life the way you want it?
Resources
One of the most valuable truths that I can tell you in this book or in life is that it's just for now. Whatever you're going through, whatever the day brings you, I assure you it's just for now. On one of the darkest days of my life, inspired by Much Ado About Nothing, I got the lines "but then there was a star danced, and under that was I born" tattooed on my forearm, where I couldn't help but see it every day. It took me some time to remember who I was and that all of this was just for now.
Places of Worship
Most places of worship welcome seekers. You can set up an appointment with the priest, rabbi, minister, imam, pujari, monk, or priestess and bring a list of questions. Do your homework first and make sure that you're dressed appropriately. Even if these beliefs are not yet your beliefs, you need to be respectful of folks in their own spiritual space. Ask questions about the community service they do, what roles you could take on, and how closely they align to your moral compass.
Online Connections
There are a great many places online in which you can find spiritual communities. Often these can be wonderful places to meet new people and learn new things. Do use your common sense online, though. Just because someone purports to be spiritual doesn't mean they're not a snake oil salesman. Don't send people money. Don't share your personal information. Don't make promises or engage deeply until you're 100 percent sure you can trust who is on the other side of the internet.
Become a Student of Religion
Even if you aren't religious at all, you can learn so much from studying the different faiths of the world. I started when I was a kid by learning about Greek and Egyptian mythology. This led to learning about Islam, which led to Judaism, which I kind of knew about because of Catholic school but not really. The more I learn about religion, the more I know about those people who practice it. If we are truly going to connect to our brothers and sisters, we need to see their perspectives as well as our own.
Volunteer
One of the best ways that you can become connected to your community is to get involved. Research your neighborhood or your city to see where the needs lie. Is it the food banks? The animal rescue organizations? Everyone needs a hand, so assess your skills and interest and see where you can help out. Bonus—by working in your community, you can get to know people in your community.
[contents]
Chapter 8
Additional Tarot Spreads for Clarity
Okay, so let's say that you did the bracket and figured out which of the five element readings you needed. You did the reading. Aaaaaaand you still don't know what to do. This is okay. Life is complicated, and sometimes you have to get to the core of the issue before you can address it.
We're going to do an example reading from beginning to the end of the process that might help, and I'm sharing three additional spreads to help parse out some steps you can take after your elements reading to clarify.
Water Reading for Ophelia
For the first reading, I didn't have any volunteers, so we're going to borrow a lady from Shakespeare and see if we can help her out.
Ophelia wants to figure out why her life is such a sodden mess. Ophelia's situation has gotten complicated. Her boyfriend, Hamlet, comes back from abroad all moody and distant. I mean, his dad has just died (under rather suspicious circumstances) and his mother remarries pretty quickly (to his uncle, no less). You'd expect your partner to lean into you after a trauma like that, but instead he's started wandering the halls of the castle, talking to himself. Her position at court is therefore at risk because if they're going to break up (because it certainly feels that way and he's acting totally weird), she's not going to be in the royal family at all. Also, her dad is always working, it seems, and she's on her own a lot.
Now, should Ophelia have ordered this very anachronistic book from Ye Olde Llewellyn Publications and had completed the bracket, she would have found out so much about herself. It turns out that her emotional self needs attention right now. Her home is secure (if drafty), her material needs are met, she stays in good shape by swimming (sorry), and she spends a lot of her downtime in the library, so she is intellectually fulfilled. A water reading it is (heh).
She does the water reading and finds that her mother's absence in her adolescent years has caused her to seek out attention from anyone around her. She tried to adopt Gertrude as a mother figure, but Gertrude has been distant. This is disappointing, as they both have a lot in common and would seem to be a good pair. Alas, their only connection is Hamlet, so when that's gone, Ophelia is adrift. (I'm so sorry.)
Since she can't fix her mother problem, what can she do? Surely the cards have some deeper advice to help her out before she starts drowning in emotions. (I can't stop. Please send help.)
One of these two readings will help her simplify the problem further. If you do an element reading and come to an impasse, you can submerge (again, I'm so sorry) yourself in one of the three following readings to get some clarity.
Each of them will serve the same purpose, and whichever you choose is entirely up to you.
Three-Card Spread
| Card 1: The problem in front of you.
Card 2: What needs to change?
Card 3: How do you change it?
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Three-Card Reading Example: Ophelia
After doing the water reading, Ophelia found that she had a lot of emotional stuff that was still hanging out there. It was making her angry and frustrated, and she's not sure what to do.
| |
---|---|---
1 | 2 | 3
Card 1: Queen of Swords Reversed
The problem in front of you.
I think that women have a bad rap when it comes to gossip. I think that men gossip, too, but our society allows them to criticize other people and situations without having to cloak it in secrecy or meanness. This is the way it is right now, though, and you're being a gossip. Don't use your words to hurt—use them to help. If you have some criticism to hand out, do it with your head up and meet their eyes. Don't be sneaky about it. That says more about you than it does about them.
Rather than turning on other people, the Queen of Swords reversed tells us that Ophelia's first instinct is to turn on herself. After all, she is in a patriarchal system where women's only currency is their virginity and ability to breed. She's been raised her entire life, therefore, to strictly curtail herself and any emotional overwhelm that she might feel consumed by, as this would knock her resale price down.
Card 2: The Hierophant
What needs to change?
You may need to find a teacher or counselor to unlock the mysteries. Remember that you are not always the smartest person in the room. Every person has something to teach you. You have to study—people, nature, feelings, whatever—so you can open the world. The world doesn't open to the lazy, and with immersion in a subject, you can begin to own it.
What needs to change is that Ophelia doesn't have the tools she needs to fix her problems. She is sorting through so much right now and needs a professional to jump in and help sort through it. Ophelia needs to find a counselor who does have the tools and can show Ophelia all her options and how to wade through them (that one was an accident, I swear).
Card 3: Two of Pentacles
How do you change it?
You often get this card when you're keeping your balance well amid rough weather. Keep light on your feet and don't lose sight of your goals. Regardless of how crazy the world can be, as long as you maintain your balance, you'll be just fine. Remember, the Buddha said be kind to everyone. You are part of everyone. Don't involve yourself in other people's issues. Just mind your business and reach out if your business involves getting assistance.
Despite her rigid surroundings and limited upbringing, Ophelia is an independent thinker. She is still very strong and independent, but this doesn't mean she can't ask for help. There is no shame in getting counseling or medication to help with mental health issues. There is nothing wrong with asking for help. Ophelia will be able to keep her balance and maintain her strength without sacrificing her mental health. It's okay to ask for help.
Summary
Luckily, a traveling monk is also staying at the castle. He notices her lingering gaze on the brook and asks if she's okay. She breaks down and admits that she feels trapped and that she is a pawn in a game of men and can only see one way out: suicide. The good monk asks if he can disguise her as a nun and help her escape in the dead of night. She arrives at the monastery's adjacent nunnery and finds that the sisters pursue studies in history, literature, and astronomy. She lives out the rest of her life in the company of strong, independent women who have conned all the patriarchy into thinking that nuns are submissive. Suckas.
The Bridge Spread
I learned this spread from my friend Beth Maiden of www. little
redtarot.com, and it's just perfect for this "after the reading" reading. If you get to a place where you feel like you know the plan but don't know how to get started, the bridge spread will work beautifully for you. We're going to use my buddy Ramon as an example.
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Card 1: You where you are right now, at the apex of the bridge.
Card 2: The shore you left behind. This card represents the past.
Card 3: The journey's lesson. How have things been so far?
Card 4: The river. Unconscious influences on you now, what lies in your subconscious, and what your intuition wants you to know.
Card 5: Your next step. Here's where to focus in the short term.
Card 6: The shore you are heading for. This may be an indication of your overall trajectory, or if you have a specific goal in mind just now, this card speaks to that.
Card 7: A guiding star: A source of support to guide you.
Bridge Reading Example: Ramon
Ramon is having trouble figuring out his work situation. He did an air reading and knows that his current job isn't working out. He knows that he needs to go back to school but isn't sure if he should change jobs before or after school. He knows that he needs a change but isn't sure the best way to get there.
Card 1: Seven of Cups
You where you are right now.
If wishes were fishes, you'd never go hungry. If you have dreams, take the time to write them down and make an action plan to make them happen. It's a vital place to visit, but you can't live there. This is the card that shows you where you could go. You need to clarify your dream, create a plan, and manifest the hell out of it.
Ramon has been daydreaming about going back to school for so long that it seems like an unattainable goal. Where he is right now is a little disheartened. Still hopeful, but it's been so long.
Card 2: Three of Swords
The past.
Ouch. Pain. Like, searing, heart-falling-out pain. The thing about pain this severe is that it's purifying. That's literally the only good thing about this card. Sometimes you have to feel all the pain before it will go away. Remember that holding pain inside can cut you to pieces. Releasing it, even if only to yourself in a journal, is one of the most amazing things you can do. If you don't want those words lying around, write them down and burn them.
Ramon has had a series of awful jobs, at which he's had to deal with homophobia, racism, and harassment. Each time he finds a new job, he makes a little more money but eventually will run into the same crappy behavior from coworkers. The three swords represent heartbreak—no wonder he's so anxious about moving on.
Card 3: Four of Wands
The journey's lesson.
You are on the path to success. You are pointed in the right direction. You can pick up more tools on the way if you need them, but I tell you it's time to start walking up the hill. Do not stop. Do not get to the end of your life and realize you didn't finish that book because you were screwing around on Facebook. Do your Job. Not your Muggle job, your real Job. The one you were put here to do.
The Four of Wands traditionally shows four wands tied together with ribbons at the top, like a bower or a chuppah. You can see past the bower to a path that leads up a hill to a beautiful home. With each job, Ramon gets a little more money, a little more confidence, and a little closer to his goal. The journey's lesson for Ramon has been that when he expects better, he'll receive better. He didn't just happen to find jobs that paid better, he expected better from himself. His confidence has grown, and he would no longer settle for the same stuff he did before.
Card 4: Queen of Pentacles
What your intuition wants you to know.
The Queen of Pentacles welcomes you into her lap for a snuggle. Comforting, loving, and compassionate. She is calm wherever she is because she chooses to be there. She doesn't overreact. She doesn't lose her balance. She decides how she's going to feel, how she's going to act, and who she's going to allow in her life. She has the BEST BOUNDARIES EVER. Feet on the ground, steady and reliable.
Ramon's intuition wants him to know that he deserves good things. He deserves a job where people respect him and where he can make a living wage. The Queen of Pents is solid and reliable, and so is Ramon. He's a good person with a good heart, and his intuition is screaming at him to believe in himself.
Card 5: Nine of Wands
The next step.
Do not let the same people who messed with you before do it again. Protect yourself. Remember that people always tell you who they are. Your job is to listen, believe them, and then take care of yourself. You have done this before, you know. You know how to set boundaries and you know how to enforce them, but knowing how to do it and actually doing it are two very different things.
Ramon has challenges ahead of him, but he's so close to achieving his dreams. He needs to find a job that allows him to attend school. He can take online classes or evening classes. Lots of people do it, but Ramon has never considered himself able or worthy to do it. It's not going to be easy, but it's absolutely going to be worth it.
Card 6: Ace of Cups
Where you're headed.
Your cup runneth over with blessings. Be sure to be grateful. Happiness is afoot. Blessings abound. Accept them, acknowledge them, and say thank you. The universe conspires to shower you with blessings. Let it. If you're not in a place to say thank you, start with "at least." At least you have power. At least you woke up today. If things are solid, kick into the thank yous. Thank you for the little things and the big things, too.
Ramon is headed toward fulfillment. He just needs to get there. Everything will be okay.
Card 7: Ten of Swords
A source of support to guide you.
Oof. You just got stomped on quite a bit. Do us all a favor and stay down for a while. Watch Broad City and eat comfort food. You'll stop bleeding eventually. The great part about this card is that you have actually reached rock bottom. Congratulations. Only one way to go from here.
The Ten of Swords isn't always interpreted in a positive way, but I see it as excellent news for Ramon. This is as bad as it gets. Working at this "better than the old but not as the good as the dream" job will help him pay his bills while he attends college. He might even be able to find a job in the field that he's interested in. He needs to understand that he's never going to go backward.
You are Here Spread
Sometimes you will know exactly what you need to do to fix things but have no resources at all to begin the process. You are weary in mind, body, or soul or are otherwise stuck, and although the impulse for change is there, the ability just isn't. This is okay. I have a reading for that.
| Card 1: Where you are.
Card 2: What's holding you up.
Card 3: What you're dragging in from the past.
Card 4: What you should reach for.
Card 5: What you can hold on to to pull yourself forward.
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You Are Here Reading Example: Ginger
Tarot readings are not just about telling the future. Actually, they're mostly about not telling the future. They look at what you're dragging in from the past, what you're stewing in in the present, what you need to fix, and what you need to stop touching. (It's never going to heal if you pick at it!) I have a good friend—Ginger. This friend has had a rough few years. Death of a loved one, breakups, loss of employment, and painful and unexpected disability. It's been hard. Ginger is at a place right now where looking into the future won't help a bit. I can tell someone that everything is going to be okay until I lose my breath, but if you're that beat down, it's just empty promises until it happens, you know? If they can't believe in themselves, they're not going to believe that I believe in them, either.
We did a spirit reading but got the sense that it was just too soon for her to think about a new spiritual path. She was struggling thinking about tomorrow. We decided to scrap the spirit reading for a while and use this one instead.
Not looking into the future. Not trying to figure out what's next, just—What do I have? Where am I? How can I work with what I have?
1
| |
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2 | 3 | 4
5
Card 1: Ten of Swords
Where you are.
Oof. You just got stomped on quite a bit. Do us all a favor and stay down for a while. Watch Broad City and eat comfort food. You'll stop bleeding eventually. The great part about this card is that you have actually reached rock bottom. Congratulations. Only one way to go from here.
Ginger needs to understand that this is bad, but it's not forever. She's starting to get so used to bad stuff happening, that she thinks that it's normal. It's not normal. It's a string of bad luck. She should try to realign her thinking to help pull herself out of this depression a bit.
Card 2: Three of Cups
What's holding you up.
Find your community! Do what you can to surround yourself with like-minded people and enjoy the companionship. Lean on them, be silly with them. Call your bestie right now and make plans to hang out. It's good for you. Sometimes you have to dance because the real world has gotten so dark that if you don't chase down a spark and make it grow, you'll become part of the darkness. Don't join the darkness.
She has a great support system of people who actively adore her and would do anything for her. She's starting to feel guilty for asking for help lately, but she's been in need. She should remember those times that she supported her friends. She has always been there for her people and needs to let them be there for her.
Card 3: Four of Cups
What you're dragging in from the past.
Don't settle. Don't pout either, but make sure that this is what you really want. Fussy and ripe for hissy fits and eye rolling. Apathy is easy to ignore. Don't wanna. Don't want to change, don't want to take my medicine, whatever that medicine may be. You have to look deeper than the apathy. Why don't you want to change? You know these behaviors make you sad, and yet you won't put them down. Consider the why of holding on to the listlessness and figure out how that's paying off for you. There's a payoff for "meh." Whether it's the reward of not failing (you can't fail if you don't try) or the reward of being comfortable, decide if it's worth being stuck.
Ginger has never been lazy a day in her life. She's worked since she was fourteen and has always supported herself and her kiddo. Since the disability and losing her job, she feels lazy. She's forgetting that this is just for now, and that her health comes first. The shame that comes with not being able to work is part of her history and taking care of herself is not.
Card 4: The Hierophant
What you should reach for.
You may need to find a teacher or counselor to unlock the mysteries. Remember that you are not always the smartest person in the room. Every person has something to teach you. You have to study—people, nature, feelings, whatever—so you can open the world. The world doesn't open to the lazy, and with immersion in a subject, you can begin to own it.
Her doctors have a good plan for her recovery, and even though the healing has been slow, she has made progress. She should continue meeting with her health team and doing her occupational therapy, and most importantly, she needs to keep track of her progress in a journal. Having this evidence of progress is going to help her feel as if she's accomplishing something (and she totally is).
Card 5: Six of Cups
What you can hold on to to pull yourself forward.
Seek joy and don't stop until you find it. Be sure that you're taking time to live in the present. Warmth, happiness, and silliness abound. Remember to accept blessings given without being paranoid. Innocence is not always a bad thing. Trust that the universe calls you beloved. When we were small, we didn't chase happiness—we just found it.
Ginger needs more silliness in her life. Even though she's at home a lot, she can still invite friends over to watch movies, enjoy dinners together, and call her out-of-town friends. She can take up a new hobby to keep her hands busy and take time to read all those books she's got piled next to her bed. Her life will be as rich as she makes it.
Additional Exercises
I have an anxiety disorder because I never learned how to properly deal with stress, so my brain instead built up this elaborate catchall of "what to do when stuff goes sideways." My brain instructed my person that if I clean the entire house at 2 a.m., I definitely won't get laid off. If I eat a salad instead of that fantastic taco, I will suddenly feel better about my body and myself. It's almost like a catch and release program for my particular neurosis. It seems that if my family or friends are under stress or undergoing loss, I feel the need to pile more things on myself to create some sort of hyperaware force field around them. If I stress out, everyone will be just fine, right?
Well, no. That's actually really stupid, but it's the way that my brain works. Whether that's a fixed neurological pathway or really ingrained behavior, I'm not sure. What I do know is that if I need it to stop—that constant braying of fix it, fix it, fix it—then the best thing I can do is sit down.
After I sit down, I try to get back to the basics. What can I control? What is clearly and wholly outside of my control? Usually, the problem is like 99.9 percent out of my control. It feels like more, though, right? It feels like we're to blame for everything that goes sideways in our world. What ego. What absolute self-centeredness. What I can handle is me. I can control my tarot cards, too. I can handle my reactions and emotions to things. Since I'm a cardslinger, I can use my cards to help me through.
I decide which part of my life this particular problem belongs to. Since I am a witch, I use the elements to do this. Witches absolutely love their elements, by the way. They bring us right back to basics.
Home/Earth
I mentioned this earlier: if you can draw a map of your home and then draw smiley or frowny faces on it, it helps immensely. You can sit in the middle of the room and take the emotional temperature of the rooms in your home. Do you hate the paint? Does the furniture remind you of your ex? Do you hate your whole house, and it's time to move? Touch base with yourself emotionally to make your house into a home. Pull a card to tell you which direction you should move in. Did you pull Temperance? You might need some balance between your things and your partner's things. Did you pull the Devil? Maybe it's beyond rescue and time to move. Even if this involves just clearing away things you haven't touched in a year, this is your home. It's supposed to support you.
Mind/Air
Are you frustrated or disengaged at work? Some folks are the most destructive when they are bored. That "do something" energy spins around, knocking over lamps and messing with relationships. The best thing to do when you're anxious or bored about work or even to just refresh your mental acuity is to visit your brain objectively.
Hello, brain. How's it going? Ooh, not so great, huh? You've been doing this work for four years, and it's become completely tedious? And you work with a bunch of a-holes? How frustrating. Tell me, then, the following things:
1. What do you really like to do?
2. Do you need more education to do that?
3. If you're doing what you want, can you do it somewhere else?
I encourage you to pull out your cards for this exercise. Let them confirm and tease out what the problem is. And then—and this is really the important part—do something about it. Change your job. Go back to school. Find a new position. For this reading, I recommend three cards: where are you now, where you need to go, and how you do it.
Whatever you do, remember that the whole point of it—over everything, really—is to be happy. It's difficult to achieve happy when you spend forty-plus hours a week at a place you can't thrive in.
Body/Fire
A simple exercise to get to the base of your physical self is one of gratitude. Draw a map of your body and include all the scars, bumps, wrinkles, scuffs, and stretch marks that you can.
For each event—for that's what all of these were, events—I want you to pull a card telling you how that event changed your life, for good or for bad. That childbirth, that broken leg, those fifty pounds gained/lost/gained again. I want you to ask the tarot one question: How did this shape me?
Pull a card for each bump and bruise. Each scar and ache. See how they shaped you. Do you pull the Empress for your stretch marks? The Strength card for your surgery scars? They made you into something different, and sometimes it's difficult to remember to assimilate the changes into who you are.
You'll find that even the rough patches were for a cause and that those events led you to who you are today. Say, "Thank you, body, for getting me here. For protecting me. For coming out on the other side." Say thank you and safely burn the picture, letting the ashes fly away with your gratitude.
Water/Heart
A lot of times the emotional overwhelm is either from an overabundance of emotional drama or from not knowing what your baseline feels like. As I mentioned earlier, I have an anxiety disorder, so my baseline runs high. I can tell (or my partner can tell) if things are going wackadoo on me because I exhibit certain behavioral tells. I have circular thinking. I get upset over small things that normally wouldn't matter. I fidget a lot. When these things start up, I have a seat with them and pull out the suit of cups.
The cups are aligned with emotion, and I use them to tell me what is a big deal... and what is not. What I can control and what I cannot.
Make a list of four or five stressors, write them down, and pull a card for each. You can put the cards back in the deck after you've written which card goes to each stressor. When I do this, I ignore the standard meanings of the cards and go for the numbers. If I pull one through five, things are manageable, or I can put them down. If I pull six through king, I need to attend to them. After I put them in order (crisis/king first), I get to work.
These rituals are simple but powerful. It's important to remember, however, that the power comes after you start the work. It isn't enough to recognize that there is a problem. You also have to do something about it.
Soul/Spirit
A lot of my clients are surprised when I ask them what their relationship with spirit is. I suppose it is an unusual question. I see matters of religion and spirituality as hugely private, and so don't often "go there" with my clients.
I have found, however, that I can tell if there is a lack of spiritual alignment. You can have the best job, the best partner and take care of your physical self in an amazing home, but if you don't have a spiritual element to your life, there is something lacking.
This doesn't mean go to church, or temple, or synagogue every week. This means find a practice outside of your day-to-day, outside of you. You can volunteer, go for a walk in the woods, spend some time with friends once a week. Whatever it is. If you do decide to go for a more traditional spiritual exercise, though, try this:
Pull one card for each question, then listen to that card.
1. Cards, what can I focus on right now—today—to enhance my spiritual self?
2. How can I connect more thoroughly to my community?
3. What can bring me peace?
Remembering that a connection to "other," to "outside," is a spiritual connection. You can find answers in books, in a sacred space, or in a walk in the woods. You can even find it across the table from a dear friend, talking smack and cracking jokes. Spirit can be structured or loose, intense or relaxed. As long as it exists, you will see a difference in your life.
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Chapter 9
Self-Care
I talked a little at the end of each chapter in the resources section about things you can do that will help you pull the energy of the readings in and help it linger. These things are mostly pointless if you don't believe that you deserve them. If you are the kind of person who constantly takes care of other people and then are too tired to help yourself. If you work seventy hours a week and can't remember the last time you got a pedicure. If you (like me) feel guilty when you spend money on yourself rather than on a responsibility—we need to talk.
You bought this book because you want a change in your life. I could wax poetic and tell you that if you don't love yourself, you'd be hard-pressed to find someone to love you, but come on. That's not always true.
What is always true is that if you don't take care of yourself, you can't give 100 percent to your family or your job or your friends. If one of these is the reason that you can't take care of yourself, then truly understand that you're shortchanging them as well as yourself. When I was a social worker, the caretakers would often end up worse health than they started in. Sometimes worse than the person they were caring for! Heart attacks, hypertension, obesity, and diabetes. Caretakers would concentrate so much on making sure that everyone else was okay that they would completely wreck themselves.
When you're totally invested in other people or in your job, you forget to make your house into a home. You forget to buy new books and play with your dog. All the pleasure can leak out of your world if you're not careful. You could look up one day and see that years have passed you by and you forgot to pay attention.
Paying attention. Being present. These are the overall lessons that I'm trying to bring forth here. If you live in the past, you'll miss the present.
Let's think about the ways that you can take care of yourself. Not the "treat yourself" ways, such as shopping or manicures or buying a new toy. Let's talk instead about ways to introduce you to self-care in your daily life.
Giving yourself space to breathe and to get things done is so important. Rushing around every morning only to rush when you get to work and then rush again when you get home is a complete nightmare. Consider simplifying your morning routine by starting some of your tasks at night. Lay out your clothing. Make your lunch and breakfast. Look at your time as if every minute of the day needs to be scheduled. This sounds counterintuitive, but I've found that if I allot a certain amount of time for things, I'm less likely to get distracted and lose my place in my day. Getting my kids ready for school takes forty minutes. Driving them to school takes twenty minutes. If we leave by 7 o' clock in the morning, we need to wake up at 6:20. If I wake up twenty minutes before the kids do, I can let the dogs out and meditate and have my breakfast alone. Delightful.
Consider releasing those friends who drive you insane. You know those folks who blow up your phone when they're having a bad day and it's crickets the rest of the time? The ones who start fights on social media and turn every bad day into a personal tragedy. You can make a chart logging the things they ask of you and the things they ask about you. Make a tally mark every time they ask you to listen, help, and let them vent. Make a tally mark on the other side whenever they do the same for you. After a month or so, look to see which side has the most—or if they're even. If the energy in doesn't equal the energy out—or nearly—you don't have a friend. You have a client and you're an unpaid counselor.
If something makes you miserable, don't do it. I'm allergic to grass. I'm allergic to pretty much every living thing in Missouri. After my divorce, I had a yard and no lawnmower. I tried to do it once, and nearly ended up in the hospital with an allergy attack. That won't do. I also hated it. I found a kid in my neighborhood who cut the lawn every week for twenty dollars. Fantastic. I don't fix computers, but I have a friend who does and who works for dinner and beer. Figure out if these things in your life are making you miserable and, for God's sake, quit doing them. The same goes for family obligations, clubs that you belong to, social media groups, social media in general, and "traditions" that you've never really enjoyed. Just stop. I promise, the world won't stop spinning.
If you have to, schedule time in your week to do nothing. I can't tell you how important it is to just be still sometimes. I can remember sitting in a tree at my grandpa's farm, reading Laura Ingalls Wilder books and watching the sun set. I can't remember capturing a moment of relaxation like that in my adult life, but I'm going to keep trying.
Unplug your brain and watch a movie or TV. When I'm writing, I watch superhero movies pretty much nonstop. I love them, they make me really happy, and I can do something else while I watch because I've seen them a million times. There is nothing like snuggling with your kiddos on the couch and watching The Muppet Show. It's entertaining, it's easy, and it helps you disconnect for a while.
Speaking of TV, please stop watching the news all the time. It's easy to get riled up by the unrelenting media frenzy. If you need to stay current, choose a news website like BBC or Al Jazeera. Check in once a day and then just stop. Ever since the OJ Simpson trial, news has become entertainment much in the way that the gladiators were entertainment to the Romans. Truth and beauty have gone by the wayside, and most news outlets are more concerned with selling you fear than they are with sharing the actual news. Don't be there for it. Walk away.
What all these things have in common is that they have to have space and time dedicated to them. Sacred space for self is such an overlooked medicine. Sometimes it isn't enough to take an antidepressant. You have to excise those things from your life that make you miserable and inject those things that light you up.
Here are some more little things that you can introduce into your week to make it a week worth living: read a book, take a walk, meditate, go to the library, go to a concert, talk to a stranger, balance your checkbook, play a video game, play a board game, pet your cat or your dog, hug your friends, have dinner all by yourself, have dinner with a bunch of friends, learn to cook something new, take a class for fun, learn a new skill, plant a garden, fix something in your house that's broken, stretch your body. Breathe.
Another tarot exercise that is valuable for folks is to meditate on a single tarot card. If you are seeking peace in your life, choose the Star card and look at the picture. Look at the graceful bend of the woman's back and the cool water she pours out. Look at the sky that surrounds her and the grass and flowers under her. Observe the tree behind her and the streams of water that are spreading on the grass. Do you see any symbols in the card? Are there repeating shapes? Make notes about the card and then put yourself into conversation with the character. Ask her a question. See what she has to say. Put yourself in her position. Why are you there? What are you going to do next? Do you dive into the pool or sit in the grass? Is your work here finished?
You can also integrate this card into your life by drawing the Star or your impression of the card, cooking something that suits her, or journaling about your conversation. One of my friends, Karin, made tarot soap when she was learning the cards. The major arcana soap was pink with sparkles and smelled like tulips.
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Conclusion
These readings are hard. They really are. They're asking you to reach down into yourself and face those things that you've been ignoring. I think that the reason they're so difficult is because if we fix that thing—that one thing that is making our love life seem impossible and forces us to challenge ourselves at work—we get pulled out of apathy and comfort and we're forced to face new challenges. Doing these readings won't fix your life. They will give you clarity about the parts of your life that need modification.
If you find that the things you want to work on don't fit in a specific reading, see if you can modify it. I had a client who was dealing with a great deal of guilt. We talked about both the water and spirit readings, and we decided that spirit would be the right one. She was able to release that pain and guilt—which had become like a religion to her—and face life without it. Instead of looking for a spiritual community, she was looking for spiritual peace. We modified the reading just a little to make it work for her.
The best part of doing these readings is that it puts you in a position of strength in that area of your life. If you are struggling with mental health issues, and this reading motivates you to find a counselor, stay on your meds, and truly live your life, you're now in charge of your brain. If you can spend six months dating yourself instead of dating the same guy in different skinny jeans, you might find that your standards shift. That you're no longer afraid of being alone, but instead you are absolutely certain that it's only worth dating someone when they're worth your time and energy.
I started doing these readings to help one client. Then I did them for a few more clients, and then a few more. I think that people were looking for a way to use the cards in a clarifying, comforting manner, and I'm so glad that they worked out. I use them myself, when life gets life-y, and even though they make me really angry, they are true—like, capital T True—so I can't ignore them.
I hope the readings bring you clarity. I hope that they bring you comfort. I hope that can laugh at yourself a little and clear away the obstacles in your path with joy and confidence. Please remember that your habits didn't happen overnight, and they're not going to go away overnight. Give yourself time and space to grow and change. This is super important, so I'm going to repeat it in all caps: THIS WILL NOT HAPPEN OVERNIGHT. Do not place unrealistic goals on yourself. Do not do all the readings in one night and try to change your entire life at the same time. One at a time. Do the whole thing (including the homework) and give yourself space to grow and change. Some of this time might be carefully avoiding the reading and the homework for about a month until you're ready to get started. That's okay. It's actually expected. I don't know of one of my clients who finished a reading and then hopped off to make the changes. Give yourself a break, would you? And have faith in yourself.
If you can't have faith in yourself just yet, I'll believe in you until you can.
XOXO,
Melissa
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appendix
Tarot Card Meanings
The deck used in this book is Llewellyn's Classic Tarot. Some other decks that I like for beginners include the Numinous Tarot, the After Tarot, Vivid Journey Tarot, and the Dreaming Way Tarot. For folks with more experience, I recommend the Sasuraibito Tarot, the Spirit Keeper's Tarot, the Slow Holler Tarot, the Pagan Otherworlds Tarot, the Raven's Prophecy Tarot, and the Augenblick Tarot.
If you'd like a book to go along with these short descriptions, have I got a book for you. Kitchen Table Tarot, written by yours truly, an Independent Publisher Book Award winner for Best First Book (bronze!). It's pretty good. Some others are Reading for Yourself by Courtney Weber, Seventy-Eight Degrees of Wisdom by Rachel Pollack, Mary K. Greer's 21 Ways to Read a Tarot Card, The Tarot Coloring Book by Theresa Reed, and Tarot for Beginners by Barbara Moore.
Major Arcana
The Fool
Jump! Take a chance. The universe will catch you. Be bold. Jump even if you're afraid. If you keep moving forward, you're winning. Even if you're only moving by inches. Chaos behind me, and probably chaos ahead. The power of being wholly in the present, though, is what gives us the ability to say "okay then" and move forward anyway. Yes, we've screwed up. Sometimes in the most spectacular fashion. So what? We still have today and tomorrow and all the rest of our lives. Let's decide. Let's stop defining ourselves by our lowest moments and focus on the shiny parts. Let's jump.
The Fool Reversed
Just because you can just doesn't mean you should. Just because the ledge is there doesn't mean you shouldn't look over the edge and double check your parachute. The Fool reversed throws caution to the wind. And common sense, a sense of pride, decency, and intellect. Just blundering stupidly into the next mistake.
The Magician
Have confidence. You have all the tools you need to get ahead. Make it happen. You put in the time. You are experienced. You are talented. You've got secrets up your sleeves and in your pockets and behind your back. Now is the time to act like the expert you are and stop hiding behind false modesty. There is a difference between confidence and arrogance. Find it.
The Magician Reversed
A Magician flipped on his head cannot be trusted. (And his rabbit is totally stuck in his hat.) This is the sleight of hand in conversation that makes you believe that something is your fault, even though it isn't. Presenting half-truths and illusions to get away with something. Stirring up confusion and intrigue.
The High Priestess
Look into the deeper mysteries for the truth. Look within yourself for the answers. This is the card for those with the Knowing of the Ways. Ask your questions. Be ready for your answers. Be honest, because she can see right through you. Be brave, because the truth is hard. Be joyful, because you can absolutely design your destiny with her on your side.
The High Priestess Reversed
Secrets, secrets are no fun. Secrets, secrets hurt someone. You're either in the dark or you're the one who blew out the candle. Someone is making space for dishonesty and tricksiness to abound. Someone is holding room for nefarious deeds, and they might not get caught. The only thing you can see is cloaked in shadows. Don't trust it.
T
he Empress
Warmth, sex, and sensuality. This lady holds the answers in her impulses and intuition. Can you love and love fiercely? Can you promise your children that your only job is to keep them safe and mean it? Can you promise your partner that you will fight for them when they can't fight for themselves? Can you stand up for yourself with integrity and compassion?
The Empress Reversed
Things are not working out. You don't feel safe, loved, or protected. You feel the steady itch of vulnerability and anxiety. There are problems that aren't going away and that you're not able to shift or improve on your own. The connection to yourself or to your partner is worn thin and near breaking. Attend to yourself and your needs.
The Emperor
Time to dot your i's and cross your t's. Be precise. Consider briefly the emotional impact of the decision, and then decide what is practical and smart. Sometimes you have to do the right and difficult thing. Even though it might break you a little bit. The Emperor wants you to be okay but realizes that sometimes the path to okay is painful (but necessary). He is willing to go through the pain to make things happen for the greater good.
The Emperor Reversed
This indicates a puppet master who is a solid core of fear with a thick crust of bullying. Do as I say because I say it. I can do what I want because I'm in charge. The type of energy that makes you wince. This person isn't a leader, they are a tyrant. They aren't strong—they are reactive and bombastic. They aren't respected—they are feared.
The Hierophant
You may need to find a teacher or counselor to unlock the mysteries. Remember that you are not always the smartest person in the room. Every person has something to teach you. You have to study—people, nature, feelings, whatever—so you can open the world. The world doesn't open to the lazy, and with immersion in a subject, you can begin to own it.
The Hierophant Reversed
Those priests who violated trust. Those folks in power who caused the #MeToo avalanche with their callous violations of others. Those predators. Those shameful excuses for people. Those who would use their position for their own benefit and drive it over the people they were meant to protect and lead. Those partners who would dominate the people they are meant to love.
The Lovers
Passion, love, and intense feelings. The important part of this card is where the two hands are almost touching. That hot, electric spark means chemistry is present. Hot damn. This doesn't always last, but it's always memorable. Think of Thelma and Louise or Romeo and Juliet. The magnetism is absolutely there. The chemistry, physics... all the science, really, except for geology, because they're certainly not grounded. Anyway, this is that pull between two people. It's bananas and usually short lived.
The Lovers Reversed
Somebody is cheating on somebody. Somebody has lied. Someone has put their promises aside and has started putting their phone face down all the time and deleting their messages as soon as they get them. Someone is putting off the hard conversation about unhappiness and loneliness and someone catching their eye and has already left the relationship in one way or another. They've got at least one foot out the door.
The Chariot
Use your drive and devotion to get out of the rut. You've got stuff to do. Remember that you're in charge of you. Move forward. Don't stop. The two steeds in front of the chariot will do what you ask, but you have to ask. Otherwise, they'll just start wandering down the paths of habit and take you where you've already been. It's important to focus on the destination and then do everything you can to pull yourself in that direction.
The Chariot Reversed
If you don't believe in yourself and in your goals, you will drive your chariot in circles. You will spin and wander and sit down suddenly, wondering what the next step is. If you have this vehicle for change, whether it be a life decision or a new job or a new partner—you have to really focus on what's good for you and what isn't. Pay attention. You're spinning.
Justice
Things are going to be fair. You might not like the outcome, but it will be even. And it will likely suck a little bit. The weight must be shared equally. The emotions must be removed from the decision, and we can move forward knowing that at the end of the day balance must be found. It's important to remember that fair doesn't mean good or bad. It just is. Justice will consider all your past mistakes and triumphs and will balance the scales accordingly.
Justice Reversed
Patently unfair. Whether it's your treatment of others, their treatment of you, or your treatment of yourself, it's off and slanted and distorted. There are lies here, and there is danger of the disingenuity taking over and covering everything in an oily film that will rot and mold. There is trouble here. Be alert.
The Hermit
Find your quiet space. Shut out the sounds of the outside world. The star in his lantern shines brightly like the Star card. He carries a gentle, soft light with him to find his answers. He wants to go where no one can reach him. If this isn't possible for you, try to find this peace within. Close your eyes. Breathe in and out. Allow yourself to be alone, and in that solitude listen for the voice inside you to tell you what is next. Sometimes what is next is more breathing and solitude.
The Hermit Reversed
It's normal to pull away from folks when you're feeling sad. Absolutely normal. It is not normal to make it a long-term thing. If you find you are isolating yourself unhappily (some people are just fine with solitude!), you should ask yourself why. What is making you push folks away, and what are you seeking in your self-inflicted exile?
The Wheel of Fortune
Either you'll ride the top of the wheel or go to the shadow side for a while. Remember that you're in charge of how you behave regardless of where the wheel takes you. There is power in knowing where you stand. There is value in all these positions. The only constant is change, so if you're being smushed into puddin' by life right now, wait till you feel the weight to start to lift and scramble to get back on top.
The Wheel of Fortune Reversed
Absolutely going backward. Likely falling down and getting run over by your own life in the process. And then the wheel rolls again and traps you underneath it, and you can't reach your phone, and when you finally do, you see that it has also been smashed. This is a bad time. This isn't a bad life or a bad existence. It will pass, but you still need to take it seriously.
Strength
Strength takes many forms. Figure out what yours is and feed it. You have reserves you might not be aware of. Use your strengths—compassion, kindness, humor, resting bitch face, active bitch face. Use your gifts to remain strong—whatever that means for you. Sometimes it means sitting still until you can stand up under your own power. Sometimes it means faking it till you're making it. Sometimes it means saying no and making that a complete sentence. No.
Strength Reversed
Screw it, it's just one more drink. Screw it, I really need some connection and she seems nice enough. Screw it, I'll finish this project late. It doesn't really matter. If you say "screw it" often enough, you're going to be screwed. If you have an elevated view of yourself and see yourself as untouchable or unique, you'll find that you're the only one who holds that opinion. Your spine is for more than holding your head up. It's also for giving a damn about something... anything.
The Hanged Man
Sometimes you need to unstick your tongue from the roof of your mouth, relax all your muscles, and just be. You will not dissolve if you aren't moving forward for a few minutes. Take a beat. Catch your breath. Take time to assess the entire situation. Even though we are seeking opportunities, opportunity is seeking us as well. Sometimes our job is to remain still and calm so that when it finds us, we are ready.
The Hanged Man Reversed
You are your problem. You and your inflated sense of self are getting in your way so often you don't know which way is up anymore. You are rigid and unwilling to look at anyone else's point of view. This isn't going to work out well for you unless you can decide that you're not the only person in the world whose opinion matters. You have got to stop trying to be so perfect. It's making you miss everything because you're wearing gigantic, self-inflicted blinders.
Death
"But I don't want to change." Welp. Tough. If Death shows up for you, you either change or get changed. The "getting changed" part stinks. You don't get to have a voice or choose a direction. You get moved around like a chess piece. If you pull death, the universe is giving you until the count of five to stop dragging yourself around and start moving forward. Remember that you used to be afraid of now, and someday you'll forget that you were afraid of the decisions you'll make tomorrow. Just move forward, certain that you're right, and may you never find that you're wrong.
Death Reversed
This is depression and living an uninspired life. And you're living it so carelessly that you don't even notice how depressed you really are. You've gotten so used to feeling flat and listless that you think it's the new normal. Your status quo has dropped to dangerous levels and you need to get some help. Reach out to a friend to see if they've noticed a change, and then do something to fix it. Right away, please.
Temperance
Balance in all things. Mind, body, and spirit. If one is out of whack, the others will soon follow. I made a checklist in my head that I follow every day to find balance. Have I tended to my altar? Did I move my body? Am I writing or reading or having on-fire discussions? These things keep me balanced. Find your things. Also, stop letting your brain feed you a litany of have tos and calm down and do one thing at a time. Simmer down.
Temperance Reversed
I imagine the angel in this card listing to one side and falling in the water. She is ill. She is disenchanted. She has lost her balance and fallen down hard. Sometimes you see this coming and sometimes it sneaks up behind you and knocks the breath right out of you. You are falling. You have fallen. You are down.
The Devil
Listen, you know what your demons are and what tempts you. Identify your personal devil and be aware of it. Don't let it get the upper hand. Just because you can sit on your phone all day doesn't mean you should. If you do, there will be consequences, and they'll suck pretty hard. If you know that a thing/person/cyclical thought process isn't good for you, put it down. You are a precious, shining treasure, and that thing/person/cyclical thought process is dimming your shine. So quit it.
The Devil Reversed
You have severed the chain that bound you to poison. Whatever your poison is, you've drained it of its power. A divorce, a rehabilitation, or a reckoning. You have decided not to drink the poison and not to self-sabotage any more. This is very, very difficult and you should be very proud of yourself. Stay vigilant, though. Your poison knows your weak spots and will wait for them.
The Tower
Everything is going to fall down. It needs to. This doesn't mean you have to be buried under it. Sometimes things need to fall so new things can grow. Batten down the hatches, and collar those new emotions that come with change and inspect them. Make sure they're yours and you're not borrowing stress. Sometimes you need to set the past on fire and let it light the way to the future. It's not comfortable, but it's a lot more comfortable than cuddling up with decay. Get in right relations with the things that serve you so you can move forward with alacrity.
The Tower Reversed
This can be a very tense situation that feels like it's going to explode at any second. It can sustain itself for years, though. I once gave a reading to a wealthy, beautiful couple with a beautiful house and beautiful children. They were also both addicts, both cheating, and both ignoring their children (and in debt up to their eyeballs). That level of delusion can only last so long. Holding your hands against the tower to keep it from falling will only leave you with bruised and bloodied hands.
The Star
Gentle, gentle, all is well. Allow yourself to be soothed and pampered. The star tells you to rest. Be still and all will be well. Her gentle, cool hand rests on your fevered brow and she tells you to drink. When your life falls apart, sometimes you have the urge to jump back up and get into the fight. The Star tells you to be still. There is no hurry in putting your feet steadily beneath you and rising once again, slowly, to stand. You're not broken, just bent. The branches that bend have marks where the stress occurred, but they keep growing afterward.
The Star Reversed
A feeling of sickness. Of body, of mind, or of soul. The grace that you had has left you and you feel empty and powerless. You need a hand. You need help. This card comes up when the universe is doing everything it can to let you know that help is around you except for dialing 911 for you. You have to reach out to rescue yourself here. No one knows that you're screaming if it's silent.
The Moon
Sometimes you should be afraid. Be sure to get a mental health check and make sure your depression/anxiety isn't lying to you. Walk your path carefully because you are not in charge right now. Listen to your intuition. You can be a warrior when you're strong again. I promise.
The Moon Reversed
I'm afraid of spiders. One time, my dad (who is also afraid of spiders) stomped on one in front of me, and thousands of baby spiders just exploded out of it. Jesus effing Christ, Mary, and Joseph, and all their carpenter friends. I nearly fainted. What this did was give me a new thing to be afraid of and a healthy respect for letting spiders outside. Sometimes your fears get eclipsed by something new and fade into the background. This is okay. They weren't helping you anyway.
The Sun
It's all okay. The sun has a way of shining light into all the dark corners of our lives. All the shadows and mistakes and flaws are illuminated. Those broken bits of you catch the light and make you shine. Be confident that the world is waiting to shower you with blessings. Know that things will be okay even if the situation is ambiguous. Feel the support of your people and of spirit, and share this light with everyone who comes into its warmth. Remember that fire burns away the darkness and that you are a strong and gifted fire dancer.
The Sun Reversed
Eh. Even reversed this is a temporary downswing. This is a cloud passing over the sun. You'll have a bad day or a bad week. You're going to feel a little blue, but it's not long term and it's not deeply rooted. Recognizing the difference between a bad day and depression can be difficult sometimes, but just breathe through it and wait for the sunshine to peek through again.
Judgement
Let the sun shine in and burn all the shadows away. Honesty and truth are here. See yourself in a place of power, just where you are. All those nicks and bruises are part of the tapestry of your life, and they enhance the you-ness. Accepting truth puts you in a position of power. Yes, institutional racism exists. Yes, you should quit smoking. No, he's not coming back. Find a truth that you've been avoiding and make peace with it. Only then can you start to fix it. Your discomfort with the truth tells you where you need the most medicine.
Judgement Reversed
The lack of confidence that comes from being told that you're wrong and believing it. Allowing all the criticism to sink so deeply into you that it feels embedded in your skin and tattooed on your soul. You can't be right because you've never been right so you'll never be right. This self-defeating card is trying to tell you to pick your chin up.
The World
This is the card of the in-betweens. Remember that the end is the beginning and the beginning is the end. If we do not keep growing and moving forward, part of our soul dies. Endings are just the tail end. Beginnings start right after them. You'll find that the less stuff you drag into your new beginning, the happier you'll be. Take time to put things away, even if it hurts. Even if these emotional weights give you comfort, that comfort might be costing you freedom.
The World Reversed
There is no ending. There is no new beginning. There is only the timeless slog of marching in place. Working at a thankless job for years on end with no upward mobility or financial gains. Staying in a marriage in which you have no love, no touch, no affection. Allowing your life to become monotonous and beige.
Pentacles
Ace of Pentacles
The universe is handing you the keys to success. You only need to reach out and take them. This is a boost to security and feeling grounded. This doesn't necessarily mean that you get to win the lottery but that you get a good night's sleep so you can be more focused. Or that you get a raise. Or that you're able to help someone out and, in doing so, help yourself.
Ace of Pentacles Reversed
It's not going to happen. The raise, the new house, the business plan. It's not going to come to fruition right now. It could be that it's not time or that you planned poorly. It could be because someone is actively working against you. Whatever it is, you need to walk away from the project and try again another time. Banging yourself against the gate that's closed against you will only leave you bruised.
Two of Pentacles
You often get this card when you're keeping your balance well amid rough weather. Keep light on your feet and don't lose sight of your goals. Regardless of how crazy the world can be, as long as you maintain your balance, you'll be just fine. Remember, the Buddha said be kind to everyone. You are part of everyone. Don't involve yourself in other people's issues. Just mind your business and reach out if your business involves getting assistance.
Two of Pentacles Reversed
If you are pulled in too many directions, you're going to end up stuck. You want to lose weight but don't want to exercise. You want to stop drinking but still buy that bottle. You want to be a good friend but still gossip about your people. It's human nature and it's habit, but it can also be changed if you really want to change it. If you're feeling conflict, there is a reason for that. It usually means that you're on the wrong path. Here's a secret, though: you can always hop over to a new path.
Three of Pentacles
Teamwork makes the dream work. Cheesy, but true. Collaborate and join up with like-minded people to make things happen. We are stronger together. Put your heads together and make magic happen.
Three of Pentacles Reversed
You have a group project and you're the only one who does the work, but everyone takes credit. You'd think that that kind of thing would stop in middle school, but you'd be mistaken. Don't be afraid to speak out. Don't allow yourself to be caught in workplace conflicts or drama. You're there to work, not make friends. Be professional, yes, but be strong, too.
Four of Pentacles
Hold on to those things that belong to you and be generous when you can. Be sure that you have everything you need before you share. Be sure that you come first. This card also indicates that you should keep an eye on the things that belong to you. Watch your money. Watch your possessions. Be safe.
Four of Pentacles Reversed
Could you be any stingier? Honestly. The Grinch ain't got nothin' on you. Charity is a part of the human condition. Seeing someone who needs something and then facilitating that exchange is part of what makes us human. If your soul has dried up and your heart has been replaced with tar and an old cell phone battery, you can still adjust your sights and do the right thing.
Five of Pentacles
Shelter and sanctuary exist, but you have to look for them. If you keep your head down, you'll miss all the good stuff. Sometimes you get pulled so far into your own darkness that you miss the lights shining near you. Sanctuary is available. You are not alone. Stop walking into the unknown and find your resources. It's going to be okay, but you have to let it be okay.
Five of Pentacles Reversed
You are nearly finished with the hard part. You've almost made it. You persevered and you're going to be able to breathe again so very soon. Just keep going. Keep saving. Keep working. Don't stop until the shadow falls short of you and you're standing in the sunshine again. Good job.
Six of Pentacles
There is an exchange here. Giving and receiving. Be sure that the balance remains. Can you give gracefully? Can you receive with equal grace? Can you allow balance and generosity into your life? Care. We all do better when we all do better. You have to invest in yourself first, then your family, and then your community. It has to flow outward. If it sticks with just you, your riches spoil and stagnate. If they flow out to only your family, they puddle and pool. If they reach your community, however, they flow and flow and flow. A rising tide raises all boats.
Six of Pentacles Reversed
Remember that three hundred dollars you loaned to your friend? Yeah, that's gone now. You invested something in someone (love, money, time), and you're not going to see it again. This is a hard lesson to learn, but part of being an adult is realizing that you can't know for sure if someone is trustworthy until they screw you over. It sucks, but it teaches a great lesson that you shouldn't loan out money (or love or time) that you're going to need back. Ever.
Seven of Pentacles
You've worked so hard, but you're not finished yet. Keep going—don't give up! Work. Practice. Get better. Decide that the status is not quo. Work until you make it look easy. Your life goes in cycles like the harvest. You prepare the fields, plant the seeds, tend your crops, and then you get the payoff. Do your work. Get your reward.
Seven of Pentacles Reversed
This card follows the script "stupid job, stupid house, stupid town." You are working hard and are underappreciated. You are trying to pay off debt and something keeps coming up. You can't catch up, no matter how hard you try, and something's got to give before you give up entirely.
Eight of Pentacles
It's said that you have to work at something for ten thousand hours before you master it. Put in the time and work so you can succeed. Hope only gets you halfway; work will bring you home. Don't be afraid of hard work. Be afraid of mediocrity. You're going to make mistakes. You're going to hurt people and yourself. You're going to do and say stupid stuff and regret your words. And then you'll keep growing and learning, and as you know better, you'll do better. It's okay to screw up, as long as you learn.
Eight of Pentacles Reversed
You can half ass a lot of things before you get caught or tripped up. I've known of people who went to work every day and did two hours of work and got paid for eight. They did this for years before someone caught on and they got super-fired. It is possible to skate through life. It's pretty shady, though, and it never lasts long. Be careful here.
Nine of Pentacles
Appreciation of what you have and fulfillment of your wishes. A time to give thanks and swim around in your happiness for a while. Part of being comfortable in the skin you're in is moving away from a mindset of scarcity to one of enough. What you have now? Be grateful for it. If you seek more, be grateful for the ability to find it. You have worked hard and you deserve good things. There is no shame or blame in being comfortable. Just remember to help those still on the path.
Nine of Pentacles Reversed
My father always said, "Stop spending money like you have it." I need you to accept these truths: (1) credit cards are stealing money from future you; (2) if you can't afford it, you don't need it; and (3) no one cares how big your TV is. Now, fold those into your life and stop living a champagne life on a beer budget. (That's one of my dad's sayings, too. I used to suck at handling money. He had to say these things a lot.)
en of Pentacles
This is about good fortune and financial success. High fives all around. You get the thing. You get peace of mind. You get what you need. This is security. The Ten of Pentacles is a wonderful card to get, because it is the ribbon at the finish line. Be sure you take time to enjoy it and pat yourself on the back.
Ten of Pentacles Reversed
Your money has fallen away from you. This happens, and it's okay. It's something that you can recover from. The most important part of falling on hard times is that you recognize and acknowledge the turn in your fortunes. If you don't look at your bank balance because you're afraid, you're going to be in an even bigger mess down the line. You can fix this, but you have to see it first.
Page of Pentacles
Making your dreams come true. Focusing on your hopes and how to make them a reality. Also, you're probably going to drop that ball. That's okay. Just be sure to pick it up again. This is the time to make things happen. That energy and joy is absolutely contagious, and you'll be able to recover beautifully from any fumbles.
Page of Pentacles Reversed
You not only dropped the ball, but you dropped it, kicked it, and then looked away before you saw where it landed. No good, man. You've got to get back on track. If you're disillusioned or disheartened, it is absolutely okay to take a small break, but then you need to realign yourself with your goals and try again. No one is going to chase your dreams for you. Move it.
Knight of Pentacles
Absolutely focused on his goals. Moving forward relentlessly and completely dedicated to finishing what they started. Put your feet on the ground. Focus on what is right in front of you. Do the dishes. This dude is solid, dependable, and grounded. Look for that calming, Zen influence who will guide you through the situation.
Knight of Pentacles Reversed
If you are too "nose to the grindstone," you're going to wear that nose right off your face. You have got to look up and around. You have to stop the sleep/work/eat/watch TV pattern that you've fallen into. Snap out of it and start looking around at your life. Are you happy? Are you bored? Figure that out. Also, stop looking at your goddamned phone so much.
Queen of Pentacles
The Queen of Pentacles welcomes you into her lap for a snuggle. Comforting, loving, and compassionate. She is calm wherever she is because she chooses to be there. She doesn't overreact. She doesn't lose her balance. She decides how she's going to feel, how she's going to act, and who she's going to allow in her life. She has the BEST BOUNDARIES EVER. Feet on the ground, steady and reliable.
Queen of Pentacles Reversed
It's good to work hard, but you can't just work hard. It's good to be organized and focused, but if you focus too much, you'll start ignoring everything that isn't a responsibility. The pentacles tend to get heavy when they're reversed, so this card will come up when folks need to ease off the till a little bit so they can relax.
King of Pentacles
Security and reliability. He makes things happen and always has his eye on the ball. This is a hard worker. You're dedicated to your family and believe that if you're paid for eight hours of work and you don't deliver, you're a thief. He's not always forgiving but is always earnest and will always be there. You have to be aware of your weaknesses and compensate for those. When you figure that out, you'll have solid ground on which to build.
King of Pentacles Reversed
The forty-year-old man who still boasts about his college days. The woman who maintains the same haircut that she's had for thirty years because it fits the ideal self that her mother decided was appropriate. The woman who fake smiles and has fake friends because her self-protective wall is so high that no one can get in. At all. This is a person with excellent camouflage who does such a good job of obscuring their true self that they aren't really sure anymore who they really are.
Swords
Ace of Swords
Follow your great idea. You're being given inspiration. The aces are the heart of their suit. The Ace of Swords is the brain of the tarot. Allow yourself to be inspired and enlightened. Be open to influences and your muse and be sure to follow this fantastic head start to its (already lucky) conclusion.
Ace of Swords Reversed
When I get this card, I always think of a dog my friend had when I was little. My friend's dad called the dog Dummy. "Hey Dummy, you can't eat cardboard." "Hey Dummy, you can't walk through that wall." The dog was lovely but, bless her, just as dumb as a box of hair. The problem was that she didn't think she was being stupid. She was just wandering about, eating cardboard. Double check with yourself (maybe ask a friend) if you're behaving like an idiot.
Two of Swords
The "shut up so I can hear myself think" card. Listen to yourself. No one else. Lock the door of your office, get rid of distractions, and focus. Sit with your thoughts and ideas until you are calm and focused. And quit asking everybody else about their opinions. Other people's opinions are none of your business.
Two of Swords Reversed
Listen, if you don't want to decide, you don't get to complain. There are a lot of reasons to remain in stasis, and you can for a while, but if you allow your decision-making to fall to other people, you get to experience their path. Do you want that? Do you want to walk someone else's road? Even if they love you and mean the best for you, it will still be inauthentic.
Three of Swords
Ouch. Pain. Like, searing, heart-falling-out pain. The thing about pain this severe is that it's purifying. That's literally the only good thing about this card. Sometimes you have to feel all the pain before it will go away. Remember that holding pain inside can cut you to pieces. Releasing it, even if only to yourself in a journal, is one of the most amazing things you can do. If you don't want those words lying around, write them down and burn them.
Three of Swords Reversed
You've pulled the last sword out and can rest. You couldn't start healing until you got to this point, right here. Now that you're free of the swords, you need to tend to yourself. Sleep. Eat good food. Drink water. Sleep some more. Remember, you just had three swords going through your heart. You need some downtime.
Four of Swords
Put your thoughts at ease. Relax. Learn how to meditate. Remember that worry is a waste of time. It burns through the present and makes the future seem grim. Shh. It's okay. Not worrying will make you feel better. I remember reading that worrying is just praying for bad things to happen. Figure out what you can fix and leave the rest to sort itself out. Use the rest of the time you would have spent worrying on stuff like reading and love and naps and cat videos.
Four of Swords Reversed
If you try to do all the things perfectly, you'll end up either screwing each one up a little bit, or doing none of the things and locking up a little bit. Stop with the frantic activity and make a list. Then take a nap. Then do one thing on the list and probably take another nap. Seriously, you gotta chill.
Five of Swords
Stop paying attention to what other people say and pay attention to what you say. Stop bickering, gossiping, trying to convince people to change their minds. It's a waste of time. This is the tarot equivalent of "don't start nothin', won't be nothin'. " Stand up for yourself and be absolutely clear about what you will and will not allow. It starts with controlling the words that come out of your mouth and those that enter your ears.
Five of Swords Reversed
I have this ex who keeps popping up in my periphery. He makes me angry every single time I see him. It's been decades, and I still just want to slap the taste out of his mouth. Now, is this his problem? No. It's mine. It's not even about him, really. I get angry because I allowed him behave so poorly and still dated him, and the shame of that is what makes me so mad. I have to forgive myself and check that my boundaries are being enforced now. And then get over it. I'm working on it. Figure out what you're really mad about and then how to get over it. (Then get over it.)
Six of Swords
Focus on where you're going and get out of where you are now. Don't look back. You don't live there anymore. Part of going from rough waters into smooth is that you don't always trust that transition. You wait for the other shoe to drop, and in doing so, sometimes you manifest that shoe right on your head. Trust that you're going to be okay. Trust that when okay comes, you might not know what it looks like right away. Trust that if you feel threatened, you can grab one of those swords to protect yourself. We never go backward. I know it's hard, and I know you're tired. Just keep going.
Six of Swords Reversed
It's all well and good to say that you're going to move on, but sometimes you just can't. The past sticks to you like glue and won't give you peace. The answer is not to just deal with it. The answer is to confront it—in counseling, with ritual, or with prayer. Deal with the problem of yester-you. Forgive yourself and let yourself heal.
Seven of Swords
You're being sneaky. Stop avoiding those things you need to figure out. They're going to be there when you start paying attention again. Be sure of who you trust and why. This includes you, by the way. Watch your steps and confidences. Remember that everything done in the dark will come to light eventually and that if you always tell the truth, you don't have to remember what you've said.
Seven of Swords Reversed
If you've built a lie and have lived in that lie for a while, it starts to feel right. The thing about lies, though, is that they crumble around the edges and become transparent. The lie that you shelter yourself with will start leaking and then you have to deal with the mess and with the tattered bits of dishonesty that litter the floor around you. Give up the ghost and let the lie die.
Eight of Swords
I know that change can be scary, but it's a lot scarier to stay still. Better to change and have some pain than stay still and be paralyzed. Are you stuck, or do you think that you're stuck, which is then making you stuck? Fear will eat your dreams and leave you cold, still, and small. It lives in our brains and tells us we're not enough. Don't believe it. If you find yourself panicked, first control your breathing, then your mental state, and then your body. After those are still, become part of the conversation. How long will fear hold you?
Eight of Swords Reversed
Eddie Murphy has this joke about skeletons in a closet. "You got skeletons in your closet? I thought I seen a bone in your shoe. Whose skeleton was that?" It's okay to have done foolish things in the past. We all have skeletons. The elaborate charade of pretending that you're perfect is so exhausting. Wouldn't it be easier to just acknowledge your messy past and come to peace with it? Take time to forgive yourself for being human.
Nine of Swords
Anxiety lives here. Remember that stupid thing that you did in 2001? Well, so does your brain, and it's going to mess with you every now and again to remind you of it. The fix for this is remember how much further along you've come since then. If I am in an anxious place, the anxieties that rush in mostly have nothing to do with reality. It's hard to discern what is real and what is just your brain poking at you. Luckily, you can rationalize your way out of this and stab your brain with Q-tips if necessary.
Nine of Swords Reversed
A little bit of stress can motivate you. A lot of stress can cause you anxiety. Too much stress can shut you down entirely. This is where self-care comes in. This is when you ask for help. From a friend or from a doctor. From a counselor or from your higher power. This is the time to hold your hand out and ask, sweetheart.
Ten of Swords
Oof. You just got stomped on quite a bit. Do us all a favor and stay down for a while. Watch Broad City and eat comfort food. You'll stop bleeding eventually. The great part about this card is that you have actually reached rock bottom. Congratulations. Only one way to go from here.
Ten of Swords Reversed
Kill your darlings, said Faulkner. He was talking about killing the characters and ideas in writing that you loved so that you could grow as an author. This applies to life, too. Kill those things that you embrace that are keeping you from your potential. Let them die and embrace entropy as a pathway to change.
Page of Swords
This card has so much potential. It is the idea that wakes you. The lightning bolt whose afterimage lingers like a neon light behind your eyes. The 4 a.m. scribble on the piece of paper by your bed that says "moon cheese" and nothing else. You've got the seed. You just have to give it care and space to grow, and it's going to change the world.
Page of Swords Reversed
There is a twinkle in your eye, because you're just about to act on an idea that you just had and haven't thought it through even a little bit.
Knight of Swords
Be bold. Be brash. Move with alacrity and be engaging. Don't, however, knock people over on the way to your goal. You can say whatever you want, but you're probably going to hurt someone's feelings. Remember that assertive isn't the same as aggressive and that throwing your words out like knives hurts you, too.
Knight of Swords Reversed
If you have a lot of energy and no focus, you end up running around like a chicken with its head cut off. You make a lot of noise and appear to be very industrious, but you are actually just bouncing around and freaking everyone out. Stop that. Find a better outlet for your energy and channel it.
Queen of Swords
Be as direct as possible. No emotions come into this decision, just a sense of fairness and of what is the right thing to do. To quote my friend Sara Benincasa, "I gave up nice a long time ago. Nice did me no good as a woman. Niceness is a lie they teach you to keep you sweet and compliant while you're screaming inside. You know what I picked instead? Kindness. I chose to be kind. Kind means I respect your boundaries and you respect mine."
Queen of Swords Reversed
I think that women have a bad rap when it comes to gossip. I think that men gossip, too, but our society allows them to criticize other people and situations without having to cloak it in secrecy or meanness. This is the way it is right now, though, and you're being a gossip. Don't use your words to hurt—use them to help. If you have some criticism to hand out, do it with your head up and meet their eyes. Don't be sneaky about it. That says more about you than it does about them.
King of Swords
You don't always have to be nice. You just have to be honest with yourself and with others. Move forward with focus and take no prisoners. This card reminds you to get your ducks in a row or the ducks will be fired and replaced with better, stronger ducks. Not even ducks, geese. Highly trained, efficient attack geese. Get to work—it will be worth it.
King of Swords Reversed
Okay, well now you're just being mean. Honestly. You don't get to treat people like things. You don't get to push your agendas on other people. You don't get to be a bully. Whatever your tool was, you've turned it into a weapon and you're bludgeoning other people with it. Stop that. Mind your manners and stop talking until you have something nice to say.
Wands
Ace of Wands
Move. The universe is giving you a window for action. Take it. It only takes one match to light a fire. Whether that's a controlled burn or a wildfire is up to you. This is the universe's hand on the small of your back, urging you forward. Sometimes it shoves you. Move forward with confidence. Know that you're where you are for a reason. If you grow roots at the starting gate, the finish line will never come closer.
Ace of Wands Reversed
Stuck, stuckity, stuck stuck stuck. You're stuck. Whether it's writer's block or inertia in finding a job or a new partner or a new house, you're just stuck. That's okay. You can be stuck for one more week, and then you have to follow whatever plan you can put together to get yourself unstuck. Stuck is helpful sometimes and helps you heal sometimes, but you don't get to live.
Two of Wands
Stay put. Sometimes if you don't know where to go, the best thing to do is to sit down. Be thoughtful and make a plan that helps you find balance. Get some rest. Plan your next move. You're not going to turn into dust if you don't go right now. If you truly feel pulled to move forward, break your goal into tiny, achievable goals and do those steps instead. Don't put all your eggs in one basket. One egg per basket, please.
Two of Wands Reversed
You've missed something in the planning stage. Something is not quite right, and you need to reevaluate. Get your toolbox out, lay out all your tools, and really look at them. Do you have everything you need to succeed? Are they the right size or shape for the job? Be sure that your attention to detail is honed so you don't have to go back and fix something.
Three of Wands
It's good to be open to opportunities and moving outside your comfort area. Keep your head on straight and realize that a bigger, better box is still a box. Time to move forward to freedom and happiness. People outgrow their dreams all the time, but those dreams have been a fixed ideal that it's difficult to see past.
Three of Wands Reversed
This is the epitome of Murphy's Law. Getting stuck, dropping the ball, dropping the ball on your foot and then tripping over it. If Mercury retrograde had a card, this would be it. Just do yourself a favor and sit on your hands until you feel like the chaos has passed. Usually this comes up when it's not a good time to start new ventures. The universe drops blockades in front of you so you can wait for a more auspicious time.
Four of Wands
You are on the path to success. You are pointed in the right direction. You can pick up more tools on the way if you need them, but I tell you it's time to start walking up the hill. Do not stop. Do not get to the end of your life and realize you didn't finish that book because you were screwing around on Facebook. Do your Job. Not your Muggle job, your real Job. The one you were put here to do.
Four of Wands Reversed
How do you expect to take care of your business when your nose is all up in everyone else's? You have got to step away from the drama and minutiae that is Not Your Business. You are not allowed to fiddle with other people's existence. I get it—if you're involved with other people's stuff, you won't have time to deal with yours, but it's not going to last for long and then your problems will be right there in your face. And they'll be bigger.
Five of Wands
Nothing good can come of this. Decide if you want to be right or if you want to be happy. Being in conflict with other people won't get you any closer to your goals. Some debate is good and healthy, but if you start raising your voice, you've already lost. Recognize why you're in the conflict to begin with, and if it can't be solved right away, leave it. You can always revisit it later when people aren't arguing. This card is telling you that you picked the wrong time and place for discord. Regroup and try again later.
Five of Wands Reversed
If you avoid conflict at every opportunity, you're going to lead a pretty miserable life. You'll go along with ideas you disagree with. You'll cave whenever someone challenges you, and you won't have any boundaries to speak of. You will go backward on your promises—to yourself and other people. If you cave on everything, you cannot possibly have moral standards or principles. You just can't.
Six of Wands
Celebration! You did the thing! Well done. Go have a nice dinner. Be gracious, but still accept praise. Don't slap it away. Just say thank you. Remember to sit still in this moment and actually feel it. Don't speed into the next goal or spend time regretting what didn't work out. Just enjoy the thing and don't let other people cast a shadow on your sunshine, Sunshine. Shine, shine, shine.
Six of Wands Reversed
To quote Captain America from Spider-Man: Homecoming, "So, you got detention. You screwed up. You know what you did was wrong. The question is, how are you gonna make things right? Maybe you were trying to be cool. But take it from a guy who's been frozen for sixty-five years—the only way to really be cool is to follow the rules." Cap is right: if you drop the ball, you gotta figure out how to pick it up again.
Seven of Wands
Tenacious. Literally, this exchange: "Are you done being sassy?" "No." This is the honey badger card of the tarot. Not backing down. Not asking permission. Not going away or sitting down.
Seven of Wands Reversed
Don't tell me what to do! (Even though you're right.) Don't criticize me! (Because it's true and it hurts.) Don't make me feel wrong, please, because if you do, I'll have to admit that I'm human and don't know everything. It is hard and jarring to be wrong, and I don't like it or want it. Please don't make me doubt myself because if I start now, I won't stop.
Eight of Wands
Time to make a move. Don't let doubt get you trapped. Decide. If it's the wrong decision, make another one. You decide your course of action. Which obstacles will you step over and which will stop you? How badly do you want to move forward? Break free of stasis. You have to understand that if you don't decide the course of your life, the universe will decide for you.
Eight of Wands Reversed
I understand that it's very easy to be comfortable. I really do. I understand that it's less scary to stay in that relationship, at that job, or in your dream instead of making it a reality. I understand. I also understand that if you don't move away from your comfort zone you're not going to grow or change or challenge yourself. You are going to sit exactly where you are, and mold will grow on you.
Nine of Wands
Do not let the same people who messed with you before do it again. Protect yourself. Remember that people always tell you who they are. Your job is to listen, believe them, and then take care of yourself. You have done this before, you know. You know how to set boundaries and you know how to enforce them, but knowing how to do it and actually doing it are two very different things.
Nine of Wands Reversed
You have no boundaries. You are not just acting like a doormat—you're voluntarily lying down in front of people and asking them to walk on you. The martyr side effect can be nice for a while, but it quickly spoils under the weight of being chronically unhappy. Look at your life—look at the places where you are happy and unhappy and then stop doing the things that make you sad. It is just that easy. And just that hard.
Ten of Wands
You are doing it wrong. All of it. Even the things you think you're doing right are actually getting overshadowed by the things that you're doing wrong. Put down everything, get some advice from someone you love, then try to pick up just a few things and see how it works out. Baby steps. Decide what your priorities are, say no to everything else, and move forward. And please stop carrying other people's expectations around—they're surprisingly heavy.
Ten of Wands Reversed
When you end a relationship, the shadow of that person follows you around for a while. When you get hurt, the shadow of that pain lingers. This card reversed asks you to sever the connection to the things that weigh you down. Have a ceremony. Write it/them a letter and burn it. Do something that tells the universe that you are finished with that person, the booze, that habit. Make a statement and then have a funeral for the thing that no longer serves you.
Page of Wands
Use your creativity and energy to parkour out of tricky situations and into a new direction. Be nimble. Be feisty. Put this energy to good use with a home project or hike or have a good make-out session. Have fun! And be sure to plan for the next step or ask for help so you don't drop your stick. You contain multitudes and can do hard things, even if you're afraid. If you bring your whole self forward, your whole self will succeed. Think Hermione Granger.
Page of Wands Reversed
Be sure to be sure. This whole "doing things and then regretting them 80 percent of the way to the destination" is not always the best way to go. Sometimes if you really want something and you really try to do it, it still doesn't work out. It just doesn't. This does not mean that you're stupid or have bad luck or don't deserve good things. This means that this thing wasn't the right thing, or it wasn't the right time or the right place. Regroup. Pull it together and try again. Try a new thing, a new time, or a new place.
Knight of Wands
Do you even joust, bro? This guy is a bit of a blowhard. Be sure you're going in the right direction and check your sources. There is a scene in Big Hero Six in which the main heroes sneak into the bad guy's lair, and Fred, the team goofball, starts singing a song about leading his intrepid friends into danger. Fred is the Knight of Wands. He's going to be the first person to jump up and help you, and he will probably trip over his feet on the way. His intent is solid. He's got your back.
Knight of Wands Reversed
Things will stop and go and stop and go. And then they'll get delayed. And then you'll totally be on the way again! And then you'll get a flat tire. This is a card of inconsistency. Not bad, exactly, but frustrating, and a good opportunity to check your equipment, your plan, and your team. Do you have everything you need to continue on? Are you sure?
Queen of Wands
Queen of sass and motivation. Tough love is important, but remembering the "love" part is the most important thing. She offers peace and beauty in one hand and a shoe with which to smack you in the other. Go with confidence. Guide those who need your influence but don't be a jerk about it. Remember that shepherds guide their flocks with their staff—they don't club them over the head with it.
Queen of Wands Reversed
Okay, this one will club you over the head with it. The Queen of Wands reversed is the worst kind of confidence: ill placed. She believes in herself and her rightness so much that she'll do whatever she can to get what she wants. Regardless of the truth, or fairness, or what's good for all. She's bossy and pushy and sometimes duplicitous. This card comes up when you're being railroaded. Or you're railroading someone.
King of Wands
Those boundaries you are working on have to remain strong. Your job is to keep Negative Nellies away and protect those things most valuable to you. This king suggests that you stop half-assing everything and start whole-assing one thing at a time. Finish what you started. Close your circle. Recognize that even though you want to be friends with everyone, you need to have a pretty keen eye for recognizing who people really are. When they show you—with their behavior, their promises, and their accountability—who they are, you need to believe them.
King of Wands Reversed
Do you remember how Wile E. Coyote would run and run with a pole and stab it into the ground, hoping to launch himself over the canyon and finally get Road Runner? What usually happened was that he first got stuck—one of those boi-yoi-yoing, back-and-forth stucks—then he would slowly go forward, nearly hitting the ground, and then BLAM—he shot backward like a rocket.
Cups
Ace of Cups
Your cup runneth over with blessings. Be sure to be grateful. Happiness is afoot. Blessings abound. Accept them, acknowledge them, and say thank you. The universe conspires to shower you with blessings. Let it. If you're not in a place to say thank you, start with "at least." At least you have power. At least you woke up today. If things are solid, kick into the thank yous. Thank you for the little things and the big things, too.
Ace of Cups Reversed
When you feel empty emotionally, that's usually an indication that you're about to start filling up again. It doesn't do any good to bemoan your life and worry that you've lost people, peace of mind, and emotional balance without realizing that this is a cycle. There is no full without an empty. There is no safe without a challenge to that safety. When you have an awful day, time, relationship—somewhere in the back of your mind (behind the pain) you need to know that good is following it. It has to.
Two of Cups
You have support in the relationship that's most prominent in your life. Allow this person to help you find the right way. Equal partnership. Respect. Esteem. No one is lacking here. No one is wondering. This love is enough. This relationship—with work, friend, sweetheart, whatever—is enough. And you can feel it.
Two of Cups Reversed
One of the problems with getting close to people is that they become close enough to see where your insecurities are. When you let your guard down, it's down for everyone and all the time. Sometimes, letting your heart show a little makes it into a target. Be careful with yourself. Be careful with your heart. Don't be careless with it, and don't let other people be careless with it either.
Three of Cups
Find your community! Do what you can to surround yourself with like-minded people and enjoy the companionship. Lean on them, be silly with them. Call your bestie right now and make plans to hang out. It's good for you. Sometimes you have to dance because the real world has gotten so dark that if you don't chase down a spark and make it grow, you'll become part of the darkness. Don't join the darkness.
Three of Cups Reversed
Sometimes friendships turn upside down. You get a sour feeling in your stomach when the person reaches out. You text less, find yourself faking a laugh or a smile more often. It could be that you grew apart or that they've done something that chews on the back of your thoughts every time you see them. You can continue being friends, but push them into the acquaintance category. You can ghost them, which is a little shady but spares you a hard conversation. You can also have that hard break-up conversation with them, which is the worst/best call. Friends shouldn't make you feel gross.
Four of Cups
Don't settle. Don't pout either, but make sure that this is what you really want. Fussy and ripe for hissy fits and eye rolling. Apathy is easy to ignore. Don't wanna. Don't want to change, don't want to take my medicine, whatever that medicine may be. You have to look deeper than the apathy. Why don't you want to change? You know these behaviors make you sad, and yet you won't put them down. Consider the why of holding on to the listlessness and figure out how that's paying off for you. There's a payoff for "meh." Whether it's the reward of not failing (you can't fail if you don't try) or the reward of being comfortable, decide if it's worth being stuck.
Four of Cups Reversed
This always reminds me of when you've been lying on the floor for a bit and you try to sit up without using your arms. You lean forward, fall back, lean forward, fall back, lean forward, and now you can stand. Just because you've been asleep for a while doesn't mean you can't wake up. It'll take a second to knock the sleepies out of your eyes and stretch, but you'll get there.
Five of Cups
Remember to give your old heartaches respect and grieve them properly. Honor that time, and then move forward without them. It's okay to grieve, really. You can miss what you had and mourn what could have been. You can feel sorry for yourself, but you can't live there. You are seeing the world through lenses of pain and will interpret it incorrectly. If you find yourself here, make only small decisions. Drink some tea. Take a nap. Wait for the pain-shaped hole in you to heal. Let it get to a place where you can breathe without pulling strands of the past into your lungs. Don't stay in the in-betweens for too long.
Five of Cups Reversed
You've been through some rough times. Do me a favor and sit down for a bit. Then, you gotta get up. If you're coming out of depression and still living as if you're depressed, those behaviors don't serve you anymore. As you're able, go outside. Breathe fresh air. Get out of your house. Go see your friends. Move in the direction you want your life to go—not where it was when you needed to be still.
Six of Cups
Seek joy and don't stop until you find it. Be sure that you're taking time to live in the present. Warmth, happiness, and silliness abound. Remember to accept blessings given without being paranoid. Innocence is not always a bad thing. Trust that the universe calls you beloved. When we were small, we didn't chase happiness—we just found it.
Six of Cups Reversed
If you are in the middle of having a temper tantrum, it's really hard to see it unless you're seven years old. You have to look around you. Are people being really, really quiet? Shooting you furtive glances? Are you saying no to everything? Does EVERYTHING suck? All of it? Okay, well, simmer down and go back to the last time you remember hearing yes. What happened? Can you deal with it without throwing things? Give it a shot.
Seven of Cups
If wishes were fishes, you'd never go hungry. If you have dreams, take the time to write them down and make an action plan to make them happen. It's a vital place to visit, but you can't live there. This is the card that shows you where you could go. You need to clarify your dream, create a plan, and manifest the hell out of it.
Seven of Cups Reversed
Congratulations, you finally know what you want. That's a huge step. Now that you want, get to gettin'. It's really important, now that you know which direction to go, that you go now. This card shows up as the green light, and ignoring it means that you'll go back to that mushy, indistinct thing where you can't decide. It was a struggle to get here, and dawdling will only make it into a struggle again.
Eight of Cups
Take only what you need and leave the rest. Excess in anything is not a good idea. Time on your phone, watching TV, daydreaming. Make sure that you spend more time on things that serve you rather than those that just fill the space. This card is an indicator that you need to dive deeper, find what is truly meaningful to you and makes your soul sing. This the card of a seeker, trying to find diamonds among the coal.
Eight of Cups Reversed
You know that thing when you get really angry at work and start planning out exactly what you'll be wearing when you quit and exactly what you'll say and ALL their jaws will drop and you will walk out like Angela Bassett's character walking away from her husband's flaming car in Waiting to Exhale? Yeah, don't let that be a real thing. If you're unhappy, take a second to make sure that you actually want to leave the situation and aren't just really pissed off.
Nine of Cups
Being present, grateful, and satisfied with what you have. Take a moment to consider how lucky you are and move forward with your shoulders back and your head high. The fella in this card is the tarot equivalent of finger guns. All is well in his world. He has no worries, and those he does have are so insignificant they can't wipe that smile off his face.
Nine of Cups Reversed
There is an element of smugness here. Eddie Murphy has a joke about kids getting ice cream and immediately teasing the one kid who didn't get any: "You can't afford it, you can't afffooooord it." Grown-ups do this, too. Usually they're not slick about it either, and they'll gleefully rub their hands together like Mr. Burns while crooning, "Excellent." Okay, not that bad, but you know when someone is acting as if what they have is more important than who they are? That's this thing. Sometimes they're right, but it doesn't make it any less uncomfortable.
Ten of Cups
Yes. Just a page full of yeses and happiness and joy. You are there. You've made it. Leave your scarcity thinking behind, and you'll get what you need and want. Don't let the celebration stop your progress to your most authentic self. There is an absolute joy in setting your intentions, working your tail off, and then achieving your goals. Whether it's finishing the dishes before you go to bed every night or finishing that book that you've stayed up till after midnight for like a year and a half in order to finish, the joy is real. The accomplishment is there.
Ten of Cups Reversed
This card is joy interrupted. You can see the happy. You remember what happy was and how good it felt. Right now, though? Not so much. There is something between you and the rainbow, and it feels like you can run and run after it and never catch it. There is a sadness here in that most folks who get this card stop chasing the rainbow right before they would have caught it.
Page of Cups
The pages are messengers, and the Page of Cups brings good news! Romance, sweetness, and emotional availability. It's called a crush for a reason. Whether it's a piece of music or a book or a person, whatever makes your heart leap out of your chest is found here. Now, is it going to be forever? Probably not, but it's going to be a hell of a ride.
Page of Cups Reversed
He fell in love and broke his heart, but he got back up to fall in love again. Even reversed, there is a sense of optimism to the pages. These guys are so flippin' earnest. They will absolutely fall down but are just as certain to get back up again. There is an element of self-pity, of delayed recovery, and of complaining to your friends for like a week about how terrible your life is. Then you get over it.
Knight of Cups
Romantic but fickle. Use your heart but keep your head on your shoulders. This knight is sweet and emotional. He can be shy and introverted but is also a flirt. When you get this card, remember that you can keep things light in love. You don't have to get married after the first date. Have fun and relax.
Knight of Cups Reversed
Have you ever met someone who is all dark and broody, but not in a romantic, James Dean way? More like a creepy a-hole who lurks in the coffee shop, scribbling in a notebook and glaring at everyone for no reason. Instead of this love flowing out, it gets stuck and sours. He couldn't find anyone to accept his love, so he turned it on its head. The energy with this card reversed is sulky, pouty, and immature.
Queen of Cups
The queen of emotion. She's in control of her feelings and surrounded by love. You can love someone but not so much that you lose your mind about it. Can you be near drama without being in drama? Can you shed those emotions that no longer serve you? She loves everyone, but I'm 100 percent sure that she loves herself first. She holds sacred space for herself and takes care of herself without apologies.
Queen of Cups Reversed
This is that kind of cloying, sticky love that you see in movies like Mommie Dearest. It usually starts innocently enough, but then it deepens into an obsession of sorts, pinning the object of affection like a butterfly on Styrofoam. It won't grow away from her that way. It won't grow at all, and she doesn't have to worry about losing it.
King of Cups
Composed and in control, emotionally grounded and caring. You decide what will rock your boat and what may not rock your boat. You decide. The King of Cups is a benevolent dictator. He's doing the work, too, and throws in with his people. He doesn't do the work because he loves the work. He does it because it makes people happy. Say it with me, though: "Other people's drama is none of my business. Amen."
King of Cups Reversed
All the intensity and power of the king's emotion flooding out of control. This is drama, you guys. Temper tantrum, hissy fit, manipulative, Real Housewives of Wherever They Are Now drama. The worst part of this is the power that the king has. He can make things really bad for you. He can trample cities under his feet. It's not a good thing, and unless this card represents a true crisis, you should seriously take cover. Egad, the drama.
[contents]
Recommended Reading
To Clean Your House
Unf*ck Your Habitat: You're Better Than Your Mess by Rachel Hoffman
To Open Your Mind
Bossypants by Tina Fey
Real Artists Have Day Jobs: (And Other Awesome Things They Don't Teach You in School) by Sara Benincasa
To Nurture Your Body
Eight Weeks to Optimum Health: A Proven Program for Taking Full Advantage of Your Body's Natural Healing Power by Andrew Weil
To Open Your Heart
Untie the Strong Woman: Blessed Mother's Immaculate Love for the Wild Soul by Clarissa Pinkola Estés
Into the Magic Shop: A Neurosurgeon's Quest to Discover the Mysteries of the Brain and the Secrets of the Heart by James R. Doty
To Open Your Soul
The Name of the Wind by Patrick Rothfuss
The Art of Asking: How I Learned to Stop Worrying and Let People Help by Amanda Palmer
[contents]
Acknowledgements
In the most unironic fashion ever, I would like to thank everyone who broke me in some way. Whether it was physical, mental, emotional, financial, or spiritual brokenness that you contributed to, it's because of you that I was forced to put myself back together. To quote the great Hannah Gadsby, "There is no way anyone would dare test their strength out on me, because you all know there is nothing stronger than a broken woman who has rebuilt herself."
Also, to Joe, who loves me altogether. You love me when I'm too much and when I feel like I'm not enough. You are my favorite, and I'm so glad I got to marry you.
[contents]
Kitchen Table Tarot
Pull Up a Chair, Shuffle the Cards, and Let's Talk Tarot
Melissa Cynova
Learn to read with your own voice and get the answers to all of your questions.
For years, Melissa Cynova has been sitting down with friends and neighbors who are curious about the tarot. She's heard all the questions and misconceptions that can confuse newcomers (and sometimes more experienced readers, too). Kitchen Table Tarot was written as a guide for anyone looking for no-nonsense lessons with a warm, friendly, and knowledgeable teacher.
Join Melissa as she shares straightforward guidance on decks, spreads, card meanings, and symbols. Filled with real-life examples and personal explanations of what it's like to read the cards, this book tells it like it is and provides the information you need to read with confidence.
978-0-7387-5077-4, 288 pp., 5 ¼ x 8
To order, call 1-877-NEW-WRLD or visit www.llewellyn.com
Prices subject to change without notice
Your Tarot Your Way
Learn to Read with Any Deck
Barbara Moore
Discover what's its like to receive tarot lessons with a warm and encouraging teacher at your side. Join Barbara Moore as she shows you how to build tarot skills from the ground up and form your own intimate connections with the cards.
Your Tarot, Your Way invites you to honor your unique life experiences as you respond to symbols and cultivate a spirit of play. With practical, real-life examples, this book helps you explore tarot as a practice for nourishing your soul and discovering new perspectives.
978-0-7387-4924-2, 216 pp., 5 ¼ x 8
To order, call 1-877-NEW-WRLD or visit www.llewellyn.com
Prices subject to change without notice
Llewellyn's Classic Tarot
Barbara Moore and Eugene Smith
Drawing classic symbolism and meaning from the world's most popular tarot deck, Llewellyn's Classic Tarot is firmly rooted in the foundation of modern reading. Whether you are new to tarot or a seasoned reader, this easy-to-use deck is your perfect guide to the wonderful world of tarot. Discover evocative artwork based on the colors and symbols of the traditional Rider-Waite-Smith system, but with a fresh and relatable perspective. With Llewellyn's Classic Companion focusing on both the classic and twenty-first-century interpretations of tarot, you have all the tools you need for better reading and understanding.
978-0-7387-3608-2, contains a book and deck
To order, call 1-877-NEW-WRLD or visit www.llewellyn.com
Prices subject to change without notice
Shadowscapes Tarot
Stephanie Pui-Mun Law and Barbara Moore
Renowned fantasy artist Stephanie Pui-Mun Law has created a hypnotic world of colorful dragons, armored knights, looming castles, and willowy fairies dancing on air—a world of imagination and dreams.
Lovingly crafted over six years, this long-awaited deck will delight all tarot enthusiasts with its wondrous blend of fairy tales, myth, and folklore from diverse cultures around the world. Featuring breathtaking watercolor artwork that fuses Asian, Celtic, and fantasy elements within the Rider-Waite structure, each exquisitely wrought card draws upon universally recognized symbols and imagery.
A companion guide also presents evocative stories and insightful interpretations for each card.
978-0-7387-1579-7, contains a book and deck
To order, call 1-877-NEW-WRLD or visit www.llewellyn.com
Prices subject to change without notice
## Contents
1. About the Author
2. Copyright Information
3. Dedication
4. Disclaimer
5. Introduction
6. Chapter 1
7. Chapter 2
8. Chapter 3
9. Chapter 4
10. Chapter 5
11. Chapter 6
12. Chapter 7
13. Chapter 8
14. Chapter 9
15. Conclusion
16. Appendix
17. Recommended Reading
18. Acknowledgements
## Landmarks
1. Cover
2. Title-Page
3. Backmatter
# List of Pages
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Ugh, the dreaded belly bloat.
I know personally it has been my never-ending quest to de-bloat my belly. And it wasn't until the summer of 2016 that I discovered one of the main reasons why I often looked 6 months preggers was because I have a gluten-intolerance. I also discovered I am lactose too.
But what if none of those conditions apply to you. Could it still be what you are eating is making your belly bloat up like a balloon? That's what we're talking about today, foods that increase our belly bloat.
Pop, soft drinks, nectar of the gods…whatever you choose to call it, soda is B-A-D for belly bloat! This is because of the carbonation (the fizz you get when you pour a drink of soda or open a can). Carbonation draws gas into your belly and intestines, causing instant distention aka belly bloat.
Another reason is sugar. Forget that this is truly the #1 reason why you should be avoiding soda at all costs in the first place (same with diet soda–it's actually a lot worse due to the fake sugar in it aka chemicals). The high sugar content of soda instantly spikes your blood sugar levels, which in turn spikes your insulin. This then causes your body to begin storing excess sugar as fat in the most convenient location possible…that's right, your belly.
Alternatives: plain, filtered water, fruit/veg/herb infused water, seltzer water (if you really dig the fizz).
Specifically, sugar-free chewing gum. I know, this is technically not a food, but it does contribute to that dreaded belly bloat due to the sugar-free aspect. Sugar-free gum contains sugar alcohols and this distends our bellies. Sugar alcohols are partially digested sugars and are well known to increase intestinal gas and bloating, which in turn expands our mid-sections.
Another reason sugar-free chewing gum can causes belly bloat is because when you chew gum, you swallow air. Now even though swallowing air won't hurt you, the more air you swallow, the more it piles up in your digestive system which can lead to pressure and belly bloat.
Alternatives: licorice root, seeds/nuts, parsley or chopped veggies.
Cabbage is a huge cause of belly bloat as are various other cruciferous veggies like broccoli, cauliflower and even brussel sprouts (remember when Sheldon on The Big Bang Theory so delicately suffered from this "affliction"?).
During digestion these veggies create a large amount of gas in our gastrointestinal tracts. The bloat is temporary and is mostly relieved the same way Sheldon did (aka passing gas).
Alternatives: gas producing veggies are much easier to digest when they are cooked, as opposed to eating them raw. You can also try different veggies, or know that the days you do eat higher-than-normal gas producing veggies, you may want to be wearing your stretchy pants.
Who doesn't love a plate of spaghetti, espesh on a chilly Saturday night! Talk about comfort food!! Although pasta can seem pretty innocent in terms of belly bloat, it's actually a recipe for disaster.
That's because white pasta is made from white flour which is refined and has the nutrients stripped from the grain. This processing causes white pasta to be digested extremely quickly causing a huge spike in your insulin, which in turn gets stored as fat, which leads a bigger waistlines.
Plus, because the digestion process of white pasta is so quick, consuming it (even once and awhile) can acutally lead you to eating even more calories because you are hungry again soon after and eating more food.
Alternatives: whole wheat pastas or if you are IBS or gluten-intolerant (like me!) opt for rice or corn pasta, which are very common in grocery stores these days (even my small town of 16,500 pop. has gluten-free options in our tiny grocery stores).
Keep in mind you don't need to bid adieu to all these belly bloat foods all at once. Baby steps are the key to sustained and lasting weight loss. Pick 1-2 foods from this list and make the switches slowly over a 1-4 week period. This isn't a race. This is your health and as long as you are making consistent efforts to improve it, you are on your way.
You can also solidify your efforts by eating a clean diet filled with lots of water and unprocessed foods. Make sure you check out my Get Lean Eating Plan (part 1 and part 2).
Sign up for my FREE 7-Day Sugar-Free Jumpstart TODAY!
Say whaaaaaat? Sugar free gum! Had no idea! I really have to work on sodas. Great tips Tanya! I just had baby #3 so I'm definitely on a mission over here lol.
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\section*{Introduction}
At present, together with the dark matter problem, one of the most
tantalizing open questions in physics is the baryon-antibaryon
asymmetry, i.e. why are we living in a matter-dominated Universe?
Where did all the antimatter go? Different theoretical and
experimental efforts trying to address this question are ongoing,
including activities focusing on the gravitational behavior of
antimatter
\cite{TheALPHACollaboration2013,AEGISCollaboration2012,Perez2012,Kirch2014,Cassidy2014}.
No compelling theoretical argument seems to support that a
difference between the gravitational behavior of matter and
antimatter should be expected \cite{Nieto1991}, although some
attempts have been made to show the contrary
\cite{Scherk1979,Chardin1992,Chardin1993,Chardin1997,Kostelecky2011}.
Moreover observations and experiments have been interpreted as
evidence against the existence of ``antigravity'' type forces
\cite{Good1961,Pakvasa1989,Apostolakis1999,Gabrielse1999}. However
those could be argued to be model dependent and therefore a simple
free fall measurement is preferable. This justifies ongoing
experimental efforts in that direction. A first attempt in this
direction was made recently by the ALPHA collaboration
\cite{TheALPHACollaboration2013} that bounds the ratio of
gravitational mass to inertial mass of antihydrogen between $-$65
and 110.
The idea of directly measuring the gravitational force acting on
antiparticles in the Earth's field goes back many decades, from the
work of Witteborn and Fairbank, attempting to measure the
acceleration of gravity for electrons and, eventually, positrons
\cite{Witteborn1967} to the PS200 experiment at CERN in the 1980's
that included measurements on antiprotons \cite{Holzscheiter2014}.
Such measurements are extremely difficult because measuring the
force of gravity on a charged particle requires a physically
unrealistic (it would seem) elimination of stray electromagnetic
fields. The obvious solution to this problem is to use neutral
antimatter particles. However, at the present time it is not
technically feasible to do so; antineutrons cannot be produced in a
controllable manner and antineutrinos are similarly elusive to
experimenters. One may instead consider using composite systems that
are electrically neutral, in which case it is only necessary to
contend with dipole moments. Only a few systems that are composed
of, or contain some fraction of, antimatter are available for
scientific study. These are antihydrogen, muonium and positronium,
which have all been suggested as possible candidates for gravity
measurements \bibnote{See for example articles 4-6 in Antimatter and
Gravity, \emph{International Journal of Modern Physics: Conference
Series}, 30, January 2014}.
Selecting between different experimental methods, one should aim at
precision experiments as they are much more strongly motivated
theoretically. Among these, the method of quantum gravitational
spectroscopy stands out by its remarkable statistical sensitivity
and its cleanness from a systematic point of view.
Gravitational quantum states are solutions of the Schr\"odinger
equation in a gravity field above a surface. They are characterized
by the following energy ($E_n$) and spatial scale ($H_n$) :
\begin{align}
&E_n=\varepsilon_0 \lambda_n,\qquad \quad
\varepsilon_0=\sqrt[3]{\frac{\hbar^2M^2g^2}{2m}}~,
\label{eq:En}\\
&H_n=E_n/Mg ~,\label{eq:Hn} \\
&\mathop{\rm Ai}(-\lambda_n)=0,\qquad \lambda_n\approx\{2.34,4.09,
5.52, 6.79, 7.94, 9.02, 10.04...\}~.
\end{align}
Here $M$ is the gravitational mass of the particle, $m$ is its
inertial mass (we distinguish between $M$ and $m$ in view of
discussing EP tests), $g$ is the gravitational field
\emph{intensity} near the Earth surface, $\overline{g}=Mg/m$ is the
acceleration of the particle in that field and $\mathop{\rm Ai}(x)$
is the Airy function \cite{Abramowitz1972,Landau1965}. For neutrons
and antihydrogen atoms, the height of the lowest gravitational level
is 13.7 $\mu$ m. For positronium, whose mass is approximately 1000
times smaller, it extends over 1.3 mm. The frequency of transitions
between first and second quantum states equals 254 Hz for neutrons
and antihydrogen, and 26 Hz for positronium. The corresponding
characteristic times needed to form quantum states are 0.5 ms and 5
ms respectively.
Quantum gravitational states were observed for the first time with
neutrons by measuring their transmission through a slit made of a
mirror and an absorber in the GRANIT experiment
\cite{Nesvizhevsky2002nature}. If the distance between the mirror
and the absorber (which is a rough surface used as a scatterer to
mix the velocity components) is much higher than the turning point
for the corresponding gravitational quantum state, the neutrons pass
through the slit without significant losses. As the slit size
decreases the absorber starts approaching the size of the neutron
wave function and the probability of neutron loss increases. If the
slit size is smaller than the characteristic size of the neutron
wave function in the lowest quantum state, the slit is not
transparent for neutrons as was demonstrated experimentally.
Here we analyze in detail several experiments which will study the
free fall of anti-atoms and argue that the observation of
gravitational quantum states of antimatter is feasible. In section
\ref{sec:gbar} we describe the forthcoming $\overline{\mathrm{H}}$ experiment GBAR. We
explain in section \ref{sec:qrefl} the quantum reflection mechanism
which allows the formation of gravitational quantum states of $\overline{\mathrm{H}}$
above material surfaces. In section \ref{sec:shaper} we show how the
filtering scheme of the GRANIT experiment could be implemented in
GBAR and in section \ref{sec:gravstateshbar} we describe a possible
spectroscopy of gravitational quantum states of $\overline{\mathrm{H}}$. Section
\ref{sec:gravfallps} reviews the status of positronium free fall
experiment at UCL and section \ref{sec:gravstatesps} explores the
possibility of observing gravitational quantum states of
positronium.
\section{GBAR status report}\label{sec:gbar}
GBAR is an experiment in preparation at CERN. Its goal is to measure
the gravitational acceleration ($\overline{g}=Mg/m$) imparted to freely
falling antihydrogen atoms, in order to perform a direct
experimental test of the Weak Equivalence Principle with antimatter.
The objective is to reach a relative precision on $\overline{g}$ of $1\%$ in
a first stage, with the perspective to reach a much higher precision
using quantum gravitational states in a second stage, as is
described in section \ref{sec:gravstateshbar}.
The principle of the experiment is described in detail in
\cite{Chardin2011} and is briefly recalled here. It is based on an
idea proposed in \cite{Walz2004}. Antihydrogen ions $\overline{\mathrm{H}}^+$ are
produced, trapped and sympathetically cooled to around 10 $\mu$ K.
The excess positron is detached by a laser pulse, which gives the
start signal for the free fall of the ultracold antihydrogen atom
$\overline{\mathrm{H}}$. The $\overline{\mathrm{H}}$ subsequent annihilation on a plate is detected and
provides the information to measure $\overline{g}$. The choice of producing
$\overline{\mathrm{H}}^+$ ions to get ultracold antihydrogen atom is the specificity
of the GBAR experiment. It is very costly in statistics, but makes
the cooling to $\mu$ K temperatures a realistic aim.
We report in this section on three recent progresses in the
preparation of the experiment: estimations of the $\overline{\mathrm{H}}^+$ production
cross sections, accumulation of positrons, and cooling of the
$\overline{\mathrm{H}}^+$ ions.
\subsection{Production cross sections of $\overline{\mathrm{H}}^+$ ions}
The $\overline{\mathrm{H}}$ production proceeds in two steps: $\overline{p} + \mathrm{Ps} \to
\overline{\mathrm{H}} + \mathrm{e}^-$ (1) followed by $\overline{\mathrm{H}}+\mathrm{Ps} \to \overline{\mathrm{H}}^+ +
\mathrm{e}^-$ (2). The Ps symbol stands for positronium. The
cross-sections of these reactions are not well known and are very
low. The matter counterpart of the first one has been measured. It
is around $10^{-15}$~cm$^2$ (10$^9$~barn) for tens of keV protons
\cite{Merrison1997}. The second one is estimated to be around
$10^{-16}$~cm$^2$ (10$^8$~barn) \cite{Walters2007}.
New calculations of these reactions have been performed in which the
first excitations levels for the Ps (up to $n=3$) and the $\overline{\mathrm{H}}$ (up
to $n=5$ ) have been considered. The results suggest that the
production of $\overline{\mathrm{H}}^+$ can be efficiently enhanced by using either a
fraction of Ps(2p) and a 2 keV antiproton beam or a fraction Ps(3d)
and antiprotons with kinetic energy below 1 keV \cite{Comini2013}.
The product of the cross sections of reactions (1) and (2) reaches
values around $10^{-29}$~cm$^4$ (10$^{19}$~barn$^2$) for an
optimized fraction of excited Ps. Simulations are underway to
estimate the effective gain with a realistic experimental setup.
This shows that very low energy antiprotons are needed. The
Extremely Low Energy Antiproton (ELENA) ring which is in
construction at CERN and which will complement the Antiproton
Decelerator (AD), will provide 75 ns rms bunches of $5\times 10^{6}$
100 keV antiprotons every 100 s. Those have to be further
decelerated and cooled to match the GBAR requirements. The
decelerator is under construction at CSNSM (``Centre de Sciences
Nucl\'eaires et de Sciences de la Mati\`ere'') in Orsay, France.
\subsection{Positron accumulation}
In addition to a high flux of low energy antiprotons, the production
of $\overline{\mathrm{H}}^+$ via reactions (1) and (2) require to form a dense cloud
of positronium. It has been shown that Ps can be efficiently
produced by dumping few keV positrons on mesoporous silica films.
Yields of 30 to 40 $\%$ depending on the incident positron energy
(few keV) have been measured \cite{Liszkay2008, Cassidy2011}. The
accumulation of a very large number of positrons, around $2 \times
10^{10}$, between two ejections of antiprotons from ELENA is thus
necessary to produce a dense enough positronium cloud.
A demonstration facility for the production and accumulation of
positrons is currently running at CEA/Saclay. It consists of a low
energy electron linear accelerator (LINAC), a high field
Penning-Malmberg trap from the Atomic Physics Laboratory in RIKEN,
Japan, and a dedicated beam line for further studies of
positron-positronium conversion and for applications in material
science. In addition, a laser system is now being built at LKB
(``Laboratoire Kastler-Brossel'') in Paris to test the excitation of
the positronium which will be formed downstream of the trap.
The LINAC produces a 4.3 MeV electron bunched beam. The bunch length
is 200 $\mu$ s, and the LINAC runs at 200 Hz, producing a mean
current of 120 $\mu$ A. Electrons are sent onto a tungsten mesh
moderator. A flux of typically $3\times {{10}^{6}}$ slow (few 100
eV) positrons per second is driven towards the Penning trap through
a vacuum tube equipped with solenoid coils producing a 80 mT field.
They are accelerated to around 1 keV to enter the high magnetic
field (5 T) region. They reach the Penning trap which is made of 23
cylindrical electrodes, surrounded by 4 additional long electrodes
to control the admission and the trapping of incident particles.
Positrons make a round trip in less than 100 ns. In order to trap
them, it is necessary to compress the 200 $\mu$ s bunch. This is
done by applying a varying voltage (20 to 150 V) when extracting the
slow positron from the moderator. In this way, it is possible to
close the entrance of the trap before the bunch escapes. This method
allows to trap one single bunch.
In order to accumulate a large number of bunches, positrons have to
be slowed down and stored in a dedicated potential well formed by a
subset of electrodes of the trap (see Fig. \ref{fig:trapping})
before the next bunch arrives. Positrons are cooled by passing
through a preloaded electron plasma in another dedicated potential
well. This method has been set up and demonstrated in
\cite{Oshima2004} with a continuous positron beam issued from a
$^{22}$Na source. With such a beam, it is not possible to close the
entrance gate, and positrons must be slowed down in one step. This
was done with a remoderator downstream of the trap. An efficiency of
1$\%$ was obtained. With a bunched positron beam, the remoderator is
not necessary, and a much higher efficiency is expected.
The cooling time fixes the maximal LINAC frequency, and depends on
the density of the electron plasma. With ${{10}^{17}}$ e$^-$/m$^3$,
simulations show that the cooling time is around 3 ms.
\begin{figure}
\centering
\includegraphics[width=0.80\textwidth]{trapping.png}
\caption{\label{fig:trapping} The positron trapping mechanism (from
\cite{Grandemange2013}). The horizontal axis is the position along
the trap axis. The left vertical axis is the voltage seen by
particles. The magnetic field strength is shown by the green curve,
with the scale on the right vertical axis. In blue, the electron
potential well filled with electrons is drawn , reducing the
apparent voltage shown by black curve with the value on the left
vertical axis. In red is shown the positron potential well. When a
positron bunch arrives, the entrance electrode voltage is low
(dashed dotted line). It is then increased and positrons go back and
forth (it is depicted by red arrows) between this gate and the
downstream part of the trap. They pass many times through the
electron plasma, and are eventually slowed down and fall into their
well. The presence of residual $\mathrm{H} _{\mathrm{2}}
^{\mathrm{+}}$ helps the final catching of positrons.}
\end{figure}
The principle of this accumulation scheme has now been successfully
demonstrated at Saclay. The details of the experimental setup used
during accumulation are given in \cite{Grandemange2013}. The result
of a successful set of accumulation trials is shown in Fig.
\ref{fig:acc}.
\begin{figure}
\centering
\includegraphics[width=0.80\textwidth]{acc.png}
\caption{\label{fig:acc} The accumulation of positrons (from \cite{Grandemange2013}).}
\end{figure}
Given the characteristics of the demonstrator facility at Saclay, a
realistic objective is now to accumulate around 10$^8$ positrons in
the trap within 2 minutes.
\subsection{Cooling of the $\overline{\mathrm{H}}^+$ ions}
Recent progresses have been made for the design of the cooling of
$\overline{\mathrm{H}}^+$ions. The cooling proceeds in two steps: Doppler cooling at
the mK level, Raman side band cooling to reach 10 $\mu$ K.
In the first step, ions are captured in a linear Paul trap inside
which Be$^+$ ions are preloaded and laser cooled. In the original
scheme of GBAR, it was assumed that the $\overline{\mathrm{H}}^+$ ions would be cooled
by Coulomb interactions with the Be$^+$ ions (``sympathetic
cooling''). Simulations show that this process is very slow. This
is due to the large mass ratio between the ions. As a consequence,
one cannot reach the mK level in a short enough time to avoid the
destruction of the $\overline{\mathrm{H}}^+$: the laser cooling of Be$^+$ induces the
photodetachment of the excess positron in a fraction of a second.
However, the simulations show also that the addition of a third
species of ions of intermediate mass, namely HD$^+$ ions, make the
process efficient enough \cite{Hilico2014}. Starting with 1800
Be$^+$ and 200 HD$^+$ ions, cooling times of ms are achievable.
Second step: to reach the 10 $\mu$ K level necessary for the free
fall experiment, a Be$^+$/$\overline{\mathrm{H}}^+$ ion pair must be transferred to a
precision trap to undergo a ground state Raman side band cooling.
Calculations show that one may achieve the desired cooling in less
than a second. This is shown in \cite{Hilico2014} and references
therein. This method will be tested with matter ions (Ca$^+$ /Be$^+$
, H$_2 ^+$ /Be$^+$) before being implemented for the GBAR
experiment. Traps are being mounted at LKB in Paris and at Mainz
University.
Since the uncertainty on the measurement of $\overline{g}$ is fully dominated
by the initial velocity dispersion due to both the vertical velocity
after cooling and the recoil due to the positron photodetachment,
the implementation of a vertical velocity selector will allow a
drastic gain in the statistics needed to reach the 1$\%$ precision
on $\overline{g}$ as is described in section \ref{sec:shaper}.
\section{Quantum reflection of antihydrogen on material surfaces}\label{sec:qrefl}
In the ultracold regime, the interaction of antihydrogen ($\overline{\mathrm{H}}$)
atoms with a surface is governed by the phenomenon of quantum
reflection. Although the atoms are strongly attracted to the
surface, the atomic wave function can be partly reflected on the
steep atom-surface potential, leading to a non-zero probability of
classically forbidden reflection. This effect is relevant to
experiments such as GBAR where ultracold $\overline{\mathrm{H}}$ atoms are detected
by annihilation on a plate (see section \ref{sec:gbar}).
A single atom placed in vacuum near a material surface experiences
an attractive Casimir-Polder (CP) force \cite{Casimir1946,
Casimir1948}. This force is a manifestation of the electromagnetic
quantum fluctuations which are coupled to the atomic dipole. Quantum
reflection occurs if an atom impinges with low velocity on such a
rapidly varying potential \cite{Friedrich2002}. We will give a more
explicit condition later on.
In this section we first describe how the CP potential is calculated
for realistic experimental conditions. We then go on to compute the
scattering amplitudes of an atom on this potential. We show that
quantum reflection can be understood as a deviation from the
semiclassical approximation. Finally we describe materials from
which quantum reflection is enhanced and above which gravitationally
bound states of $\overline{\mathrm{H}}$ could be observed.
\subsection{Calculation of the Casimir-Polder potential}
We use the scattering approach to Casimir forces
\cite{Lambrecht2006} to give a realistic estimation of the
atom-surface interaction energy. In this approach the interacting
objects are described by reflection matrices for the electromagnetic
field. Reflection on a plane is described by Fresnel coefficients,
while reflection on the atom is treated in the dipolar approximation
and depends on the dynamic polarizability \cite{Messina2009}. This
allows an evaluation of the CP potential for any material when its
optical properties are known. Those used here are detailed in
\cite{Dufour2013qrefl}. Note that since the typical length scale for
quantum reflection ($\sim$100 nm) is below the thermal wavelength at
300 K ($\sim$1 $\mu$ m), we carried out all calculations at null
temperature.
\begin{figure}
\centering
\includegraphics[width=0.80\textwidth]{potential.png}
\caption{\label{fig:CPpotential} Casimir-Polder (CP) potential for
antihydrogen in the vicinity of a material bulk; from top to
bottom, perfect conductor (full line), silicon (dashed line), silica
(dotted line); (inset:ratio $V/V^*$ to the retarded potential $V^*$
for a perfectly conducting mirror, see text).}
\end{figure}
The CP potential for $\overline{\mathrm{H}}$ at a distance $x$ of a perfectly
conducting plane and thick silicon and silica slabs are presented in
Fig. \ref{fig:CPpotential}. For a perfectly conducting mirror in the
long-distance regime we recover the historic result of Casimir and
Polder \cite{Casimir1946,Casimir1948}:
\begin{equation}
V(x) \underset{x \gg \lambda}{\approx} V^*(x) = - \frac{3 \hbar c}{8
\pi x^4} \frac{\alpha(0)}{4\pi\epsilon_0}~,
\end{equation}
where $\alpha(0)$ is the static polarizability of the atom.
For real mirrors, the potential is reduced but it shows the same
power law dependence in the van der Waals (short distance) and
retarded (long distance) regimes:
\begin{equation}
V(x) \underset{x \ll \lambda}{\approx} - \frac{C_3}{x^3} ~, \qquad
\qquad V(x) \underset{x \gg \lambda}{\approx} -\frac{C_4}{x^4}~,
\end{equation}
where $\lambda$ is a typical wavelength associated with the optical
response of atom and plane.
\subsection{Scattering on the Casimir-Polder potential}
We now solve the Schr\"odinger equation for an atom of energy $E>0$
scattering on the CP potential $V(x)$ :
\begin{equation}
\frac{\textrm{d}^2 }{\textrm{d} x^2} \psi(x) +
\frac{p(x)^2}{\hbar^2} \psi(x) = 0 ~,
\end{equation}
with $p(x) = \sqrt{2 m (E - V(x))}$ the classical momentum. We write
the exact wave function as a sum of counter-propagating WKB waves
whose coefficients are allowed to vary:
\begin{equation}\label{eq:wkbbasis}
\psi(x) = \frac{c_{in}(x)}{\sqrt{p(x)}}
\exp \left(-\frac{i}{\hbar} \int^x p(x ') \mathrm{d} x ' \right)
+ \frac{c_{out}(x)}{\sqrt{p(x)}} \exp \left(\frac{i}{\hbar}
\int^x p(x ') \mathrm{d} x ' \right)~.
\end{equation}
Upon insertion in the Schr\"odinger equation we obtain coupled first
order equations for the coefficients $c_{in}(x),c_{out}(x)$
\cite{Berry1972}. The annihilation of $\overline{\mathrm{H}}$ on the material surface
translates as a fully absorbing boundary condition on the surface:
$c_{out}(x=0)=0$. This is in contrast with matter atoms, for which
more complicated surface physics is involved in the boundary
condition. Close to the surface, the energy becomes negligible
compared with the potential, which takes the van der Waals form and
$c_{in}(x),c_{out}(x)$ can be solved for analytically
\cite{Dufour2013qrefl}. The equations are then integrated
numerically until $c_{in}(x),c_{out}(x)$ become constants.
The reflection probability
$|r|^2=\underset{x\to\infty}{\lim}|c_{out}(x)/c_{in}(x)|^2$ is
plotted against the energy $E$ in Fig. \ref{fig:reflectivity} for
various semi-infinite media. Note that the quantum reflection
probability is larger for materials with a weaker CP interaction,
such as silica.
\begin{figure}
\centering
\includegraphics[width=0.80\textwidth]{reflectivity.png}
\caption{\label{fig:reflectivity} Quantum reflection probability
$|r|^2$ as a function of the energy for antihydrogen atoms on bulk
mirrors; from bottom to top, perfect conductor (full line), silicon
(dashed line), silica (dotted line).}
\end{figure}
\subsection{The badlands function}
To understand this surprising result we look more closely at what
distinguishes the exact solution of the Schr\"odinger equation from
the reflectionless WKB approximation. If $c_{in},c_{out}$ are no
longer allowed to vary, one can show that the wave function
\eqref{eq:wkbbasis} obeys a modified Schr\"odinger equation where
$p(x)^2$ is replaced by $\tilde p(x)^2 = p(x)^2(1+Q(x))$
\cite{Berry1972}. $Q(x)$ is known as the badlands function since the
WKB approximation is not valid in regions where it is
non-negligible:
\begin{equation}
Q(x) = \frac{\hbar^2}{2 p(x)^2} \left(\frac{p''(x)}{p(x)}
- \frac{3}{2} \frac{p'(x)^2}{p(x)^2} \right)~.
\end{equation}
For the CP potential, the badlands function exhibits a peak in the
region where $|V(x)|=E$ but goes to zero both as $x\to\infty$ (where
the potential cancels) and as $x\to0$ (where the classical momentum
diverges).
\begin{figure}
\centering
\includegraphics[width=0.80\textwidth]{badlands.png}
\caption{\label{fig:badlands} The badlands function $Q(x)$ for an
antihydrogen atom with energy $E=10$~neV; from bottom to top,
perfect conductor (full line), silicon (dashed line), silica (dotted
line).}
\end{figure}
As the energy is decreased, the semiclassical approximation breaks
down and the badlands function's peak becomes larger. But for a
given energy, the peak is larger and closer to the surface when the
potential is weak, as shown in Fig. \ref{fig:badlands}. The
difference between exact and WKB solutions is larger in weaker CP
potentials, leading to enhanced quantum reflection.
\subsection{Enhancing quantum reflection}
Quantum reflection first appears as a bias in the context of the
GBAR experiment, since it tends to exclude low energy atoms from the
statistics. However this phenomenon opens perspectives for the
storage and guiding of antimatter with material walls. With this in
mind we consider materials which couple weakly to the
electromagnetic field and are therefore good mirrors for atoms, as
we have seen in the previous paragraph.
A simple strategy is to remove matter from the reflective medium, by
using thin slabs or porous materials for example. Our versatile
approach allowed us to compute the CP interaction near thin slabs,
an undoped graphene sheet \cite{Dufour2013qrefl} and nanoporous
materials \cite{Dufour2013porous}. The latter consist in a solid
matrix which forms an array of nanometric pores. Aerogels, which are
obtained by supercritically drying a silica gel, are a well known
example. We also consider porous silicon and powders of diamond
nanoparticles formed by explosive shock.
From a distance larger than the typical pore size, such materials
can be modeled as homogeneous effective media with properties
averaged between that of vacuum and of the solid matrix. In
consequence their effective dielectric constant is extremely low, as
a result of which quantum reflection is exceptionally efficient. In
table \ref{tab:lifetimes} we show the lifetime of an antihydrogen
atom in the first gravitationally bound state above a surface (see
section \ref{sec:gravstateshbar} for more details).
\begin{table}
\centering
\begin{tabular}{c|c}
Surface (porosity) & Lifetime (s)\\
\hline\hline
perfect conductor & 0.11\\ \hline
bulk silicon & 0.14\\ \hline
bulk silica & 0.22\\ \hline
nano-diamond powder (95\%) & 0.89\\ \hline
porous silicon (95\%) & 0.94\\ \hline
silica aerogel (98\%) & 4.6
\end{tabular}
\caption{\label{tab:lifetimes} The lifetime of the first
gravitationally bound state of antihydrogen above various surfaces.}
\end{table}
Note that this approach does not take into account the possible
presence of stray charges on the surface, a question that would have
to be addressed to observe the predicted reflection probabilities.
Moreover, the effective medium approximation is applicable only for
low atom velocities, such that the atom is reflected far enough from
the surface. With these caveats, nanoporous materials are an
outstanding candidate for the manipulation and study of antihydrogen
and its gravitationally bound states, with lifetimes above the
second.
\section{Shaping of vertical velocity components of antihydrogen atoms for GBAR}
\label{sec:shaper}
The main source of uncertainty on the determination of $\overline{g}$ in the
GBAR experiment is the width of the vertical velocity distribution
of the atom at the beginning of the free fall. This spread in
velocities is due to the quantum uncertainty on the momentum of
$\overline{\mathrm{H}}^+$ in the ground state of the harmonic Paul trap and to the
additional recoil associated with the photo-detachment of the extra
positron (see section \ref{sec:gbar}).
In this section we give an estimation of the uncertainty on the
arrival time associated with the initial vertical velocity spread
and show how it can be reduced by filtering out the fastest atoms.
Since slow antihydrogen atoms bounce on material surfaces thanks to
quantum reflection (see previous section), the filtering scheme used
in GRANIT with ultracold neutrons \cite{Nesvizhevsky2002nature} can
also be applied in GBAR.
\subsection{Width of the arrival time distribution}
We consider a wave-packet falling in a linear gravitational
potential and want to determine the arrival time distribution on a
fixed horizontal plane, supposing there is no reflection from that
(ideal) detector. In this case classical and quantum calculations
give identical results, as can be seen by noticing that the Wigner
quasi-distribution function obeys the classical equations of motion
if the potential is at most quadratic. Therefore a given initial
phase-space distribution simply propagates along the classical
trajectories.
For a wave-packet initially centered at a height $H$ above the
detector, with zero mean velocity and uncorrelated vertical position
and velocity distributions of width $\Delta z$ and $\Delta v$
respectively, the spread of the arrival time distribution is
\begin{equation}
\frac{\Delta t}{t_H}= \sqrt{{\left(\frac{\Delta z}{2H}\right)}^2 +
{\left(\frac{\Delta v}{\sqrt{2 \overline{g} H}}\right)}^2}~,
\label{eq:deltat}
\end{equation}
with $t_H=\sqrt{2 H/\overline{g}}$ the classical free fall time. This
translates as a statistical uncertainty $\Delta \overline{g}/\overline{g}=2\Delta
t/\sqrt{N}t_H$ on the determination of $\overline{g}$ after $N$ independent
measurements.
If the particle is initially in the ground state of a harmonic trap,
the distribution is Gaussian and saturates the Heisenberg
inequality: $\Delta z\Delta v=\hbar/2m$. Then the time uncertainty
is minimal for
\begin{equation} \Delta v=\Delta v_{opt} =
\sqrt{\frac{\hbar}{2m} \sqrt{\frac{\overline{g}}{2 H}}}~.
\end{equation}
For $H=30$~cm and $\overline{g}=g$ this evaluates to $\Delta v_{opt} \approx
3.6\times 10^{-4}$~m/s, and the relative uncertainty on the arrival
time is $2\times 10^{-4}$. However the current expected value for
GBAR is three orders of magnitude larger $\Delta v_0\approx
0.5$~m/s, which leads to a relative uncertainty of $0.2$.
The uncertainty in GBAR is largely dominated by the vertical
velocity dispersion. If the initial velocity dispersion can be
reduced from $\Delta v_0$ to $\Delta v$ by filtering out the hottest
atoms, the single-shot precision and the number of atoms are both
reduced by a factor $\Delta v/\Delta v_0$. Despite the loss in
statistics, this results in a net reduction of the statistical
uncertainty on $\overline{g}$.
\subsection{Shaping of the vertical velocity distribution}
Our proposal \cite{Dufour2014} to realize this filtering is to let
the atoms pass through a horizontal slit between two disks. The
bottom disk has a smooth top surface on which atoms reflect with
high probability whereas the top disk has a rough bottom surface
which effectively acts as an absorber for the atoms (see Fig.
\ref{fig:shaper}). Antihydrogen is initially trapped in the center
of the two disks (openings are made in the center of the disks to
allow operation of the trap). If its vertical velocity is high
enough to reach the rough surface, it is reflected non-specularly
and remains inside the device until it annihilates with high
probability. On the contrary, if it cannot reach the top disk the
atom will exit the device with high probability after bouncing on
the bottom mirror a few times. It then falls freely to a detector a
height $H$ below. Since the horizontal velocity is conserved, the
knowledge of the total time between photo-detachment and
annihilation and of the total horizontal $L$ distance traveled,
allows one to correct for the time spent inside the device before
the free fall.
If $h$ is the height of the slit, the velocity spread at the output
is $\Delta v\approx \sqrt{2 \overline{g} h}$ and the proportion of atoms that
exit the device is $N/N_0 \approx \Delta v/\sqrt{2\pi}\Delta v_0$.
Using the shaping device therefore reduces the statistical
uncertainty on $\overline{g}$ by a factor which scales as $h^{1/4}$.
\begin{figure}
\centering
\includegraphics[width=0.80\textwidth]{shaper.png}
\caption{\label{fig:shaper} Scheme of the proposed device to reduce the vertical velocity spread of the falling wavepacket (see text).}
\end{figure}
Classically going to ever smaller slit heights leads to arbitrarily
good precision. For example for $h=1$~mm, $\Delta v \approx
0.14$~m/s, $N/N_0\approx 5\%$ and the accuracy is improved by a
factor 2, whereas for $h=50$~$\mu$ m, $\Delta v \approx 0.03$~m/s,
$N/N_0\approx 1\%$ and the accuracy improved by a factor 4.
There are however two limits: the number of repetitions of the
experiment must be large enough that at least some atoms make it
through the filter, but more fundamentally, the wave function of the
atom must fit inside the slit. Indeed for slit sizes below 50 $\mu$
m the discrete spectrum of states bound by gravity must be taken
into account. For a slit size of 20 $\mu$ m only the ground state
can travel through the guide, below that the transmission drops to
zero.
This fact has been used to demonstrate the existence of
gravitationally bound states for neutrons
\cite{Nesvizhevsky2002nature}. The next section explores the
possibilities of similar experiments with antihydrogen to further
increase the precision of equivalence principle tests on antimatter.
\section{Resonance spectroscopy of gravitational states of
antihydrogen near material surface}
\label{sec:gravstateshbar}
In this section we will study a motion of an $\overline{\mathrm{H}}$ atom, localized
in a gravitational state near a horizontal plane mirror. The
existence of such states though counterintuitive is explained by the
phenomenon of quantum reflection of ultracold (anti-)atoms from a
steep attractive Casimir-Polder atom-surface potential. Such states
have similar properties with those discovered for neutrons
\cite{Nesvizhevsky2002nature,Nesvizhevsky2000,Nesvizhevsky2003,Nesvizhevsky2005,Voronin2006}.
To account for the interaction of $\overline{\mathrm{H}}$ with a material wall, the
gravitational quantum states \eqref{eq:En} receive a complex energy
shift $\varepsilon_0 \Delta$, with $\Delta\simeq-i0.005$ for a
perfectly conducting wall \cite{Voronin2011}. All states therefore
acquire equal width, which is a function of a material surface
substance $\Gamma=2|\Delta|\varepsilon_0$. This width corresponds to
the lifetime of $0.1$~s in case of a perfectly conducting surface
and is twice longer for silica
\cite{Voronin2005pra,Dufour2013qrefl,Dufour2013porous} for instance.
The interest to study gravitational quasi-stationary states of $\overline{\mathrm{H}}$
is due to their comparatively long life-time on one hand and easy
identification of certain state because of it's mesoscopic spatial
scale. This opens an interesting perspective to apply potentially
very precise resonance spectroscopy method to establish the
gravitational properties of anti-atoms. These methods are based on
inducing an observation of resonance transitions between
gravitational states. One of possible approach is to use an
alternating inhomogeneous magnetic field for such a purpose.
The interaction of a magnetic field with a ground state $H$ atom
moving through the field \cite{Lamb1952,Gorkov1968,Lozovik2004} is
dominated by the interaction of an average magnetic moment of the
atom \cite{Landau1965} in a given hyperfine state with the magnetic
field. We are going to focus on an alternating magnetic field with a
gradient in the vertical direction. This condition is needed for
coupling the field and the center of mass (c.m) $\overline{\mathrm{H}}$ motion in the
gravitational field of the Earth. It allows one to induce resonant
transitions between quantum gravitational states of $\overline{\mathrm{H}}$
\cite{Voronin2011}.
We will consider the magnetic field in the following form:
\begin{equation}\label{Magn1}
\vec{B}(z,x,t)=B_0 \vec{e}_z+ \beta \cos(\omega t) \left(z
\vec{e}_z-x \vec{e}_x \right)~.
\end{equation}
Here $B_0$ is the amplitude of a constant, vertically aligned,
component of magnetic field, $\beta$ is the value of magnetic field
gradient, $z$ is a distance measured in the vertical direction, $x$
is a distance measured in the horizontal direction, parallel to the
surface of a mirror. A time-varying magnetic field (\ref{Magn1}) is
accompanied with an electric field ($[\vec{\nabla} \vec{E}] =
-\frac{1}{c} \partial \vec{B} / \partial t$). However, for the
velocities of ultracold atoms, corresponding interaction terms are
small and thus will be omitted.
An inhomogeneous magnetic field couples the spin and the spatial
degrees of freedom. A $\overline{\mathrm{H}}$ wave function is described in this case
using a four-component column (in a non-relativistic treatise) in
the spin space, each component being a function of the c.m.
coordinate $\vec{R}$, relative $\overline{p}-\overline{e}$ coordinate $\vec{\rho}$ and
time $t$. The corresponding Schr\"{o}dinger equation is:
\begin{equation} \label{Schr}
i \hbar\frac{\partial \Phi_{\alpha}(\vec{R},\vec{\rho},t)}{\partial
t} = \sum_{\alpha'}\left[ -\frac{\hbar^2}{2m}\Delta_R+Mgz+V_{CP}(z)
+ \widehat{H}_{in}+\widehat{H}_m \right]_{\alpha, \alpha'}
\Phi_{\alpha'}(\vec{R},\vec{\rho},t)~.
\end{equation}
A subscript $\alpha$ in this equation indicates one of four spin
states of the $\overline{p}-\overline{e}$ system. The meaning of the interaction terms
is the following. $V_{CP}(z)$ is an atom-mirror interaction
potential, which turns into the Casimir-Polder potential at an
asymptotic atom-mirror distance (see
\cite{Voronin2005pra,Voronin2005} and references therein).
$\widehat{H}_{in}$ is the Hamiltonian of the internal motion, which
includes the hyperfine interaction:
\begin{equation}\label{Hc}
\widehat{H}_{in} = -\frac{\hbar^2}{2
\mu}\Delta_\rho-e^2/\rho+\frac{\alpha_{HF}}{2}\left(\hat{F}^2-3/2\right)~.
\end{equation}
Here $\mu=m_1m_2/m$, $m_1$ is the antiproton mass, $m_2$ is the
positron mass, $m=m_1+m_2$, $\alpha_{HF}$ is the hyperfine constant,
$\hat{F}$ is the operator of the total spin of the antiproton and
the positron. We will treat only $\overline{\mathrm{H}}$ atoms in a $1S$-state (below
we will show that the excitation of other states in the studied
process is improbable). The term
$\frac{\alpha_{HF}}{2}\left(\hat{F}^2-3/2\right)$ is a model
operator, which effectively accounts for the hyperfine interaction
and reproduces the hyperfine energy splitting correctly. The term
$\widehat{H}_m$ describes the field-magnetic moment interaction:
\begin{equation}\label{Hm}
\widehat{H}_m = -2\vec{B}(z,x,t)\left( \mu_{\overline{e}}\hat{s}_{\overline{e}}\times
\hat{I}_{\overline{p}}+ \mu_{\overline{p}}\hat{s}_{\overline{p}}\times \hat{I}_{\overline{e}}\right)~.
\end{equation}
Here $\mu_{\overline{e}}$ and $\mu_{\overline{p}}$ are magnetic moments of the
positron and the antiproton respectively, $\hat{s}_{\overline{e}}$,
$\hat{s}_{\overline{p}}$ is a spin operator, acting on spin variables of
positron (antiproton), $\hat{I}_{\overline{e}}$, $\hat{I}_{\overline{p}}$ is a
corresponding identity operator. As far as the field
$\vec{B}(z,x,t)$ changes in space and in time, this term couples the
spin and the c.m. motion.
We will assume that in typical conditions of a spectroscopy
experiment the $\overline{\mathrm{H}}$ velocity component $v$ parallel to the mirror
surface (directed along $x$-axis) is of the order of a few $m/s$ and
is much larger than a typical vertical velocity in lowest
gravitational states (which is of the order of $cm/s$). We will
treat the motion in a frame moving with the velocity $v$ of the
$\overline{\mathrm{H}}$ atom along the mirror surface. Thus we are going to consider
the x-component motion as a classical motion with a given velocity
$v$, and we will substitute a $x$-dependence by a $t$-dependence. We
will also assume that $B_0\gg \beta L$, where $L\sim 30$~cm is a
typical size of an experimental installation of interest. This
condition is needed for "freezing" the magnetic moment of an atom
along the vertical direction; it provides the maximum transition
probability.
We will be interested in the weak field case, such that the Zeeman
splitting is much smaller than the hyperfine level spacing $\mu_B
B_0\ll \alpha_{HF}$. The hierarchy of all mentioned above
interaction terms could be formulated as follows:
\begin{equation}\label{hierar}
m_2e^2/\hbar^2 \gg \alpha_{HF}\gg \mu_{\overline{e}} |B_0 |\gg E_n~,
\end{equation}
and thus it justifies the use of the adiabatic expansion for solving
Eq. (\ref{Schr}); it is based on the fact that an internal state of
an $\overline{\mathrm{H}}$ atom follows adiabatically the spatial and temporal
variations of an external magnetic field. Neglecting non-adiabatic
couplings, an equation system for the amplitude $C_n(t)$ of a
gravitational state $\psi_n(z)$ has the form:
\begin{equation}\label{Adiab}
i \hbar \frac{d C_{n}(t)}{dt} = \sum_{k} C_{k}(t) V_{n,k}(t)\exp
\left(-i\omega_{n k} t\right )~.
\end{equation}
The transition frequency $\omega_{n k}=(E_k-E_n)/\hbar$ is
determined by the gravitational energy level spacing. This fact is
used in the proposed approach to access the gravitational level
spacing by means of scanning the applied field frequency, as will be
explained in the following.
Within this formalism the role of the coupling potential $V(z,t)$ is
played by the energy of an atom in a fixed hyperfine state thought
of as a function of (slowly varying) distance $z$ and time $t$.
\begin{equation}
V_{n,k}(t)=\int_0^\infty \psi_n(z) \psi_{k}(z) E(t,z)dz~.
\end{equation}
Here $\psi_n(z)$ is the gravitational state wave function, which is
known in terms of the Airy function \cite{Voronin2011}.
The energy $E(z,t)$ is the eigenvalue of the internal and magnetic
interactions $\widehat{H}_{in}+\widehat{H}_m$, where the c.m.
coordinate $\vec{R}$ and time $t$ are treated as slow-changing
parameters. Corresponding expressions for the eigen-energies of a
$1S$ manifold are:
\begin{eqnarray}\label{Ea}
E_{a,c}&=& E_{1s} -
\frac{\alpha_{HF}}{4}\mp\frac{1}{2}\sqrt{\alpha_{HF}^2 +
|(\mu_B-\mu_{\overline{p}})B(z,t)|^2},
\\ \label{Eb} E_{b,d}&=& E_{1s}+\frac{\alpha_{HF}}{4}\mp
\frac{1}{2}|(\mu_B+\mu_{\overline{p}})B(z,t)|.
\end{eqnarray}
Subscripts $a,b,c,d$ are standard notations for hyperfine states of
a $1S$ manifold in a magnetic field. The presence of a constant
field $B_0$ produces the Zeeman splitting between states $b$ and
$d$. As far as the energy of states $b,d$ depends on magnetic field
linearly, while for states $a,c$ it depends quadratically, only
transition between $b,d$ states take place in case of a weak field.
In the following we will consider only transitions between
gravitational states in a $1S( b,d)$ manifold.
A qualitative behavior of the transition probability is given in the
Rabi formula, which can be deduced by means of neglecting the high
frequency terms compared to the resonance couplings of only two
states, initial $i$ and final $f$, in case the field frequency
$\omega$ is close to the transition frequency
$\omega_{if}=(E_f-E_i)/\hbar$:
\begin{equation}\label{Rabbi}
P = \frac{1}{2}\frac{(V_{if})^2}{(V_{if})^2 +
\hbar^2(\omega-\omega_{if})^2}
\sin^2\left(\frac{\sqrt{(V_{if})^2+\hbar^2(\omega-\omega_{if})^2}}{2\hbar}t\right)
\exp(-\Gamma t)~.
\end{equation}
The factor $1/2$ appears in front of the right-hand side of the
above expression due to the fact that only two $(b,d)$ of four
hyperfine states participate in the magnetically induced
transitions.
It is important that the transition frequencies $\omega_{if}$ do not
depend on the anti-atom-surface interaction up to the second order
in the splitting $\Delta$. This is a consequence of the already
mentioned fact that all energies of gravitational states acquire
equal shift due to the interaction with a material surface.
A resonant spectroscopy of $\overline{\mathrm{H}}$ gravitational states could consist
of observing $\overline{\mathrm{H}}$ atoms localized in the gravitational field above
a material surface at a certain height as a function of the applied
magnetic field frequency. A ``flow-through type'' experiment,
analogous to the one discussed for the spectroscopy of neutron
gravitational states \cite{Kreuz2009}, includes three main steps. A
sketch of a principle scheme of an experiment proposed in
\cite{Dufour2014} is shown in Fig. \ref{FigSketch} (see also
section \ref{sec:shaper}).
\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{sketchRes3.png}
\caption{A sketch of the principle scheme of an experiment on
magnetically induced resonant transitions between $\overline{\mathrm{H}}$
gravitational states. 1 - a source of ultracold antihydrogen, 2 - a
mirror, 3 - an absorber, 4 - a magnetic field, 5 - a
detector.}\label{FigSketch}
\end{figure}
First, an atom of $\overline{\mathrm{H}}$ is shaped in a ground gravitational state.
This is achieved by means of passing $\overline{\mathrm{H}}$ through a slit, formed by
a mirror and an absorber, which is placed above the mirror at a
given height $H_a$. The mirror and the absorber form a waveguide
with a state-dependent transmission \cite{Voronin2006}. The choice
of $H_a=H_1\simeq 13.6$~$\mu$ m implies that only $\overline{\mathrm{H}}$ atoms in
the ground gravitational state pass through the slit. Second, $\overline{\mathrm{H}}$
atoms are affected by an alternating magnetic field (\ref{Magn1})
while they are moving parallel to the mirror. An excited
gravitational state is resonantly populated. Third, the number of
$\overline{\mathrm{H}}$ atoms in an excited state is measured by means of counting the
annihilation events in a detector, which is placed at a height $H_d$
above the mirror. The value of $H_d$ is chosen to be larger than the
spatial size of the gravitational ground state and smaller than the
spatial size of the final state (\ref{eq:Hn}), $ H_1\ll H_d<H_f$, so
that the ground state atoms pass through, while atoms in the excited
state are detected.
We present a simulation of the number of detected annihilation
events as a function of the field frequency in Fig. \ref{FigTrans}
for the transition from the ground to the $6$-th excited state,
based on a numerical solution of the equation system Eq.
(\ref{Adiab}). The corresponding resonance transition frequency is
$\omega=972.46$~Hz. The value of the field gradient, optimized to
obtain the maximum probability of $1\rightarrow 6$ transition during
the time of flight $t_{fl}=\tau=0.1$~s, turned to be equal
$\beta=27.2$~Gs/m, the corresponding guiding field value, which
guarantees the adiabaticity of the magnetic moment motion, is
$B_0=30$~Gs.
\begin{figure}
\centering
\includegraphics[width=0.8\textwidth]{ResLine1.png}
\caption{The transition probability as a function of the magnetic
field frequency for the transition from the ground state to $6$-th
gravitational state.}\label{FigTrans}
\end{figure}
It follows from (\ref{eq:En}) that the $\overline{\mathrm{H}}$ gravitational mass
could be deduced from the measured transition frequency
$\omega_{nk}$ as follows:
\begin{equation}
M=\sqrt{\frac{2m\hbar \omega_{nk}^3 }{g^2(\lambda_k-\lambda_n)^3}}~.
\end{equation}
Let us mention that $g$ in the above formula means the gravitational
field intensity near the Earth surface, a value which characterizes
properties of the field and is assumed to be known with a high
precision. At the same time all the information about gravitational
properties of $\overline{\mathrm{H}}$ is included in the gravitational mass $M$.
Equality of the gravitational mass $M$ and the inertial mass $m$,
imposed by the Equivalence principle, results in the following
expression:
\begin{equation}
M=\frac{2\hbar \omega_{nk}^3 }{g^2(\lambda_k-\lambda_n)^3}~.
\end{equation}
Estimation of the accuracy of the above expression requires account
of different effects, including dynamical Stark shift of the
resonance line, non-adiabatic corrections to the transition
probability, interaction of alternating magnetic field with a
mirror, etc. The detailed study of different systematic effects is
under way. Assuming that the spectral line width is determined by
the lifetime $\tau\approx 0.1$~s of gravitational states, we
estimate that the gravitational mass $M$ can be deduced with the
relative accuracy $\epsilon_M\sim 10^{-3}$ for $100$ annihilation
events for the transition to the $6$-th state.
\section{Gravitational free fall of cold positronium}\label{sec:gravfallps}
Antihydrogen, muonium and positronium are the possible candidates
for gravity measurements on antimatter, with various pros and cons.
Antihydrogen and muonium \cite{Kirch2014} are extremely difficult to
produce, requiring large facilities (i.e., PSI, CERN), whereas
positronium is relatively easy to produce in smaller university
laboratories. However, Ps has an inconvenient propensity to
self-annihilate; the triplet ground state vacuum lifetime of only
142 ns, would seem to preclude using this system for a free fall
measurement. As has been pointed out by various authors, in
particular A. P. Mills, Jr., \cite{MillsJr.1989}, this is not the
case, since one need only excite Ps atoms into long-lived Rydberg
states to prevent self-annihilation. Indeed, for any Ps state with
$n>1$ the radiative lifetime is always less than the annihilation
lifetime\footnote{The only excited state for which this is not true
is the metastable 2s state.}. That is to say, for excited states the
overlap of the positron and electron wave functions is sufficiently
low that annihilation can be considered to be negligible (see Fig.
\ref{fig:lifetimesps}).
\begin{figure}
\centering
\includegraphics[width=0.80\textwidth]{PsLifetime.png}
\caption{\label{fig:lifetimesps} Radiative lifetimes of various Ps
states as a function of the principal quantum number $n$. The
lifetimes were calculated by summing the Einstein $A$ coefficients
of all electric-dipole-allowed decay channels from each Rydberg
state. For each $A$ coefficient the appropriate radial integrals
were determined using analytic expressions for the radial wave
functions in a pure Coulomb potential \cite{Bethe1957}. The dashed
line is the annihilation lifetime of $n$s states. After Ref.
\cite{Cassidy2014}. }
\end{figure}
The radiative lifetimes of excited Ps states, shown in Fig.
\ref{fig:lifetimesps} , are almost twice those of the corresponding
states in hydrogen. For practical reasons the smallest Ps beam
deflections one can expect to observe will be 10's of micrometers or
more. Therefore, if Ps falls with the usual gravitational
acceleration, it would be necessary to produce states with lifetimes
of the order of a few ms to observe such deflections. As is evident
from Fig. \ref{fig:lifetimesps}, achieving such long radiative
lifetimes requires either exciting low $l$ Rydberg levels (i.e., s
or d) to extremely high principal quantum numbers, or going to lower
$n$ states (perhaps around $n$ = 30 or so) and then transferring the
atoms to circular states \bibnote{For a discussion of the properties
of circular states, and methods for producing them, see
\cite{Gallagher1994}} (or, if not true circular states, at least
states with higher angular momentum).
Aside from the creation of sufficiently long-lived Rydberg levels,
conducting a Ps free fall experiment will require solving many other
problems. In order to accomplish an experiment of the type first
outlined by Mills and Leventhal \cite{MillsJr.2002} it will be
necessary to produce a small (10-50 micron) ``point'' source of slow
positronium in a cryogenic environment. The resulting long-lived
Rydberg atoms will then have to be formed into a beam, perhaps by
electrostatic manipulation (focusing, and deceleration) via their
electric dipole moments \cite{Seiler2011}, and finally detected with
good spatial resolution (as a function of flight time) in order to
observe a deflection due to gravity. Possible methods to accomplish
some of these tasks are considered elsewhere
\cite{Cassidy2014,MillsJr.2002}.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{PsScheme.png}
\caption{\label{fig:expscheme} A schematic representation of a
Mills-Leventhal type of Ps free fall experiment. A real experiment
will undoubtedly be significantly different from this illustration,
which is intended only to highlight some of the different steps
involved. Of distinct practical concern will be the need to keep the
apparatus at low temperatures to mitigate effects of black body
radiation, as well as minimizing the Ps speed, which will determine
the length of the flight path, and hence the experiment. }
\end{figure}
The production of Ps Rydberg states with principle quantum numbers
around 30 can be accomplished using a two-step process
(1s$\to$2p$\to n$d), and has already been experimentally
demonstrated using broad-band ($\sim$ 100 GHz) lasers to accommodate
the large Doppler-broadened width of the transitions
\cite{Jones2014}. However, this methodology is not well suited to
the requirement that these atoms are subsequently transferred to
higher angular momentum states, and in order to achieve the required
state selectivity it may be necessary to use a different excitation
mechanism; i.e., a Doppler-free two-photon transition from the
ground state directly to a well-defined Rydberg Stark state
\bibnote{This has not yet been demonstrated for positronium; the
feasibility of doing so is considered in T. E. Wall, D. B. Cassidy
and S. D. Hogan, to be published.}.
As is well known, Rydberg atoms exhibit exaggerated properties
\cite{Gallagher1994} (see table \ref{tab:rydbergs}). In the present
case this is critical, since we seek to produce Ps atoms with very
long lifetimes, and also to take advantage of the large electric
dipole moments of Rydberg atoms to create and control an atomic
beam. However, insofar as we are compelled to make use of
electrically neutral systems to measure the weak gravitational force
acting on antimatter particles without extraneous electromagnetic
fields dominating their motion, excitation to states with very large
dipole moments brings us back to the original problem of extraneous
field effects. The situation is considerably less dire when dealing
with electric dipoles (and, to a much lesser extent, magnetic
dipoles) since in this case only field gradients give rise to
forces. Nevertheless, in an experiment designed to probe the weak
force of gravity with Rydberg atoms, forces due to stray fields must
be taken into account.
\begin{table}
\centering
\begin{tabular}{c|c|c|c|c}
& $n$-scaling & Ps & H & He \\ \hline\hline Binding energy (meV) &
$n^{-2}$ & -7.56 & -15.11 & -15.12\\ \hline State separation (meV) &
$n^{-3}$ & 0.48 & 0.96 & 0.96\\ \hline Orbital radius ($a_0$) &
$n^2$ & 2694 & 1347 & 1347 \\ \hline Radiative lifetime ($\mu$ s) &
$n^3$ & 28.4 & 14.2 & 14.2\\ \hline Dipole moment/mass ($e
a_0$/amu) & $n^2$ & 2.2 $\times 10^6$ & 1206 & 304
\end{tabular}
\caption{\label{tab:rydbergs}The $n$-dependence of several
properties of Rydberg atoms, with examples shown for the 30d state
of Ps, H and He. The state separation is calculated for 30d $\to$
31d. The orbital radius is defined here as the expectation value
$\braket{r} = \frac{1}{2} (3 n_{eff}^2 -l(l+1))$, where $n_{eff}$
includes the relevant quantum defect. The electric dipole
moment-to-mass ratios are calculated for the outermost state of the
$n$ = 30, $m$ = 2 Stark manifold. The radiative lifetime
$n$-dependence applies only to low $l$ states: for circular states
the scaling is closer to $n^5$ (see Fig. \ref{fig:lifetimesps}). }
\end{table}
When an atom is placed in an external electric field of strength $F$
and direction $z$, the field mixes the atom's angular momentum
states. To first order the state $\ket{n,l,m}$ is mixed with states
of adjacent $l$ but the same $n$ and $m$, \cite{Gallagher1988}. The
resulting Stark states repel each other, causing them to spread out
as the electric field strength is increased, as shown in Fig.
\ref{fig:starkstates}. Following the example of hydrogen, where the
first order Stark shift is analytically calculable, the
Schr\"odinger equation for an atom in an electric field can be
written in cylindrical coordinates, where the relevant quantum
numbers are $n$, $m$ and the parabolic quantum numbers $n_1$ and
$n_2$, which together satisfy the condition $n= n_1 + n_2 + |m| +1$.
The Stark states in a given $\ket{n,m}$ manifold are described by
the index $k=n_1-n_2$, where $k$ has values in the range from
$k_{min}=-(n-|m|-1)$ up to $k_{max}=n-|m|-1$ (with $\Delta k = 2$).
The first order Stark shift in Ps is given by $E_S=-\mu \times F$,
where the electric dipole moment has magnitude $
|\mu|=\frac{3}{2}n|k| a_{\mathrm{Ps}}$ (where the Ps Bohr radius
$a_{\mathrm{Ps}}$ is (almost) twice that of hydrogen, i.e., $2
a_0$). For a high-$n$ Rydberg state with low angular momentum, for
example the 30d state with $m = 2$, the value of the electric dipole
moment can be very large. The Stark state with $k_{max}=27$ has an
electric dipole moment of $2430 e a_0$. This large electric dipole
moment arises because within this $n$-state there are many
degenerate angular momentum states with the same value of $m$ that
are coupled by the electric field. While this is extremely useful
for atomic control \cite{Seiler2011,Vliegen2006,Hogan2008} it
presents a significant problem for gravity measurements, since the
electric field gradient experienced by a Ps atom in this state that
would result in a force equal to that of ``normal'' gravity ($\sim
2\times10^{-29}$~N ) is only $\sim 10^{-3}$~V/m$^2$. Although this
is by no means insignificant, it does compare favorably with the
$\sim 5\times 10^{-11}$~V/m electric field that would apply a
$g$-like force to a bare positron (or electron).
States with the maximum absolute values for the orbital and magnetic
quantum numbers for a given $n$, the so-called circular states,
experience no first-order Stark shift. For these states $m=|n|-1$,
meaning that there is only one Stark state associated with this
value of $m$, which has $k=0$, and thus, to first order, \emph{no
electric dipole moment} (see Fig. \ref{fig:starkstates}). The
explanation for this is that, within a given $n$-manifold, the
circular states have unique values of $m$, and thus are not coupled
to any other degenerate angular momentum states. In the classical
limit these states correspond to circular orbits, in which the
average $z$-position of the electron is zero, resulting in no
electric dipole moment, unlike the lower angular momentum Rydberg
states where, in the classical limit, the electronic orbit is highly
anisotropic, with the electron having a large average displacement
from the atomic core. Although there is no atomic core or nucleus in
the case of Ps, the wave function is nevertheless hydrogenic, and
the same arguments apply. The circular states do experience a
second-order Stark shift, from coupling of adjacent $n$-states,
however, this shift is extremely weak. It should, therefore, be
possible to produce Ps states with high angular momentum and
minimize the effects of stray fields while simultaneously extending
the lifetimes to useful levels. However, any manipulation techniques
that rely on large dipole moments will obviously have to be
performed after the optical excitation to the relevant $n$ states,
but before transferring the atoms to states with high angular
momentum.
\begin{figure}
\centering
\includegraphics[width=0.80\textwidth]{StarkShift.png}
\caption{\label{fig:starkstates} Stark states of $n$ = 30 and 31
states of Ps, with $m$ = 2 (grey dashed) and $m$ = 29 (black). In
the $n$ = 30 level the $m$ = 29 state is a circular state and
experiences no first-order Stark shift and only a very weak
second-order shift, as explained in the text.}
\end{figure}
Performing a gravity measurement on any system containing antimatter
is clearly very challenging, and many of the potential obstacles are
currently being investigated. The ability to produce controllable
beams of Ps atoms may also open the door to other types of
experiments, such as interferometry
\cite{Phillips1997,Oberthaler2002}, which could provide an
alternative route to an antimatter gravity measurement.
\section{Can we observe gravitational quantum states of positronium~? }
\label{sec:gravstatesps}
Positronium is about 1000 times lighter than a neutron or
antihydrogen. Therefore the expected height of the gravitational
quantum state is 100 larger corresponding to a macroscopic size of
$H_1 =1.3$~mm while the energy is 10 times smaller, $E_1=0.13$~peV
(see Eqs. \ref{eq:En}-\ref{eq:Hn}). The observation time to resolve
a quantum gravitational state can be estimated using the Heisenberg
uncertainty principle to be of the order of $\hbar/E_1 \simeq
4.5$~ms. This value is much larger than the long lived triplet
positronium lifetime in the ground state which is 142 ns (the Ps
singlet state only lives 125 ps and thus in the following we will
only consider the triplet state and refer to it as Ps). Hence, as
for the case of a measurement of the gravitational free fall of Ps
described in the previous section, the Ps lifetime can be increased
by excitation to a higher level. A possible scheme to observe the Ps
gravitational quantum states could employ the flow-through technique
used for the first observation of this effect with neutrons (see
Fig. \ref{fig:SchemeQMBounce}). Greater detail of the proposed
experimental set-up and technique are described in a dedicated
contribution to this workshop \bibnote{P. Crivelli, V.V.
Nesvizhevsky, and A.Yu. Voronin. Can we observe the gravitational
quantum states of Positronium? \emph{Advances in High Energy
Physics}, this issue, 2014.}. Here we describe the main idea.
\begin{figure}
\centering
\includegraphics[width=\textwidth]{SchemeQMBounce.png}
\caption{\label{fig:SchemeQMBounce} Possible scheme for the
observation of the gravitational quantum states of positronium.}
\end{figure}
Positronium is formed by implanting keV positrons from a
re-moderated pulsed slow positron beam in a positron-positronium
converter. To observe the quantum mechanical behavior of Ps in the
gravitational field its vertical velocity should be of the same
order of the gravitational energy levels and thus $v_y<0.15$~m/s.
Furthermore to resolve the quantum state the Ps atom has to interact
long enough with the slit and therefore it has to be laser excited
to a Rydberg state with $n>30$ and maximum $l$ quantum number (see
previous section). To keep a reasonable size of the experimental
setup (i.e a slit size of the order of 0.5~m) and minimize the
number of detectors the velocities in the horizontal plane should be
smaller than $v_{x,z}<100$~m/s. Similar to neutrons a collimator
could be used to select the velocity components $v_x,v_y$ of the
positronium distribution. However since no reliable thermal cold
source of positronium exists the velocity component perpendicular to
the surface $v_z$ has to be lowered by some other means. Relying on
the fact that atoms in Rydberg states have a large dipole moment
Stark deceleration can be used for this purpose. This method has
been demonstrated for different atomic species (including hydrogen)
\cite{Hogan2008} and molecules \cite{Hogan2009}. Atoms in Rydberg
states have large dipole moments thus electric field gradients can
be used to manipulate them. The acceleration/deceleration $a$
imparted to the Rydberg atoms is given by:
\begin{equation}
a=76 \nabla F \frac{1}{m} n k ~,
\end{equation}
where $\nabla F$ is the gradient of the electric field in
V~cm$^{-2}$, $m$ the mass of the decelerated particles in atomic
units, $n$ and $k$ the Stark state quantum numbers. H atoms in
$n=25$ and an initial velocity of 700 m/s can be brought at rest in
3 mm \cite{Hogan2008}.
For Ps being 1000 times lighter decelerations exceeding $10^9$~m/s
could be realized and therefore the vertical velocity of Ps emitted
from thin silica films with initial velocities of the order of
$10^5$~m/s \cite{Cassidy2010,Crivelli2010} could be reduced to below
100 m/s. Since one is interested only in decelerating the
distribution that is almost perpendicular to the surface of the Ps
target, one can expect for those atoms an efficiency close to 100
\%. This is confirmed by preliminary simulations \bibnote{Private
communication with Dr. C. Seiler.}. The collimator will be placed
after the deceleration stage and the microwave region where
circularly polarized radiation will spin up the Ps to the maximum
$l$ so that kicks to the momentum imparted to the atoms in the
vertical direction during these processes will be accounted for.
The fraction of atoms with $v_y<0.15, v_x,v_z<100$~m/s is estimated
to be of the order of $2 \times 10^{-9}$. After the collimator the
Ps will fly through the slit made of a mirror and the absorber. If
the distance between them is smaller than the first expected
gravitational state (i.e $<1$~mm) this will not be transparent and
therefore no signal will be detected above the expected background
in the detectors. If the width of the slit is increased to a value
lying between the first and the second gravitational state (i.e.
$<2$~mm) the Ps wave function can propagate and a signal is expected
to be detected via field-ionization and subsequent detection with
MCPs. This quantum jump would provide the unambiguous indication of
the observation of a quantum gravitational state of positronium.
As a mirror for Ps we propose to exploit a gradient of magnetic
field created using wires arranged parallel to each other with a
constant current to create a uniform gradient of the magnetic field.
Only the Ps triplet atoms with $m=0$ have a non-zero net magnetic
moment. For the $m=\pm1$ the electron and the positron magnetic
moments cancel and therefore those are insensitive to the magnetic
field. Therefore, only one third of the initial population will be
reflected. To equate the $E_y = 0.1$~peV a field of few mG at the
wire surface will be sufficient.
Because of the large spacial size of gravitational quantum states
and the very large characteristic length of the mirror needed to
form the gravitational states that is much larger than a
characteristic inter-wire distance, we expect that the very weak
magnetic gradient will not perturb the gravitational states. The
strict theoretical analysis of this clearly mathematically defined
problem is ongoing. A matter mirror could also be considered. Due to
the large spacial size of the gravitational quantum state, the
surface potential is expected to be very sharp and therefore result
in efficient quantum reflection (see section \ref{sec:qrefl}).In
both cases (magnetic or material mirror) we expect to have
effectively (quasi-classically) only a few collisions with the
surface. Nevertheless, the transitions rates due to quenching and
ionization caused by the electric or magnetic fields have to be
calculated. The absorber as for the neutrons is a rough surface on
which the impinging Ps will mix its velocity components and
therefore be lost.
With such a scheme assuming a mono-energetic slow positron beam flux
of $9\times 10^8$ e$^+$/s (this the highest intensity reported so
far reached at the FMR~II NEMOPUC source in Munich
\cite{Hugenschmidt2008}) an event rate of 0.8 events/day with a
background 0.05 events/day might be achievable with a realistic
extrapolation of current technologies. Possible losses due to
spurious effects like stray electric or magnetic fields or black
body radiation seems to be negligible but as for the case of a free
gravity fall further calculations and preliminary experiments should
be done to confirm this assumption and that all the required
efficiencies (e.g. Ps excitation in the $n=33$, $l=32$ state) can be
attained.
To note that the expected height of the gravitational state is
related to the gravitational mass $M$ by Eq. \eqref{eq:Hn}. This
means that for an uncertainty in the determination of $H_1$ of
$\delta H_1$ one can get an accuracy in the determination of $M$ at
the level of $\delta M/M= 3 \delta H_1/H_1\sqrt{N}$ where $N$ is the
number of detected signals. Assuming an uncertainty of $\delta H_1 =
0.1$~mm which is mainly determined by the finite source size the
value of $\delta M/M$ can be determined to 3\% in three months. This
is comparable to the accuracy that is aimed for antihydrogen
experiments at CERN
\cite{Kellerbauer2008,Vanderwerf2014,Zhmoginov2013}. Therefore,
observation of Ps gravitational quantum states offers a
complementary approach to test the effect of gravity on a pure
leptonic system. Most of the techniques required for such an
experiment are under development for the ongoing free gravity fall
experiment of Ps (see section \ref{sec:gravfallps}) and Rydberg Ps
deceleration experiments are being considered at ETH Zurich where
Prof. B. Brown's (Marquette University) buffer gas trap is being
commissioned. The advantage of using gravitational quantum states is
that unpredicted perturbations of the Ps atoms will not result in a
systematic effect for the experiment but will only affect the signal
rate. Therefore as for the case of antihydrogen this approach seems
promising to provide a much higher accuracy than a free fall
experiment.
\section*{Conclusion}
In this review, we have reported the progress of ongoing experiments
to measure gravitational free fall of antimatter. The GBAR
experiment will produce antihydrogen atoms in the ultracold regime
where quantum reflection from surfaces takes place. Quantum
reflection will allow the observation of gravitational quantum
states of antimatter that promise to lead to a very sensitive probe
of the effect of gravity on anti-atoms (2 orders of magnitude
improvement compared to the free fall experiments).
The techniques developed in experiments designed to produce a cold
beam of Ps for a free fall measurement will also eventually find
application in creating ultracold Ps atoms, as required for
observing gravitational quantum states. They will also enable a wide
variety of other experimental areas, such as precision spectroscopy.
Antimatter atoms in gravitational quantum states also provide a
unique opportunity to constrain experimentally extra short-range
forces between the mirror and the anti-atom with about the same
sensitivity as we do for normal matter \cite{Antoniadis2011}.
\section*{Acknowledgements}
The authors wish to thank the GRANIT collaboration and the GBAR
collaboration (\href{http://gbar.in2p3.fr}{gbar.in2p3.fr}) for
providing excellent possibilities for discussions and exchange. In
addition to the involved institutes and laboratories, preparatory
work for GBAR has been funded by the ``Conseil G\'en\'eral de
l'Essonne'', by the ``Agence Nationale de la Recherche'' in France,
and by the LABEX P2IO.
P.C. acknowledges the support by the Swiss National Science
Foundation (grant PZ00P2\_132059) and ETH Zurich (grant
ETH-47-12-1). DBC and TEW gratefully acknowledge funding from the
EPSRC (EP/K028774/1), the Leverhulme Trust (RPG-2013-055) and the
ERC (CIG 630119).
\bibliographystyle{unsrt}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 9,973
|
Q: Programmatically apply Knockout bindings to specific tags within a text block I have a block of html markup that I'd like to databind to a div using Knockout JS. This is simple enough with the knockout html binding.
What I'd like to do is conditionally bind child elements within that html markup, based on their class. These elements can appear anywhere within the html markup, as they are inline elements (like span, a tags, etc).
The purpose of this is to databind click events to specific words or phrases within this text block, so that I can call a function (in this case, a modal with a definition of the target word or phrase).
Does anyone know how to do this? From what I can see, I have two main avenues to explore:
*
*Use jQuery to grab the elements by class, and somehow databind to then after the fact. I don't know if this will work for various
reasons - not having the DOM ready, not being able to databind in
that method, etc.
*Use knockout's template binding, and add an afterRender template callback that parses the text block for the elements and databinds
them accordingly.
*A combination of the two above. (Yeah, I know, that's three avenues.)
Has anyone done anything like this before? I'd like to get your advice and feedback, if possible. Thank you!
A: I don't recommend this approach unless you are sure that the html you are binding to is safe. The html binding looks like this.
ko.bindingHandlers['html'] = {
'init': function() {
// Prevent binding on the dynamically-injected HTML
// (as developers are unlikely to expect that, and it
// has security implications)
return { 'controlsDescendantBindings': true };
},
'update': function (element, valueAccessor) {
var value = ko.utils.unwrapObservable(valueAccessor());
ko.utils.setHtml(element, value);
}
};
Looking at this you could write a html2 binding
ko.bindingHandlers['html2'] = {
'init': function() {
return { 'controlsDescendantBindings': false };
},
'update': function (element, valueAccessor) {
var value = ko.utils.unwrapObservable(valueAccessor());
ko.utils.setHtml(element, value);
}
};
http://jsfiddle.net/madcapnmckay/LDtuF/1/
Be careful with this, you want to make sure the html cannot be used for insecure purposes.
Hope this helps.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,843
|
"Quid pro quo" means "this for that" in Latin. So most trainers will explain that it's inappropriate for managers to condition job benefits on the employee providing sexual favors or making the rejection of a sexual advance the basis for an adverse employment decision. But in the real world, "quid pro quo" harassment doesn't usually result from a manager's calculated decision about how to bargain his or her authority for sexual favors. In the real world, it's often the result of a man misperceiving whether a female subordinate is sexually interested.
Some psychologists think there's an evolutionary basis for this. Carin Perilloux, a psychologist at Williams College, hypothesizes that overconfident men were more likely to "go for it" (my words, not hers) and therefore had more opportunities to pass on their genes. As a result, as reported in Discovery News, "men are more likely to walk away from an interaction with a woman thinking that she's into him, while the woman thinks, 'Well that was a nice friendly conversation.'"
This situation isn't hard to imagine. A male supervisor is thinking about his interactions with his younger, female subordinate. He thinks about how pleasant she is towards him, how she appears to pay close attention to what he says and even laughs at the jokes that his wife tells him are stupid. He's wondering, is it because he's her boss and she has a powerful economic incentive to stay in his good graces? Or is it because she thinks he's hot?
This is where men's innate tendency to overestimate women's interest comes into play. If he pursues a relationship and she rejects his advances, is he going to treat her the same? Because, even if it's subconscious, if he becomes more critical of her work, less inclined to give her desirable assignments, or looks for ways to get her out of the organization, then that's pretty blatant quid pro quo harassment. The motive was not a calculated effort to use his authority to get sexual favors. It was his mistaken belief that she would be receptive to his advances.
Training on quid pro quo harassment needs to help managers understand this dynamic. As a start, perhaps we should teach that "quid pro quo" is Latin for "No. I won't sleep with you for a raise. You're gross. Now leave so I can call my lawyer."
|
{
"redpajama_set_name": "RedPajamaC4"
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| 131
|
Q: Android: Get user-provided information sent to a private email How would I get user-provided information (e.g. product feedback) from an android app to be sent to a private email account?
I currently run a website and we would like to develop an android app that allows users to input product feedback. We would like this feedback to then be sent or stored in a way that is only accessible to certain people. It doesn't necessarily have to be sent to email, it just needs to be private. We just want the product feedback to be available for us for review and comment.
Is there any place I could go to get started thinking about this?
A: Assume you have a EditText in your layout file, you get the text in your activity like this, where the et_some_user_input is the id of the EditText in your layout file.
EditText etSomeUserInput = (EditText)findViewById(R.id.et_some_user_input);
String someUserInput = etSomeUserInput.getText().toString();
Now you have the user input in someUserInput, you can send this string like this:
Intent i = new Intent(Intent.ACTION_SEND);
i.setType("message/rfc822");
i.putExtra(Intent.EXTRA_EMAIL , new String[]{"recipient@private.com"});
i.putExtra(Intent.EXTRA_SUBJECT, "subject of email");
i.putExtra(Intent.EXTRA_TEXT , someUserInput);
try {
startActivity(Intent.createChooser(i, "Send mail..."));
} catch (android.content.ActivityNotFoundException ex) {
Toast.makeText(MyActivity.this, "There are no email clients installed.", Toast.LENGTH_SHORT).show();
}
Reference of the send email code.
This will make the user to fire up an email client such as google email, or yahoo email or any other email client the user choose. In case, you want the user input to be sent to your private email or somewhere you can access to without going through an email client. One suggestion is to set up a server that accepts user messages, so in your android app, you can send this message to your server, and you server can save that message into some sort of database, and you can then set up an admin web app to connect to that database to see what's being submitted by the user, or can you query that database directly.
If you are not sure what I am talking about, you will need to do some research to
How to set up a server to accept REST service calls in Java, PHP, Ruby, Python,
Node.js or whatever programming language you prefer?
How to make REST service calls in Android?
If you don't have the resources to set up the server, then you can try to ask the internet if it's possible to send an email in Android without going through an email client.
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 80
|
Great ResignationInflationSupply ChainsLeadership
Retail ·Department Stores
J.C. Penney Shares Hit All-Time Low as Turnaround Fails
BYPhil Wahba
J.C. Penney's (JCP) turnaround has officially turned around.
The department store's shares fell 22% on Friday morning to less than $4—heading for a new all-time low when trading opens—after it reported a fourth straight quarter of comparable sales declines and a wider net loss hurt by how cheaply it has sold items in liquidation from the dozens of stores it has closed.
The company tried to put a brave face on it with CEO Marvin Ellison touting "significant acceleration" in its kids' apparel and a purportedly strong start to back-to-school, but investors were having none of it, concerned by Penney's ability to get any leverage from its manifold efforts to revive the once iconic chain's fortunes.
Penney, which has closed 127 of its 1,000-plus stores in the last year, saw its net loss net loss widen to $62 million, or 20 cents per share, in the second quarter ended July 29, from $56 million, or 18 cents per share, a year earlier. So far this year, Penney has had a net loss of $242 million, almost twice what it was at the end of the second quarter last year.
The results come on the heels of difficult reports for rivals like Macy's (M), which on Thursday announced a 10th straight quarter of sales declines, Dillard's (DDS) and to a lesser extent Kohl's. (KSS) Department stores have been grappling with consumer indifference, weak traffic at many malls and merchandise overlap.
But given Penney's high level of indebtedness—long term debt is $3.8 billion, enormous for an unprofitable company generating less than $12 billion in annual sales—these results are particular worrisome to investors concerned Penney is unable to fix its business, and they have punished the stock accordingly. A decade ago, Penney was worth $19 billion on the stock market; today it is barely worth $1.5 billion.
Penney has not been sitting idly by: it has reconfigured its home good area to focus on appliances and home services to become less reliant on apparel, it has staked a claim on the plus-size apparel market with new brands and it recently launched a new loyalty program.
Yet its business is failing to capitalize on those initiatives or the free fall of mall rival Sears (SHLD). Ellison, who became CEO two years ago, called out home goods, jewelry, shoes, handbags, Sephora beauty shops and its salon business as better performers, but made no mention of appliances, a big bet by the former Home Depot (HD) executive, in his statement. He will discuss Penney's results later Friday morning on a conference call.
Looking ahead, Penney said it still expects comparable sales, a key metric that strips out the impact of newly closed stores, would range from a 1% drop to a 1% increase for the full year but did warn investors cost of goods sold would be higher, a pressure point on its earnings.
Net sales in the second quarter climbed 1.5 percent, to $2.96 billion, slightly above forecasts in a rare bright spot in the report.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,304
|
Nant (toponimo francese) è una frazione di 231 abitanti del comune svizzero di Mont-Vully, nel Canton Friburgo (distretto di Lac).
Geografia fisica
Storia
Formata dai villaggi di Nant-Dessous e Nant-Dessus, nel 1850 è stata unita alle altre località di Chaumont, Praz e Sugiez, chiamate collettivamente fino al 1831 Commune générale des quatre villages de La Rivière, per formare il comune di Vully-le-Bas (dal 1977 Bas-Vully), il quale a sua volta il 1º gennaio 2016 è stato accorpato all'altro comune soppresso di Haut-Vully per formare il nuovo comune di Mont-Vully, del quale Nant è il capoluogo.
Note
Collegamenti esterni
Frazioni di Mont-Vully
Località del Canton Friburgo
|
{
"redpajama_set_name": "RedPajamaWikipedia"
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| 7,250
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{"url":"https:\/\/gigaom.com\/2011\/06\/17\/tech-companies-to-begin-cleaning-water-at-japan-nuclear-plants\/","text":"# Tech companies to begin cleaning water at Japan nuclear plants\n\nA group of nuclear tech companies are poised to begin cleaning the contaminated water in the turbine buildings at the Fukushima nuclear power plants in Japan that suffered damage in the wake of the earthquake and tsunami. Japanese conglomerates Hitachi and Toshiba, French nuclear provider AREVA, and Silicon Valley startup Kurion plan to start cleaning the massive amount of water \u2014 tens of millions of gallons \u2014 that have been pumped into the turbine buildings and has now been contaminated by nuclear materials, and is filled with debris, oil and salt (from seawater).\n\nThe group has been racing to officially start the cleansing process as soon as possible, hoping to officially begin at noon on Friday Japan time (8 p.m. Thursday night PDT). Speed is essential, because the rainy season just started in Japan, and if the contaminated water overflows, it could damage the area environmentally even more. There\u2019s also the concern that more earthquakes could occur in the area, which could also cause the water to overflow. \u201cWe delivered the technology to TEPCO in five weeks, which is a fraction of the time it took to start cleaning water at Three Mile Island,\u201d Kurion\u2019s CEO John Raymont, told me in an interview.\n\nFor three-year-old Kurion, working on TEPCO\u2019s nuclear cleanup is a game-changing deal. The company, which is backed by Lux Capital, and Firelake Capital, is the only American company and the only startup tackling the problem. Kurion has already run two tests on the contaminated water at the Japanese nuclear plants using its cleanup material (they call it ion specific media). One test it ran on its own, and one in collaboration with the complete tech cleanup crew. The tests were successful, \u201cmet the criteria, the performance and the flow rates,\u201d said Raymont.\n\nNow the colossal process of cleaning some 90 million gallons will begin imminently. First Toshiba\u2019s tech will remove the oil and debris, then Kurion\u2019s material comes in and soaks up radioactive cesium and iodine and then AREVA\u2019s technology soaks up radioactive strontium. Kurion\u2019s \u201cmedia\u201d contains the radioactive waste, and shrinks it down to a small enough size so that it can be turned into glass, a process called vitrification. Vitrification permanently encapsulates the nuclear waste so it can be stored and transported more easily, and is the standard way that nuclear waste is dealt with.\n\nKurion\u2019s business model is based on making vitrification modular, which makes it cheaper, faster and more efficient. Often, the standard vitrification process requires the contaminated materials to be moved to a centralized plant, but Kurion\u2019s process brings the technology to the contaminated materials.\n\nNuclear waste management is a problem that hasn\u2019t seen a whole lot of innovation over the past few decades. According to some estimates, $1 out of every$4 from the Department of Energy\u2019s budget goes toward nuclear waste management, so there is a sizable opportunity to help the DOE cut that expense. Now with the Japanese nuclear disaster, there\u2019s an immediate market.\n\nImages courtesy of TEPCO.","date":"2021-09-19 23:19:55","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 1, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.23023450374603271, \"perplexity\": 3400.1000321351703}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-39\/segments\/1631780056902.22\/warc\/CC-MAIN-20210919220343-20210920010343-00317.warc.gz\"}"}
| null | null |
package org.apache.beam.sdk.io.cassandra;
import com.datastax.driver.core.Cluster;
import com.datastax.driver.core.ConsistencyLevel;
import com.datastax.driver.core.PlainTextAuthProvider;
import com.datastax.driver.core.QueryOptions;
import com.datastax.driver.core.ResultSet;
import com.datastax.driver.core.Row;
import com.datastax.driver.core.Session;
import com.datastax.driver.core.policies.DCAwareRoundRobinPolicy;
import com.datastax.driver.core.policies.RoundRobinPolicy;
import com.datastax.driver.core.policies.TokenAwarePolicy;
import com.datastax.driver.core.querybuilder.QueryBuilder;
import com.datastax.driver.core.querybuilder.Select;
import com.datastax.driver.mapping.Mapper;
import com.datastax.driver.mapping.MappingManager;
import com.google.common.annotations.VisibleForTesting;
import java.io.IOException;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.NoSuchElementException;
import org.apache.beam.sdk.io.BoundedSource;
import org.slf4j.Logger;
import org.slf4j.LoggerFactory;
/**
* An implementation of the {@link CassandraService} that actually use a Cassandra instance.
*/
public class CassandraServiceImpl<T> implements CassandraService<T> {
private static final Logger LOG = LoggerFactory.getLogger(CassandraServiceImpl.class);
private static final long MIN_TOKEN = Long.MIN_VALUE;
private static final long MAX_TOKEN = Long.MAX_VALUE;
private static final BigInteger TOTAL_TOKEN_COUNT =
BigInteger.valueOf(MAX_TOKEN).subtract(BigInteger.valueOf(MIN_TOKEN));
private class CassandraReaderImpl<T> extends BoundedSource.BoundedReader<T> {
private final CassandraIO.CassandraSource<T> source;
private Cluster cluster;
private Session session;
private ResultSet resultSet;
private Iterator<T> iterator;
private T current;
public CassandraReaderImpl(CassandraIO.CassandraSource<T> source) {
this.source = source;
}
@Override
public boolean start() throws IOException {
LOG.debug("Starting Cassandra reader");
cluster = getCluster(source.spec.hosts(), source.spec.port(), source.spec.username(),
source.spec.password(), source.spec.localDc(), source.spec.consistencyLevel());
session = cluster.connect();
LOG.debug("Query: " + source.splitQuery);
resultSet = session.execute(source.splitQuery);
final MappingManager mappingManager = new MappingManager(session);
Mapper mapper = mappingManager.mapper(source.spec.entity());
iterator = mapper.map(resultSet).iterator();
return advance();
}
@Override
public boolean advance() throws IOException {
if (iterator.hasNext()) {
current = iterator.next();
return true;
}
current = null;
return false;
}
@Override
public void close() {
LOG.debug("Closing Cassandra reader");
if (session != null) {
session.close();
}
if (cluster != null) {
cluster.close();
}
}
@Override
public T getCurrent() throws NoSuchElementException {
if (current == null) {
throw new NoSuchElementException();
}
return current;
}
@Override
public CassandraIO.CassandraSource<T> getCurrentSource() {
return source;
}
}
@Override
public CassandraReaderImpl<T> createReader(CassandraIO.CassandraSource<T> source) {
return new CassandraReaderImpl<>(source);
}
@Override
public long getEstimatedSizeBytes(CassandraIO.Read<T> spec) {
try (Cluster cluster = getCluster(spec.hosts(), spec.port(), spec.username(), spec.password(),
spec.localDc(), spec.consistencyLevel())) {
if (isMurmur3Partitioner(cluster)) {
try {
List<TokenRange> tokenRanges = getTokenRanges(cluster,
spec.keyspace(),
spec.table());
return getEstimatedSizeBytes(tokenRanges);
} catch (Exception e) {
LOG.warn("Can't estimate the size", e);
return 0L;
}
} else {
LOG.warn("Only Murmur3 partitioner is supported, can't estimate the size");
return 0L;
}
}
}
/**
* Actually estimate the size of the data to read on the cluster, based on the given token
* ranges to address.
*/
@VisibleForTesting
protected static long getEstimatedSizeBytes(List<TokenRange> tokenRanges) {
long size = 0L;
for (TokenRange tokenRange : tokenRanges) {
size = size + tokenRange.meanPartitionSize * tokenRange.partitionCount;
}
return Math.round(size / getRingFraction(tokenRanges));
}
@Override
public List<BoundedSource<T>> split(CassandraIO.Read<T> spec,
long desiredBundleSizeBytes) {
try (Cluster cluster = getCluster(spec.hosts(), spec.port(), spec.username(), spec.password(),
spec.localDc(), spec.consistencyLevel())) {
if (isMurmur3Partitioner(cluster)) {
LOG.info("Murmur3Partitioner detected, splitting");
return split(spec, desiredBundleSizeBytes, getEstimatedSizeBytes(spec));
} else {
LOG.warn("Only Murmur3Partitioner is supported for splitting, using an unique source for "
+ "the read");
String splitQuery = QueryBuilder.select().from(spec.keyspace(), spec.table()).toString();
List<BoundedSource<T>> sources = new ArrayList<>();
sources.add(new CassandraIO.CassandraSource<T>(spec, splitQuery));
return sources;
}
}
}
/**
* Compute the number of splits based on the estimated size and the desired bundle size, and
* create several sources.
*/
@VisibleForTesting
protected List<BoundedSource<T>> split(CassandraIO.Read<T> spec,
long desiredBundleSizeBytes,
long estimatedSizeBytes) {
long numSplits = 1;
List<BoundedSource<T>> sourceList = new ArrayList<>();
if (desiredBundleSizeBytes > 0) {
numSplits = estimatedSizeBytes / desiredBundleSizeBytes;
}
if (numSplits <= 0) {
LOG.warn("Number of splits is less than 0 ({}), fallback to 1", numSplits);
numSplits = 1;
}
LOG.info("Number of splits is {}", numSplits);
double startRange = MIN_TOKEN;
double endRange = MAX_TOKEN;
double startToken, endToken;
endToken = startRange;
double incrementValue = endRange - startRange / numSplits;
String splitQuery;
if (numSplits == 1) {
// we have an unique split
splitQuery = QueryBuilder.select().from(spec.keyspace(), spec.table()).toString();
sourceList.add(new CassandraIO.CassandraSource<T>(spec, splitQuery));
} else {
// we have more than one split
for (int i = 0; i < numSplits; i++) {
startToken = endToken;
endToken = (i == numSplits) ? endRange : (startToken + incrementValue);
Select.Where builder = QueryBuilder.select().from(spec.keyspace(), spec.table()).where();
if (i > 0) {
builder = builder.and(QueryBuilder.gte("token($pk)", startToken));
}
if (i < (numSplits - 1)) {
builder = builder.and(QueryBuilder.lt("token($pk)", endToken));
}
sourceList.add(new CassandraIO.CassandraSource(spec, builder.toString()));
}
}
return sourceList;
}
/**
* Get a Cassandra cluster using hosts and port.
*/
private Cluster getCluster(List<String> hosts, int port, String username, String password,
String localDc, String consistencyLevel) {
Cluster.Builder builder = Cluster.builder()
.addContactPoints(hosts.toArray(new String[0]))
.withPort(port);
if (username != null) {
builder.withAuthProvider(new PlainTextAuthProvider(username, password));
}
if (localDc != null) {
builder.withLoadBalancingPolicy(
new TokenAwarePolicy(new DCAwareRoundRobinPolicy.Builder().withLocalDc(localDc).build()));
} else {
builder.withLoadBalancingPolicy(new TokenAwarePolicy(new RoundRobinPolicy()));
}
if (consistencyLevel != null) {
builder.withQueryOptions(
new QueryOptions().setConsistencyLevel(ConsistencyLevel.valueOf(consistencyLevel)));
}
return builder.build();
}
/**
* Gets the list of token ranges that a table occupies on a give Cassandra node.
*
* <p>NB: This method is compatible with Cassandra 2.1.5 and greater.
*/
private static List<TokenRange> getTokenRanges(Cluster cluster, String keyspace, String table) {
try (Session session = cluster.newSession()) {
ResultSet resultSet =
session.execute(
"SELECT range_start, range_end, partitions_count, mean_partition_size FROM "
+ "system.size_estimates WHERE keyspace_name = ? AND table_name = ?",
keyspace,
table);
ArrayList<TokenRange> tokenRanges = new ArrayList<>();
for (Row row : resultSet) {
TokenRange tokenRange =
new TokenRange(
row.getLong("partitions_count"),
row.getLong("mean_partition_size"),
row.getLong("range_start"),
row.getLong("range_end"));
tokenRanges.add(tokenRange);
}
// The table may not contain the estimates yet
// or have partitions_count and mean_partition_size fields = 0
// if the data was just inserted and the amount of data in the table was small.
// This is very common situation during tests,
// when we insert a few rows and immediately query them.
// However, for tiny data sets the lack of size estimates is not a problem at all,
// because we don't want to split tiny data anyways.
// Therefore, we're not issuing a warning if the result set was empty
// or mean_partition_size and partitions_count = 0.
return tokenRanges;
}
}
/**
* Compute the percentage of token addressed compared with the whole tokens in the cluster.
*/
@VisibleForTesting
protected static double getRingFraction(List<TokenRange> tokenRanges) {
double ringFraction = 0;
for (TokenRange tokenRange : tokenRanges) {
ringFraction = ringFraction + (distance(tokenRange.rangeStart, tokenRange.rangeEnd)
.doubleValue() / TOTAL_TOKEN_COUNT.doubleValue());
}
return ringFraction;
}
/**
* Measure distance between two tokens.
*/
@VisibleForTesting
protected static BigInteger distance(long left, long right) {
if (right > left) {
return BigInteger.valueOf(right).subtract(BigInteger.valueOf(left));
} else {
return BigInteger.valueOf(right).subtract(BigInteger.valueOf(left)).add(TOTAL_TOKEN_COUNT);
}
}
/**
* Check if the current partitioner is the Murmur3 (default in Cassandra version newer than 2).
*/
@VisibleForTesting
protected static boolean isMurmur3Partitioner(Cluster cluster) {
return cluster.getMetadata().getPartitioner()
.equals("org.apache.cassandra.dht.Murmur3Partitioner");
}
/**
* Represent a token range in Cassandra instance, wrapping the partition count, size and token
* range.
*/
@VisibleForTesting
protected static class TokenRange {
private final long partitionCount;
private final long meanPartitionSize;
private final long rangeStart;
private final long rangeEnd;
public TokenRange(
long partitionCount, long meanPartitionSize, long rangeStart, long
rangeEnd) {
this.partitionCount = partitionCount;
this.meanPartitionSize = meanPartitionSize;
this.rangeStart = rangeStart;
this.rangeEnd = rangeEnd;
}
}
/**
* Writer storing an entity into Apache Cassandra database.
*/
protected class WriterImpl<T> implements Writer<T> {
private final CassandraIO.Write<T> spec;
private final Cluster cluster;
private final Session session;
private final MappingManager mappingManager;
public WriterImpl(CassandraIO.Write<T> spec) {
this.spec = spec;
this.cluster = getCluster(spec.hosts(), spec.port(), spec.username(), spec.password(),
spec.localDc(), spec.consistencyLevel());
this.session = cluster.connect(spec.keyspace());
this.mappingManager = new MappingManager(session);
}
/**
* Write the entity to the Cassandra instance, using {@link Mapper} obtained with the
* {@link MappingManager}. This method use {@link Mapper#save(Object)} method, which is
* synchronous. It means the entity is guaranteed to be reliably committed to Cassandra.
*/
@Override
public void write(T entity) {
Mapper<T> mapper = (Mapper<T>) mappingManager.mapper(entity.getClass());
mapper.save(entity);
}
@Override
public void close() {
if (session != null) {
session.close();
}
if (cluster != null) {
cluster.close();
}
}
}
@Override
public Writer createWriter(CassandraIO.Write<T> spec) {
return new WriterImpl(spec);
}
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,299
|
Prinsip kerja dari spatial multiplexing adalah mengirim TekTrasSel-Modul#6 Prinsip kerjadari spatial multiplexing adalah mengirim by Luis M. Correia.
Find this Pin and more on Curiosidades by espetacular. A correia transportadora maislonga do mundo Indonesia Kerja Nyata ! #hutri71 #kemerdekaan.
catalogo link belt crusher; amount of ore used in industry for copper production; doesthe magnetic ore separator hve any limitations; Chat for Free.
lebar belt conveyor 280 mm. LOADING CONVEYOR Correia Transportadora,Máquina detrituração efektifitas tenaga kerja di proses crushing pertambangan batubar.
Transportador Por Correia; Prinsip Kerja Grinder Mill; SMALL Plant Grinder In UK. smallgrinding mills sri lanka – CGM Grinding Plant.
GRINDING MILL SERIES shibangchina. In the lizenithne crushing plant, GRINDING MILL SERIES.Ball mill is an efficient tool for grinding materials into fine powder.
paper mill in the philippines mill for sale/ Paper Mills in Metro Manila Prinsip kerjamesin hammer mill / my google api appliion hammer mill for feed mills in .
» prinsip kerja belt conveyor minerao correia transportadores industria chinesa. minerao e industria de transformao de manganes.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 8
|
Brian Hooks plays a character who is just released from jail. And the state adopts a "3 strikes" rule for felons that involves serious penalties. Hooks has 2 strikes, and wants to change his life for the better. When a friend picks him up, they are pulled over, and his friend shoots at police officers, and Hooks escapes. Now Hooks, a wanted man, must clear his name of having nothing to do with the shooting.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 2,406
|
Boguszowice may refer to the following places in Poland:
Boguszowice, Cieszyn in Silesian Voivodeship (south Poland);
Boguszowice, former town in Silesian Voivodeship (south Poland), now two districts of Rybnik:
Boguszowice Osiedle
Boguszowice Stare
|
{
"redpajama_set_name": "RedPajamaWikipedia"
}
| 3,125
|
You seek to express at work the real you, and yet, struggle with communicating with others and dealing with emotionally charged situations?
You aspire to be a successful manager or leader, and yet, relationships with you colleagues and teams could be more fulfilling?
You are a manager and/or leader, and are struggling to lead both with your head and your heart?
What kind of leadership do we need to reach new levels of performance, which is ecological and sustainable for all parties?
What can we do to help others feel more motivated and engaged in what they do at work?
What new type of communication do we need to develop?
How can we contribute to our colleagues' full potential development and growth?
What new type of organizations and governance can we put in place to facilitate our teams' full expression?
How can we lead with both our head and our heart?
In my blog, you will find my perspective on leadership, management, non violent communication, mindfulness, team engagement and performance, more agile organizational structures and more.
Welcome and please leave your comments and share ideas and suggestions!
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,459
|
import json
import os
import sys
import urllib
import urllib2
from ConfigParser import ConfigParser, NoOptionError, NoSectionError
from version import __version__
from urllib2 import HTTPError, URLError
API_VERSION = 1
API_URL_BASE_TEMPLATE = 'https://api.geoloqi.com/%d/%s'
class Geoloqi:
"""
A simple interface wrapper for the Geoloqi API.
"""
config = None
api_key = None
api_secret = None
access_token = None
session = None
def __init__(self, api_key=None, api_secret=None, access_token=None):
"""
Initializes a new instance of the Geoloqi API wrapper.
Application credentials can be provided as kwargs or in a config
file. Clients *must* provide either a user access_token or an
application api_key and api_secret. Application credentials can be
obtained from the Geoloqi developer site.
Credentials can be provided in a config file in the user's home
directory named `.geoloqi` or in the system directory
`/etc/geoloqi/geoloqi.cfg`. In either case, config files should
use the following format.
::
[Credentials]
user_access_token = <your_user_access_token>
application_access_key = <client_api_key>
application_secret_key = <client_api_secret>
If both a user access_token *and* a set of client application
credentials are provided, the user access token will be used.
Args:
api_key: Your application's Geoloqi API key.
api_secret: Your application's Geoloqi API secret.
access_token: Your personal user access token.
Raises:
ValueError: If the proper client credentials were not provided.
"""
self.api_key = api_key
self.api_secret = api_secret
self.access_token = access_token
# Parse any config files
self.config = ConfigParser()
self.config.read(['/etc/geoloqi/geoloqi.cfg', os.path.expanduser('~/.geoloqi')])
if not self.api_key:
try:
self.api_key = self.config.get('Credentials', 'application_access_key')
except NoOptionError:
pass
except NoSectionError:
pass
if not self.api_secret:
try:
self.api_secret = self.config.get('Credentials', 'application_secret_key')
except NoOptionError:
pass
except NoSectionError:
pass
if not self.access_token:
try:
self.access_token = self.config.get('Credentials', 'user_access_token')
except NoOptionError:
pass
except NoSectionError:
pass
# Determine if Credentials exist
if not self.access_token and not (self.api_key and self.api_secret):
raise ValueError('Missing application credentials or a valid user access token!')
# Create our session
self.session = Session(self.api_key, self.api_secret, self.access_token)
def get(self, path, args=None, headers=None):
"""
Make a GET request to the Geoloqi API server.
Args:
path: Path to the resource being requested (example: 'account/profile')
args: An optional dictonary of GET arguments.
headers: An optional dictonary of extra headers to send with the request.
Returns:
The JSON response as a dictionary.
"""
return self.session.get(path, args, headers)
def post(self, path, data=None, headers=None):
"""
Make a POST request to the Geoloqi API server.
Args:
path: Path to the resource being requested (example: 'account/profile')
data: An optional dictonary to be sent to the server as a POST.
headers: An optional dictonary of extra headers to send with the request.
Returns:
The JSON response as a dictionary.
"""
return self.session.post(path, data, headers)
def run(self, path, data=None, headers=None):
"""
Make a request to the Geoloqi API server.
Args:
path: Path to the resource being requested (example: 'account/profile')
data: An optional dictonary to be sent to the server as a POST.
headers: An optional dictonary of extra headers to send with the request.
Returns:
The JSON response as a dictionary.
"""
return self.session.run(path, data, headers)
class Session:
"""
This class represents a session with the Geoloqi API.
"""
api_key = None
api_secret = None
access_token = None
auth = None
retry_attempt = 0
def __init__(self, api_key=None, api_secret=None, access_token=None):
"""
Create a new Geoloqi API session.
Args:
api_key: Your application's Geoloqi API key.
api_secret: Your application's Geoloqi API secret.
access_token: Your personal user access token.
Raises:
ValueError: If the proper client credentials were not provided.
"""
self.api_key = api_key
self.api_secret = api_secret
self.access_token = access_token
# Verify the Session has the needed Credentials
if not self.access_token and not (self.api_key and self.api_secret):
raise ValueError('A Session requires a set of application credentials'\
+ ' or a valid user access token!')
# Get an access token
if not self.access_token:
self.get_access_token()
def get(self, path, args=None, headers=None):
"""
Make a GET request to the Geoloqi API server.
Args:
path: Path to the resource being requested (example: 'account/profile')
args: An optional dictonary of GET arguments.
headers: An optional dictonary of extra headers to send with the request.
Returns:
The JSON response as a dictionary.
"""
if args:
path = "%s?%s" % (path, urllib.urlencode(args))
return self.run(path, None, headers)
def post(self, path, data=None, headers=None):
"""
Make a POST request to the Geoloqi API server.
Args:
path: Path to the resource being requested (example: 'account/profile')
data: An optional dictonary to be sent to the server as a POST.
headers: An optional dictonary of extra headers to send with the request.
Returns:
The JSON response as a dictionary.
"""
if not headers:
headers = {}
headers.update({
'Content-Type': 'application/json',
})
return self.run(path, data, headers)
def run(self, path, data=None, headers=None):
"""
Make a request to the Geoloqi API server.
Args:
path: Path to the resource being requested (example: 'account/profile')
data: An optional dictonary to be sent to the server as a POST.
headers: An optional dictonary of extra headers to send with the request.
Returns:
The JSON response as a dictionary.
"""
if not headers:
headers = {}
# Update the request headers
headers.update({
'User-Agent': self.get_user_agent_string(),
})
# Authorize the request
if self.access_token:
headers.update({'Authorization': 'OAuth %s' % self.access_token,})
# Execute request
f = self.execute(path, data, headers)
raw = f.read()
# Parse response
response = json.loads(raw)
if response.has_key('error'):
error = response.get('error')
if error == 'expired_token':
# Our access token has expired
if self.retry_attempt < 1:
self.renew_access_token()
# Retry the request
self.retry_attempt += 1
return self.run(path, data, headers)
else:
# TODO: Failed to refresh the access token!
pass
else:
# TODO: Throw or log the error?
pass
# Reset our retry counter
self.retry_attempt = 0
return response
def execute(self, path, data=None, headers=None):
"""
Makes a low-level request to the Geoloqi API server. Does no
processing of the response.
Args:
path: Path to the resource being requested (example: 'account/profile')
data: An optional dictonary to be sent to the server as a POST.
headers: An optional dictonary of extra headers to send with the request.
Returns:
A file like object returned from `urllib2.urlopen`.
"""
if data:
data = json.dumps(data)
if not headers:
headers = {}
request = urllib2.Request(API_URL_BASE_TEMPLATE % (API_VERSION, path),
data, headers=headers)
# Execute the request
try:
return urllib2.urlopen(request)
except (HTTPError, URLError), e:
return e
def establish(self, data):
"""
Used to retrieve the access token from the Geoloqi OAuth2 server. This
is used internally and you shouldn't need to call it manually.
Args:
data: A dictionary of data to be included with the request. You should
include the OAuth2 'grant_type' and other data like your
refresh token.
Returns:
None
"""
data.update({
'client_id': self.api_key,
'client_secret': self.api_secret,
})
self.auth = self.post('oauth/token', data)
self.access_token = self.auth.get('access_token')
def renew_access_token(self):
"""
Renew the access token using the stored refresh token. This method is
called automatically when the server returns an expired_token response,
so you shouldn't need to call it manually.
Returns:
None
"""
self.establish({
'grant_type': 'refresh_token',
'refresh_token': self.auth.get('refresh_token'),
})
def get_access_token(self):
"""
Retrieve an access token for this session. This token is used in
the same manner as the user access token, it simply allows the
application to make requests on the behalf of itself instead of a
user within the app.
This call makes a request to the API server once for each instantiation
of the object, then cache the result on the object.
Returns:
The current access token as a String.
"""
if not self.access_token:
self.establish({
'grant_type': 'client_credentials',
})
return self.access_token
def get_user_agent_string(self):
"""
Retrieve a 'User-Agent' string to be used when making API requests.
Returns:
The 'User-Agent' string.
"""
return 'geoloqi-python %s' % __version__
if __name__ == "__main__":
# TODO: Write a `main` method that enables simple command-line
# access to the Geoloqi API.
sys.exit()
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,619
|
\chapter{Inferring the singlet state from polarisation statistics}
\label{sec:appendix_a}
We assume that Alice and Bob possess a shared reference frame of photon polarisation. They perform random measurements in two basis, one of which is the linear basis $B_1$ spanned by $\ket{h}$ and $\ket{v}$ and the other, the circular basis $B_2$ spanned by $\ket{\pm}=\frac{1}{\sqrt{2}}(\ket{h}\pm\ket{v})$. We assume that the 2-photon state to be measured by Alice and Bob is in the general form $\rho$.
We can write the POVM elements $E_{1s}$ and $E_{1d}$ for $B_1$ as
\begin{equation}
E_{1s}=\ket{hh}\bra{hh}+\ket{vv}\bra{vv} \, \, \, , E_{1d}=\ket{hv}\bra{hv}+\ket{vh}\bra{vh} \, ,
\end{equation} and the POVM elements $E_{2s}$ and $E_{2d}$ for $B_2$ as
\begin{equation}
E_{2s}=\ket{++}\bra{++}+\ket{--}\bra{--} \, \, \, , E_{2d}=\ket{+-}\bra{+-}+\ket{-+}\bra{-+} \, .
\end{equation} Note that $E_{1s}+E_{1d}=E_{2s}+E_{2d}=1$.
If Alice and Bob always detects orthogonal polarisations in $B_1$ and $B_2$, we wish to show that the only state consistent with this observation is the singlet state(which is maximally entangled) given by $\rho=\ket{\Phi_-}\bra{\Phi_-}$\footnote{I thank Jens Eisert for his help in this problem.} where $\ket{\Phi-}=\frac{1}{\sqrt{2}}(\ket{hv}-\ket{vh})$. This approach does not require a full tomographic measurement and may shed new perspectives to the notion of optimal measurements in quantum tomography.
We therefore have
\begin{equation} \label{Povm1}
{\rm Tr}(E_{1d}\rho)=1
\end{equation} and
\begin{equation}\label{Povm2}
{\rm Tr}(E_{2d}\rho)=1 \, .
\end{equation}
We know that $\rho$ must be positive semi-definite. This is to allow a valid probability interpretation should we choose to arbitrarily reduce the degree of freedom specifying $\rho$. This restriction together with Eq.~(\ref{Povm1}) implies that $\rho$ can only be of the form $\rho=a\ket{hv}\bra{hv}+c\ket{hv}\bra{vh}+c^*\ket{vh}\bra{hv}+b\ket{vh}\bra{vh}$ where $a+b=1$ and $a$,$b$ are positive real numbers.
Adding condition (\ref{Povm2}) yields the further constraint $a=b=\frac{1}{2}$ as well as $c=-\frac{1}{2}$. This implies that $\rho=\ket{\Phi_-}\bra{\Phi_-}$.
\chapter{Distributed Photon Entanglement on Demand} \label{demand}
In this short chapter, we highlight a useful result from Chapters \ref{firework}
and \ref{minsk} and show that distributed photon
entanglement generation is possible on demand. In particular, we also
demonstrate a duality relation that arises from our previous study
of multiports which may give new perspectives in multiport design for quantum
information processing.
\section{Multiatom entanglement and multiphoton entanglement on demand}
The key to multiphoton entanglement on demand lies in the initial creation of
multiatom entanglement. We denote this step as the initialisation. Recalling Chapter \ref{firework}
where we have mainly considered the Bell multiport, we place no restriction
on the multiports considered in this chapter. We first allow
each atom to be entangled with a
photon and study
what happens if the photons are passed through a
multiport.
We assume that each $i$th atom is specified by two states $\ket{\pm}$ notated by
$g^{\dagger}_{i\pm}\ket{0}$ using a second-quantised notation.
Each of the
atoms should first be maximally entangled with a photon feeding into each input
of the $N \times N$ multiport such that we can write the total combined initial state as
\begin{equation}
\ket{\Psi_{in}}= \frac{1}{\sqrt{2^N}} \prod_{i=1}^N \Big(
\sum_{\mu=+,-}
g_{i \mu}^\dagger \, a_{i \mu}^\dagger \Big) \, \ket{0}.
\end{equation}
Such a state preparation can be performed
deterministically \cite{Gheri98,LimSPIE04} by an atom-cavity system as described
in Chapter \ref{minsk}. The general
atom-photon state $\ket{\Psi_{in}}$ after passing the photons through the
multiport and upon
collecting one photon per output port is then, up
to normalisation and by analogy to Eq. (\ref{firework:output2}), given by
\begin{equation} \label{firework:output3}
\ket{\Psi_{out}}=\frac{1}{\sqrt{2^N}}\sum_{\sigma} \Bigg[ \prod_{i=1}^N
U_{\sigma(i) i}
\Big(
\sum_{\mu=+,-} g_{i \mu}^\dagger b_{\sigma (i) \mu}^{\dagger} \Big)
\Bigg] \, |0 \rangle \, .~
\end{equation}
Now, the next step is to choose a detection syndrome of the photons. Let the
detection syndrome be defined by the postselected state
\begin{equation}
\ket{{\rm S}}=\prod_{j=1}^N \sum_{\mu}\alpha_{j \mu}^\ast b_{j \mu}^\dagger
\ket{0}.
\end{equation} For example, $\alpha_{j \mu}^\ast b_{j \mu}^\dagger
\ket{0}$ defines the state of a photon detected in output port $j$ with
polarisation $\mu$. Note that we
see a direct correspondance or analogy of $\ket{{\rm S}}$ to the input photon state
(\ref{firework:in}).
Applying the relevant projector, the multiatomic state $\ket{\rm A}$ can be
shown to be projected onto
\begin{equation} \label{dual1}
\ket{{\rm A}}=\frac{1}{\sqrt{2^N}}\sum_{\sigma} \Bigg[ \prod_{i=1}^N
U_{\sigma(i) i} \Big(
\sum_{\mu=+,-} \alpha_{\sigma(i) \mu} g_{i \mu}^{\dagger} \Big) \Bigg] \, |0
\rangle
\end{equation} Hence, a multiatomic state can be prepared by choosing an appropriate detection syndrome. Provided that the photons detectors have negligible dark counts, photon loss and inefficient detectors do not decrease the fidelity of state preparation. This is an advantage of choosing a coincidence (one photon per output port) detection syndrome. To get further insight, we can next substitute $i=\sigma^{-1}(j)$ and obtain
\begin{eqnarray} \label{dual2}
\ket{{\rm A}}&=&\frac{1}{\sqrt{2^N}}\sum_{\sigma^{-1}} \Bigg[ \prod_{j =1}^{N}
U_{j \sigma^{-1}(j)} \Big(
\sum_{\mu=+,-} \alpha_{j \mu} g_{\sigma^{-1}(j) \mu}^{\dagger} \Big) \Bigg] \,
|0 \rangle \nonumber \\
&=&\frac{1}{\sqrt{2^N}}\sum_{\sigma} \Bigg[ \prod_{i=1}^N U_{i \sigma(i)} \Big(
\sum_{\mu=+,-} \alpha_{i \mu} g_{\sigma(i) \mu}^{\dagger} \Big) \Bigg] \, |0
\rangle \, .
\end{eqnarray} since $\sigma^{-1}$ is just a dummy index for a permutation.
A close comparision of (\ref{dual2})
with (\ref{firework:output2}) shows that this multiatomic state is of a similar\footnote{For simplicity, we have assumed bosonic statistics in the atoms for the discussion of the analogy with (\ref{firework:output2}). In any case, (\ref{dual1}) is the projected multiatomic state.}
form to
that of prepared multiphoton state given the input photon
state (\ref{firework:in}),
except that the input and output of the multiport are swapped. This suggests a
duality relation of designing optical
circuits aimed at preparing entangled atoms by examining how entangled
photons(in analogy to the desired entangled atoms) can be prepared from product states of photons(analogous to the detection syndrome).
An example of producing a three atom W-state was demonstrated using a $3 \times 3$
multiport
\cite{Zou04a}. It was then conjectured, but not proven,
that an $N \times N$ multiport may be used to prepare an $N$ atom W-state. The
work in this chapter qualifies this result rigourously for a
Bell multiport.
\begin{figure}\label{demand:setup}
\begin{center}
\begin{tabular}{c}
\includegraphics[height=3.5cm]{setup}
\end{tabular}
\end{center}
\caption{
Experimental setup. (a) During the final ``{\em push button}'' step, the
entanglement of $N$ atom-cavity systems is mapped onto the state of $N$ newly
generated photons. (b) The initialisation of the system requires postselective
measurements on the photon emission from the $N$ cavities through a
multiport beamsplitter.}
\end{figure}
In general, one can prepare or initialise a wide variety of atomic states with
this method. One only needs to design the multiport that
yields the desired mulitphoton state postselectively as described in Chapter
\ref{firework}. An alternative and more general way to
create multiatom states by using universal two-qubit gates with insurance is
described in Chapter \ref{minsk}.
Once the atomic qubits have been initialised, $N$ photons in
exactly the same state can be
created by simply mapping the state of the sources onto the state of $N$ newly
generated photons whenever required \cite{LimSPIE04}\footnote{ A very recent
proposal by Kok {\em et al.}
\cite{Kok05} built on the same point (i.e. initialise and map) to implement a
multiphoton entanglement on demand source with a slightly
different physical setup and procedure. The double-heralding step they used in the initialisation process can have the same advantages as coincidence detection. In addition, a cluster state is prepared offline and arbitrary multiatomic qubits can then be prepared by single qubit measurements to complete the initialisation process. Our scheme employs either the multiport approach for direct preparation of the atomic states or a series of universal two-qubit gates for atomic state initialisation.} (See Fig.~\ref{demand:setup}). To
accomplish this, the
state of each photon source
should first be encoded as in Chapter \ref{minsk}. Afterwards, the atomic
qubits can be decoupled from the flying qubits by measuring again in a mutually
unbiased basis
with respect to the computation basis (i.e. the encoding basis) and
performing a local operation on the photon whenever necessary. The generation of
multiphoton entanglement on
demand superficially resembles a remote state preparation of
the state of $N$ newly
created photons by the multiatomic state. This mapping can also be accomplished
more efficently without measurement by choosing atomic
levels similar to that in Lim {\em et al.} \cite{LimSPIE04}. For example, one
can use a 5-level atom with ground states $\ket{0}$ and
$\ket{1}$ constituting the logical qubit states as well as another ground state
$\ket{2}$. The excited states are $\ket{e_0}$ and
$\ket{e_1}$ where the cavity couples the transition $e_i-2$ with a cavity photon
of polarisation $i$. The exciting laser couples to the
transition $e_i-i$ and drives an initialised atom similar to the description in
Chapter~\ref{minsk}, for example in the state
$(\alpha\ket{0}+\beta\ket{1}) \otimes \ket{\rm vac}$ to the state $\ket{2} \otimes
(\alpha a_0^\dagger+\beta a_1^\dagger) \ket{\rm vac}$
where $a_i^\dagger \ket{\rm vac}$ denotes a photon with polarisation $i$ in the
external cavity field.
We can in principle create any arbitrary distributed $N$-photon state on demand
in comparision with the schemes of Gheri {\em et al.}
\cite{Gheri98} and Sch{\"o}n {\em et al.}\footnote{I thank Christian Sch{\"o}n for stimulating discussions. } \cite{Schon05}, where a restricted set of
states on the same spatial mode can be created efficiently from a
single atom-cavity system.
Before we conclude our thesis, we observe that linear optics resources have been a
crucial component in most parts of this thesis. We
remove this resource next and consider entanglement generation with distant
sources without cavities, i.e. in free space.
\chapter{Multiphoton Entanglement through a Bell Multiport Beam Splitter using
Independent Photons} \label{firework}
\section{Introduction}
In this chapter, we are concerned with the practical generation of multiphoton entanglement. It is
not possible to
create a direct interaction between photons and hence they are difficult to
entangle as already highlighted in Chapter \ref{overview}. One way to overcome this problem is to create polarisation or time-bin
entanglement via photon pair creation within the same source as in atomic
cascade and parametric down-conversion experiments. This has already been
demonstrated experimentally by many groups
\cite{Aspect82,Kwiat95,Brendel99,Thew02,Riedmatten04}. Other, still theoretical
proposals employ certain features of the combined level structure of atom-cavity
systems \cite{Gheri98,Lange00,Schon05}, photon emission from atoms in free space (described in Chapter \ref{photon})
or suitably initialised distant single photon sources (to be demonstrated in Chapter \ref{demand}).
Alternatively, highly entangled multiphoton states can be prepared using
independently generated single photons with no entanglement in the initial
state, linear optics and postselection. This method shall be the main focus of this chapter. In general, the photons should enter the
linear optics network such that all information about the origin of each photon
is erased. Afterwards postselective measurements are performed in the output
ports of the network \cite{Lapaire03}. Using this approach, Shih and Alley
verified the generation of maximally entangled photon pairs in 1988 by passing
two photons simultaneously through a 50:50 beam splitter and detecting them in
different output ports of the setup \cite{Shih88}. For a recent experiment based
on this idea using quantum dot technology, see Ref.~\cite{Fattal04}.
Currently, many groups experimenting with single photons favour parametric down
conversion because of the quality of the output states produced. However, these
experiments cannot be scaled up easily, since they do not provide efficient
control over the arrival times of the emitted photons. It is therefore
experimentally challenging to interfere more than two photons successfully.
Interesting experiments involving up to five photons have nevertheless been performed
\cite{Eibl03,Bourennane04,Zhao03,Zhao04,Zhao05} but going to higher photon numbers
might require different technologies. To find alternatives to parametric down
conversion, a lot of effort has been made over the last years to propose
experimentally realisable sources for the generation of single photons on demand
\cite{Law97,Kuhn99,Duan03,Jeffrey04}. Following these proposals, a variety of
experiments has already been performed, demonstrating the feasibility and
characterising the quality of sources based on atom-cavity systems
\cite{Kuhn00,Kuhn02,Lange04,Mckeever04}, quantum dots \cite{Benson00,Pelton02}
and NV color centres \cite{Kurtsiefer00,Beveratos02}. Before we proceed further, it is appropriate to give a more detailed survey of the
above mentioned single photon sources.
\subsection{Photon sources}
Photon sources can be generally subdivided into sources that give strictly antibunched
photons, (i.e. the normalised intensity time correlation, also known as $g^{(2)}(\tau)$, is smaller than unity
for zero time separation) or sources that yield otherwise. True
single photon sources yield only antibunched photons with $g^{(2)}(0)=0$. Furthermore, an ideal turnstile
single photon source should consistently yield exactly one photon in the same
pure quantum state whenever required. Particularly for applications
\cite{Knill01} relying on Hong-Ou-Mandel two-photon type interference, it is
important for photons to be indistinguishable and of high purity. An
example of such a candidate source is an atom-like system which includes quantum
dots, diamond NV-color centers and atom-cavity systems. These systems
also afford push-button photon generation, which is an ideal requirement for
experiments requiring single photons such as quantum cryptography or linear
optics based quantum computing. When a photon is
required, the source can be triggered to yield a photon. There also exist
approximate single photon sources that cannot be directly triggered on
demand. In principle, even these sources can simulate an on-demand single photon
source with the help of photon memory and non-demolition
measurement, a currently challenging experimental requirement that has undergone
much interest and development. Two prominent examples of
pseudo single photon sources are a weak coherent laser pulse and the parametric
downconversion source. We review below a selection of single photon sources that
are currently in use.
\subsection {Weak Coherent laser pulse}
A laser pulse can be modelled to a good approximation as a equal weighted
mixture of coherent states of the same amplitude $\alpha$ but different phase
$\phi$ \cite{vanEnk02,Molmer97}. This is equivalent to a mixture\footnote{In fact, the relative and not the absolute phase in quantum optics experiments turns out to be the crucial parameter. So, it is equally valid and may be more useful to model the laser pulse as an effective pure coherent state instead of a mixture of coherent states as in (\ref{fallacy}). One should however be careful to ascribe realism to such an interpretation. This issue has been a source of hot debate. See Ref. \cite{Bartlett05} and references therein for further discussion.} of photon Fock states weighted with a
Poissonian distribution,
\begin{equation}\label{fallacy}
\rho_{\rm laser}=\int d\phi \ket{\alpha {\rm e}^{{\rm i} \phi}}\bra{\alpha {\rm
e}^{{\rm i} \phi}}=\sum_n \frac{{\rm e}^{-\alpha^2} \alpha^{2
n}}{n!}\ket{n}\bra{n}.
\end{equation} The probability weight of an $n$-photon Fock state is thus given by
$P(n)=\frac{{\rm e}^{-\alpha^2} \alpha^{2 n}}{n!}$. When $\alpha \ll 1$, then $P(0)
\gg P(1) \gg P(2)$. This implies that a weak laser pulse can indeed function as a
pseudo single photon source. This necessarily implies low count
rate for single photons which is due to the necessity to use a weak pulse to suppress any
multiphoton component weighted by Poissonian statistics. Furthermore, any single
photon pulse generated must be detected postselectively and cannot be heralded(except with the help of a photon non-demolition
measurement) since $P(0) \gg P(1)$ therefore implying a necessarily
large vacuum component. The weak coherent laser pulse finds its application in
quantum key distribution (QKD) for example. It was once thought that photon-number splitting attack would be a strong impediment to
achieve a high key rate in the presence of channel loss.
However, in the light of some recent advancement of secure QKD protocols robust against photon-number splitting attack, such as the
decoy-state \cite{Hwang03} and strong phase-reference pulse \cite{Koashi04} protocol, the weak coherent laser pulse is likely to remain
an important tool for QKD.
\subsection{Parametric downconversion}
A useful photon source arises from the process of spontaneous parametric downconversion(SPDC).
Such a
source is used widely in a large number of quantum optics experiments such as the famous Hong-Ou-Mandel effect \cite{Hong87}. Similar to
the coherent laser pulse
source, it is also not a true single photon source. It is, however
able
to yield a wide variety of multiphoton states postselectively. If only a single
photon is desired, it can act as a heralded source where
a trigger allows one to infer the emission of a photon in a certain mode. On the other hand, it is widely used
to generate entangled photon pairs \cite{Kwiat95,Tittel98,Brendel99,Thew02,Riedmatten04} in various encodings such as polarisation,energy-time, time-bin etc. and a wide variety of experiments ranging from fundamental test of quantum
mechanics to linear optics quantum computation have been performed with it.
SPDC can generally yield quite a high count rate of entangled photon pairs, for example about $10^5-10^6 s^{-1}$ \cite{Kurtsiefer01,Kumar04}. However,
experiments for multiphoton interferometery typically yield,
for example for $N=4$ photons, a coincidence count rate of $10^{-2} s^{-1}$\cite{Pan98}. This low
count rate is partially due to both the random nature of photon emission as well as the need for frequency filters to erase the
time-stamp of
the generated photons
for experiments such as entanglement swapping with Bell measurements \cite{Zukowski93}. The reason is due to the strong temporal correlation of the signal and idler photons emitted.
Due to this, only
quantum optics experiment in the few photons level utilising the above states
($N\leq 5$) are viable.
Although there exist quasi-deterministic schemes, for
example in Ref. \cite{Pittman02a,Jeffrey04} for photon
generation, they require photon recycling circuits or photon memories, both still
experimentally challenging. On the other hand, parametric downconversion is
useful
for generating squeezed states (see for example Ref.~\cite{Wu86}), which are useful for applications in continuous
variable quantum information processing.
The SPDC process can be roughly understood in terms of a higher
energy
photon being converted by an energy conserving process to two lower energy photons, traditionally known as the
signal and idler photons. If the signal and idler photons are of the same polarisation, this is known as a
Type-I process. If their polarisation are mutually orthogonal, this is known as a Type-II process. To generate
a photon pair, a birefringent noncentrasymmetric nonlinear crystal is pumped by a laser, either in cw mode or
pulsed mode. Phase matching conditions determine the direction and frequencies of the signal and idler photon pair
generated.
We now denote the photon creation operator with frequency $\omega$, polarisation $\lambda$ and direction
$\hat{\bf k}$ as $a^\dagger_{\hat{\bf k},\lambda}(\omega)$. We denote the emitted directions(polarisation) of
a signal and idler photon as $\hat{\bf k}_s$($\lambda_s$) and $\hat{\bf k}_i$($\lambda_i$) respectively and
we assume a type-II process. As in Kwiat {\em et al.}\cite{Kwiat95}, we assume the presence of two photon
collection directions, $\hat{\bf k}_A$ and $\hat{\bf k}_B$ such that when $\hat{\bf k}_A=\hat{\bf k}_i$ or
$\hat{\bf k}_s$, $\hat{\bf k}_B=\hat{\bf k}_s$ or $\hat{\bf k}_i$ respectively. In these two directions,
together with frequency filters, the postselected 2-photon state $\ket{\psi}$ generated by SPDC
\cite{Zukowski95} is then given by
\begin{eqnarray}
\ket{\psi} &=&
\int d\omega_p \int d\omega_i \int d\omega_s F(\omega_p,\omega_i,\omega_s) \nonumber \\
&&[a^\dagger_{\hat{\bf k}_A,\lambda_i}(\omega_i) a^\dagger_{\hat{\bf k}_B,\lambda_s}(\omega_s)+
a^\dagger_{\hat{\bf k}_A,\lambda_s}(\omega_s) a^\dagger_{\hat{\bf k}_B,\lambda_i}(\omega_i)]\ket{\rm vac}
\end{eqnarray} where $\omega_p$ is the pump frequency and $F(\omega_p,\omega_i,\omega_s)$ is a function
dependent on the phase matching condition as well as the frequency envelope of the pump and the frequency
filters. Under suitable phase matching condition, spatial pin-hole filtering, and or frequency filtering,
$F(\omega_p,\omega_i,\omega_s)$ can be highly peaked at $F(\omega_p,\omega_p/2,\omega_p/2)$ and $\ket{\psi}$
therefore reduces approximately to a polarisation Bell state \cite{Kwiat95,Zukowski95} that is widely used in
quantum optics experiments.
\subsection{Atom-like systems for the generation of single photons on demand}
Candidate systems that could yield single photons on demand include
mainly atom-like systems such as atoms, quantum dots, NV (Nitrogen-Vacancy) color centers or even
molecules. These proposals are mainly based on the ability to excite the photon source which
then decays back to a ground state as a result yielding a photon. Due to the
fact that every photon generated by this method requires an excitation time
overhead, this results in naturally antibunched photon production. These systems are more recent developments, compared to SPDC sources
and weak coherent laser pulses. They benefit from recent technological advancements such as semiconductor processing, laser cooling
and trapping etc. and are still an exciting development avenue. Quantum information processing has further served as an important
motivating factor, as is investigated in this thesis, for the continual development of these sources.
The quantum dot single photon source is operated by performing
a sharp laser pulse excitation to an excited level representing the creation of a so-called excited
exciton which rapidly decays non-radiatively to the lowest excited
state of the exciton. A subsequent slower decay back to the ground state yields
a photon. In practice, biexcitonic excitation is usually preferable, due to the ability to spectrally isolate the
last but one photon \cite{Santori00}. With the quantum dot integrated in monolithic cavity structures, the
spontaneous emission rate can be increased substantially with the
emission mainly into the cavity mode which results in directed photon emission.
Due to the non-radiative decay in the excitation process, there is a slight
uncertainty in the photon emission time. Even with this and all other effects
contributing to decoherence, photon pulses of sufficient purity and
indistinguishability can be generated consistently to observe a Hong-Ou-Mandel
2-photon interference\cite{Santori02} at low temperatures. If spectral purity is not needed, single photon
generation can still be performed at room temperature \cite{Michler00}. The quantum dot also allows for
coherent
Raman excitation schemes \cite{Kiraz04} and may lead to an attractive solid-state
alternative to photon guns based on atom-cavity systems. A good review on the
physics of photon generation through quantum dots is found in Santori {\em et
al.} \cite{Santori04}. It is worth mentioning also the quantum dot can be excited electronically via a
Coloumb blockade and Pauli effect \cite{Kim99} leading to an electronic turnstile single photon device.
NV color center, an optically active defect inside a diamond nanocrystal, is an alternative atom-like system
for photon generation. Unlike the quantum dot, for applications in quantum
cryptography where the purity of photons generated is not too important, NV
color centers can be maintained at room temperature during operation. Their key advantage lies in the fact that
they boast of extremely stable operation even at room temperature. The
excitation of an NV color center to generate a single photon is similar to that
of the quantum dot. To date, photon antibunching has been observed with this
method \cite{Kurtsiefer00,Beveratos02}. The demonstration of Hong-Ou-Mandel
2-photon interference would probably require cyrogenic temperature operation.
Before moving to atom-cavity systems, we mention that single molecules are yet another attractive atom-like system capable of yielding
single photons. In fact, the most recent experiment with a TDI (Terrylenediimide) molecule at cryogenic temperature have yielded photons
demonstrating the Hong-Ou-Mandel 2-photon interference \cite{Kiraz05}.
The atom-cavity system consists of an atom ideally trapped in a high-finesse
cavity. A laser and cavity-driven Raman process which is described in more
detail in Chapter \ref{minsk} transfers a photon in the cavity which subsequently
leaks out. This has been experimentally demonstrated
\cite{Kuhn02,Mckeever04,Lange04} and photons generated from such systems have a
sufficient purity and consistency to observe the Hong-Ou-Mandel 2-photon
interference \cite{Legero04}. In principle, barring imperfections such as photon absorption, weak
cavity coupling, the photon generation probability approaches unity. The
atom-cavity system allows also for generation of entangled photons on demand \cite{Schon05,Gheri98}.
More generally, it also allows for the state of an atom to be redundantly
encoded to the photon it generates which is described in Chapter \ref{minsk}.
Current experimental achievements of all these sources have admittedly not yet achieved photon production on demand. The best reported
photon production efficiency \cite{Mckeever04} is still less than $70 \%$ although there is no limit in principle to achieving near unit
efficiency. Compared to SPDC, these sources generally demand greater experimental complexity at the present. Moreover, these sources are generally not as wavelength tunable as SPDC sources, although this need not be a real disadvantage. However, with
strong motivations for scalability in linear optics quantum computation and distributed quantum computing as well as quantum cryptography, much effort to the development of a robust true
photon on demand source is ongoing.
\subsection{Single photons and multiport} \label{photonmultiport}
Motivated by the above recent developments, several authors studied the creation of
multiphoton entanglement by passing photons generated by a single photon source
through a linear optics network
\cite{Zukowski97,Lee02,Fiurasek02,Zou02,Wang03,Sagi03,Pryde03,Mikami04,Shi05}. A variety
of
setups has been considered. Zukowski {\em et al.} showed that the $N \times N$
Bell multiport beam splitter (see below) can be used to produce higher dimensional EPR (Einstein-Podolsky-Rosen)
correlations \cite{Zukowski97}. Shi and Tomita \cite{Shi05} for example studied 3 and 4-photon {\em W}-state preparation with multiports
which led to high generation efficency. They also conjectured but did not prove that a symmetric $N \times N$ multiport may be used to
generate an $N$-photon {\em W}-state. Mikami {\em et al.} studied the generation of $N$-photon states through parametric
downconversion, coherent laser states and multiports with photon number-resolving detectors. Such multiports have an important
application in boosting the success probability of linear optics teleportation to
near unity \cite{Knill01} using a special highly entangled multiphoton ancilla.
Special attention has been paid to the optimisation of schemes for the
generation of the so-called NOON state with special applications in lithography
\cite{Lee02,Fiurasek02,Zou02,Pryde03}. Wang studied the event-ready generation
of maximally entangled photon pairs without photon number-resolving detectors
\cite{Wang03} and Sagi proposed a scheme for the generation of $N$-photon
polarisation entangled GHZ states \cite{Sagi03}. It is also possible to prepare
arbitrary multiphoton states \cite{Fiurasek03} using for example probabilistic
but universal linear optics quantum gates, like the one described in
Refs.~\cite{Pittman02,Knill01} or using a large enough optical cluster state
\cite{Yoran03,Nielsen04,Browne05} which still remains an experimental challenge.
However, these approaches are not always the most favourable and often require a
large number of entangled photon ancillas.
Here we are interested in the generation of highly entangled qubit states of $N$
photons using only a single photon source and a symmetric $N \times N$ Bell
multiport beam splitter, which can be realised by combining single beam
splitters into a symmetric linear optics network with $N$ input and $N$ output
ports \cite{Zukowski97,Torma95}. In the two-photon case, the
described scheme
simplifies to the experiment by Shih and Alley \cite{Shih98}. To entangle $N$
photons, every input port $i$ of the Bell multiport should be entered by a
single photon prepared in a state $|\lambda_i \rangle$. The photons then
interfere with each other before leaving the setup (see Fig.~\ref{scheme}). We
consider the state preparation as successful under the condition of the
collection of one photon per output port, which can be relatively easily
distinguished from cases with at least one empty output port. In general, this can be done with photon number-resolving non-demolition
detectors \cite{Munro05,Nemoto04} and we obtain preselected multiphoton entanglement. Otherwise, the entangled state is postselected
without the need of photon number-resolving detectors. Postselected photon state preparation is nevertheless useful if one can
accomplish a non-trivial task. Examples
are teleportation \cite{Joo03}, quantum secret sharing, secure quantum key distribution \cite{Chen05}, testing entanglement with
witnesses and observing a violation of Bell's inequality \cite{Bourennane04a,Toth05} .
\begin{figure}
\begin{minipage}{\columnwidth}
\begin{center}
\resizebox{\columnwidth}{!}{\rotatebox{0}{\includegraphics{fire2}}}
\end{center}
\caption{Experimental setup for the generation of multiphoton entanglement by
passing $N$ single photons through an $N \times N$ Bell multiport beam splitter.
The state preparation is considered successful under the condition of the
collection of one photon per output.} \label{scheme}
\end{minipage}
\end{figure}
One advantage of using a Bell multiport beam splitter for the generation of
multiphoton entanglement is that it redirects the photons without changing their
inner degrees of freedom, like polarisation, arrival time and frequency. The
described setup can therefore be used to generate polarisation, time-bin and
frequency entanglement. Especially, time-bin entanglement can be very robust
against decoherence and has, for example, applications in long-distance fibre
communication \cite{Gisin02}. Moreover, the preparation of the input product
state does not require control over the relative phases of the incoming photons,
since the phase factor of each photon contributes at most to a global phase of
the combined state with no physical consequences.
It is the purpose of this chapter to explore some novel properties of a Bell
multiport beam splitter. This chapter is organised as follows. In Section
\ref{firework:scatter} we introduce the notation for the description of photon scattering
through a linear optics setup. Section \ref{fourphoton} shows that a wide range
of highly entangled photon states can be obtained for $N=4$, including the {\em
W}-state, the GHZ-state and a double singlet state. Afterwards we discuss the
generation of $W$-states for arbitrary photon numbers $N$ and calculate the
corresponding probabilities for a successful state preparation. We observe an
interesting non-monotonic decreasing trend in the success probability as $N$
increases owing to quantum interference. Finally we
conclude our results in Section \ref{firework:conclusions}.
\section{Photon scattering through a linear optics setup} \label{firework:scatter}
Let us first introduce the notation for the description of the transition of the
photons through the $N \times N$ multiport beam splitter. In the following,
$|+ \rangle$ and $|- \rangle$ denote the state of a photon with polarisation
``$+$'' and ``$-$'' respectively. Alternatively, $|+ \rangle$ could describe a
single photon with an earlier arrival time or a higher frequency than a photon
prepared in $|- \rangle$. As long as the states $|\pm \rangle$ are orthogonal
and the incoming photons are in the same state with respect to all other degrees
of freedom, except of course their input spatial positions, the calculations presented in this paper apply throughout. Moreover,
we assume that each input port $i$ is entered by one independently generated
photon prepared in $|\lambda_i \rangle = \alpha_{i+} |+\rangle_i + \alpha_{i-}
|-\rangle_i$, where $\alpha_{i \pm}$ are complex coefficients with $|\alpha_{i+}
|^2 + |\alpha_{i-}|^2 = 1$. If $a_{i \mu}^\dagger$ denotes the creation operator
for one photon with mode $\mu$ in input port $i$, the $N$-photon input
state can be written as
\begin{eqnarray} \label{firework:in}
|\phi_{\rm in} \rangle &=& \prod_{i=1}^N \Big( \sum_{\mu=+,-}
\alpha_{i \mu} \, a_{i \mu}^\dagger \Big) \, |0 \rangle
\end{eqnarray}
with $|0 \rangle$ being the vacuum state with no photons in the setup.
Let us now introduce the unitary $N \times N$-multiport transformation operator,
namely the scattering matrix $S$, that relates the input state of the system to
the corresponding output state
\begin{eqnarray} \label{fin}
|\phi_{\rm out} \rangle &=& S \, |\phi_{\rm in} \rangle \, .
\end{eqnarray}
Using Eq.~(\ref{firework:in}) and the relation $S^\dagger S = I \!\! I$ therefore yields
\begin{eqnarray} \label{fin2}
|\phi_{\rm out} \rangle &=& S \, \Big( \sum_{\mu=+,-} \alpha_{1 \mu} \,
a_{1 \mu}^\dagger \Big)
\, S^\dagger S \, \Big( \sum_{\mu=+,-} \alpha_{2 \mu} \, a_{2
\mu}^\dagger \Big) \cdot \, . \, . \, . \, \cdot S^\dagger S \, \Big(
\sum_{\mu=+,-} \alpha_{N \mu} \,
a_{N \mu}^\dagger \Big) \, S^\dagger S \, |0 \rangle \nonumber \\
&=& \prod_{i=1}^N \, \Big( \, \sum_{\mu=+,-} \alpha_{i \mu} \, S \, a_{i
\mu}^\dagger \, S^\dagger \, \Big) \, |0 \rangle \, .
\end{eqnarray}
In the following, the matrix elements $U_{ji}$ of the unitary transformation
matrix $U$ denote the amplitudes for the redirection of a photon in input $i$
to output $j$. Generally speaking, an $N \times N$ multiport described by any arbitrary transfer matrix $U$
may be constructed by a pyramidal arrangement of beamsplitters and phase plates
as shown in Fig.~\ref{construction}.
The most familiar example of a multiport is the $2 \times 2$ beamsplitter that has
2 input and 2 output ports. It can be described by a unitary $2 \times 2$ matrix $B(R,\phi)$
given by
\begin{equation}
B(R,\phi)=\left( \begin{array}{cc} \sqrt{T} & {\rm e}^{{\rm i}\phi}\sqrt{R} \\
\sqrt{R} &
-{\rm e}^{{\rm i}\phi} \sqrt{T} \end{array} \right),
\end{equation} where the $R$ denotes the reflectivity and $T=1-R$ denotes the
transmittivity of the beamsplitter. The phase $\phi$ is obtained by placing a phase
shifter at one of the input ports.
\begin{figure}\label{pyramid}
\begin{center}
\includegraphics[scale=0.75]{multiport1}
\end{center}
\caption{ Pyramidal construction of a $N \times N$ multiport consisting of
beamsplitters and phase plates } \label{construction}
\end{figure}
Reck \cite{Reck94,Reck96} (see also \cite{Sun01}) has shown this using similar methods as used in Gaussian
elimination. The key to his proof is to factorize the matrix $U$ into a product
of block matrices describing only $2 \times 2$ beam splitter matrices together with phase
shifts. We begin by defining the $N \times N$ matrix $T_{pq}(R_{pq}, \phi_{pq})$
which is essentially an identity matrix except for possibly four elements indexed by $pp$, $pq$, $qp$ and $qq$
which denote effectively a $2 \times 2$ unitary matrix. Note
that $T_{pq}(R_{pq}, \phi_{pq})$ represents a $2 \times 2$ beamsplitter with
matrix $B(R_{pq},\phi_{pq})$ where the two input ports are $p$ and $q$ input
ports of the $N \times N$ multiport.
For the inverse matrix of $U$ denoted as $U^{-1}$, it is possible to find
appropriate $T_{pq}(R_{pq}, \phi_{pq})$ such that $U^{-1} \prod^{1}_{i=N-1}
T_{Ni}(\omega_{Ni}, \phi_{Ni})$ is another unitary matrix where the last row and
column contains only zeros except on the diagonal element which contains only a
phase factor. Defining $\prod^{1}_{i=N-1} T_{Ni}(\omega_{Ni}, \phi_{Ni})=L(N)$,
we can systematically reduce $U^{-1}$ to a diagonal matrix $D^{-1}$ that
contains only phase factors in the diagonal elements by the following operation,
\begin{equation}
U^{-1} L(N)L(N-1)...L(2)=D^{-1}.
\end{equation}
It is clear that $U=L(N)L(N-1)...L(2)D$ can be built by a series of $2 \times 2$
beamplitters with phase shifters in each of the $N$ output ports corresponding
to the diagonal elements of $D^{-1}$. Indeed, the pyramidal construction shown in
Fig.~\ref{pyramid} corresponds precisely to such a decomposition operated in reverse. From this
construction, the maximum number of $2 \times 2$ beamsplitters needed for the
construction for a $N \times N$ multiport is given by $N(N-1)/2$.
Since the multiport beam splitter does not contain any elements
that change the inner degrees of freedom of the incoming photons, the transition
matrix $U$ does not depend on $\mu$. Denoting the creation operator for a single
photon with parameter $\mu$ in output port $j$ by $b_{j \mu}^\dagger$ therefore
yields
\begin{eqnarray} \label{tran}
S \, a_{i \mu}^\dagger \, S^\dagger = \sum_j U_{ji} \, b_{j \mu}^\dagger \, .
\end{eqnarray}
Inserting this into Eq.~(\ref{fin}) we can now calculate the output state of the
$N \times N$ multiport given the initial state (\ref{firework:in}) and obtain
\begin{eqnarray} \label{firework:output1}
|\phi_{\rm out} \rangle &=& \prod_{i=1}^N \, \Bigg[ \, \sum_{j=1}^N \, U_{ji} \,
\Big( \sum_{\mu=+,-} \alpha_{i \mu} \, b_{j \mu}^\dagger \Big) \,
\Bigg] \, |0 \rangle \, . ~
\end{eqnarray}
This equation describes the independent redirection of all photons to their
potential output ports. Conservation of the norm of the state vector is
guaranteed by the unitarity of the transition matrix $U$. It is also important to note that any multiplication of phase factors in any of
the input or output ports as well as any relabelling of the input or output ports constitutes a multiport which is essentially equivalent
to the original multiport. In this sense, the original multiport is defined up to an equivalence class.
The state preparation is considered successful under the condition of the
collection of one photon per output port. To calculate the final state, we apply
the corresponding projector to the output state (\ref{firework:output1}) and find that
the thereby postselected $N$-photon state equals, up to normalisation,
\begin{equation} \label{firework:output2}
|\phi_{\rm pro} \rangle=\sum_{\sigma} \Bigg[ \prod_{i=1}^N U_{\sigma(i) i} \Big(
\sum_{\mu=+,-} \alpha_{i \mu} b_{\sigma (i) \mu}^{\dagger} \Big)
\Bigg] \, |0 \rangle \, .~
\end{equation}
Here $\sigma$ are the $N!$ possible permutations of the $N$ items $\{1,\, 2, \,
..., \, N\}$. Note that the bosonic statistics of photons has been taken into account inherently in the formulation. A further
elaboration on this is due in Chapter \ref{fusion}. Moreover, the norm of the state (\ref{firework:output2}) squared, namely
\begin{equation} \label{firework:suc}
P_{\rm suc} = \| \, |\phi_{\rm pro} \rangle \, \|^2 \, ,
\end{equation}
is the success rate of the scheme and probability for the collection of one
photon in each output $j$.
\subsection{The Bell multiport beam splitter}
Motivated by a great variety of applications, we are particularly interested in
the generation of highly entangled photon states of a high symmetry, an example
being {\em W}-states. This suggests to consider symmetric multiports as described in \ref{photonmultiport}, which
redirect each incoming photon with equal probability to all potential output
ports. A special example for such an $N \times N$ multiport is the Bell
multiport beam splitter. Its transformation matrix
\begin{eqnarray}\label{fourier}
U_{ji} &=& {\textstyle{1 \over \sqrt{N}} \, \omega_N^{(j-1)(i-1)}}
\end{eqnarray}
is also known as a discrete Fourier transform matrix and has been widely
considered in the literature \cite{Zukowski97,Torma95,Torma98}. Indeed, the Bell multiport is a linear optical realisation of a quantum Fourier transform. Here $\omega_N$
denotes the $N$-th root of unity,
\begin{equation} \label{root}
\omega_N \equiv \exp \left( 2{\rm i} \pi /N \right) \, .
\end{equation}
Proceeding as in Section II.D of Ref.~\cite{Zukowski97}, it can easily be
verified that $U$ is unitary as well as symmetric. Especially for $N=2$, the
transition matrix (\ref{fourier}) describes a single 50:50 beam splitter.
\section{The generation of 4-photon states} \label{fourphoton}
Before we discuss the $N$-photon case, we investigate the possibility of preparing
highly entangled 4-photon states using specially prepared photons and a $4
\times 4$ Bell multiport beam splitter. For $N=4$, the transition matrix
(\ref{fourier}) becomes
\begin{eqnarray} \label{firework:matrix4}
U &=& {\textstyle {1 \over 2}} \left( \begin{array}{rrrr}
1 & 1 & 1 & 1\\
1 & \omega_4 & \omega_4^2 & \omega_4^3\\
1 & \omega_4^2 & \omega_4^4 & \omega_4^6\\
1 & \omega_4^3 & \omega_4^6 & \omega_4^9 \end{array} \right)
= {\textstyle {1 \over 2}} \left( \begin{array}{rrrr}
1 & 1 & 1 & 1\\
1 & {\rm i} & -1 & -{\rm i}\\
1 & -1 & 1 & -1\\
1 & -{\rm i} & -1 & {\rm i} \end{array} \right) \, .~ \nonumber \\ &&
\end{eqnarray}
The following analysis illustrates the richness of the problem as well as
motivating possible generalisations for the case of arbitrary photon numbers.
\subsection{Impossible output states} \label{pose}
Let us first look at the seemingly trivial situation, where every input port of
the multiport beamspliter is entered by one photon in the same state, let us say
in $|+ \rangle$, so that
\begin{equation} \label{inni}
|\phi_{\rm in} \rangle = a_{1 +}^{\dagger}a_{2 +}^{\dagger}a_{3 +}^{\dagger}a_{4
+}^{\dagger} \, |0 \rangle \, .
\end{equation}
Using Eqs.~(\ref{firework:output2}) and (\ref{firework:matrix4}), we then find that the collection
of one photon per output port prepares the system in the postselected state
\begin{equation}
|\phi_{\rm pro} \rangle = \sum_{\sigma} \Bigg[ \prod_{i=1}^4 U_{\sigma (i)i} \,
b_{i +}^{\dagger} \Bigg] \,
|0 \rangle = 0 \, .
\end{equation}
This means, that it is impossible to pass four photons in the same state through
the considered setup with each of them leaving the multiport in a different
output port. More generally speaking, the state with four photons in the same
state does not contribute to the event of collecting one photon per output port.
It is therefore impossible to prepare any superposition containing the states
$b_{1 +}^{\dagger}b_{2 +}^{\dagger}b_{3
+}^{\dagger}b_{4 +}^{\dagger} \, |0 \rangle$ or $b_{1 -}^{\dagger}b_{2
-}^{\dagger}b_{3 -}^{\dagger}b_{4 -}^{\dagger} \, |0 \rangle$, respectively. The
reason is destructive interference of certain photon states within the linear
optics setup, which plays a crucial role for the generation of multiphoton
entanglement via postselection. This effect is further studied and generalised in Chapter \ref{fusion}.
\subsection{The 4-photon {\em W}-state} \label{W4}
We now focus our attention on the case, where input port 1 is entered by a
photon prepared in $|+ \rangle$ while all other input ports are entered by a
photon in $|- \rangle$, i.e.
\begin{equation} \label{Win}
|\phi^W_{\rm in} \rangle = a_{1 +}^{\dagger}a_{2 -}^{\dagger}a_{3
-}^{\dagger}a_{4 -}^{\dagger} \, |0 \rangle \, .
\end{equation}
Using again Eqs.~(\ref{firework:output2}) and (\ref{firework:matrix4}), we find that the
collection of one photon per output port
corresponds to the postselected 4-photon state
\begin{eqnarray} \label{proni}
|\phi_{\rm pro}^W \rangle &=& \sum_{j=1}^4 U_{j1} \, b_{j +}^{\dagger} \,
\sum_{\sigma_j} \Bigg[
\prod_{i=2}^4 U_{\sigma_j(i) i} \, b_{\sigma_j (i) -}^{\dagger} \Bigg] \,
|0 \rangle \, , \nonumber \\&&
\end{eqnarray}
where the $\sigma_j$ are the $3!$ permutations that map the list $\{2, \, 3, \,
4\}$ onto the list $\{1, \, ..., \, (j-1), \, (j+1), \, ..., \, 4 \}$. If
$|j_{\rm out} \rangle$ denotes the state with one photon in $|+ \rangle$ in
output port $j$ and one photon in $|-\rangle$ everywhere else,
\begin{equation} \label{z}
|j_{\rm out} \rangle \equiv b_{N -}^{\dagger} \, . \, . \, . \, b_{(j+1)
-}^{\dagger} b_{j +}^{\dagger} b_{(j-1) -}^{\dagger} \, . \, . \, . \, b_{1
-}^{\dagger} \, |0\rangle \, ,
\end{equation}
and $\beta_j$ is a complex probability amplitude, then the output state
(\ref{proni}) can be written as
\begin{eqnarray} \label{pro3}
|\phi_{\rm pro}^W \rangle &=& \sum_{j=1}^4 \beta_j \, |j_{\rm out} \rangle \, .
\end{eqnarray}
Furthermore, we introduce the reduced transition matrices $U_{\rm red}^{(j)}$,
which are obtained by deleting the first column and the $j$-th row of the
transition matrix $U$. Then one can express each $\beta_j$ as the permanent\footnote{Note that the permanent of matrix $U$ is ${\rm perm}(U)=\sum_{\sigma}\prod_{i=1}^N U_{i\sigma(i)}.$}
\cite{Horn85,Scheel04,Minc78} of a matrix,
\begin{equation} \label{c}
\beta_j = U_{j 1} \sum_{\sigma_j} \prod_{i=2}^4 U_{\sigma_j (i)i}
= U_{j 1} \, {\rm perm} \, \left( U_{\rm red}^{(j) {\rm T}} \right) \, .
\end{equation}
The output state (\ref{pro3}) equals a {\em W}-state, if the coefficients
$\beta_j$ are all of the same size and differ from each other at most by a phase
factor.
To show that this is indeed the case, we calculate the reduced matrices $U_{\rm
red}^{(j)}$ explicitly\footnote{The reason for not simplifying these matrices
is that the following equations
provide the motivation for the proof of the general case in Section
\ref{doubleW}.} and obtain
\begin{eqnarray} \label{sing}
&& \hspace*{-0.5cm} U_{\rm red}^{(1)} = {\textstyle {1 \over 2}}
\left( \begin{array}{rrrr} \omega_4 & \omega_4^2 & \omega_4^3 \\ \omega_4^2 &
\omega_4^4 &
\omega_4^6 \\ \omega_4^3 & \omega_4^6 & \omega_4^9 \end{array} \right) \, , ~
U_{\rm red}^{(2)} = {\textstyle {1 \over 2}}
\left( \begin{array}{rrrr} 1 & 1 & 1 \\ \omega_4^2 & \omega_4^4 & \omega_4^6 \\
\omega_4^3
& \omega_4^6 & \omega_4^9 \end{array} \right) \, , \nonumber \\
&& \hspace*{-0.5cm} U_{\rm red}^{(3)} = {\textstyle {1 \over 2}}
\left( \begin{array}{rrrr} 1 & 1 & 1 \\ \omega_4 & \omega_4^2 & \omega_4^3 \\
\omega_4^3 &
\omega_4^6 & \omega_4^9 \end{array} \right) \, , ~
U_{\rm red}^{(4)} = {\textstyle {1 \over 2}}
\left( \begin{array}{rrrr} 1 & 1 & 1 \\ \omega_4 & \omega_4^2 & \omega_4^3 \\
\omega_4^2 &
\omega_4^4 & \omega_4^6 \end{array} \right) \, . \nonumber \\
\end{eqnarray}
The coefficients $\beta_j$ differ at most by a phase factor, if the norm of the
permanents of the transpose of these reduced matrices is for all $j$ the same.
To show that this is the case, we now define the vector
\begin{equation}
{\bf v} = (\omega_4, \omega_4^2, \omega_4^3) \, ,
\end{equation}
multiply each row of the matrix $U_{\rm red}^{(1)}$ exactly $(j-1)$ times with
${\bf v}$ and obtain the new matrices
\begin{eqnarray}
&& \hspace*{-0.5cm} \tilde{U}_{\rm red}^{(1)} = U_{\rm red}^{(1)} \, , ~
\tilde{U}_{\rm red}^{(2)} = {\textstyle {1 \over 2}}
\left( \begin{array}{rrrr} \omega_4^2 & \omega_4^4 & \omega_4^6 \\ \omega_4^3 &
\omega_4^6 & \omega_4^9 \\ 1 & 1 & 1 \end{array} \right) \, , ~ \nonumber \\
&& \hspace*{-0.5cm} \tilde{U}_{\rm red}^{(3)} = {\textstyle {1 \over 2}}
\left( \begin{array}{rrrr} \omega_4^3 & \omega_4^6 & \omega_4^9 \\ 1 & 1 & 1 \\
\omega_4 &
\omega_4^2 & \omega_4^3 \end{array} \right) \, , ~
\tilde{U}_{\rm red}^{(4)} = U_{\rm red}^{(4)} \, .
\end{eqnarray}
The above described multiplication amounts physically to the multiplication of
the photon input state with an overall phase factor and
\begin{equation}
\left| \, {\rm perm} \left( U_{\rm red}^{(1) {\rm T}} \right) \, \right| =
\left| \, {\rm perm} \left( \tilde{U}_{\rm red}^{(j) {\rm T}} \right) \, \right|
\, .
\end{equation}
Moreover, using the cyclic symmetry of permanents \cite{Horn85}, we see that
\begin{equation}
{\rm perm} \left( U_{\rm red}^{(j) {\rm T}} \right) = {\rm perm} \left(
\tilde{U}_{\rm red}^{(j) {\rm T}} \right) \, .
\end{equation}
This implies together with Eq.~(\ref{c}) that the norm of the coefficients
$\beta_j$ is indeed the same for all $j$. Furthermore, using the above argument
based on the multiplication of phase factors to the photon input state, one can
show that
\begin{equation} \label{phase}
\beta_j= \beta_1 \left( \prod_{k=0}^3 \omega_4^k \right)^{j-1} \, .
\end{equation}
Inserting this into Eq.~(\ref{proni}), we find that the postselected state with
one photon per output port equals, after normalisation\footnote{In the
following we denote normalised states by marking them with the $\hat{~}$
symbol.}, the {\em W}-state
\begin{equation} \label{beta}
|\hat{\phi}_{\rm pro}^W \rangle
= {\textstyle {1 \over 2}} \, \big[ \, b_{1 +}^{\dagger}b_{2 -}^{\dagger}b_{3
-}^{\dagger}b_{4 -}^{\dagger}-b_{1 -}^{\dagger}b_{2
+}^{\dagger}b_{3 -}^{\dagger}b_{4 -}^{\dagger}
+ b_{1 -}^{\dagger} b_{2 -}^{\dagger} b_{3 +}^{\dagger} b_{4 -}^{\dagger}
-b_{1 -}^{\dagger} b_{2 -}^{\dagger} b_{3 -}^{\dagger} b_{4
+}^{\dagger} \, \big] \, |0 \rangle \, .
\end{equation}
In analogy, we conclude that an input state with one photon in $|-\rangle$ in
input port 1 and a photon in $|+ \rangle$ in each of the other input ports,
results in the preparation of the {\em W}-state
\begin{equation} \label{beta2}
|\hat{\phi}_{\rm pro}^{W'} \rangle
= {\textstyle {1 \over 2}} \, \big[ \, b_{1 -}^{\dagger}b_{2 +}^{\dagger}b_{3
+}^{\dagger}b_{4 +}^{\dagger}-b_{1 +}^{\dagger}b_{2 -}^{\dagger}b_{3
+}^{\dagger}b_{4 +}^{\dagger}
+b_{1 +}^{\dagger}b_{2 +}^{\dagger}b_{3 -}^{\dagger}b_{4 +}^{\dagger}-b_{1
+}^{\dagger}b_{2 +}^{\dagger}b_{3 +}^{\dagger}b_{4 -}^{\dagger} \, \big] \,
|0 \rangle
\end{equation}
under the condition of the collection of one photon per output port. Both
states, (\ref{beta}) and (\ref{beta2}), can be generated with probability
\begin{equation} \label{s}
P_{\rm suc}={\textstyle {1 \over 16}} \, .
\end{equation}
Transforming them into the usual form of a $W$-state with equal coefficients of
all amplitudes \cite{Dur00} only requires further implementation of a Pauli
$\sigma_z$ operation (i.e.~a state dependent sign flip) on either the first and
the third or the second and the fourth output photon, respectively.
Although symmetry considerations may suggest that one can obtain a {\em W}-state given the described input, it is not obvious from a
rigorous point of view that this is the case. We have therefore performed explicit calculations to obtain the output state. In fact,
naive application of a symmetry argument may lead to an incorrect predicted state which we will show in the next 2 subsections.
\subsection{The 4-photon GHZ-state} \label{fun}
Besides generating {\em W}-states, the proposed setup can also be used to
prepare 4-photon GHZ-states. This requires, feeding each of the input ports 1
and 3 with one photon in $|+ \rangle$ while the input ports 2 and 4 should each
be entered by a photon in $|- \rangle$ such that
\begin{equation}
|\phi_{\rm in}^{\rm GHZ} \rangle = a_{1 +}^{\dagger}a_{2 -}^{\dagger}a_{3
+}^{\dagger}a_{4 -}^{\dagger} \, |0 \rangle \, .
\end{equation}
Calculating again the output state under the condition of collecting one photon
per output port, we obtain
\begin{equation}
|\phi_{\rm pro}^{\rm GHZ} \rangle
= \sum_{\sigma}U_{\sigma (1)1}U_{\sigma (2)2}U_{\sigma (3)3}U_{\sigma (4)4}
b_{\sigma (1)+}^{\dagger}b_{\sigma (2) -}^{\dagger}b_{\sigma (3)
+}^{\dagger}b_{\sigma (4) -}^{\dagger} \, |0 \rangle \ ,
\end{equation}
where the $\sigma$ are the $4!$ permutations that map the list $\{1, \, 2, \, 3,
\, 4 \}$ onto itself. On simplification, one finds that there are only two
constituent states with non-zero coefficients and $|\phi_{\rm pro}^{\rm GHZ}
\rangle$ becomes after normalisation
\begin{equation} \label{outGHZ}
|\hat{\phi}_{\rm pro}^{\rm GHZ} \rangle = {\textstyle {1 \over \sqrt{2}}} \,
\big[ \, b_{1 +}^{\dagger}b_{2
-}^{\dagger}b_{3 +}^{\dagger}b_{4 -}^{\dagger}-b_{1
-}^{\dagger}b_{2+}^{\dagger}b_{3 -}^{\dagger}b_{4 +}^{\dagger} \, \big] \, |0
\rangle
\end{equation}
which equals the GHZ-state up to local operations. Transforming (\ref{outGHZ})
into the usual form of the GHZ-state requires changing the state of two of the
photons, for example, from $|+ \rangle$ into $|- \rangle$. This can be realised
by applying a Pauli $\sigma_x$ operation to the first output port as well as a
$\sigma_y$ operation to the third output.
Finally, we remark that the probability for the creation of the GHZ-state
(\ref{ss}) is twice as high as the probability for the generation of a {\em
W}-state (\ref{s}),
\begin{equation} \label{ss}
P_{\rm suc} = {\textstyle {1 \over 8}} \, .
\end{equation}
Unfortunately, the experimental setup shown in Fig.~\ref{scheme} does not allow
for the preparation of GHZ-states for arbitrary photon numbers $N$. For a
detailed description of polarisation entangled GHZ states using a different
network of $50:50$ and polarising beam splitters, see Ref.~\cite{Sagi03}.
\subsection{The 4-photon double singlet state}
For completeness, we now ask for the output of the proposed state preparation
scheme, given that the input state equals
\begin{equation}
|\phi_{\rm in} \rangle = a_{1 +}^{\dagger} a_{2 +}^{\dagger} a_{3 -}^{\dagger}
a_{4 -}^{\dagger} |0 \rangle \, .
\end{equation}
Proceeding as above, we find that this results in the preparation of the state
\begin{equation}
|\phi_{\rm pro}^{\rm DS} \rangle
= \sum_{\sigma}U_{\sigma (1)1}U_{\sigma (2)2}U_{\sigma (3)3}U_{\sigma (4)4}
b_{\sigma (1)+}^{\dagger}b_{\sigma (2) +}^{\dagger}b_{\sigma
(3)-}^{\dagger}b_{\sigma (4)-}^{\dagger}\, |0 \rangle
\end{equation}
under the condition of the collection of one photon per output port. Here the
permutation operators $\sigma$ are defined as in Section \ref{fun}, which yields
\begin{eqnarray} \label{DS}
|\hat{\phi}_{\rm pro}^{\rm DS} \rangle
&=& {\textstyle {1 \over 2}} \, \big[ \, b_{1 +}^{\dagger}b_{2 +}^{\dagger}b_{3
-}^{\dagger}b_{4 -}^{\dagger}+b_{1 -}^{\dagger}b_{2
-}^{\dagger}b_{3 +}^{\dagger}b_{4 +}^{\dagger}
-b_{1 +}^{\dagger}b_{2 -}^{\dagger}b_{3 -}^{\dagger}b_{4 +}^{\dagger}-b_{1
-}^{\dagger}b_{2 +}^{\dagger}b_{3 +}^{\dagger}b_{4 -}^{\dagger} \, ] \, |0
\rangle \nonumber \\
&=& \frac{1}{\sqrt{2}}[ b_{1 +}^{\dagger}b_{3 -}^{\dagger}-b_{3 +}^{\dagger}b_{1 -}^{\dagger}] \otimes \frac{1}{\sqrt{2}}[ b_{2
+}^{\dagger}b_{4 -}^{\dagger}-b_{2 +}^{\dagger}b_{4 -}^{\dagger}]\ket{0} \, .
\end{eqnarray}
This state can be prepared with probability
\begin{equation}
P_{\rm suc} = {\textstyle {1 \over 16}} \, .
\end{equation}
The state (\ref{DS}) is a double singlet state, i.e.~a tensor product of two
2-photon singlet states, with a high robustness against decoherence
\cite{Eibl03,Bourennane04}. In this 2 subsections, naive symmetry considerations may suggest a state with equal
superpositions of all permutations of the given input state as the output. We have seen here clearly that this is not the case.
\subsection{The general 4-photon case}
Finally, we consider the situation where the input state is of the general form
(\ref{firework:in}). Calculating Eq.~(\ref{firework:output2}), we find that the unnormalised
output state under the condition of one photon per output port equals in this
case
\begin{eqnarray} \label{final}
|\phi_{\rm pro} \rangle &=&
{\textstyle {{\rm i} \over 4}} \big( \, \gamma_1+\gamma_2-\gamma_3-\gamma_4 \,
\big) \, |\hat{\phi}_{\rm pro}^{\rm DS} \rangle
+ {\textstyle {1 \over 2 \sqrt{2}}} \big( \, \gamma_6-\gamma_5\, \big) \,
|\hat{\phi}_{\rm pro}^{\rm GHZ} \rangle \nonumber \\
&& + {\textstyle {1 \over 4}} \big( \, \gamma_8 +\gamma_{10}-\gamma_7-\gamma_9
\, \big) \, |\hat{\phi}_{\rm pro}^{W} \rangle
+ {\textstyle{1 \over 4}} \big( \,
\gamma_{12}+\gamma_{14}-\gamma_{11}-\gamma_{13} \, \big) \, |\hat{\phi}_{\rm
pro}^{W'} \rangle ~~~
\end{eqnarray}
with the coefficients
\begin{eqnarray} \label{coefficients}
&\gamma_{1}=\alpha_{1+}\alpha_{2+}\alpha_{3-}\alpha_{4-} \, , ~
& \gamma_{2}=\alpha_{1-}\alpha_{2-}\alpha_{3+}\alpha_{4+} \, ,\nonumber \\
& \gamma_{3}=\alpha_{1-}\alpha_{2+}\alpha_{3+}\alpha_{4-} \, , ~
& \gamma_{4}=\alpha_{1+}\alpha_{2-}\alpha_{3-}\alpha_{4+} \, ,\nonumber \\
& \gamma_{5}=\alpha_{1+}\alpha_{2-}\alpha_{3+}\alpha_{4-} \, , ~
& \gamma_{6}=\alpha_{1-}\alpha_{2+}\alpha_{3-}\alpha_{4+} \, , \nonumber \\
& \gamma_{7}=\alpha_{1+}\alpha_{2-}\alpha_{3-}\alpha_{4-} \, , ~
& \gamma_{8}=\alpha_{1-}\alpha_{2+}\alpha_{3-}\alpha_{4-} \, , \nonumber \\
& \gamma_{9}=\alpha_{1-}\alpha_{2-}\alpha_{3+}\alpha_{4-} \, , ~
& \gamma_{10}=\alpha_{1-}\alpha_{2-}\alpha_{3-}\alpha_{4+} \, , \nonumber \\
& \gamma_{11}=\alpha_{1-}\alpha_{2+}\alpha_{3+}\alpha_{4+} \, , ~
& \gamma_{12}=\alpha_{1+}\alpha_{2-}\alpha_{3+}\alpha_{4+} \, , \nonumber \\
& \gamma_{13}=\alpha_{1+}\alpha_{2+}\alpha_{3-}\alpha_{4+} \, , ~
& \gamma_{14}=\alpha_{1+}\alpha_{2+}\alpha_{3+}\alpha_{4-} \, .~~~~~~
\end{eqnarray}
The form of the coefficients (\ref{coefficients}) reflects the full symmetry of
the transformation of the input state. Each of the entangled states
$|\hat{\phi}_{\rm pro}^{\rm DS} \rangle$, $|\hat{\phi}_{\rm pro}^{\rm GHZ}
\rangle$, $|\hat{\phi}_{\rm pro}^{W} \rangle$ and $|\hat{\phi}_{\rm pro}^{W'}
\rangle$ are generated independently from the different constituent parts of the
input (\ref{firework:in}). Besides, Eq.~(\ref{final}) shows that the output state is
constrained to be of a certain symmetry, namely the symmetry introduced by the $N \times N$ Bell multiport and the postselection criteria of
finding one photon per output port.
\section{The generation of $N$-photon {\em W}-states} \label{doubleW}
Using the same arguments as in Section \ref{W4}, we now show that the $N \times
N$ Bell multiport beam splitter can be used for the generation of {\em W}-states
for arbitrary photon numbers $N$. Like Bell states, {\em W}-states are highly
entangled but their entanglement is more robust \cite{Dur00}. Moreover, as $N$
increases, {\em W}-states perform better than the corresponding GHZ states
against noise admixture in experiments to violate local realism \cite{Sen(De)03}
and are important for optimal cloning protocols \cite{Buzek96}.
In analogy to Eq.~(\ref{Win}), we assume that the initial state contains one
photon in $|+ \rangle$ in the first input port, while every other input port is
entered by a photon prepared in $|- \rangle$ so that
\begin{eqnarray}
|\phi_{\rm in} \rangle &=& a_{1 +}^{\dagger} \prod_{i=2}^N a_{i -}^{\dagger} \,
|0 \rangle \, .
\end{eqnarray}
Using Eq.~(\ref{firework:output2}), we find that the state of the system under the
condition of the collection of one photon per output port equals
\begin{eqnarray} \label{output3}
|\phi_{\rm pro} \rangle &=& \sum_{j=1}^N U_{j1} \, b_{j +}^{\dagger} \,
\sum_{\sigma_j} \Bigg[
\prod_{i=2}^N U_{\sigma_j(i) i} \, b_{\sigma_j (i) -}^{\dagger} \Bigg] \,
|0 \rangle \, , \nonumber \\&&
\end{eqnarray}
where the $\sigma_j$ are the $(N-1)!$ permutations that map the list $\{2, \, 3,
\, ..., \, N\}$ onto the list $\{1,\, 2, \, ..., \, (j-1), \, (j+1), \, ..., \,
N \}$. As expected, the output is a superposition of all states with one photon
in $|+ \rangle$ and all other photons prepared in $|- \rangle$.
To prove that Eq.~(\ref{output3}) describes indeed a {\em W}-state, we use again
the notation introduced in Eqs.~(\ref{z}) and (\ref{pro3}) and write
\begin{eqnarray} \label{pro4}
|\phi_{\rm pro} \rangle & \equiv& \sum_j \beta_j \, |j_{\rm out} \rangle \, .
\end{eqnarray}
To show that the coefficients $\beta_j$ differ from $\beta_1$ at most by a phase
factor, we express the amplitudes $\beta_j$ as in Eq.~(\ref{c}) using the
permanents of the reduced transition matrices and find
\begin{equation} \label{ccc}
\beta_j = U_{j 1} \sum_{\sigma_j} \prod_{i=2}^N U_{\sigma_j (i)i} = U_{j 1} \,
{\rm perm} \, \left( U_{\rm red}^{\rm T}(j) \right) \, .
\end{equation}
Inserting the concrete form of the transition matrix $U$, this yields
\begin{equation}
\beta_j = \frac{1}{\sqrt{N^N}}\sum_{\sigma_j} \prod_{i=2}^N \omega_N^{(\sigma_j
(i)-1)(i-1)} \, .~~
\end{equation}
Proceeding as in Section \ref{W4}, we now multiply $\beta_j$ with the phase
factor
\begin{equation}
v_j \equiv \left( \prod_{k=0}^{N-1} \omega_N^k \right)^{-(j-1)}
\end{equation}
and obtain
\begin{eqnarray} \label{generalproof}
v_j \, \beta_j &=& {\textstyle {1 \over \sqrt{N^N}}} \sum_{\sigma_j}
\prod_{i=2}^N \omega_N^{(\sigma_j (i)-j)(i-1)} \nonumber \\
&=& {\textstyle {1 \over \sqrt{N^N}}} \sum_{\sigma_j} \prod_{i=2}^N
\omega_N^{\big({\rm mod}_N (\sigma_j (i)-j) \big)(i-1)} \, .~~
\end{eqnarray}
The expression ${\rm mod}_N(\sigma_j (i)-j)+1$ represents a set of $(N-1)!$
permutations that map $\{2, \, 3, \, ..., \, N\}$ onto the list $\{2, \, 3, \,
..., \, N\}$. It is therefore equivalent to the permutations $\sigma_1(i)$,
which allows us to simplify Eq.~(\ref{generalproof}) even further and to show
that
\begin{equation} \label{close}
v_j \, \beta_j = {\textstyle {1 \over \sqrt{N^N}}}
\sum_{\sigma_1} \prod_{i=2}^N \omega_N^{(\sigma_1 (i)-1)(i-1)} = \beta_1 \, .~~
\end{equation}
From this and the fact that $1+2+...+(N-1)={1 \over 2} N(N-1)$ we finally arrive
at the relation
\begin{eqnarray}
\beta_j &=& \left( \prod_{k=0}^{N-1} \omega_N^k \right)^{j-1} \beta_1 \nonumber
\\
&=& \left\{ \begin{array}{cl} \beta_1 \, , & {\rm if}~N~{\rm is~odd} \, , \\
(-1)^{j-1} \, \beta_1 \, , & {\rm if}~N~{\rm is~even} \, . \end{array} \right.
\end{eqnarray}
This shows that the amplitudes $\beta_j$ are all of the same size and the Bell
multiport can indeed generate $N$-photon {\em W}-states. If one wants the
coefficients $\beta_j$ to be exactly the same, one can remove unwanted minus
signs in case of even photon numbers by applying a $\sigma_z$ operation in each
output port with an even number $j$.
The logic of the described proof exploits the symmetry of a Bell multiport and
avoids calculating the coefficients of the constituent states of the output
photon. Indeed, there exist no known efficient method \cite{Scheel04,Minc78} to
calculate these coefficients in general.
In the case $N=2$, the above described state preparation scheme reduces to the
familiar example, where two photons prepared in the two orthogonal states $|+
\rangle$ and $|- \rangle$ pass through a 50:50 beam splitter. The collection of
one photon in the each output port prepares the system in this case in the state
${1 \over \sqrt{2}} \, [ \, b_{1+}^\dagger b_{2-}^\dagger - b_{1-}^\dagger
b_{2+}^\dagger \, ] \, |0 \rangle$, which can be transformed into ${1 \over
\sqrt{2}} \, [ \, b_{1+}^\dagger b_{2-}^\dagger + b_{1-}^\dagger b_{2+}^\dagger
\, ] \, |0 \rangle$ by flipping the sign of the state, i.e.~depending on
whether the photon is in $|+ \rangle$ or $|-\rangle$, in one of the output
ports.
\subsection{Success probabilities}
Let us finally comment on the success rate of the proposed {\em W}-state
preparation scheme. Computing the probability (\ref{firework:suc}) can be done by finding
the amplitude $\beta_1$ with the help of Eq.~(\ref{ccc}). Although the
definition of the permanent of a matrix resembles the definition of the
determinant, there exist only few theorems that can be used to simplify their
calculation \cite{Horn85,Scheel04,Minc78}. In fact, the computation of the permanent is an
NP-complete problem compared to that of a determinant which is only of complexity P. We therefore calculated $P_{\rm suc}$ numerically
(see
Fig.~\ref{fireplot}).
\begin{figure}
\begin{minipage}{\columnwidth}
\begin{center}
\resizebox{\columnwidth}{!}{\rotatebox{0}{\includegraphics{fireplot1}}}
\end{center}
\caption{The success rate for the generation of $N$-photon {\em W}-states
$P_{\rm suc}$ as a function of $N$. The solid line approximates the exact
results via the equation $P_{\rm suc}={\rm e}^{a-bN}$ with $a=1.35 \pm 1.32$ and
$b=1.27 \pm 0.10$} \label{fireplot}
\end{minipage}
\end{figure}
As it applies to linear optics schemes in general, the success probability
decreases unfavourably as the number of qubits involved increases. Here the
probability of success drops on average exponentially with $N$. We observe the
interesting effect of a non-monotonic decreasing success probability as $N$
increases. For example, the
probabilty of success for $N=13$ is higher than for $N=9$. Moreover, for $N=6$
and $N=12$, {\em W}-state generation is not permitted due to destructive
interference. This may lead one to speculate that this is the case for all multiples of $N=6$. Unfortunately, this does not apply to
$N=18$ and precludes an easy explanation of this effect.
\section{Conclusions} \label{firework:conclusions}
We analysed the generation of multiphoton entanglement with the help of
interference and postselection in a linear optics network consisting of an $N
\times N$ Bell multiport beam splitter. Each input port should be entered by a
single photon prepared in a certain state $|\lambda_i \rangle$. As long as the
photons are the same with respect to all other degrees of freedom and it can be
guaranteed that photons prepared in the same state overlap within their
coherence time inside the linear optics network, the described scheme can be
implemented using only a single photon source
\cite{Kuhn02,Mckeever04,Lange04,Benson00,Pelton02,Kurtsiefer00,Beveratos02}. We
believe that the described approach allows one to entangle much higher photon
numbers than what can be achieved in parametric down conversion experiments.
In general, a highly entangled output state is obtained under the condition of
the collection of one photon per output port. The motivation for this
postselection criteria is that distinguishing this state from other output
states does not require photon number resolving detectors, and can also accommodate lossy photon production. Ideally, the detectors should
have negligible dark counts which is possible with current technology \cite{Rosenberg05}. For
simplicity of discussion, we would take this to be the assumption in the rest of the thesis. Moreover, the photons
can easily be processed further and provide a resource for linear optics quantum
computing and quantum cryptographic protocols.
Firstly, we analysed the case $N=4$ and showed that the $4 \times 4$ Bell multiport
allows for the creation of a variety of highly-symmetric entangled states
including the {\em W}-state, the GHZ-state and double singlet states. It was
found that some states are easier to prepare than others. A straightforward
generalisation of the 4-photon case yields a scheme for the creation of
$N$-photon {\em W}-states. We calculated the rates for successful state
preparations and showed that they decrease in a non-monotonic fashion and on
average exponentially with $N$.
The motivation for considering a Bell multiport beam splitter was that it only
redirects the photons without affecting their inner degrees of freedom. The
proposed setup can therefore be used to produce polarisation, time-bin and
frequency entanglement, respectively. To generate, for example, polarisation
entangled photons, the initial photon states may differ in polarisation but
should otherwise be exactly the same. The high symmetry of the Bell multiport
beam splitter allows for the generation of a variety of highly entangled
symmetric states. Furthermore, except for interferometric stability being required for the multiport, the scheme is highly robust to
slow external and unknown phase fluctutation as this contributes to only a trivial global phase in the scheme.
The results in this chapter need not be limited only to postselected photon entanglement generation. As a foretaste, we will highlight an even more important application based on this chapter in
Chapter \ref{demand}. We continue our study on multiports in
the next chapter with the aim of studying multiparticle interference, which is the crucial underlying mechanism in much of this thesis.
\chapter{Abstract}
\clearpage \pagestyle{empty}
\begin{center}
{\Huge Abstract}\\[1cm]
\end{center}
Photons are natural carriers of quantum information due to their ease of distribution and long lifetime. This thesis concerns various
related aspects of quantum information processing with single photons.
Firstly, we demonstrate $N$-photon entanglement
generation through a generalised $N \times N$ symmetric beam splitter
known as the Bell multiport. A wide variety of 4-photon entangled states as well as the $N$-photon
W-state can be generated with an unexpected non-monotonic decreasing probability of success with $N$. We also show how the same setup can
be used to generate multiatom entanglement. A further study of multiports also leads us to a multiparticle generalisation of the
Hong-Ou-Mandel dip
which holds for all Bell multiports of even number of input ports.
Next, we demonstrate a generalised linear optics based photon filter that has a constant success
probability regardless of the number of photons involved. This filter has the
highest reported success probability and
is interferometrically robust. Finally, we demonstrate
how repeat-until-success quantum computing can be performed with two distant nodes
with unit success probability using only linear optics resource. We further show that using non-identical photon sources, robustness can
still be achieved, an illustration of the nature and advantages of measurement-based quantum computation. A
direct application to the same setup leads naturally to arbitrary multiphoton
state
generation on demand. Finally, we demonstrate how
polarisation entanglement of photons can be detected from the emission of two atoms in a Young's double-slit type experiment without
linear optics, resulting in both atoms being also maximally entangled.
\begin{publications}
\\
1. Y. L. Lim and A. Beige, {\em Photon polarisation entanglement from distant
dipole sources}, J. Phys. A {\bf 38}, L7 (2005), quant-ph/0308095 \\
2. Y. L. Lim and A. Beige, {\em Push button generation of multiphoton
entanglement}, Proc. SPIE {\bf 5436}, 118 (2004), quant-ph/0403125 \\
3. Y. L. Lim and A. Beige, {\em An efficient quantum filter for multiphoton
states}, J. Mod. Opt. {\bf 52}, 1073 (2005), quant-ph/0406008 \\
4. Y. L. Lim and A. Beige, {\em Multiphoton entanglement through a Bell
multiport beam splitter}, Phys. Rev. A {\bf 71}, 062311 (2005), quant-ph/0406047 \\
5. Y. L. Lim and A. Beige and L. C. Kwek, {\em Repeat-Until-Success linear optics quantum computing}, Phys. Rev. Lett. {\bf 95}, 030505
(2005), quant-ph/0408043 \\
6. Y. L. Lim and A. Beige, {\em Generalised Hong-Ou-Mandel Experiments with
Bosons and Fermions}, New J. Phys {\bf 7}, 155 (2005), quant-ph/0505034 \\
7. Y. L. Lim, S. Barrett, A. Beige, P. Kok and L. C. Kwek, {\em Repeat-until-success distributed quantum computing with
stationary and flying qubits},(submitted to Phys. Rev. A), quant-ph/0508218 \\
\end{publications}
\begin{acknowledgements}
I thank both Dr Almut Beige and Prof Sir Peter Knight for being my PhD supervisors. Almut has patiently guided me from being a novice to
the stage where I can hopefully say interesting and new things about physics. A valuable lesson which I have learnt from her is the
importance of being dogged and persevering in research. She has also taught me to believe in the impossible. Had it not been for her encouragement, I might have given up too easily on a difficult problem! Peter has been very encouraging and supportive to my education and have helped to ensure that I get the support and opportunities to attend
summer schools and conferences, all of which have proven to be crucial to this PhD experience. I also thank both of them for guiding me in my thesis and giving me many
valuable comments. Furthermore, many of the results in this thesis have been inspired from my interactions with them.
I thank Dan Browne for kindly ploughing through my thesis and offering many valuable scientific feedback, as well as suggesting improvements to the English. I also thank him for our many discussions that has greatly enriched me, and his patience in answering my many curious questions. In addition, I also thank Shash for reading my introductory chapter and also patiently explaining things like stabilizers and POVMs to me. I also thank Terry Rudolph for providing me with many inspiring and provocative thoughts that has helped to shape my research.
I thank Jim Franson for many stimulating discussions and encouragement on many of the work in this thesis, when he was visiting Imperial College. I also thank Geoff Pryde and Marek {\. Z}ukowski for their encouragement and interest in my work. I also appreciate Martin Plenio for his encouragement.
I thank Jesus Roger-Salazar for kindly giving me the latex template for this thesis. I also thank Jens Eisert for his kind help with Appendix A.
I thank my collaborators Sean Barrett and Pieter Kok for their friendship and sharing their valuable experience and knowledge in research
and physics. I also thank another collaborator Kwek Leong Chuan from Quantum LAH for funding a few of my visits to my homeland Singapore to facilitate research collaborations,
and also helping me to open up opportunities for future research. My appreciation goes also to Christian Kurtsiefer for getting me fired up with excitement on things that can be done back in Singapore after my PhD.
Special thanks goes to Hugo Cable for being the best of buddies to me in QOLS. We have both "grown up" much together through this PhD experience and I wish him all the best after his PhD.
I also thank Jae, Rachele, Jeremy and Adele for their friendship.
I want to specially thank Huang Sen and Emily for such a special and dear friendship that cannot be expressed with words.
I thank a special brother, Bae Joon Woo. By divine circumstances, our paths had crossed twice - once in IQING 2002 and the other time in Cargese 2004, and we have since maintained close contact. He has been a special encouragement and inspiration to me and I thank him for his friendship.
I would like to thank David Oliver, Ros, Carmel, Ladi and Iyiola of Riverpark Church who had so warmly welcomed me when I first arrived in London and eased my settling in. Also many thanks to all members of the church for their love and prayer support during my three years in London.
I thank David Ong, Adrian Ying, Patrick Wong as well as Melvin Tan for their friendship and encouragement, and for always remembering me in their prayers even when I was away from Singapore.
I also thank Kin Seng and Kien Boon who have been responsible for encouraging me to apply for the DSO postgraduate scholarship program and facilitating the application process. I thank Kin Seng in particular for his encouragement and belief in me. I also thank Yuqing and Ernest for their prayers and support. I thank Poh Boon for helping me to get into the Applied Physics Lab in DSO, where my love for physics was rekindled. Of course, many thanks to DSO for granting me the funding to complete the PhD program in Imperial College!
I thank everyone in my family, especially Daddy, Mummy, Granny, Aunty Lucia, Aunty Letitia, my dear sister Yuan Ping, brother Yuan Sing, as well as my parents-in-law for all their love, support and encouragement. My love of physics was first ignited when my dad bought me the book, {\em At the opening of the atomic era}, when I was only a child. Ever since, I had always wanted to be a physicist!
I thank my wife Puay-Sze for her neverending patience, support and love which have been so crucial to me. I also want to make special mention of my soon-to-be-born son, Isaac, for the joy and additional motivation that he brings to me during the last lap of my thesis-writing!
I thank God who has loved me, always guided me and taken care of me.
\end{acknowledgements}
\tableofcontents
\chapter{Generalised Hong-Ou-Mandel Effect for \\ Bosons and Fermions}
\label{fusion}
\section{Introduction}
The 2-photon Hong-Ou-Mandel (HOM) dip has been demonstrated first in 1987
\cite{Hong87}. In their experiment, Hong, Ou and Mandel sent two identical
photons through the separate input ports of a $50:50$ beam
splitter. Each output port contained a photon detector. No
coincidence detections within the temporal coherence length of the photons,
i.e.~no simultaneous clicks in both detectors, were recorded when there is no
relative delay of the input photons\footnote{The term ``HOM
dip" refers to the ``dip" of the coincidence counts in both detectors under zero
relative time delay of the input photons or photon
detection.}. Crucial for the
observation of this effect was the indistinguishability of the pure quantum
states of
the input photons, which differed only in the directions of their wave vectors.
This allowed the photons to interfere within the
setup. The detectors could not resolve the origin of each observed photon.
Due to the nature of this experiment, the HOM dip was soon employed for quantum
mechanical tests of local realism and for the generation of
postselected entanglement between two photons \cite{Shih88}. Linear optics Bell
measurements on photon pairs rely intrinsically on the HOM dip
\cite{Braunstein95,Mattle96}, which has also been a building block for the
implementation of linear optics gates for quantum information processing with
photonic qubits \cite{Knill01}. Shor's factorisation algorithm \cite{Shor94},
for
example, relies on multiple path interference to achieve massive parallelism
\cite{Ou99} and multiphoton interference has to play a crucial role in any
implementation of this algorithm using linear optics.
Since it requires temporal and spatial mode-matched photons, observing the HOM
dip for two photons is also a good test of their indistinguishability. HOM
interference has been applied to characterise recently introduced sources for
the
generation of single photons on demand by testing the identicalness of
successively generated photons \cite{Fattal04,Legero04,Kiraz05}. Another
interesting test based on the HOM dip has been studied by Bose and Home, who
showed that it can reveal whether the statistics of two identical particles
passing through a $50: 50$ beam splitter is fermionic or bosonic \cite{Bose02}.
Motivated by the variety of possible applications of the 2-photon HOM dip,
this
chapter investigates generalised HOM experiments. We consider a straightforward
generalisation of the scattering of two particles through a $50:50$ beam
splitter, namely the scattering of $N$ particles through a symmetric $N \times
N$
Bell multiport beam splitter. While numerous studies on $N$ photon interference
in the {\em constructive} sense, i.e.~resulting in the enhancement of a certain
photon detection syndrome, have been made (see e.g.~Refs.~\cite{Ou99}), not much
attention has been paid to multiple path interference in the {\em destructive}
sense. Mattle {\em et al.} \cite{Mattle95} has studied both constructive and
destructive detection syndromes for two photons scattering
through an $N \times N$ Bell multiport. Recently, Walborn {\em et al.} studied
so-called multimode HOM effects for
photon pairs with several inner degrees of freedom, including the spatial and
the
polarisation degrees of freedom \cite{Walborn03}. A notable example for
destructive HOM interference has been given by Campos \cite{Campos00}, who
studied certain triple coincidences in the output ports of an asymmetric $3
\times 3$ multiport beam splitter, which is also known as a tritter.
We consider {\em bosons} as well as the simultaneous scattering of {\em
fermions}. The difference between both classes of particles is most elegantly
summarised in the following commutation rules. While the annihilation and
creation operators $a_i$ and $a_i^\dagger$ for a boson in mode $i$ obey the
relation
\begin{equation} \label{boson}
[a_i,a_j^\dagger] \equiv a_i a_j^\dagger - a_j^\dagger a_i = \delta_{ij} ~~~
{\rm
and} ~~~ [a_i^\dagger,a_j^\dagger]=[a_i,a_j]=0 ~~ \forall ~ i, \, j
\end{equation}
with $\delta_{ij}=0$ for $i \neq j$ and $\delta_{ii} = 1$, the annihilation and
creation operators $a_i$ and $a_i^\dagger$ of fermionic particles obey the
anticommutation relation
\begin{equation} \label{fermion}
\{a_i,a_j^\dagger\} \equiv a_i a_j^\dagger + a_j^\dagger a_i =\delta_{ij} ~~~
{\rm and} ~~~ \{a_i^\dagger,a_j^\dagger\}=\{a_i,a_j\}=0 ~~ \forall ~ i, \, j \,
.
\end{equation}
Here $i$ and $j$ refer to the inner degrees of freedom of the particles, like
their respective path, polarisation, spin, frequency or energy.
Multiport beam splitters exist in general for a wide variety of fermionic
and bosonic particles. Possible realisations of a photonic multiport have been
discussed in Chapter \ref{firework}. For example,
multiports for bosonic or fermionic atoms can
consist of a network of electrode wave guide beam splitters on an atom chip
\cite{Cassettari00}. Multiports for electrons, which behave like fermions, can be
realised by fabricating a network of quantum point contacts acting as
2-electron beam splitters \cite{Samuelsson04}. Specially doped optical fibres
have recently been introduced in the literature and are expected to constitute
beam splitters for ``fermion-like" photons \cite{Franson04}.
As in the original HOM experiment \cite{Hong87}, we assume in the following
that
a particle detector is placed in each output port of the scattering beam
splitter array. The incoming particles should enter the different input ports
more or less simultaneously and in such a way that there is one particle per input port.
Moreover, we assume that the particles are identical. We will show that it is
impossible to observe a particle in each output port for even numbers $N$ of
bosons. We denote this effect of zero coincidence detection as the {\em
generalised HOM dip}. We will also show that fermions always leave the setup
separately exhibiting perfect coincidence detection. Since the interference
behaviour of both types of particles is very different, the Bell multiport can
be
used to reveal their quantum statistics.
This chapter is organised as follows. In Section \ref{scatterfusion}, we introduce the
theoretical description of particle scattering through a symmetric Bell
multiport. Section \ref{twofusion} describes the scattering of two particles through a
$50 : 50$ beam splitter as an example. In Section \ref{many}, we derive the
condition for the generalised HOM dip for bosons and analyse the scattering of
fermions through the same setup for comparison. Finally we conclude our results
in Section \ref{conclusions}.
\section{Scattering through a Bell multiport beam splitter} \label{scatterfusion}
The description of particle scattering through a multiport is essentially the
same as the previous Chapter \ref{firework}. Suppose each
input port $i$ is entered by a
particle with creation operator $a_i^\dagger$. Then the input state of the
system
equals
\begin{eqnarray} \label{infusion}
|\phi_{\rm in} \rangle &=& \prod_{i=1}^N a_i^\dagger \, |0 \rangle \, ,
\end{eqnarray}
where $|0 \rangle$ is the vacuum state with no particles in the setup.
If $b_j^\dagger$ denotes the creation operator for a single particle in output
port $j$, similar to Chapter \ref{firework}, we obtain for the output state of
the photons given the input state (\ref{infusion}),
\begin{eqnarray} \label{output1}
|\phi_{\rm out} \rangle &=& \prod_{i=1}^N \, \Bigg( \, \sum_{j=1}^N \, U_{ji} \,
b_j^\dagger \, \Bigg) \, |0 \rangle \, .
\end{eqnarray} Again, $U_{ji}$ denotes the matrix element representing the
transition amplitude of the $i$th input port to the $j$th output port of the matrix
$U$ defining the multiport. Specially for a Bell multiport, $U$ is a discrete
fourier transform matrix defined in Chapter \ref{firework}.
Note that up to now, we
have not invoked any assumptions about the nature of the particles. The
formalism
in this section applies to bosons and fermions equally.
\section{HOM interference of two particles} \label{twofusion}
\begin{figure}
\begin{minipage}{\columnwidth}
\begin{center}
\vspace*{0cm}
\resizebox{\columnwidth}{!}{\rotatebox{0}{\includegraphics{beamsplit}}}
\end{center}
\vspace*{-1.5cm}
\caption{(a) HOM dip for two bosons scattering through a $50:50$ beam splitter.
(b) Perfect coincidence in the output ports for fermion scattering.} \label{hom}
\end{minipage}
\end{figure}
Before analysing the general case, we motivate our discussion by considering two
identical particles entering the different input ports of a $50 : 50$ beam
splitter. For $N=2$, the transition matrix (\ref{fourier}) becomes the Hadamard
matrix\footnote{We remind the reader that the transition matrix chosen here is not unique. It represents rather an equivalence class of $50:50$ beam splitters with can be transformed to each other via phase shifts. Our discussion on the bunching or antibunching of particles apply to this equivalence class.}
\begin{eqnarray}
U &=& {\textstyle {1 \over \sqrt{2}}} \left( \begin{array}{rr} 1 & 1 \\ 1 & -1
\end{array} \right)
\end{eqnarray}
and the input state (\ref{in}) becomes $|\phi_{\rm in} \rangle= a_1^\dagger
a_2^\dagger \, |0 \rangle $. Note that local measurements on this input state
cannot reveal any information about the bosonic or fermionic nature of the two
particles. However, using Eq.~(\ref{output1}), we find that the beam splitter
prepares the system in the state
\begin{equation} \label{n2}
|\phi_{\rm out} \rangle = {\textstyle {1 \over 2}} \,
\big(b_1^\dagger+b_2^\dagger \big) \big(b_1^\dagger-b_2^\dagger \big) \, |0
\rangle = {\textstyle {1 \over 2}} \left[ (b_1^\dagger)^2 - b_1^\dagger
b_2^\dagger + b_2^\dagger b_1^\dagger - (b_2^\dagger)^2 \right] |0 \rangle \, .
\end{equation}
This state no longer contains any information about the origin of the particles,
since each incoming one is equally likely
transferred to any of the two output ports. Passing through the setup, the input
particles become indistinguishable by detection (see Fig.~ \ref{hom}). Their
quantum statistics can now be revealed using local measurements.
Bosons obey the commutation law (\ref{boson}). Using this, the output state
(\ref{n2}) becomes
\begin{equation}
|\phi_{\rm out} \rangle = {\textstyle {1 \over 2}} \left[ (b_1^{\dagger})^2 -
(b_2^{\dagger})^2 \right] |0 \rangle \, ,
\end{equation}
which implies a zero-coincidence count rate at the output ports. The particles
bunch together in the same output port and exhibit the famous HOM dip (see Fig.~
\ref{hom}(a)). In contrast, fermions obey the anticommutation relation
(\ref{fermion}) and their output state
\begin{equation}
|\phi_{\rm out} \rangle = b_1^\dagger b_2^\dagger \, |0 \rangle
\end{equation}
implies perfect particle coincidence. This means that the fermions always arrive
in separate output ports and never bunch together (see Fig.~\ref{hom}(b)). A
$50 : 50$ beam splitter can therefore be used to distinguish bosons and
fermions indeed \cite{Bose02}.
\section{Multiparticle HOM interference} \label{many}
We now consider the general case of $N$ particles passing through an $N \times
N$
Bell multiport beam splitter. As in the $N=2$ case, the setup redirects each
incoming particle with equal probability to any of the possible output ports,
thereby erasing the information about the origin of each particle and making
them
indistinguishable by detection. For even numbers of bosons, this results in the
generalised HOM dip and zero coincidence detection. In contrast, fermions leave
the setup always separately, thus demonstrating maximum coincidence detection.
Observing this extreme behaviour can be used, for example, to verify the quantum
statistics of {\em many} particles experimentally.
\subsection{Bosonic particles} \label{bos}
In order to derive the necessary condition for the appearance of the generalised
HOM dip for even numbers of bosons, we calculate the output state
(\ref{output1})
of the system under the condition of the collection of one particle per output
port. Each term contributing to the projected conditional output state
$|\phi_{\rm pro} \rangle$ can be characterised by a certain permutation, which
maps the particles in the input ports $1, \, 2, \, ..., \, N$ to the output
ports
$1, \, 2, \, ..., \, N$. In the following, we denote any of the $N!$
permutations
by $\sigma$ with $\sigma (i)$ being the $i$-th element of the list obtained when
applying the permutation $\sigma$ onto the list $\{1,\, 2, \, ..., \, N\}$.
Using
this notation, $|\phi_{\rm pro} \rangle$ equals up to normalisation
\begin{equation} \label{pro}
|\phi_{\rm pro} \rangle = \sum_{\sigma} \Bigg[ \prod_{i=1}^N U_{\sigma (i)i} \,
b_{\sigma(i)}^{\dagger} \Bigg] \, |0 \rangle \, .
\end{equation}
The norm of this state has been chosen such that
\begin{equation} \label{suc}
P_{\rm coinc} = \| \, |\phi_{\rm pro} \rangle \, \|^2
\end{equation}
is the probability to detect one particle per output port. It is therefore also
the probability for observing coincidence counts in all $N$ detectors as in Chapter \ref{firework}.
Up to now, the nature of the particles has not yet been taken into account.
Using
the commutation relation (\ref{boson}) for bosons, the conditional output state
(\ref{pro}) becomes
\begin{equation} \label{properm}
|\phi_{\rm pro} \rangle = {\rm perm} \, U \cdot \prod_{i=1}^N b_i^{\dagger} \,
\ket{0}
\end{equation}
with the permanent of the square matrix $U$ defined as
\cite{Scheel04,Horn85,Minc78}
\begin{equation} \label{perm}
{\rm perm} \, U \equiv {\rm perm} \, U^T \equiv \sum_{\sigma} \prod_{i=1}^N
U_{\sigma (i) \, i} \, .
\end{equation}
The permanent of a matrix is superficially similar to the determinant. However,
there exist hardly any mathematical theorems that can simplify the calculation
of
the permanent of an arbitrary matrix.
To derive a condition for the impossibility of coincidence detections, we have
to
see when the probability (\ref{suc}) equals zero. Using Eq.~(\ref{properm}), we
find
\begin{equation} \label{almost}
P_{\rm coinc} = | \, {\rm perm} \, U \, |^2 \, .
\end{equation}
The key to the following proof is to show that the transition matrix $U$ of the
Bell multiport possesses a certain symmetry such that its permanent vanishes in
certain cases. Suppose the matrix $U$ is multiplied by a diagonal matrix
$\Lambda$ with matrix elements
\begin{equation}
\Lambda_{jk} \equiv \omega_N^{j-1} \, \delta_{jk} \, .
\end{equation}
This generates a matrix $\Lambda U$ with
\begin{eqnarray} \label{ma}
(\Lambda U)_{ji} = \sum_{k=1}^N \Lambda_{jk}U_{ki} = \Lambda_{jj}U_{ji}
= {\textstyle{1 \over \sqrt{N}}} \, \omega_N^{(j-1)i} \, .
\end{eqnarray}
We now introduce the modulus function defined as ${\rm mod}_N (x)=j$, if $x-j$
is
dividable by $N$ and $0\leq j<N$. Since $\omega_N^ N = \omega_N^0 =1$, the
matrix
elements (\ref{ma}) can be expressed as
\begin{eqnarray}
(\Lambda U)_{ji} = {\textstyle{1 \over \sqrt{N}}} \, \omega_N^{(j-1)({\rm
mod}_N (i)+1-1)} \, .
\end{eqnarray}
Note that the function $\tilde \sigma(i)={\rm mod}_N (i)+1$ maps each element of
the list $\{1,2,...N-1,N \}$ respectively to the list $\{2,3,...N,1\}$. A
comparison with Eq.~(\ref{fourier}) therefore shows that
\begin{equation}
(\Lambda U)_{ji} = U_{j \, \tilde \sigma(i)} \, .
\end{equation}
In other words, the multiplication with $\Lambda$ amounts to nothing more than a
cyclic permutation of the columns of the matrix $U$. Taking the cyclic
permutation symmetry of the permanent of a matrix (see definition (\ref{perm}))
into account, we obtain
\begin{equation} \label{proof1}
{\rm perm} \, U ={\rm perm} \, (\Lambda U) \, .
\end{equation}
However, we also have the relation
\begin{equation} \label{proof2}
{\rm perm} \, (\Lambda U) ={\rm perm} \, \Lambda \cdot {\rm perm} \, U
\end{equation}
with the permanent of the diagonal matrix $\Lambda$ given by
\begin{eqnarray} \label{relative}
{\rm perm} \, \Lambda = \prod_{k=1}^N \omega_N^{k-1} = \omega_N^{\sum_{k=1}^N
k}
= \omega_N^{N(N+1)/2} = {\rm e}^{ {\rm i} \pi (N+1)} = \left\{ \begin{array}{rl}
1 \, , & {\rm if}~N~{\rm is~odd} \, , \\ -1 \, , & {\rm if}~N~{\rm is~even} \,
. \end{array} \right.
\end{eqnarray}
For $N$ being even, a comparison of Eqs.~(\ref{proof1}) - (\ref{relative})
reveals that
\begin{equation} \label{last}
{\rm perm} \, U = - {\rm perm} \, U = 0 \, .
\end{equation}
As a consequence, Eq.~(\ref{almost}) implies that $P_{\rm coinc} =0$.
Coincidence
detection in all output ports of the setup is impossible for even numbers of
bosons. This is not necessarily so, if the number of particles is odd. For
example, for $N=3$ one can check that there is no HOM dip by calculating ${\rm
perm} \, U$ explicitly. Campos showed that observing a HOM dip for $N=3$ is
nevertheless possible with the help of a specially designed asymmetric multiport
beam splitter \cite{Campos00}.
Furthermore, even if the number of particles is even, the HOM dip does not
appear to hold for all symmetric multiports. For example, it
is known that all symmetric $4 \times 4$ multiport can be represented generally
by the transition matrix $U$ given in the form as \cite{Zukowski97}
\begin{eqnarray} \label{fiber}
U &=& {\textstyle {1 \over 2}} \left( \begin{array}{rrrr} 1 & 1 & 1 & 1 \\ 1 &
{\rm e}^{{\rm i}\phi} & -1 & -{\rm e}^{{\rm i}\phi} \\ 1 &
-1 & 1 & -1 \\ 1 & -{\rm e}^{{\rm i}\phi} & -1 & {\rm e}^{{\rm i}\phi}
\end{array} \right) \, ,
\end{eqnarray} where each choice of $\phi$ in the range between $0$ and $\pi$
parameterize an equivalence class. Note that the Bell
multiport coincides with the choice of $\phi=\frac{\pi}{2}$. As before, one can
compute the probability of coincidence detection and it
is given by
\begin{equation}
P_{\rm coinc} =\frac{1}{8}(1+\cos (2 \phi)) \, .
\end{equation} This suggest that by performing a HOM experiment on coincidence
detection, one can characterise an unknown symmetric $4
\times 4$ multiport. This may find new application in symmetric multiports made
by fiber splicing \cite{Pryde03}\footnote{For example,
it was communicated to me by Geoff Pryde that the phase factor $\phi$ in
Eq.~(\ref{fiber}) is not a parameter easily controllable in
fiber splicing. }. In the case of a Bell multiport, one
recovers the HOM dip.
\subsection{Fermionic particles}
Fermions scattering through a Bell multiport show another extreme
behaviour. Independent of the number $N$ of particles involved, they always
leave
the setup via different output ports, thereby guaranteeing perfect coincidence
detection. As expected, particles obeying the quantum statistics of fermions
cannot populate the same mode.
Again, we assume that each input port is simultaneously entered by one particle
and denote the creation operator of a fermion in output port $i$ by
$b_i^\dagger$. Proceeding as in Section \ref{bos}, one finds again that the
output state of the system under the condition of the collection of one particle
per output port is given by Eq.~(\ref{pro}). To simplify this equation, we now
introduce the sign function of a permutation with $\rm{sgn}(\sigma) = \pm 1$,
depending on whether the permutation $\sigma$ is even or odd. An even (odd)
permutation is one, that can be decomposed into an even (odd) number of
interchanges. Using this notation and taking the anticommutation relation for
fermions (\ref{fermion}) into account, we find
\begin{equation}
|\phi_{\rm pro} \rangle = \sum_{\sigma}{\rm sgn}(\sigma) \Bigg( \prod_{i=1}^N
U_{\sigma (i) \, i} \, b_i^{\dagger} \Bigg) \, |0 \rangle \, .
\end{equation}
A closer look at this equation shows that the amplitude of this state relates to
the determinant of the transformation matrix given by
\begin{equation}
{\rm det} \, U =\sum_{\sigma}{\rm sgn}(\sigma) \prod_{i=1}^N U_{\sigma (i) \,
i}
\, .
\end{equation}
Since $U$ is unitary, one has $|{\rm det} \, U| =1$ and therefore also, as
Eq.~(\ref{suc}) shows,
\begin{equation}
P_{\rm coinc} = | \, {\rm det} \, U \, |^2 = 1 \, .
\end{equation}
This means that fermions leave the system separately indeed, i.e.~with one
particle per output port. In the above, we only used the unitarity of the
transition matrix $U$ but not its concrete form. Perfect coincidence detection
therefore applies to any situation where fermions pass through an $N \times N$
multiport, i.e.~independent of its realisation.
\section{Conclusions} \label{conclusions}
We analysed a situation, where $N$ particles enter the $N$ different input ports
of a symmetric Bell multiport beam splitter simultaneously. If these particles
obey fermionic quantum statistics, they always leave the setup independently
with one particle per output port. This results in perfect coincidence
detection,
if detectors are placed in the output ports of the setup. In contrast to this,
even numbers $N$ of bosons have been shown to never leave the setup with one
particle per output port. This constitutes a generalisation of the 2-photon
HOM
dip to the case of arbitrary even numbers $N$ of bosons. The generalised HOM dip
is in general not observable when $N$ is odd.
The proof exploits the cyclic symmetry of the setup. We related the coincidence
detection in the output ports to the permanent or the determinant of the
transition matrix $U$ describing the multiport, depending on the bosonic or
fermionic nature of the scattered particles. The NP complexity of computing the
permanent compared to the determinant has been discussed
in Chapter~\ref{firework}.
Experimental setups involving the scattering of bosons through a multiport
therefore have important applications in quantum information processing.
For example, part of the linear optics quantum computing scheme by Knill,
Laflamme and Milburn \cite{Knill01} is based on photon scattering through a Bell
multiport beam splitter. In contrast to this, the scattering of non-interacting
fermions through the same corresponding circuit, can be efficiently simulated on
a classical computer \cite{Terhal02,Knill01a}. Moreover, the quantum statistics
of particles has been used for a variety of quantum information processing tasks
such as entanglement concentration \cite{Paunkovic02} and entanglement transfer
\cite{Omar02}. Completely new perspectives might open when using setups that can
change the quantum behaviour of particles and convert, for example, photons into
fermions \cite{Franson04}.
Finally, we remark that observing HOM interference of many particles is
experimentally very robust. Our results can therefore also be used to verify the
quantum statistics of particles experimentally as well as to characterise or
align an experimental setup. Testing the predicted results does not require
phase
stability in the input or output ports nor detectors with maximum efficiency.
The
reason is that any phase factor that a particle accumulates in any of the input
or output ports contributes at most to an overall phase factor of the output
state $|\phi_{\rm out} \rangle$. However, the coincidence statistics are
sensitive
to the phase factors accumulated inside the multiport beam splitter as they
affect the form of the transition matrix $U$.
In the next chapter, we propose a scheme for an entanglement assisted
photon manipulation. The required entangled photon ancillas
can be either generated on demand (see Chapter \ref{demand}) or postselectively
(see Chapter \ref{firework}).
\chapter{An Efficient Quantum Filter for Multiphoton States} \label{hummingbird}
\section{Introduction}
Much effort has been made to find efficient schemes for the realisation of useful operations between photons contributing to quantum information processing. For example, we have discussed the process of entangling photons in Chapter \ref{firework}. In this chapter, we discuss a very useful operation, namely the {\em parity} or {\em
quantum filter} \cite{Pan98a,Franson01,Hofmann02,Grudka02,Zou02a}. The application of parity filters is diverse, ranging from quantum non demolition
measurements of entanglement to the generation of multiphoton quantum codes \cite{Hofmann02} and the generation of multipartite
entanglement \cite{Zou02a}. Moreover, it has been shown that the parity filter can constitute a crucial component for the generation of
cluster states for one-way quantum computing \cite{Verstraete04,Browne05}. Furthermore, Nemoto and Munro\cite{Nemoto04} applied the parity filter based on weak nonlinearity to achieve nearly deterministic linear optics quantum computing. Together with single qubit rotations and measurements, the
parity
filter constitutes a universal set of gate operations \cite{Browne05}.
Applied to two photons, the parity filter projects their state onto the 2-dimensional subspace of states where the photons have
identical polarisation in the $\ket{H}$ and $\ket{V}$ basis\footnote{This is also known as the states of even parity}. We denote the corresponding operator as $P_2$ and define
\begin{equation} \label{par}
P_2 = \sqrt{p_2} \, \Big( \, \ket{HH}\bra{HH}+\ket{VV}\bra{VV} \, \Big) ~,~
\end{equation}
where $H$ and $V$ describe a horizontally and a vertically polarised photon, respectively. Besides, $p_2$ is the success probability for
the performance of the parity projection on an arbitrary input state. This means, even when applied to a parity eigenstate, the photons
only pass through the filter with probability $p_2$. Here, the term {\em success probability} denotes the projection efficiency
of a given setup.
In the original proposal of a linear optics implementation of the 2-photon parity filter \cite{Hofmann02}, Hofmann and Takeuchi obtained a success
probability of $p_2={1 \over 16}$ after passing the photons through several beam splitters and performing postselective measurements.
Two other proposals yield a higher success probability of $p_2 = {1 \over 4}$ \cite{Grudka02,Zou02a}. Grudka and Wojcik achieve this by
using
the idea of teleportation \cite{Knill01} and by employing ancilla states containing six photons. Zou and Pahlke use a
single
mode quantum filter that separates the 1-photon state from the vacuum and the 2-photon state. By combining two such single mode filters,
a parity filter can be realised that requires a 4-photon ancilla state as a resource \cite{Zou02a}.
In direct analogy to the 2-photon parity filter (\ref{par}), a quantum filter for $N$ photons can be defined
by the operator
\begin{equation} \label{PN}
P_N=\sqrt{p_N} \, \Big( \, \ket{HH \, . \, . \, . \, H}\bra{HH \, . \, . \, . \, H}+\ket{VV \, . \, . \, . \, V}\bra{VV \, . \, . \, .\,
V} \, \Big) ~.~
\end{equation}
Applied to an arbitrary input state with $N$ photons, this filter projects the system with probability $p_N$ onto the 2-dimensional
subspace where all photons have the same polarisation in the $\ket{H}$ and $\ket{V}$ basis. One way to implement this gate is to pass
the input state through $(N-1)$ 2-photon parity filters, which succeeds with overall probability $p_N= p_2^{N-1}$. This approach
presents a steep challenge for large photon number $N$, given the above mentioned success probabilities of a single 2-photon parity
check.
In this chapter, we describe a potential implementation of the $N$-photon quantum filter (\ref{PN}) with a success rate as high as $p_N
= {1\over 2}$, which is much more effective than performing operation (\ref{PN}) with the previously proposed 2-photon parity filters.
As a resource we require the presence of the $N$-photon GHZ-state
\begin{equation} \label{GHZ}
|A^{(N)} \rangle = {\textstyle {1 \over \sqrt{2}}} \, \Big( \, \ket{HH \, . \, . \, . \, H} + \ket{VV \, . \, . \, . \, V} \, \Big) ~.~
\end{equation}
In principle, this photon state can be prepared on demand \cite{Gheri98,Lange00,LimSPIE04}. Furthermore, we require a photon-number
resolving
detector that can distinguish between 0, 1 and 2 photons. To implement the quantum filter (\ref{PN}), we use ideas that have been
inspired by a recently performed entanglement purification protocol \cite{Pan03}. Indeed, the same setup can be reconfigured and interpreted as a 2-photon parity filter. It should also be emphasised, with some changes in the definition of the photon basis in our setup, our quantum filter also maps to the CNOT gate proposed by Pittman {\em et al.} \cite{Pittman01} with a $\frac{1}{4}$ probability of success.
\section{A multipartite quantum filter} \label{humfilter}
The most important component of our scheme is the polarising beam splitter, which redirects a photon depending on its polarisation to
one of the output modes. In the following, $\ket{\lambda_i}$ describes a photon with polarisation $\lambda$ travelling in mode $i$.
Besides, we denote the input modes $i=1$ and $2$ and the output modes $i=1'$ and $2'$ such that a $V$ polarised photon entering input
mode $1$ and an $H$ polarised photon entering input mode $2$ leave the setup through output $1'$. Suppose two photons enter the setup
in different modes. Then the effect of the beam splitter can be summarised in the transformation
\begin{equation} \label{comp}
\ket{\lambda_1 \mu_2} \otimes \ket{0_{1'} 0_{2'}} \longrightarrow
\left\{ \begin{array}{ll} \ket{0_1 0_2} \otimes \ket{H_{1'}H_{2'}} ~,~ & {\rm if} ~~ \lambda = \mu = H ~,~ \\[0.1cm]
\ket{0_1 0_2} \otimes \ket{V_{1'}V_{2'}} ~,~ & {\rm if} ~~ \lambda = \mu =V ~,~ \\[0.1cm]
\ket{0_1 0_2} \otimes \ket{(HV)_{1'} 0_{2'}} ~,~ & {\rm if} ~~ \lambda = V ~~ {\rm and} ~~ \mu = H ~, \\[0.1cm]
\ket{0_1 0_2} \otimes \ket{0_{1'} (HV)_{2'}} ~,~ & {\rm if} ~~ \lambda = H ~~ {\rm and} ~~ \mu = V ~.~ \end{array} \right.
\end{equation}
We show now that this operation can be used to realise a filter which compares the polarisation $\lambda$ of a target photon with the
polarisation of an ancilla photon prepared in $\ket{\mu_2}$. With $\mu$ being either $V$ or $H$, the filter operation corresponds to the
projector $\ket{\mu} \bra{\mu}$ and can be implemented with {\em unit} efficiency.
Suppose a photon number resolving detector is placed in one of the output modes, say output $2'$, and the target photon enters the
system prepared in $\ket {\lambda_1} =\alpha \, \ket{H_1} + \beta \, \ket{V_1}$. Using Eq.~(\ref{comp}), one can calculate the
unnormalised output state after a click in the detector corresponding to polarisation $\mu$. It is either $\alpha \, \ket{H_{1'}}$ or $\beta \, \ket{V_{1'}}$, depending on whether $\mu$
equals $H$ or $V$. Note that the probability for a 1-photon detection ($|\alpha|^2$ or $|\beta|^2$, respectively) is exactly what one
would expect after applying the filter operation $\ket{\mu} \bra{\mu}$ with efficiency 1 to the incoming photon. Remarkably, the target
photon is effectively not destroyed in the process. The reason is that it does not matter whether the detector absorbs the target photon
or the ancilla photon, if both have the same polarisation and are anyway indistinguishable.
\subsection{The 2-photon case} \label{two}
Let us now describe how the polarising beam splitter (\ref{comp}) can be used for the implementation of a 2-photon parity filter. The
setup we consider here contains two polarising beam splitters and two polarisation sensitive detectors (see Fig.~\ref{fig1}). The target
state
enters the setup via the input modes $1$ and $3$. We further require the presence of the 2-photon ancilla state $\ket{A^{(2)}}$, which
is a 2-photon Bell state. The ancilla photons should enter the setup via the input modes $2$ and $4$. The two detectors are placed in
the output modes $2'$ and $4'$. If they both receive a photon each, the filter operation is deemed a success. Output modes $1'$
and $3'$ are designated the filter output.
\begin{figure}
\begin{center}
\includegraphics{humfig1}
\end{center}
\caption{Experimental setup for the realisation of a 2-photon parity filter. The target photons enter the setup via the input
modes 1 and 3, while the ancilla photons enter the setup via inputs 2 and 4. Within the setup, each photon has to pass one polarising
beam splitter. Under the condition of the detection of one photon in each of the outputs $2'$ and $4'$, the filter succeeded and the
projected output state leaves the system via the modes $1'$ and $3'$.} \label{fig1}
\end{figure}
In the following, we consider the general input pure state
\begin{equation} \label{huminput}
\ket{\psi_{\rm in}^{(2)}} = \alpha \, \ket{H_1 H_3} +\beta \, \ket{V_1 V_3} +\gamma \, \ket{H_1 V_3} + \delta \, \ket{V_1 H_3} ~.~
\end{equation}
Our aim is to eliminate the components, where the photons are of different polarisation. Together with the ancilla state $|A^{(2)}
\rangle$, the setup in Fig.~\ref{fig1} is entered by the 4-photon state
\begin{eqnarray} \label{in}
\ket{\tilde \psi_{\rm in}^{(2)}} &=& \ket{\psi_{\rm in}^{(2)}} \otimes \ket{A^{(2)}} \nonumber \\
&=& {\textstyle {1 \over \sqrt{2}}} \, \Big( \, \alpha \, \ket{H_1H_2H_3H_4} + \alpha \, \ket{H_1V_2H_3V_4} + \beta \,
\ket{V_1V_2V_3V_4} + \beta \, \ket{V_1H_2V_3H_4} \nonumber \\
&& + \gamma \, \ket{H_1H_2V_3H_4} + \gamma \, \ket{H_1V_2V_3V_4} + \delta \, \ket{V_1V_2H_3V_4} + \delta \, \ket{V_1H_2H_3H_4} \,
\Big) ~.~ \nonumber \\
\end{eqnarray}
We now show that the system can act like a parity filter, if one photon is collected in output mode $2'$ and another one is collected in
output mode $4'$. Using Eq.~(\ref{comp}), one can show that the 4-photon states (\ref{in}) becomes in this case, the unnormalised state
\begin{equation} \label{bus}
\ket{\tilde \psi_{\rm out}^{(2)}} = {\textstyle {1 \over \sqrt{2}}} \, \Big( \, \alpha \, \ket{H_{1'}H_{2'}H_{3'}H_{4'}} +
\beta \, \ket{V_{1'}V_{2'}V_{3'}V_{4'}} \, \Big) ~.~
\end{equation}
We further assume that the detectors measure the polarisation of the incoming photons in the rotated basis defined by the 1-photon
states
\begin{equation} \label{tractor}
\ket{\pm} \equiv {\textstyle {1 \over \sqrt{2}}} \, \Big( \, \ket{H} \pm \ket{V} \, \Big) ~.~
\end{equation}
It is important that the detectors distinguish the polarisation of each incoming photon in this basis (opposed to just absorbing the
photon), since this approach guarantees that the output becomes the expected pure state. Using the definition (\ref{tractor}), we can
rewrite the state (\ref{bus}) as
\begin{eqnarray}
\ket{\tilde \psi_{\rm out}^{(2)}} &=& {\textstyle {1 \over 2}} \, \Big( \, \alpha \, \ket{H_{1'}H_{3'}} + \beta \,
\ket{V_{1'}V_{3'}} \, \Big) \otimes {\textstyle {1 \over \sqrt{2}}} \, \Big( \, \ket{+_{2'}+_{4'}} + \ket{-_{2'}-_{4'}} \, \Big) \nonumber \\
&& + {\textstyle {1 \over 2}} \, \Big( \, \alpha \, \ket{H_{1'}H_{3'}} - \beta \, \ket{V_{1'}V_{3'}} \, \Big) \otimes {\textstyle
{1 \over \sqrt{2}}} \, \Big( \, \ket{+_{2'}-_{4'}} + \ket{-_{2'}+_{4'}} \, \Big) ~.~
\end{eqnarray}
Suppose the photons in output ports $2'$ and $4'$ are absorbed in the measurement process. Then the output state of the system equals in
case of a single click in each of the detectors
\begin{equation}
\ket{\psi_{\rm out}^{(2)}} = {\textstyle {1 \over 2}} \, \Big( \, \alpha \, \ket{H_{1'} H_{3'}} \pm \beta \, \ket{V_{1'} V_{3'}}
\, \Big) ~.~
\end{equation}
The ``$+$" sign applies when both detectors measure the same polarisation (which happens with probability ${1 \over 2}$); the ``$-$"
sign applies when both detectors measure different polarisations (which also happens with probability ${1\over 2}$). More generally, every measurement of the state $\ket{-}$ yields a phase flip error on the output state. Therefore, measuring even numbers of $\ket{-}$(or in this case, the same polarisations) yield no phase flip error or identity operation on the output state. The implementation
of the parity filter only needs a correction of this phase flip error in the event of measuring odd numbers of $\ket{-}$ (in this case, different polarisations) which can be implemented with the help of a Pauli $\sigma_z$ operation on any of the output photons. In any case, the 4-photon state (\ref{in}) can be reduced by measurement in the $\ket{\pm}$ with appropriate $\sigma_z$
correction to the following 2-photon state,
\begin{equation} \label{car}
\ket{\psi_{\rm out}^{(2)}} = {\textstyle {1 \over \sqrt{2}}} \Big( \, \alpha \, \ket{H_{1'} H_{3'}} + \beta \, \ket{V_{1'} V_{3'}} \,
\Big) ~,~
\end{equation} with unit efficiency. We have taken into account all appropriate measurement syndromes which explains the normalisation.
A closer look at the normalisation of this state tells us that the parity filter shown in Fig.~\ref{fig1} works with
efficiency $p_2 = {1 \over 2}$. If the success probability of the scheme would be 1, the output state (\ref{car}) would be $\alpha \,
\ket{H_{1'} H_{3'}}
+ \beta \, \ket{V_{1'} V_{3'}}$. It can be shown that the filter can also be operated with mixed states as inputs.
\subsection{The $N$-photon case}
\begin{figure}
\begin{center}
\includegraphics{humfig2}
\end{center}
\caption{Experimental setup for the realisation of the $N$-photon quantum filter (\ref{PN}). The $N$ polarising beam splitters
each compare the state of one of the target photons with the state of one of the ancilla photons, which are initially prepared in the
GHZ state $|A^{(N)} \rangle$. Besides, $N$ detectors perform photon measurements in the polarisation basis (\ref{tractor}). The output
photons leave the system via the odd numbered output ports.} \label{fig2}
\end{figure}
The generalisation of the above described 2-photon parity filter to the $N$-photon quantum filter (\ref{PN}) is straightforward and
requires $N$ polarising beam splitters and $N$ polarisation sensitive detectors (see Fig.~\ref{fig2}). One side of the setup is
entered by the $N$-photon input state $|\psi_{\rm in}^{(N)} \rangle$ with one photon in each odd-numbered input mode, while the other
side is entered by an $N$-photon ancilla state $|A^{(N)} \rangle$ with one photon in each even numbered input mode. In the following we
denote the modes containing the detectors by $2'$, $4'$, ..., $(2N)'$, while the modes $1'$, $3'$, ..., $(2N-1)'$ contain the output
state.
Again, the successful operation of the quantum filter is indicated by a single click in each of the detectors. Suppose $\alpha$ denotes the amplitude
of the state $|H_1H_3 \, . \, . \, . \, H_{2N-1} \rangle$ while $\beta$ is the amplitude of the state $\ket{V_1V_3 \, . \, . \, . \,
V_{2N-1}}$ with respect to the target state $\ket{\psi_{\rm in}^{(N)}}$. Then we find, using Eq.~(\ref{comp}) and in analogy to
Eq.~(\ref{bus}), that the collection of one photon in each of the detector output ports transforms the total input state $\ket{\tilde
\psi_{\rm in}^{(N)}} = \ket{\psi_{\rm in}^{(N)}} \otimes \ket{A^{(N)}}$ into
\begin{equation} \label{bicycle}
\ket{\tilde \psi_{\rm out}^{(N)}} = {\textstyle {1 \over \sqrt{2}}} \, \Big( \, \alpha \, \ket{H_{1'}H_{2'} \, . \, . \, . \, H_{(2N)'}}
+ \beta \, \ket{V_{1'}V_{2'} \, . \, . \, . \, V_{(2N)'}} \, \Big) ~.~
\end{equation}
For the same reason as in the 2-photon case, we assume that the detectors measure the polarisation of the incoming photons in the
polarisation basis (\ref{tractor}) by absorption. Suppose $J$ is the number of photons found in the $\ket{-}$ state, then one can show
using Eq.~(\ref{bicycle}) and proceeding as in Section \ref{two} that the output state of the remaining $N$ photons equals
\begin{equation}
\ket{\psi_{\rm out}^{(N)}} = {\textstyle {1 \over 2}} \, \Big( \, \alpha \, \ket{H_{1'}H_{3'} \, . \, . \, . \, H_{(2N-1)'}} +
(-1)^{J} \, \beta \, \ket{V_{1'}V_{3'} \, . \, . \, . \, V_{(2N-1)'}} \, \Big) ~.~
\end{equation} Note that the probability of $J$ being an odd number, which incurs a phase flip error on the output state in analogy to the 2-photon filter, is $1 \over 2$.
As before, we can transform with unit efficiency the state (\ref{bicycle}) with the help of a phase flip correction to the final state given by
\begin{equation} \label{tram}
\ket{\psi_{\rm out}^{(N)}} = {\textstyle {1 \over \sqrt{2}}} \, \Big( \, \alpha \, \ket{H_{1'}H_{3'}...H_{(2N-1)'}} + \beta \,
\ket{V_{1'}V_{3'}...V_{(2N-1)'}} \, \Big) ~.~
\end{equation}
This is exactly the output state that one expects after the application of the quantum filter (\ref{PN}) to the input state $|\psi_{\rm
in}^{(N)} \rangle$ with success probability $p_N = {1 \over 2}$, which is the highest that has been predicted so far without the use of
universal two-qubit quantum gate operation such as the CNOT or CZ gates\footnote{Alternatively, a straightforward way of implementing the quantum filter (\ref{PN}) is to replace each polarising beam splitter in the setup (see
Fig.~\ref{fig2}) by a CNOT gate(which is difficult to realise with linear optics alone). Furthermore, the detectors in all even numbered output modes should perform a polarisation
sensitive measurement in the $H/V$ basis. The projection efficiency of such a scheme would only be limited by the success probability
$p$ of a single controlled-NOT operation and would scale like $p^N$. For sufficiently large photon numbers $N$, this might decrease
below ${1 \over 2}$. Therefore the use of polarising beam splitters, which can operate with a very high fidelity, should be favoured
\cite{Pan03}.}.
Naively, one might expect that the efficiency of the filter decreases with the number of photons in the setup. However, this is not the case here.
Furthermore, we remark that the described quantum filter also works for mixed $N$-photon input states.
Taking into account real detector efficiencies and dark count rates will diminish both the success probability and fidelity of the
above described filter. In general, the success probability and fidelity depend on the nature of the input state as well as the ancilla.
Here, we focus on a simple example of analysing the
error probability of a 2-photon parity filter by assuming imperfect photon detectors but perfect ancilla state. It is shown in
\cite{Saavedra00} that the ancilla states considered here can be prepared with high fidelity and success probability. As in
\cite{Hofmann02}, we assume that the dark count rate can be reduced by time gating and consider the effect of detector inefficiencies causing
an error due to a mistake of registering a 2-photon detection event as a single photon event. This is known as preselective error. If
we also postselect the output state, then such an error can in principle be eliminated. Without loss of generality, we analyse the
case where the detectors each register a click for an alleged photon in the state $\ket{+}$. This can be represented by a POVM(Positive
operator valued measure) element $E_{i'}$ given by \cite{Lee04}
\begin{equation}
E_{i'}=p_d\ket{+_{i'}}\bra{+_{i'}}+2p_d(1-p_d)\ket{(++)_{i'}}\bra{(++)_{i'}} \, ,
\end{equation} where $p_d$ is the single photon detection efficiency. From Kok and Braunstein \cite{Kok01,Barnett98}, we know that the reduced projected state is $\rho_{1'3'}=\frac{{\rm Tr}_{2'4'}(E_{2'}E_{4'}\rho_{1'2'3'4'})}{{\rm Tr}_{1'3'}(\cdot)}$ where $\rho_{1'2'3'4'}$ is the state after passing $\ket{\tilde \psi_{\rm in}^{(2)}}$ through the 2 polarising beam splitters. We also fix
$|\alpha|^2=|\beta|^2=|\gamma|^2=|\delta|^2=\frac{1}{4}$ to compute for the most typical input state to obtain the average fidelity.
One can show that the fidelity\footnote{This is analogous to the true QND or preselective fidelity discussed in Ref \cite{Kok05a}. Clearly, if we define our fidelity based on coincidence counting, where a photon is detected in all outputs 1', 2' 3' 4', the postselective fidelity can be much higher.} of the quantum filter is given by $F=\bra{\psi_{\rm out}^{(2)}}\rho_{1'3'}\ket{\psi_{\rm out}^{(2)}}/\langle \psi_{\rm out}^{(2)}\ket{\psi_{\rm out}^{(2)}}=(5-6p_d+2p_d^2)^{-1}$. For example, given a $p_{d}$ of 0.88
(\cite{Takeuchi99,Rosenberg05}), the maximum error rate $1-F$ would be 0.19 in the light of current technology. Especially for the
recent work by Rosenberg {\em et al.} \cite{Rosenberg05}, superconducting transition-edge sensors are expected to have photon-number
resolution with negligible dark counts at arbitrary high efficiency in the future.
\section{Conclusions}
We described the realisation of a 2-photon parity filter that requires only two polarising beam splitters, two photons prepared in a
maximally entangled Bell state and two polarisation sensitive detectors. The success rate of the scheme $p_2 = {1 \over 2}$ is the
highest that has been predicted so far without the help of universal two-qubit quantum gate operations and is reached here due to employing an
entangled ancilla state as a resource. A generalisation of the proposed scheme to the $N$-photon case is straightforward. We showed that
the quantum filter (\ref{PN}) can be implemented with the help of $N$ polarising beam splitters and an $N$-photon GHZ state as a
resource. Remarkably, the success rate of the filter remains ${1 \over 2}$, irregardless of the size of the input state.
To implement the quantum filter (\ref{PN}), the $N$ polarising beam splitters compare the state of the incoming photons pairwise with
the state of the ancilla photons. In Section \ref{humfilter}, we showed that a single polarising beam splitter can be used to realise a
filter, which measures polarisation $H$ or $V$, respectively, with unit efficiency. Preparing the ancilla photons, for example, in the
state $\ket{HH \, . \, . \, . \, H}$, would result in a filter that measures whether all target photons are prepared in $\ket{H}$.
However, since we compare the input state with a GHZ state, which contains two terms, namely $\ket{HH \, . \, . \, . \, H}$ and $\ket{VV
\, . \, . \, . \, V}$, the probability of the described filter is only as high as ${1 \over2}$. Indeed, the highly entangled $N$-photon
GHZ state acts as a ``mask" for the filter.
A straightforward extension of the ideas of this chapter is to consider a different form of the ``mask" or ancilla state $\ket{A^{CZ}}$
given by $\frac{1}{2}(\ket{H_2H_4}+\ket{H_2V_4}+\ket{V_2H_4}-\ket{V_2V_4})$. Under the condition that the photons pass the filter,
heralded by single photon detection in both output detectors in $\ket{\pm}$, the output state, with correction of sign errors, would
instead be given by
\begin{equation}
\ket{\psi^{(CZ)}_{\rm out}}=\frac{1}{2}(\alpha \ket{H_{1'}H_{3'}}+\beta \ket{V_{1'}V_{3'}}+\gamma \ket{H_{1'}V_{3'}}-\delta
\ket{V_{1'}H_{3'}}) \, .
\end{equation} This is the same as the application of a CZ filter or gate $P_{CZ}$
\begin{equation}
P_{CZ}=\frac{1}{2}(\ket{H_{1'}H_{3'}}\bra{H_1H_3}-\ket{V_{1'}V_{3'}}\bra{V_1V_3}+\ket{H_{1'}V_{3'}}\bra{H_1V_3}+\ket{V_{1'}H_{3'}}\bra{V_
1H_3})
\end{equation} with efficiency $\frac{1}{4}$ to the input state (\ref{huminput}). This is analogous to the CNOT gate proposed by Pittman
{\em et al.} \cite{Pittman01} with success probability $\frac{1}{4}$.
We have seen an example of how quantum computing with photons assisted with entangled ancillas can result in a more efficient
implementation. However, the scheme is still necessarily probabilistic as are all known linear optics based schemes where the input
state is not already
necessarily encoded offline. We move to the next chapter where we add just one more ingredient, a special single
photon source with encoding
ability, and show how quantum computing with linear optics can become effectively deterministic.
\chapter{Photon Polarisation Entanglement from Distant Sources in Free
Space}\label{photon}
\section{Introduction}\label{photon:intro}
In this chapter, we attempt to develop new perspectives to photon generation through distant sources in free
space. There exist roughly
two general approaches to create entangled photon pairs. Firstly, entangled photon pairs can be
created within the {\em
same} source as in atomic cascades \cite{Aspect82}, in
parametric down conversion schemes \cite{Kwiat95} and in the biexciton emission of
a single quantum dot in a cavity \cite{Stace03}. If
the entanglement is not created within the same source, single photons can be
brought together to overlap within their coherence time on
a beamsplitter where a postselective entangling measurement can be performed on
the output ports \cite{Shih98}. A more detailed survey of single photon sources and entanglement
generation is given in Chapter \ref{firework}.
\begin{figure}
\begin{center}
\includegraphics[scale=1.0]{lambda}
\end{center}
\caption{$\Lambda$-level configuration of the dipole source with the two degenerate ground states $|0\rangle$ and $|1 \rangle$, the excited state
$|2 \rangle$ and optical transitions corresponding to the
two orthogonal polarisations ``$+$" and ``$-$"} \label{lambda}
\end{figure}
In contrast to this, we show that polarisation entanglement can also be obtained postselectively
when the photons are created by {\em distant} sources in
free space without having to control their photon collection times. As an example,
we analyse the photon emission from two dipole sources
that might be realised in the form of trapped atoms, diamond NV color centres,
quantum dots or by using single atoms doped onto a
surface. An interaction between the sources is not required. Each source should
possess a $\Lambda$-type three-level configuration with
the two degenerate ground states $|0\rangle$ and $|1 \rangle$, the excited state
$|2 \rangle$ and optical transitions corresponding to the
two orthogonal polarisations ``$+$" and ``$-$" along a well defined axis (see Fig. \ref{lambda}). Polarisation entanglement arises
under the condition of the emission of two photons in
different but carefully chosen directions independent from the initial state of
the sources. Furthermore in our scheme, this leaves the
dipoles in a maximally entangled state. Therefore, we can obtain
both usable postselected 2-photon entanglement\footnote{For example, postselected photon entanglement can be used for quantum
cryptography} and
preselected dipole-dipole entanglement.
In order to understand how the scheme works, it is important to note that
fluorescence from two distant dipole sources can produce an
interference pattern on a far away screen, if the distance between
the sources \cite{Scully82,Eichmann93,Schon01} is comparable to the wavelength of the emitted
photons. This can be understood as both
sources contributing {\em coherently} to the creation of each photon.
Consequently, the emission of one photon leaves a trace in the
states of {\em all} its potential sources, depending on its polarisation and the
direction of its wave vector \cite{Schon01,Schon02}, and
can thus affect the state of the subsequently emitted photon. Such a picture is
seen most directly using the quantum jump formalism
\cite{Hegerfeldt93,Dalibard92,Carmichael93}.
The described interference pattern has already been observed \cite{Eichmann93}
in the intensity profile due to the flourescence of two
four-level atoms scattered by laser light. Various attempts
\cite{Wong97,Itano98,Schon01,Agarwal02} to elucidate this have been made with the central theme that
interference can only be
observed when the which-way information is in principle
absent. In addition, work aimed at investigating aspects of second-order photon or intensity-intensity correlations at perfect
photon detection coincidences (i.e. at the same time) has also been made
\cite{Mandel83,Schon01,Agarwal02}.
The modulation depth of such intensity-intensity
correlations of the same polarisation is shown to be reachable to 100 \% even when
the intensity interference pattern may disappear. In other
words, there exist a strong spatial antibunching of emitted photons of the same
polarisation in free space where the detection of one
photon does not permit the detection of another photon in certain directions at the
same instant. Here, we exploit this feature for the
generation of entangled photon pairs.
In this chapter, the detectors of Alice and Bob are placed such that all wave
vector amplitudes contributing to the creation of a second
photon with the same polarisation as the first one interfere destructively. In
case of the collection of two photons (one by Alice and
one by Bob) the shared pair has to be in a superposition of the state where Alice
receives a photon with polarisation ``$+$" and Bob a
photon with polarisation ``$-$" and the state where Alice receives a photon with
polarisation ``$-$" and Bob a photon with
polarisation ``$+$". Both share a maximally entangled pair, if the amplitudes for
these two states are of the same size. In summary,
polarisation entanglement is obtained with the help of postselection and
interference effects. Related mechanisms have been proposed in
the past to create atom-atom entanglement \cite{Cabrillo99,Plenio99,Protsenko02,Simon03}.
The pair creation scheme proposed in this paper is feasible with present
technology and might offer several advantages to quantum
cryptography. In contrast to parametric down conversion, the setup
guarantees antibunching between subsequent photon pairs
since the creation of a new pair is not possible without reexcitation of both
sources. Furthermore, the scheme is robust with respect to the possible phase fluctuation in the exciting laser\footnote{Axel Kuhn first
brought this to my attention on discussion of this scheme.}. The final
photon state does not depend on the initial state of the sources in case of a
successful collection.
Finally, the scheme may offer the possibility to generate {\em multiphoton}
entanglement by incorporating more than two radiators in the
setup.
\section{Theory}\label{photon:theory}
Let us now discuss the creation of such an entangled photon pair in detail. We describe
the interaction of the dipole sources with the
surrounding free radiation field by the Schr\"odinger equation. The annihilation
operator for a photon with wave vector ${\bf k}$,
polarisation $\lambda$ with polarisation vector\footnote{In this thesis, the
notation is chosen such that $\hat {\bf x} \equiv {\bf x}/\|
{\bf x} \| $.} defined as ${\bf \epsilon}_{\hat {\bf k}\lambda}$ is $a_{{\bf
k}\lambda}$. The two dipole sources considered here are placed at
the fixed positions ${\bf r}_1$ and ${\bf r}_2$ and should be identical in the
sense that they have the same dipole moment ${\bf D}_{2j}=\bra{2}{\bf D}\ket{j}$
for the 2-$j$ transition ($j=0,1$). The energy separation between the degenerate
ground states and level 2 is $\hbar \omega_0$ while
$\omega_k=kc$ and $L^3$ is the quantisation volume of the free radiation field. In
addition, we define the $i$th atomic lowering and
raising operator as
\begin{equation}
S^-_{i,j}=\ket{j}_{ii}\bra{2}, \,\,\, S^+_{i,j}=\ket{2}_{ii}\bra{j}.
\end{equation}
Using this notation, the system Hamiltonian becomes
within the rotating wave approximation and with respect to the interaction-free
Hamiltonian in the interaction picture,
\begin{eqnarray} \label{21}
H_{\rm I} &=& \sum_{i=1,2} \sum_{j=0,1} \sum_{{\bf k},\lambda} \hbar g_{{\bf
k}\lambda}^{(j)} \,
{\rm e}^{ -{\rm i} (\omega_0 - \omega_k) t} \, {\rm e}^{ -{\rm i} {\bf k} \cdot
{\bf r}_i} \,
a_{{\bf k}\lambda}^\dagger \, S^-_{i,j}+ {\rm H.c.} ~, \nonumber \\
&=& \sum_{j=0,1} H_{{\rm I}1}^j+H_{{\rm I}2}^j
\end{eqnarray}
which can be decomposed into terms $H_{{\rm I}i}^j$ relating only to each of the $i$th
atom and
\begin{eqnarray}
g_{{\bf k}\lambda}^{(j)}
&=& {\rm i} e \, \left[ \frac{\omega_k}{2 \epsilon_0 \hbar L^3 } \right]^{1/2}
\! ({\bf D}_{2j} , {\bf \epsilon}_{\hat {\bf k}\lambda})
\end{eqnarray}
is the coupling constant for the field mode $({\bf k},\lambda)$ to the 2-$j$
transition of each source. With $H_{\rm I}$, we can associate the unitary operator describing the evolution of the combined system from
time $t_1$ to $t_2$ as $U_{\rm I}(t_2,t_1)$. The rotating wave approximation
corresponds to neglecting the non-energy conserving terms that describe the
excitation of atoms combined with the creation of a photon or
the deexcitation of atoms combined with the annihilation of a photon. These
effects are not unphysical \cite{Knight73} but their
contribution to the time evolution of the described system can be shown to be very
small and almost impossible to observe.
\subsection{Entangled photon and entanged dipole generation}
To describe the effect of an emission on the state of the sources, we introduce
the spontaneous decay rate of the 2-$j$ transition
$\Gamma_j \equiv (e^2 \omega_0^3 \, |{\bf D}_{2j}|^2)/(3\pi \epsilon_0 \hbar c^3)$
and the reset or collapse operator $R_{{\bf \hat k},\lambda}$ which is associated with the quantum jump formalism \cite{Hegerfeldt93,
Dalibard92,
Carmichael93}. A good review of the quantum jump approach can be found in Ref. \cite{Plenio98}. For the
sake of simplicity, we set $\Gamma_0=\Gamma_1=\Gamma$ in this chapter.
The quantum jump formalism is an instance of a type of unravelling of the master equation describing the evolution of the dipole sources
in an open environment such as the free radiation field. The source follows a so-called quantum state trajectory based on knowledge
obtained from a real or ficticious continuous\footnote{More precisely, the measurement is not truly continuous but coarsed-grained at a
timescale of $\Delta t$ much larger than the transition optical period but also much smaller than the average timescale of atomic evolution. A
truly continuous measurement will instead freeze the system due to the quantum Zeno effect.} time-resolved measurement that yields
generally two types of observables. One of them is the no-photon observation and the other is a photon detection observation that can be
direction and (or) polarisation specific. We first denote the
free radiation field in the vacuum state $\ket{0_{\rm ph}}$ and define the reduced density operator of the dipole sources at time $t$ as
$\rho_{a}(t)$. We furthermore denote the 1-photon state of wave vector
${\bf k}=k{\bf \hat k}$ and
polarisation $\lambda$ by $\ket{1_{k{\bf \hat k} \lambda}}$. In the theory of quantum evolution of an open system, under the Born-Markovian approximation, the
evolution of $\rho_{a}(t)$ given that at time $t$, the combined state is $\rho(t)=\ket{0_{\rm ph}} \rho_{a}(t) \bra{0_{\rm ph}}$, can be
described by a superoperator ${\cal L}(\Delta t)$ that yields a Kraus operator sum representation given as
\begin{equation}
\rho_{a}(t) \to {\cal L}(\Delta t) \rho_{a}(t) = \rho_{a}(t+\Delta t)= \sum_{\mu} M_{\mu} \rho_{a}(t) M^\dagger_{\mu} \, .
\end{equation} Here $M_{\mu}=\bra{\mu}U_{\rm I}\ket{0_{\rm ph}}$ is associated with an observable $\mu$ and is also known as a Kraus operator. The above evolution is valid
if we have no information of $\mu$ and puts $\rho_{a}(t+\Delta t)$ into a generally mixed state. This can be intepreted as an environment-induced measurement \cite{Schon01} where the results of the measurement is not known. The situation changes if we perform a
measurement and have full information on $\mu$. The evolution is now described by
\begin{equation}
\rho_{a}(t) \to {\cal L}(\Delta t) \rho_{a}(t) = \rho_{a}(t+\Delta t)= \frac{M_{\mu} \rho_{a}(t) M^\dagger_{\mu}}{{\rm Tr}(M_{\mu}
\rho_{a}(t) M^\dagger_{\mu})} \, ,
\end{equation} if the measurement in $\Delta t$ yields an observable $\mu$ with probability ${\rm Tr}(M_{\mu} \rho_{a}(t)
M^\dagger_{\mu})$.
The evolution of the source is thus generally stochastic leading to a quantum trajectory and the average of all stochastic evolutions
yields the density operator obtained on solving the master equation, which gives an ensemble description. An instance of a stochastic
evolution of the state is defined as a quantum trajectory of the quantum jump formalism.
If a photon of polarisation $\lambda$ and a wave
vector pointing in the $\hat {\bf k}$ direction is detected within a coarse-grained time $\Delta t$ small compared to the average timescale of
the system evolution and large compared to the optical period, the evolution of the state of the source (see \cite{Wong97,Schon01,Beige97}
for more details) is given by
\begin{equation}
\rho_a(t+\Delta t)=\sum_{k} M_{\mu_k} \rho_{a}(t) M^\dagger_{\mu_k}/{\rm Tr}(\cdot) \approx R_{{\bf \hat k},\lambda}\rho_{a}(t_0)
R^{\dagger}_{{\bf \hat k},\lambda} \Delta t/{\rm Tr}(\cdot) \, ,
\end{equation} with $M_{\mu_k}=\bra{1_{k{\bf \hat k} \lambda}} U_I(\Delta t +t,t) \ket{0_{\rm ph}}$ and\footnote{Our analysis here apply only for degenerate levels $\ket{0}$ and $\ket{1}$ or in the case of non-degeneracies, when their frequency split $\Delta \omega$ is small enough such that $|\Delta \omega|\Delta t \approx 0$. For a more general discussion, see the formulation in Ref. \cite{Hegerfeldt93}. }
\begin{eqnarray} \label{R21}
R_{{\bf \hat k},\lambda} & \equiv & \sum_{i,j} \left[ {3 \Gamma \over 8 \pi}
\right]^{1/2}
(\hat{\bf D}_{2j} , {\bf \epsilon}_{\hat{\bf k} \lambda}) \,
{\rm e}^{-{\rm i} k_0 \hat {\bf k} \cdot {\bf r}_i} \, S^-_{i,j} ~.~~
\end{eqnarray} One can see that the detection of a photon within a short time $\Delta t$ is always accompanied with a lowering or
jump of the source within the same time $\Delta t$ hence motivating the name ``quantum jump" approach. Note that the probability density
for the described emission is given by ${\rm Tr}
(R_{{\bf \hat k},\lambda}\rho_{a}(t) R^{\dagger}_{{\bf \hat k},\lambda})$.
The no-photon time evolution of the system say between $t_2$ and $t_1$ is associated with a Kraus operator $M_0=\bra{0_{\rm ph}}
U_I(t_2,t_1) \ket{0_{\rm ph}}=U_{\rm cond}(t_2,t_1)$. More precisely, the state of the sources at $t_2$ after a
no-photon event from $t_1$ is given by
\begin{equation}
\rho_{a}(t_2)=U_{\rm cond}(t_2,t_1)\rho_a(t_1)U^{\dagger}_{\rm cond}(t_2,t_1) /{\rm Tr}(\cdot) \,
\end{equation} where $U_{\rm cond}(t_2,t_1)={\rm e}^{-{\rm i}H_{\rm cond}(t_2-t_1)/\hbar}$.
This approach provides a
non-Hermitian conditional Hamiltonian $H_{\rm cond}$
given by the following
relation
\begin{equation} \label{Hcondrelation}
I-\frac{\rm i}{\hbar} H_{\rm cond} \Delta t \approx U_{\rm cond}(\Delta t+t_0,t_0)=\bra{0_{\rm ph}} U_I(\Delta t+t_0,t_0)
\ket{0_{\rm ph}}
\end{equation} where the r.h.s is evaluated by second-order perturbation theory
for a coarse-grained time $\Delta t$ similar to that of the timescale used in the
derivation of the reset operator. In Eq.~(\ref{Hcondrelation}), $\bra{0_{\rm ph}} U_I(\Delta t+t_0,t_0) \ket{0_{\rm ph}}$ is
given by
\begin{eqnarray}
&& \bra{0_{\rm ph}} U_I(\Delta t+t_0,t_0) \ket{0_{\rm ph}} \\ \nonumber
&=& \bra{0_{\rm ph}} I-\frac{\rm i}{\hbar} \int_{t_0}^{\Delta t+t_0} dt'H_{\rm I}(t')
-\frac{1}{\hbar^2}\int_{t_0}^{\Delta t+t_0}dt' \int_{t_0}^{t'} dt''
H_{\rm I}(t')H_{\rm I}(t'') +O(\Delta t^2) \ket{0_{\rm ph}} \\ \nonumber
&=&I-\frac{1}{\hbar^2}\int_{t_0}^{\Delta t+t_0}dt' \int_{t_0}^{t'} dt'' \bra{0_{\rm ph}} \Big[
\sum_m H_{{\rm I}1}^m(t')H_{{\rm I}1}^m(t'')+H_{{\rm
I}2}^m(t')H_{{\rm I}2}^m(t'') \\ \nonumber
&&+\sum_{m \neq n} H_{{\rm I}1}^m(t')H_{{\rm I}1}^n(t'')+H_{{\rm
I}2}^n(t')H_{{\rm I}2}^m(t'') \\ \nonumber
&&+\sum_{m,n} H_{{\rm I}1}^m(t')H_{{\rm I}2}^n(t'')+H_{{\rm I}2}^n(t')H_{{\rm
I}1}^m(t'') \Big] \ket{0_{\rm ph}}+O(\Delta t^2)
\end{eqnarray}
We further define the relative position vector ${\bf r}={\bf r}_1-{\bf r}_2$ where
$r=\|{\bf r}\|$ is the distance between the atoms and denote $k_0=\frac{\omega_0}{c}$. For
the setup considered here, one finds in the absence of laser driving \cite{Beige97,Wong97},
\begin{eqnarray} \label{hcondgrand}
H_{\rm cond} &=& \frac{\hbar}{2{\rm i}} \Big[ \Gamma \sum_m
(S^+_{1,m}S^-_{1,m}+S^+_{2,m}S^-_{2,m}) \\ \nonumber
&&+\sum_{m,n}C_{1m,2n}S^+_{1,m}S^-_{2,n}+C_{1n,2m}S^+_{1,n}S^-_{2,m} \Big]\, ,
\end{eqnarray} where $C_{in,jm}$ arises from dipole-dipole interaction and is
given by
\begin{eqnarray} \label{dipoledipole}
C_{in,jm} &=& \frac{3 \Gamma}{2} {\rm e}^{{\rm i}k_0r} \Big[ \frac{1}{{\rm i}
k_0r} ((\hat{\bf D}_{2m},\hat{\bf D}_{2n})-(\hat{\bf
D}_{2m},\hat{\bf r})(\hat{\bf r},\hat{\bf D}_{2m})) \nonumber \\
&&+ \Big( \frac{1}{(k_0r)^2}-\frac{1}{{\rm i} (k_0r)^3} \Big) ((\hat{\bf
D}_{2m},\hat{\bf D}_{2n})-3(\hat{\bf
D}_{2m},\hat{\bf r})(\hat{\bf r},\hat{\bf D}_{2m})) \Big].
\end{eqnarray}
We consider only the cases where $C_{in,jm}$ is very small, or in other words,
where the dipole-dipole interaction is insignificant. Without
calculating the terms $C_{in,jm}$ explicitly, one can see that relative to the
rate of spontaneous decay $\Gamma$, $C_{in,jm}$ scales as
$(k_0r)^{-1}$ in the strongest possible dipole-dipole coupling scenario. This
occurs when both dipoles are parallel with each other and orthogonal to the
line joining both atoms. Therefore, in the limit of large $k_0r$, for example,
$r>25 \lambda_0$ with $\lambda_0=\frac{2 \pi}{k_0}$,
dipole-dipole coupling becomes insignificant.
The two-atom double slit experiment
performed by Eichmann {\em et al.} \cite{Eichmann93}
also operates at this regime.
We can thus simplify (\ref{hcondgrand}) and get
\begin{equation} \label{hcond}
H_{\rm cond} = \frac{\hbar \Gamma}{2{\rm i}} \sum_m
(S^+_{1,m}S^-_{1,m}+S^+_{2,m}S^-_{2,m}).
\end{equation} This Hamiltonian can also be derived by assuming that each atom
couples to its own separate radiation field.
We now determine the state of the system under the condition of the collection
of two photons, the first one at $t_1$ in the $\hat{\bf
k}_{\rm X}$ direction with polarisation ${\bf \epsilon}_{\hat {\bf k}_{\rm X} \lambda}$ and the second one at $t_2$ in the $\hat{\bf
k}_{\rm Y}$
direction with polarisation ${\bf \epsilon}_{\hat {\bf k}_{\rm Y} \lambda'}$. If the initial state of the dipole sources at
$t=0$ is $|\varphi_0 \rangle$, whilst the free radiation field is in its vacuum
state, the unnormalised state of the dipole sources \cite{Schon01} after the collection of the second photon is given by
\begin{eqnarray}
\ket{\psi( {\bf \epsilon}_{\hat {\bf k}_{\rm Y} \lambda'}t_2| {\bf
\epsilon}_{\hat {\bf k}_{\rm X} \lambda}t_1)}&=& R_{\hat{\bf k}_{\rm Y},\lambda'} \, U_{\rm
cond}(t_2,t_1) \, R_{\hat{\bf k}_{\rm X},\lambda} U_{\rm cond}(t_1,0) |\varphi_0 \rangle \nonumber \\
&=& N(t_1,t_2)\langle 22| \varphi_0
\rangle \sum_{i,j=0}^1 ( (\hat{\bf D}_{2i} , {\bf \epsilon}_{\hat {\bf k}_{\rm Y}
\lambda'})(\hat{\bf D}_{2j} , {\bf \epsilon}_{\hat
{\bf k}_{\rm X} \lambda}){\rm e}^{-{\rm i} k_0 \hat {\bf k}_{\rm Y} \cdot {\bf
r}_1}{\rm e}^{-{\rm i} k_0 \hat {\bf k}_{\rm X} \cdot {\bf
r}_2} \nonumber \\
&& +(\hat{\bf D}_{2j} , {\bf \epsilon}_{\hat {\bf k}_{\rm Y}
\lambda'})(\hat{\bf D}_{2i} , {\bf \epsilon}_{\hat {\bf k}_{\rm X}
\lambda}){\rm e}^{-{\rm i} k_0 \hat {\bf k}_{\rm Y} \cdot {\bf r}_2}{\rm e}^{-{\rm
i} k_0 \hat {\bf k}_{\rm X} \cdot {\bf r}_1}) \ket{ij} \, ,
\end{eqnarray} with
\begin{equation}
N(t_1,t_2)={3\over 8 \pi} \, \Gamma {\rm e}^{ -\Gamma (t_1+t_2) } \, .
\end{equation}
Note that $\| \, |\psi (\hat{\bf k}_{\rm Y}, t_2 | \hat{\bf k}_{\rm X}, t_1)
\rangle \, \|^2$ yields the probability density for the
corresponding event \cite{Plenio98}.
We now calculate the polarisation correlation $C_{\hat {\bf k}_{\rm A}\lambda,
\hat {\bf k}_{\rm B} \lambda'}$ (i.e. the joint probability where Alice and Bob get a $
\lambda$ and $\lambda'$ polarised photon respectively if Alice and Bob
collect a photon each). It is simply given by
\begin{equation} \label{correlation}
C_{\hat {\bf k}_{\rm A} \lambda, \hat {\bf k}_{\rm B}
\lambda'}=|\ket{\psi({\bf \epsilon}_{\hat {\bf k}_{\rm B} \lambda'}t_2|
{\bf \epsilon}_{\hat {\bf k}_{\rm A} \lambda}t_1})|^2 /\sum_{\lambda_1,
\lambda_2}|\ket{\psi( {\bf \epsilon}_{\hat {\bf k}_{\rm B}
\lambda_1}t_2| {\bf \epsilon}_{\hat {\bf k}_{\rm A} \lambda_2}t_1)}|^2 \, .
\end{equation} One can easily check that this is independent of $t_1$ and $t_2$ as all the time
dependence cancels out in the
normalising factor $N(t_1,t_2)$.
We can use (\ref{correlation}) to calculate the probability $C_{\pm}$ $(C_{hv})$ that both Alice and Bob get orthogonal
polarisation if they each collect a
photon in the circular (linear) basis. The importance of such a calculation lies in the fact that the circular and linear basis are mutually
unbiased. This corresponds closely to the quantum cryptographic BB84 protocol where Alice and
Bob perform measurements in a set of
mutually unbiased bases. The existence of polarisation correlations in a set of mutually unbiased bases is a signature of entanglement
(See Appendix~\ref{sec:appendix_a}.).
To assure that Alice and Bob can receive a polarisation entangled pair, they
should place their detectors in directions $\hat {\bf
k}_{\rm A}$ and $\hat {\bf k}_{\rm B}$ with
\begin{equation} \label{hold}
{\rm e}^{ -{\rm i} k_0 (\hat{{\bf k}}_{\rm A}-\hat{{\bf k}}_{\rm B}) \cdot ({\bf r}_1-{\bf r}_2) } = -1 \, .
\end{equation}
One can see that these positions are in general not unique. They have the physical interpretation of
corresponding to a half-fringe interval
in the far
field of a double-slit experiment, in which the two atoms are replaced by pinholes which are symmetrically irradiated by a laser.
With condition (\ref{hold}), one obtains
\begin{eqnarray} \label{hurray}
&& \ket{\psi( {\bf \epsilon}_{\hat {\bf k}_{\rm B} \lambda'}t_2| {\bf
\epsilon}_{\hat {\bf k}_{\rm A} \lambda}t_1)} = \ket{\psi( {\bf \epsilon}_{\hat {\bf k}_{\rm A} \lambda}t_2| {\bf
\epsilon}_{\hat {\bf k}_{\rm B} \lambda'}t_1)} \nonumber \\
&& = N(t_1,t_2)2^{1/2}
\, {\rm e}^{ -{\rm i} k_0 (\hat {\bf k}_{\rm A} \cdot {\bf r}_1 + \hat {\bf k}_{\rm B} \cdot {\bf r}_2 )} \, \langle 22| \varphi_0
\rangle \nonumber \\
&& \left[ \big( \hat{\bf D}_{20} ,
{\bf \epsilon}_{\hat {\bf k}_{\rm B} \lambda'} \big)
\big( \hat{\bf D}_{21} , {\bf \epsilon}_{\hat {\bf k}_{\rm A} \lambda} \big)
- \big( \hat{\bf D}_{21} , {\bf
\epsilon}_{\hat{\bf k}_{\rm B} \lambda'} \big) \big( \hat{\bf D}_{20} , {\bf
\epsilon}_{\hat{\bf k}_{\rm A} \lambda} \big) \right] \otimes |a_{01} \rangle
\end{eqnarray}
with $|a_{01} \rangle \equiv (|01 \rangle - |10 \rangle)/\sqrt{2}$. After two
emissions, the dipole radiators are left in a maximally
entangled state which is completely disentangled from the free radiation field.
A coordinate
system is introduced where the $\hat {\bf z}$-axis points in
the direction of the line connecting the two sources and the $\hat {\bf x}$-axis
coincides with the quantisation axis. In addition, we
choose $\hat {\bf k}_{\rm B} = (1,0,0)^{\rm T}$, $ {\bf \epsilon}_{\hat{\bf
k}_{\rm B} +} = \hat{\bf D}_{20} = (0,1,{\rm i})^{\rm
T}/\sqrt{2}$ and $ {\bf \epsilon}_{\hat{\bf k}_{\rm A} -} = \hat{\bf D}_{21} =
\hat{\bf D}^*_{20}$. Using the spherical coordinates
$(\vartheta,\varphi)$ for Alice's detector position, one can write $ {\bf
\epsilon}_{\hat{\bf k}_{\rm A} \pm} =\frac{1}{\sqrt{2}}( {\bf
\epsilon}_{\hat{\bf k}_{\rm A} h}\pm {\rm i} {\bf
\epsilon}_{\hat{\bf k}_{\rm A} v}) $ with linear polarisations $ {\bf
\epsilon}_{\hat{\bf k}_{\rm A} h}=( - \sin \varphi, \cos \varphi , 0)^{\rm T}$ and $ {\bf
\epsilon}_{\hat{\bf k}_{\rm A} v}=( - \cos \vartheta \cos \varphi, - \cos \vartheta \sin \varphi , \sin \vartheta)^{\rm T}$ (see
\cite{Itano98}).
Using (\ref{hurray}), we have
\begin{equation}
C_{\pm}=C_{\hat {\bf k}_{\rm A} +, \hat {\bf k}_{\rm B} -}+C_{\hat {\bf k}_{\rm A} -, \hat {\bf k}_{\rm B} +}=\frac{(\cos
\varphi+\sin \vartheta)^2+(\cos \vartheta \sin \varphi)^2}{2(1+(\sin \vartheta \cos \varphi)^2)}
\end{equation} and
\begin{equation}
C_{hv}=C_{\hat {\bf k}_{\rm A} h, \hat {\bf k}_{\rm B} v}+C_{\hat {\bf k}_{\rm A} v, \hat {\bf k}_{\rm B} h}=\frac{(\cos
\varphi)^2+(\sin \vartheta)^2}{(1+(\sin \vartheta \cos \varphi)^2)} \, .
\end{equation} A straightforward evaluation of both equations for the range $\frac{\pi}{2}-0.5< \vartheta < \frac{\pi}{2}+0.5$ and
$-0.5<\varphi<0.5$ shows that\footnote{Note that $\vartheta = \pi/2$ should be excluded and intepreted as a limit point since we demand that condition (\ref{hold}) is fulfilled. At this limit where also $\varphi=0$, then $C_{\pm} = C_{hv} = 1$ and we have a maximally entangled postselected state (See also Appendix~\ref{sec:appendix_a}). }
\begin{equation}
C_{\pm} \approx C_{hv} \approx 1 \,.
\end{equation}
Therefore, having $\hat {\bf k}_{\rm A}$ pointing in a direction
relatively close to the quantisation axis ($\vartheta=\pi/2
, \varphi=0$) which is the $\hat {\bf x}$-axis with a tolerance of $\pm 0.5$ radians for both angles $\vartheta
, \varphi$ together with condition (\ref{hold}) fulfilled will guarantee that Alice and Bob
obtain an approximate postselected maximally entangled photon pair state which is maximally entangled in the ideal limit when $\hat {\bf k}_{\rm A} \to \hat {\bf x}$.
For illustration, we fix Alice's azimuthal angle $\varphi=0$ and consider the case where $\theta \approx \frac{\pi}{2}$. Fig.~\ref{photonerror1}
illustrates this case for both dipole separations
at 25 and 26 wavelengths. As a
comparision, Fig.~\ref{photonerror2} illustrates the case for both dipole separations at 25 and 27
wavelengths.
\begin{figure}
\begin{center}
\psfrag{a}{$\vartheta$}
\includegraphics[scale=0.5]{photonerrorc1}
\includegraphics[scale=0.5]{photonerrorv1}
\end{center}
\caption{Photon-photon polarisation correlation for orthogonal polarisation as a function
of the spherical coordinate $\vartheta$ of Alice's
detector location while Bob collects photons in the $\hat {\bf
x}$-direction in the circular(left)
and vertical(right) basis for $r= 25
\lambda_0$ (solid curve), and $r= 26 \lambda_0$ (dotted curve). } \label{photonerror1}
\end{figure}
\begin{figure}
\begin{center}
\psfrag{a}{$\vartheta$}
\includegraphics[scale=0.5]{photonerrorc2}
\includegraphics[scale=0.5]{photonerrorv2}
\end{center}
\caption{Photon-photon polarisation correlation for orthogonal polarisation as a function
of the spherical coordinate $\vartheta$ of Alice's
detector location while Bob collects photons in the $\hat {\bf
x}$-direction in the circular(left)
and vertical(right) basis for $r= 25
\lambda_0$ (solid curve), and $r= 27 \lambda_0$ (dotted curve).}\label{photonerror2}
\end{figure}
We observe that only when (\ref{hold}) is fufilled, Alice and Bob always collect photons of orthogonal
polarisation each be it in the circular or linear basis, which therefore agrees with a postselected 2-photon entangled state in the
singlet form. It can also be seen
from both figures that even when an error of the dipole separation occurs within 2 wavelengths,
the orthogonal polarisation correlation for collecting a photon pair can
still above $90\%$. Therefore, strict Lamb-Dicke localisation of the dipole source is not essential. One may estimate the order of
magnitude of the probability of an entangled photon pair collection with the help of Fig.~\ref{photonerror1}. One can then obtain the
maximum count rate for detectors with solid angles $\Delta_A$ and $\Delta_B$ with the approximate formula $\frac{9}{64 \pi^2} \Delta_A
\Delta_B$ \cite{LimJPA05}. For example, for two detectors each of solid
angular extent of $0.0225$ steradians\footnote{Each detector consist of an array of slit detectors, with 15 slits each of $0.002 \times
0.75$ steradians at intervals fulfiling (\ref{hold}) for $r=25 \lambda_0$.} yielding a minimum orthogonal polarisation correlation of $0.96$, the order of
magnitude for the collection probability $P_{c}$ is approximately $10^{-6}$. This is comparable to the scenario considered by Duan {\em
et al.} \cite{Duan04a} where he estimated the probability (also about $10^{-6}-10^{-7}$) of entangling two distant ions in free space
with the aid of a beamsplitter based on a similar scheme by Simon and Irvine \cite{Simon03}.
\section{Experimental Implementation}
As an example\footnote{I acknowledge Phillip Grangier for his kind discussion on experimental issues concerning this scheme during the 2003
summer school in Les Houches, Session 79, {\em Quantum Information and Entanglement}.} we describe now a setup for entangled photon pair creation with two
trapped $^{87}$Rb atoms that is feasible with present
technology \cite{Schlosser01}. The ground states $|0 \rangle$ and $|1 \rangle$ are
obtained from the $5^2S_{1/2}$ levels with $F=1$ and
have the quantum numbers $m_{\rm F}= -1$ and $m_{\rm F}= 1$. The excited state $|2
\rangle$ is provided by the $5^2P_{3/2}$ level with
$F=0$. Suppose the atoms are initially in the $5^2S_{1/2}$ ground state with $F=1$
and $m_{\rm F}=0$ and a $\pi$ polarised laser field is
applied to excite to level 2 by a sharp $\pi$ -pulse. After spontaneous emission into the ground
states
$|0\rangle$ and $|1 \rangle$, another $\pi$ polarised laser
reinitialises the system by coupling these states to the $5^2P_{3/2}$ states with
$F=1$. From there the atoms return into the initial
state via spontaneous decay. Due to their differences in polarisation and because
of the detector locations, ``$+$" ($\sigma^+$) and
``$-$" ($\sigma^-$) polarised signal photons are distinguishable from the laser
photons and spontaneously emitted $\pi$ polarised photons. With a typical spontaneous decay time of order $10^{-8}$ s and assuming a
rapid excitation with efficiency $90\%$ and recycling time of order $10^{-7}$ with detection efficiency of $0.88$
\cite{Takeuchi99,Rosenberg05} and taking the estimate for collection efficiency $P_{c}$, the estimated count rate of entangled photons
from this setup is expected to be $10^2s^{-1}$. Compared to the yield possible in parametric downconversion being $10^6s^{-1}$
\cite{Kumar04}, this scheme has relatively low yield. However, it does not require frequency filters for entangled photon detection
and it also offers entanglement of the dipole sources as an attractive byproduct.
\section{Conclusion}
In conclusion, we proposed a scheme for the creation of polarisation entangled
photon pairs by using two distant dipole radiators in free
space. The entanglement is obtained by carefully choosing the detector positions
with respect to the sources and arises under the
condition of the collection of two photons independent of their emission times and
the initial state of the sources. This also results in the source being maximally entangled. It is
important to note that the photon
entanglement detected can be used for quantum cryptography or Bell's inequality test. The scheme introduced in this chapter has the advantage of not requiring any linear optics and cavities compared to schemes in the previous chapters.
Another application of the scheme would be to merely prepare two distant dipole sources
in the maximally entangled ground state $|a_{01}
\rangle$. In this case, no degeneracy of the atomic ground levels $\ket{0}$ and $\ket{1}$ is
required. As in the case of Simon and Irvine \cite{Simon03}, this 2-photon detection protocol
for preparing an entangled dipole state is robust against random laser phase fluctuations during the
atomic excitation process as it
contributes to just a trivial global phase factor. Furthermore, the 2-photon protocol can yield high fidelity of entangled state
preparation more easily compared to the 1-photon protocol originally proposed by Cabrillo {\em et al.} \cite{Cabrillo99}. This is due to the
fact that in the 1-photon protocol, photons have to be gathered around the entire solid angle of emission to rule
out the possibility of an undetected 2-photon emission which ruins the entanglement. This problem may be solved at the cost of a
very weak excitation on the photon sources. The presented idea might find interesting
applications in quantum
computing with trapped atoms, diamond NV color centres,
quantum dots or single atoms doped onto a surface and opens new possibilities for
the creation of antibunched polarisation entangled
photon pairs and even multiphoton entanglement by including more than two
radiators in the setup.
Finally, we remark that the free-radiation field can be perceived roughly as a type of continuous
beamsplitter, similar to a discrete multiport with infinite inputs and outputs. This leads generally to low entangled photon pair
collection efficiency of $10^{-6}$ if we only gather photons in a
well-defined directional spatial mode as explained earlier.
Photon entanglement
schemes are therefore generally more realistic in the long-run with linear optics resources and single photon sources emitting on demand in well
directed spatial modes as demonstrated in the rest of the thesis, owing to the higher success probability that can be obtained. Linear
optics also offers flexibility in generating a wider variety of
entangled states compared to the free space approach described in this chapter. It is now time to conclude this thesis.
\chapter{Distributed Quantum Computing with Distant Single Photon
Sources}\label{minsk}
\section{Introduction}
Practical implementations of quantum computing to solve non-trivial problems
require a scalable architecture, i.e. the ability to process,
address and store many qubits. This is particularly challenging if all the
interactions between qubits are controlled locally and coherently. Particular
advances in this aspect have been made in ion traps \cite{Kielpinski02} and atoms
trapped in optical lattices \cite{Jaksch99}. Even with optical
lattices, with the inherent capability to store many qubits, controlled
addressibility and manipulation of individual qubits still
remains an experimental challenge despite advances to alleviate these
requirements \cite{You00,Kay04,Calarco04} through the help of marker atoms. In
ion traps, while addressibility is not an issue, interaction between distant
qubits still requires some form of ion transport to the range
where coherent interaction is possible between two ions
\cite{Duan04,Kielpinski02}.
An attractive alternative approach is the concept of distributed quantum
computing \cite{Eisert00,Grover96,Cirac99}. This consists of a network of
nodes with each node processing and storing a small number of qubits, which is
comparatively easy to realise. The qubits in each node are
stationary qubits, i.e. qubits that are not transported, with long decoherence
times and serve as a quantum memory. The stationary qubits
in each node communicate with distant nodes through the means of flying qubits,
i.e. qubits that are transported. Distributed quantum
computing can lead to a more efficient implementation of the phase estimation
problem compared to a classical computer in the presence of
decoherence \cite{Grover96,Cirac99}. Furthermore, distributed quantum computing
allows distant users to share quantum resources.
Traditionally, the stationary qubit of a certain node maps its state to a flying
qubit which leaves the node. On the arrival at the
target node, the flying qubit maps its state to a stationary qubit in the target
node. It is thus assumed that interconvertability of
stationary and flying qubits are required. Schemes related to this have been
proposed based on atom-cavity as stationary qubits and
photons as flying
qubits \cite{Enk97,Cirac97,Sorensen98,Xiao04,Cho04b,Zhou05,Duan05b}. All these
schemes involve single photon sources with direct transmisions of photons
through cavities. In all these cases, such transmissions occur one or several
times to complete the gate operation protocol. Another
scheme by Mancini {\em et al.} involves engineering a direct interaction between 2
distant coupled cavities via fibers \cite{Mancini04}. All these schemes
demand a high level of precision and might pose a great experimental challenge
\cite{Browne03} if one requires a high success probability.
In contrast to this, we avoid all these challenges by not requiring any form of
photon transmission through cavities. Note that we do
not really require the interconversion of stationary qubits and flying qubits
for quantum computation in a network. The unidirectional
encoding of stationary to flying qubits is already sufficient for distributed
quantum computation. Schemes along these lines
\cite{Protsenko02,Schlosser03,Zou05,Barrett04} have already been
proposed\footnote{Very recently, after this work has been submitted for
publication in August 2004, Benjamin {\em et al.}
\cite{Benjamin05} reported a scheme on creating graph states by optical
excitation in stationary qubits. They also obtain the insurance
scenario reported in this chapter with a $4 \times 4$ multiport at the cost of
having two distant stationary qubits encoding a qubit.
Furthermore, interferometric stability is not inherent in their scheme.}. To
implement a two-qubit universal gate between two distant
stationary
qubits, the basic idea is to redundantly encode the pair of stationary qubits to
a pair of flying qubits. Following that, a maximally
entangling or Bell measurement, which is normally accomplished with linear
optics, is performed on the pair of flying qubits. A universal
two-qubit gate is accomplished between the stationary qubits if the entangling
measurement is successful. Another related scheme based on
trapped ions has been proposed \cite{Duan04a} which uses ancilla ions in which they
have to be pre-entangled.
In this chapter, we will demonstrate that
scalable quantum computing between distant stationary qubits, where the
stationary qubits are single photon sources which generate the photons naturally
as flying qubits, can be made deterministic even if the
entangling measurement does not succeed.\footnote{We remind the reader here that a never-failing complete Bell measurement on two photons is not possible with linear optics \cite{Lutkenhaus99}.} We {\em do not} require any ancilla
stationary qubits nor any photon transmission through
cavities to achieve this. The ability to encode the state of the atom unto the
photon is all that is required. We use linear optics to
perform the Bell measurements on the photon. Generally, the measurement basis we
choose does not yield any information about the
stationary qubits and therefore cannot destroy the qubits in any case. As above,
for a successful Bell measurement, a two-qubit gate is
accomplished. If not, the state of the stationary qubit is not destroyed and
this allows us to repeat the encoding and subsequent
measurement until it succeeds. We have shown that this can be done by carefully
choosing the measurement basis in the entangling
measurement with linear optics. A related idea to protect a photon state
against
gate failure has been proposed in the past by Knill {\em et al.} \cite{Knill01} in the context
of
photon gate implementation by a two-qubit quantum code in their
teleportation-based gate. Such ideas are closely related to quantum error
correction \cite{Shor95,Calderbank96,Steane96}. We however use a form of
redundant encoding
natural to diverse kinds of single photon sources and show that distributed
quantum
computation between stationary qubits can require similar experimental
resources
as linear optics computation, i.e. single photon sources, optical elements and
photon detectors. At the same time, it can be performed much more
efficiently\footnote{We require no prepared ancillas nor any photon storage
and feedforward
operations.} as compared to conventional linear optics computation \cite{Knill01}.
The single photon sources that we use can take the form of atom-cavity systems
\cite{Law97,Kuhn99}, quantum dots, diamond NV colour centers or
even atomic ensembles \cite{Matsukevich04}. In principle, any photon source that
allows redundant encoding of the state of the source to the
photon it generates is a viable candidate for our scheme.
Having generated the photons, the photons must subsequently travel to the linear
optics apparatus that performs the entangling or partial Bell
measurement. Finally, the partial Bell measurements
on the encoded photons are performed and based on measurement results, we either
halt the scheme upon a heralded success or repeat the
scheme until success.
This chapter is organised as follows. The next section details the general
principle of a remote two-qubit gate
implementation with our scheme. We also show how teleportation with insurance
can be accomplished with minimal change to the setup.
Following that, in Section \ref{photonencoder} and \ref{photonmeasurement} we
describe the two ingredients of the scheme, photon encoding
and measurements. Finally, we conclude in the last section with a short
discussion on possible applications to cluster state buildup for
robust computing.
\section{Basic Idea of a remote two-qubit phase gate}
One of the requirements for universal quantum computing is the ability to
perform a universal two-qubit gate
operation, like a controlled
phase gate. Here we describe the general concept for the implementation of such
an entangling two-qubit phase
gate between two distant
single photon sources. Note that our method of distributed quantum computing
only allows the realisation of non-local
phase gates, since the
measurement on a photon pair can imprint a phase on the state of the
corresponding sources but cannot change
the distribution of their
populations. This is however sufficient for universal quantum computation. The
first step for the implementation of a two-qubit gate is
the generation of a photon within
each respective source, which
encodes the information of the stationary qubit.
\subsection{Encoding}
Let us denote the states of the photon sources, which encode the logical qubits
$|0 \rangle_{\rm L}$ and $|1
\rangle_{\rm L}$ as $|0
\rangle$ and $|1 \rangle$, respectively. For example, for atom-like single photon sources, the stable ground states can be chosen as the logical qubits. An arbitrary pure state of two
stationary qubits can be written
as
\begin{equation} \label{original}
\ket{\psi_{\rm in}}=\alpha \, \ket{00} + \beta\, \ket{01} + \gamma \, \ket{10} +
\delta \, \ket{11} \, ,
\end{equation}
where $\alpha$, $\beta$, $\gamma$ and $\delta$ are the corresponding complex
coefficients with $|\alpha|^2 +
|\beta|^2 + |\gamma|^2 +
|\delta|^2=1$. Suppose a photon is now generated in each of the two sources,
whose state (i.e. ~polarisation,
frequency or generation
time) depends on the state of the source. As we see below, it is helpful to
assume that the encoding is for
both sources different. In
the following, we assume that source 1 prepared in $|i \rangle$ leads to the
creation of one photon in state
$|{\sf x}_i \rangle$, while
source 2 prepared in $|i \rangle$ leads to the creation of one photon in state
$|{\sf y}_i \rangle$, such
that
\begin{equation} \label{enc}
\ket{i}_1 \rightarrow \ket{i;{\sf x}_i}_1 \, ,
~~ \ket{i}_2 \rightarrow \ket{i;{\sf y}_i}_2 \, .
\end{equation}
The simultaneous creation of a photon in both sources then transfers the initial
state (\ref{original}) into
\begin{equation} \label{theencoding}
\ket{\psi_{\rm enc}} = \alpha \, \ket{00;{\sf x}_0{\sf y}_0} + \beta \,
\ket{01;{\sf x}_0{\sf y}_1} + \gamma
\,
\ket{10;{\sf x}_1{\sf y}_0} +\delta \, \ket{11;{\sf x}_1{\sf y}_1} \, .
\end{equation}
The way this encoding step can be realised experimentally using either emission
time or
polarisation degrees of freedom to
encode the stationary qubits is discussed in Section \ref{photonencoder}.
\subsection{Mutually Unbiased Basis}
Once the photons have been created, an entangling phase gate can be implemented
by performing an absorbing
measurement on the photon
pair. Therefore, it is important to choose the photon measurement such that none
of the possible outcomes
reveals any information about the
coefficients $\alpha$, $\beta$, $\gamma$ and $\delta$. That such measurements
exists is well
known \cite{Wootters89}. The corresponding
measurement basis forms a so-called {\em mutually unbiased basis} (MUB) with
respect to the computational basis.
Here we are interested in
photon pair measurements in a MUB\footnote{In this chapter, our MUB basis is
always defined with respect to the computational basis.}
given the computational basis $\{\ket{{\sf x}_0{\sf
y}_0}, \, \ket{{\sf x}_0{\sf y}_1}, \, \ket{{\sf x}_1{\sf y}_0}, \, \ket{{\sf
x}_1{\sf y}_1} \}$.
More concretely, the potential outcomes of the photon measurement should all be of the
form
\begin{equation} \label{unbiased}
\ket{\Phi} = {\textstyle {1 \over 2}} \big[ \ket{{\sf x}_0{\sf y}_0} + {\rm
e}^{{\rm i} \varphi_1} \,
\ket{{\sf x}_0{\sf y}_1} + {\rm
e}^{{\rm i} \varphi_2} \, \ket{{\sf x}_1{\sf y}_0} + {\rm e}^{{\rm i} \varphi_3}
\, \ket{{\sf x}_1{\sf y}_1}
\big] \, .
\end{equation} Indeed, this is possible with linear optics as we will show in this thesis.
Detecting this state and absorbing the two photons in the process transfers the
encoded state
(\ref{theencoding}) into
\begin{equation} \label{out}
\ket{\psi_{\rm out}} = \alpha \, \ket{00} + {\rm e}^{-{\rm i} \varphi_1} \,
\beta \, \ket{01} + {\rm e}^{-{\rm
i} \varphi_2} \,
\gamma \, \ket{10} + {\rm e}^{-{\rm i} \varphi_3} \, \delta \, \ket{11} \, ,
\end{equation}
and therefore does not reveal any information about the input state
(\ref{original}) indeed. It is thus
equivalent to a phase gate implementation on the stationary qubit.
Here we are especially interested in the implementation of an entangling phase
gate with maximum entangling
power. This requires
detecting the photons in one of the four Bell states. If
\begin{equation}
\varphi_3 = \varphi_1 + \varphi_2 \, ,
\end{equation}
the state $\ket{\Phi}$ is a product state and the output (\ref{out}) differs
from the initial state
(\ref{original}) only by local
operations. However, the state (\ref{unbiased}) becomes a maximally entangled one if and only if
\begin{equation}
\varphi_3 = \varphi_1 + \varphi_2 \pm (2n-1)\pi \, .
\end{equation}
.
\subsection{A deterministic entangling gate}
Let us denote the states of the measurement basis, i.e. the mutually unbiased
basis, in the following by $\{
\ket{\Phi_i} \}$. In order to find a complete Bell basis with all
states of the form (\ref{unbiased}), we introduce the following notation,
\begin{eqnarray} \label{base}
\ket{\Phi_1} &\equiv & {\textstyle {1 \over \sqrt{2}}} \big[ \ket{{\sf a}_1 {\sf
b}_1}+\ket{{\sf a}_2 {\sf
b}_2} \big] \, \, , \,
\ket{\Phi_2} \equiv {\textstyle {1 \over \sqrt{2}}} \big[ \ket{{\sf a}_1 {\sf
b}_1}-\ket{{\sf a}_2 {\sf b}_2}
\big] \, , \nonumber \\
\ket{\Phi_3} &\equiv & {\textstyle {1 \over \sqrt{2}}} \big[ \ket{{\sf a}_1 {\sf
b}_2}+\ket{{\sf a}_2 {\sf
b}_1} \big] \, \, , \,
\ket{\Phi_4} \equiv {\textstyle {1 \over \sqrt{2}}} \big[ \ket{{\sf a}_1 {\sf
b}_2}-\ket{{\sf a}_2 {\sf b}_1}
\big] \, ,
\end{eqnarray}
where the states $|{\sf a}_i \rangle$ describe photon 1 and the states $ |{\sf b}_i
\rangle $ describe photon 2 and
$\langle {\sf a}_1 | {\sf a}_2 \rangle =0$ and $\langle {\sf b}_1 | {\sf b}_2
\rangle = 0$. One can then write
the photon states on the
right hand side of Eq.~(\ref{base}) without loss of generality as
\begin{eqnarray} \label{ab}
\ket{{\sf a}_1} &=& \cos{\theta_1} \, \ket{{\sf x}_0} + {\rm e}^{{\rm i}
\vartheta_1} \sin{\theta_1} \,
\ket{{\sf x}_1} \, , \,
\ket{{\sf a}_2} = {\rm e}^{-{\rm i} \xi_1}( {\rm e}^{-{\rm i} \vartheta_1}
\sin{\theta_1}\, \ket{{\sf x}_0} -
\cos{\theta_1} \, \ket{{\sf x}_1}) \nonumber \\
\ket{{\sf b}_1} &=& \cos{\theta_2} \, \ket{{\sf y}_0} + {\rm e}^{{\rm i}
\vartheta_2} \sin{\theta_2}
\,\ket{{\sf y}_1} \, , \,
\ket{{\sf b}_2} = {\rm e}^{-{\rm i} \xi_2}({\rm e}^{-{\rm i} \vartheta_2}
\sin{\theta_2} \, \ket{{\sf y}_0} -
\cos{\theta_2} \, \ket{{\sf y}_1}) \, .\nonumber \\
\end{eqnarray}
Inserting this into Eq.~(\ref{base}), we find
\begin{eqnarray} \label{base2}
\ket{\Phi_1} &=& {\textstyle {1 \over \sqrt{2}}} \big[ \big
(\cos{\theta_1}\cos{\theta_2}+{\rm e}^{-{\rm
i}(\vartheta_1+\vartheta_2)} {\rm e}^{-{\rm i}(\xi_1+\xi_2)}
\sin{\theta_1}\sin{\theta_2} \big) \ket{{\sf
x}_0{\sf y}_0} \nonumber \\
&& + \big( {\rm e}^{{\rm i}\vartheta_2}\cos{\theta_1}\sin{\theta_2} - {\rm
e}^{-{\rm i}\vartheta_1} {\rm
e}^{-{\rm i}(\xi_1+\xi_2)} \sin{\theta_1}\cos{\theta_2} \big)
\ket{{\sf x}_0{\sf y}_1} \nonumber \\
&&+ \big( {\rm e}^{{\rm i}\vartheta_1}\sin{\theta_1}\cos{\theta_2}-{\rm
e}^{-{\rm i}\vartheta_2} {\rm
e}^{-{\rm i}(\xi_1+\xi_2)} \cos{\theta_1}\sin{\theta_2} \big) \ket{{\sf x}_1{\sf
y}_0} \nonumber \\
&&+ \big( {\rm e}^{{\rm
i}(\vartheta_1+\vartheta_2)}\sin{\theta_1}\sin{\theta_2}+ {\rm e}^{-{\rm
i}(\xi_1+\xi_2)} \cos{\theta_1}\cos{\theta_2} \big) \ket{{\sf x}_1{\sf
y}_1} \big], \nonumber \\
\ket{\Phi_2} &=& {\textstyle {1 \over \sqrt{2}}} \big[
\big(\cos{\theta_1}\cos{\theta_2}-{\rm e}^{-{\rm
i}(\vartheta_1+\vartheta_2)} {\rm e}^{-{\rm i}(\xi_1+\xi_2)}
\sin{\theta_1}\sin{\theta_2} \big)\ket{{\sf
x}_0{\sf y}_0} \nonumber \\
&& + \big({\rm e}^{{\rm i}\vartheta_2}\cos{\theta_1}\sin{\theta_2}+{\rm
e}^{-{\rm i}\vartheta_1} {\rm
e}^{-{\rm i}(\xi_1+\xi_2)} \sin{\theta_1}\cos{\theta_2} \big) \ket{{\sf x}_0{\sf
y}_1} \nonumber \\
&& + \big( {\rm e}^{{\rm i}\vartheta_1}\sin{\theta_1}\cos{\theta_2}+{\rm
e}^{-{\rm i}\vartheta_2} {\rm
e}^{-{\rm i}(\xi_1+\xi_2)} \cos{\theta_1}\sin{\theta_2} \big) \ket{{\sf x}_1{\sf
y}_0} \nonumber \\
&&+ \big( {\rm e}^{{\rm
i}(\vartheta_1+\vartheta_2)}\sin{\theta_1}\sin{\theta_2}-{\rm e}^{-{\rm
i}(\xi_1+\xi_2)} \cos{\theta_1}\cos{\theta_2} \big) \ket{{\sf x}_1{\sf y}_1}
\big], \nonumber \\
\ket{\Phi_3} &=&{\textstyle {1 \over \sqrt{2}}}\big[ \big({\rm e}^{-{\rm
i}\vartheta_2} {\rm e}^{-{\rm
i}\xi_2}\cos{\theta_1}\sin{\theta_2}+{\rm e}^{-{\rm i}\vartheta_1} {\rm
e}^{-{\rm i}\xi_1}
\sin{\theta_1}\cos{\theta_2} \big)\ket{{\sf x}_0{\sf y}_0} \nonumber \\
&&- \big({\rm e}^{-{\rm i}\xi_2} \cos{\theta_1}\cos{\theta_2}-{\rm e}^{-{\rm
i}(\vartheta_1-\vartheta_2)} {\rm
e}^{-{\rm i}\xi_1}\sin{\theta_1}\sin{\theta_2} \big) \ket{{\sf x}_0{\sf y}_1}
\nonumber \\
&&+ \big({\rm e}^{{\rm i}(\vartheta_1-\vartheta_2)} {\rm e}^{-{\rm i}\xi_2}
\sin{\theta_1}\sin{\theta_2}- {\rm
e}^{-{\rm i}\xi_1} \cos{\theta_1}\cos{\theta_2} \big) \ket{{\sf x}_1{\sf y}_0}
\nonumber \\
&& - \big({\rm e}^{{\rm i}\vartheta_1} {\rm e}^{-{\rm i}\xi_2}
\sin{\theta_1}\cos{\theta_2}+{\rm e}^{{\rm
i}\vartheta_2} {\rm e}^{-{\rm i}\xi_1}\cos{\theta_1}\sin{\theta_2} \big)
\ket{{\sf x}_1{\sf y}_1} \big] ,
\nonumber \\
\ket{\Phi_4} &=&{\textstyle {1 \over \sqrt{2}}}\big[ \big({\rm e}^{-{\rm
i}\vartheta_2} {\rm e}^{-{\rm
i}\xi_2} \cos{\theta_1}\sin{\theta_2}-{\rm e}^{-{\rm i}\vartheta_1} {\rm
e}^{-{\rm
i}\xi_1}\sin{\theta_1}\cos{\theta_2} \big) \ket{{\sf x}_0{\sf y}_0} \nonumber \\
&& - \big({\rm e}^{-{\rm i}\xi_2} \cos{\theta_1}\cos{\theta_2}+{\rm e}^{-{\rm
i}(\vartheta_1-\vartheta_2)}
{\rm e}^{-{\rm i}\xi_1} \sin{\theta_1}\sin{\theta_2} \big) \ket{{\sf x}_0{\sf
y}_1} \nonumber \\
&& + \big({\rm e}^{{\rm i}(\vartheta_1-\vartheta_2)} {\rm e}^{-{\rm i}\xi_2}
\sin{\theta_1}\sin{\theta_2}+
{\rm e}^{-{\rm i}\xi_1} \cos{\theta_1}\cos{\theta_2} \big) \ket{{\sf x}_1{\sf
y}_0} \nonumber \\
&&- \big({\rm e}^{{\rm i}\vartheta_1} {\rm e}^{-{\rm i}\xi_2}
\sin{\theta_1}\cos{\theta_2}-{\rm e}^{{\rm
i}\vartheta_2} {\rm e}^{-{\rm i}\xi_1} \cos{\theta_1}\sin{\theta_2} \big)
\ket{{\sf x}_1{\sf y}_1} \big] .
\end{eqnarray}
These states are of the form (\ref{unbiased}), if the amplitudes are all of the
same size, which yields the
conditions
\begin{eqnarray} \label{constraint1}
&& \hspace*{-0.6cm} \big| \cos{\theta_1}\cos{\theta_2}\pm{\rm e}^{-{\rm
i}(\vartheta_1+\vartheta_2+\xi_1+\xi_2)}\sin{\theta_1}\sin{\theta_2} \big|
\nonumber
\\
&=& \big|\cos{\theta_1}\sin{\theta_2}\pm{\rm e}^{-{\rm
i}(\vartheta_1+\vartheta_2+\xi_1+\xi_2)}\sin{\theta_1}\cos{\theta_2} \big| =
\textstyle{1 \over \sqrt{2}} \, ,
\end{eqnarray}and
\begin{eqnarray} \label{constraint2}
&& \hspace*{-0.6cm} \big|\cos{\theta_1}\sin{\theta_2}\pm{\rm e}^{-{\rm
i}(\vartheta_1 - \vartheta_2+\xi_1-\xi_2)}\sin{\theta_1}\cos{\theta_2} \big|
\nonumber \\
&=& \big|\cos{\theta_1}\cos{\theta_2}\pm{\rm e}^{-{\rm
i}(\vartheta_1-\vartheta_2+\xi_1-\xi_2)}\sin{\theta_1}\sin{\theta_2}| =
\textstyle{1 \over \sqrt{2}} \, .
\end{eqnarray}
The constraints (\ref{constraint1}) and (\ref{constraint2}) can be fulfilled by the condition,
\begin{equation}
\label{con}
\cos(2\theta_1)\cos(2\theta_2)=\cos(\vartheta_1\pm\vartheta_2+\xi_1 \pm \xi_2)=0
\, .
\end{equation}
The $\pm$ sign in Eq.~(\ref{con}) apply to Eq.~(\ref{constraint1}) and
(\ref{constraint2}), respectively,
provided that neither
$\cos(2\theta_1)$ or $\cos(2\theta_2)$ equal 1. In the special case, where
either $\cos(2\theta_1)=1$ or
$\cos(2\theta_2)=1$, condition (\ref{con})
simplifies to $\cos(2\theta_1)\cos(2\theta_2)=0$ with no restrictions in the
angles $\vartheta_1$,
$\vartheta_2$, $\xi_1$ and $\xi_2$\footnote{An example of this special case can be found in Ref. \cite{Zou05}. However, this case does not yield a gate operation with insurance when a partial Bell measurement on the photons is performed.}. One particular way to fulfil
these restrictions is to set
\begin{equation} \label{angels}
\xi_2=-{\textstyle {1 \over 2}} \pi \, , ~\xi_1=\vartheta_1=\vartheta_2=0 ~~
{\rm and} ~~\theta_1 = \theta_2
= {\textstyle {1 \over 4}} \pi \, ,
\end{equation}
which corresponds to the choice
\begin{eqnarray} \label{angelencoding}
\ket{{\sf a}_{1}} &=& {\textstyle {1 \over \sqrt{2}}} ( \ket{{\sf x}_0} + \,
\ket{{\sf x}_1}) \, ,
\ket{{\sf a}_{2}} = {\textstyle {1 \over \sqrt{2}}} ( \ket{{\sf x}_0} - \,
\ket{{\sf x}_1}) \, , \nonumber \\
\ket{{\sf b}_{1}} &=& {\textstyle {1 \over \sqrt{2}}} ( \ket{{\sf y}_0} +
\ket{{\sf y}_1}) \, ,
\ket{{\sf b}_{2}} = {\textstyle {{\rm i} \over \sqrt{2}}} ( \ket{{\sf y}_0} -
\ket{{\sf y}_1}) \, .
\end{eqnarray}
Therefore, the Bell states (\ref{base}) have the following form which satisfies
the form of the mutually
unbiased basis states (\ref{unbiased}),
\begin{eqnarray}
\ket{\Phi_1} &= & {\textstyle {1 \over 2}}{\rm e}^{{\rm i}\pi/4} \big[ \ket{\sf
x_0y_0}-{\rm i}\ket{\sf
x_0y_1}-{\rm i}\ket{\sf x_1y_0}+\ket{\sf x_1y_y} \big] \, , \nonumber \\
\ket{\Phi_2} &= & {\textstyle {1 \over 2}}{\rm e}^{-{\rm i}\pi/4} \big[ \ket{\sf
x_0y_0}+{\rm i}\ket{\sf
x_0y_1}+{\rm i}\ket{\sf x_1y_0}+\ket{\sf x_1y_y} \big] \, , \nonumber \\
\ket{\Phi_3} &= & {\textstyle {1 \over 2}}{\rm e}^{{\rm i}\pi/4} \big[ \ket{\sf
x_0y_0}-{\rm i}\ket{\sf
x_0y_1}+{\rm i}\ket{\sf x_1y_0}-\ket{\sf x_1y_y} \big] \, , \nonumber \\
\ket{\Phi_4} &= & -{\textstyle {1 \over 2}}{\rm e}^{-{\rm i}\pi/4} \big[
\ket{\sf x_0y_0}+{\rm i}\ket{\sf
x_0y_1}-{\rm i}\ket{\sf x_1y_0}-\ket{\sf x_1y_y} \big] \, .
\end{eqnarray}
To find out which gate operation the detection of the corresponding maximally
entangled states (\ref{base})
combined with a subsequent
absorption of the photon pair results into, we decompose the encoded state
(\ref{theencoding}) into a state of
the form
\begin{equation} \label{bla}
\ket{\psi_{\rm enc}} = {\textstyle {1 \over 2}} \sum_i^4 \ket{\psi_i,\Phi_i}
\end{equation} and determine the states $\ket{\psi_i}$ of the stationary qubits.
Using the notation
\begin{equation} \label{CZ}
U_{\rm CZ} \equiv \ket{00}\bra{00} + \ket{01}\bra{01} + \ket{10}\bra{10} -
\ket{11}\bra{11}
\end{equation}
for the controlled two-qubit phase gate and the notation
\begin{equation}
Z_i(\phi) \equiv \ket{0}_{\rm ii}\bra{0}+{\rm e}^{-{\rm i}\phi}\ket{1}_{\rm
ii}\bra{1}
\end{equation}
for the local controlled-Z gate on photon source $i$ \footnote{This gate can be
accomplished by
applying a strongly detuned laser
field for a certain time $t$.}, we find
\begin{eqnarray} \label{totalsuccess}
\ket{\psi_1} &=& \exp \big({- \textstyle {1 \over 4}} {\rm i} \pi \big) \, Z_2
\big(- {\textstyle {1 \over 2}}
\pi \big) \, Z_1 \big(- {\textstyle {1 \over 2}} \pi \big) \,
U_{\rm CZ} \, \ket{\psi_{\rm in}} \, , \nonumber \\
\ket{\psi_2} &=& \exp \big({\textstyle {1 \over 4}} {\rm i} \pi \big) \, Z_2
\big({\textstyle {1 \over 2}}
\pi \big) \,Z_1 \big( {\textstyle {1 \over 2}} \pi \big) \, U_{\rm CZ} \,
\ket{\psi_{\rm in}} \, , \nonumber \\
\ket{\psi_3}&=& \exp \big({- \textstyle {1 \over 4}} {\rm i} \pi \big) \, Z_2
\big(-{\textstyle {1 \over 2}}
\pi \big) \, Z_1 \big({\textstyle {1 \over 2}} \pi \big) \,
U_{\rm CZ} \, \ket{\psi_{\rm in}} \, , \nonumber \\
\ket{\psi_4} &=& -\exp \big({\textstyle {1 \over 4}} {\rm i} \pi \big) \, Z_2
\big({\textstyle {1 \over 2}}
\pi \big) \, Z_1 \big(-{\textstyle {1 \over 2}} \pi \big) \, U_{\rm CZ}
\, \ket{\psi_{\rm in}} \, .
\end{eqnarray}
From this we see that one obtains the CZ gate operation (\ref{CZ}) up to local
unitary operations upon the
detection of any of the four
Bell states $\ket{\Phi_{i}}$.
\subsection{Gate implementation with insurance} \label{insurance}
When implementing distributed quantum computing with photons as flying qubits
and single photon sources as
stationary qubits, the problem
arises that it is impossible to perform a complete Bell measurement on the
photons using only linear optics
elements. As it has been
shown in the past \cite{Lutkenhaus99}, in the best case, one can only
distinguish two of the four Bell states on average.
The construction of
efficient non-linear optical elements remains a difficult problem
experimentally.
The above described phase
gate could therefore be operated at most with success rate ${1 \over 2}$.
Therefore, we choose the photon pair measurement basis $\{ |\Phi_i \rangle \}$
such that two of
the basis states are maximally entangled
while the other two basis states are product states. This is also naturally motivated from the fact that such a measurement basis can be
easily implemented using a linear optics setup \cite{Braunstein95,Mattle96}. In the following, we choose
$\ket{\Phi_3}$ and $\ket{\Phi_4}$ as in Eq.~(\ref{base}) and $\ket{\Phi_1}$ and
$\ket{\Phi_2}$ as
\begin{eqnarray} \label{partialproduct}
\ket{\Phi_1} =\ket{{\sf a_1b_1}} \, , ~~
\ket{\Phi_2}=\ket{{\sf a_2b_2}} \, .
\end{eqnarray}
As long as the states $\{ |\Phi_i \rangle \}$ constitute a MUB, the
implementation of an
eventually deterministic entangling
phase gate remains possible. In this way, we obtain quantum computing {\em with
insurance}. In case of the failure of
the gate implementation,
a product state is detected and the system remains, up to a local phase gate, in
the original qubit state.
This means that the original
qubit state (\ref{original}) can be restored and the described protocol can be
repeated, thereby eventually
resulting in the performance
of the universal controlled phase gate (\ref{CZ}). The probability for the
realisation of the gate operation
within one step equals ${1
\over 2}$ and the completion of the gate requires, on average, only two steps.
Let us now determine the conditions under which the states $\{ |\Phi_i \rangle
\}$ constitute a MUB. Proceeding as
above, we find that $\ket{\Phi_3}$ and $\ket{\Phi_4}$ are of form
(\ref{unbiased}) if the angles
$\vartheta_i$, $\xi_i$ and $\theta_i$ in
Eq.~(\ref{ab}) fulfil, for example, Eq.~(\ref{angels}). In analogy to
Eqs.~(\ref{constraint1}) and
(\ref{constraint2}), we find that
$\ket{\Phi_1}$ and $\ket{\Phi_2}$ belong to a MUB, if
\begin{equation} \label{magic}
\big|\cos{\theta_1}\cos{\theta_2} \big| = \big |\cos{\theta_1}\sin{\theta_2} \big|
= \big|\sin{\theta_1}\cos{\theta_2} \big| = \big|\sin{\theta_1}\sin{\theta_2} \big| = \textstyle{1 \over 2} \, ,
\end{equation}
which also holds for the parameter choice in Eq.~(\ref{angels}). Note that
Eq.~(\ref{magic}) is general and applies
for any product state detection in a Partial Bell basis measurement. One can
easily verify with the above
choice that the product states $\ket{\Phi_1}$ and $\ket{\Phi_2}$ are given by
\begin{eqnarray}
\ket{\Phi_1}&=& {\textstyle {1 \over 2}} \big[ \ket{\sf x_0y_0}+\ket{\sf
x_0y_1}+\ket{\sf x_1y_0}+\ket{\sf
x_1y_1} \big] \, , \nonumber \\
\ket{\Phi_2}&=& {\textstyle {{\rm i} \over 2}} \big[ \ket{\sf x_0y_0}-\ket{\sf
x_0y_1}-\ket{\sf
x_1y_0}+\ket{\sf x_1y_1} \big] \, ,
\end{eqnarray} which fulfils (\ref{unbiased}).
This means that choosing the states $|{\sf a}_i \rangle$ and $|{\sf b}_i
\rangle$ as in Eq.~(\ref{angelencoding}) allows to implement the gate operation
(\ref{CZ}) with insurance\footnote{The term insurance
was first coined by Bose {\em et al.} in the context of teleportation between
atoms in different cavities with the aid of a backup atom
in one of the cavities \cite{Bose99}.}.
Finally, we determine the gate operations corresponding to the detection of a
certain measurement outcome
$|\Phi_i \rangle$. To do this,
we decompose the encoded state (\ref{theencoding}) again into a state of the
form (\ref{bla}). Proceeding as in
the previous subsection we
find
\begin{eqnarray}
\ket{\psi_1} &=& \ket{\psi_{\rm in}} \, , \nonumber \\
\ket{\psi_2} &=& -{\rm i} \, Z_2 \big( \pi \big) \, Z_2 \big( \pi \big) \,
\ket{\psi_{\rm in}} \, , \nonumber
\\
\ket{\psi_3}&=& \exp \big({- \textstyle {1 \over 4}} {\rm i} \pi \big) \, Z_2
\big(-{\textstyle {1 \over 2}}
\pi \big) \, Z_1 \big({\textstyle {1 \over 2}} \pi \big) \,
U_{\rm CZ} \, \ket{\psi_{\rm in}} \, , \nonumber \\
\ket{\psi_4} &=& -\exp \big({\textstyle {1 \over 4}} {\rm i} \pi \big) \, Z_2
\big({\textstyle {1 \over 2}}
\pi \big) \, Z_1 \big(-{\textstyle {1 \over 2}} \pi \big) \, U_{\rm CZ}
\, \ket{\psi_{\rm in}} \, .
\end{eqnarray}
From this, we see that one obtains indeed the CZ gate operation (\ref{CZ}) up to
local unitary operations upon
the detection of either
$\ket{\Phi_{3}}$ or $\ket{\Phi_{4}}$ as in (\ref{totalsuccess}). In case of the detection of the product
states $\ket{\Phi_1}$ or
$\ket{\Phi_2}$, the initial state
can be restored with the help of one-qubit phase gates, which then allows to
repeat the operation.
It should be emphasized that there are other possible photon pair measurement bases that yield a
universal two-qubit phase gate
upon the detection of a Bell-state
but where the original state is destroyed upon the detection of a product state
(see e.g.~\cite{Zou05}). The
reason is that, while the
detected Bell states might result in a universal gate operation, the
corresponding product states are not
mutually unbiased and their
detection erases the qubit state in the photon sources. To achieve the effect of
an {\em insurance}, the photon pair
measurement basis should be
chosen as described in this Section, as an example.
\subsection{Teleportation with insurance}\label{teleportation}
Here, we first show that the setup can be directly used to realise a quantum
filter operation with insurance.
This would lead us naturally to teleportation. Particularly, we describe a scheme
for the implementation of the
parity filter operation(see Chapter \ref{hummingbird})
\begin{equation} \label{filter}
P^{1\pm}_{\rm filter} = |00 \rangle \langle 00| \pm |11 \rangle \langle 11| \,
,
\end{equation}
which projects the initial qubit state $\ket{\psi_{\rm in}}$ with probability
$|\alpha|^2+|\delta|^2$ onto the even-parity state,
\begin{equation} \label{fin3}
\ket{\psi_{\rm fin}}= \big( \alpha \, \ket{00} \pm \delta \, \ket{11}
\big)/\sqrt{|\alpha|^2+|\delta|^2} \, .
\end{equation} or
\begin{equation}
P^{2\pm}_{\rm filter} = |01 \rangle \langle 01| \pm |10 \rangle \langle 10| \,
,
\end{equation}
which projects $\ket{\psi_{\rm in}}$ with probability
$|\beta|^2+|\gamma|^2$ onto the odd-parity state,
\begin{equation} \label{fin4}
\ket{\psi_{\rm fin}}= \big( \beta \, \ket{01} \pm \gamma \, \ket{10}
\big)/\sqrt{|\beta|^2+|\gamma|^2} \, .
\end{equation}
Again, this can be achieved if the photon states are detected in the
desired form $\frac{1}{\sqrt{2}}(\ket{\sf x_0y_0}\pm{\rm e}^{{\rm
i}\delta_{1\pm}}\ket{\sf x_1y_1})$
or $\frac{1}{\sqrt{2}}(\ket{\sf x_0y_1}\pm{\rm e}^{{\rm
i}\delta_{2\pm}}\ket{\sf x_1y_0})$ where $\delta_{i\pm}$ is any arbitrary phase
angle. Looking
again at Eq. (\ref{base2}), all the
basis states $\ket{\Psi_i}$ will be the desired form by setting
\begin{equation}\label{teleportationcondition}
\sin(\theta_1\mp\theta_2)=\cos(\theta_1\pm\theta_2)=0 \, ,
\end{equation}and
\begin{equation}
\sin(\vartheta_1\pm \vartheta_2+\xi_1
\pm \xi_2)=0 \, ,
\end{equation} provided that $\sin(2\theta_1)\sin(2\theta_2) \neq 0$. In the
special case where $\sin(2\theta_1)\sin(2\theta_2) =0$, then
the only constraint would be $\sin(2\theta_1)=\sin(2\theta_2)=0$ with no
restriction on the angles $\vartheta_1$, $\vartheta_2$, $\xi_1$
and $\xi_2$.
However now, we redefine $\ket{\Psi_1}$ and $\ket{\Psi_2}$ as product states
defined in Eq.~(\ref{partialproduct}), collectively forming
a partial Bell-measurement basis. Combining with the condition of insurance in
Eq.~(\ref{magic}), we see that the choice
\begin{equation}
\theta_1=\theta_2=\frac{\pi}{4}, \, \vartheta_1=\vartheta_2=\xi_1=\xi_2=0,
\end{equation} allows us to implement a parity filter with
insurance. This is the choice where
\begin{eqnarray} \label{demonencoding}
\ket{{\sf a}_{1}} &=& {\textstyle {1 \over \sqrt{2}}} ( \ket{{\sf x}_0} + \,
\ket{{\sf x}_1}) \, , \,
\ket{{\sf a}_{2}} = {\textstyle {1 \over \sqrt{2}}} ( \ket{{\sf x}_0} - \,
\ket{{\sf x}_1}) \, , \nonumber \\
\ket{{\sf b}_{1}} &=& {\textstyle {1 \over \sqrt{2}}} ( \ket{{\sf y}_0} +
\ket{{\sf y}_1}) \, , \,
\ket{{\sf b}_{2}} = {\textstyle {1 \over \sqrt{2}}} ( \ket{{\sf y}_0} -
\ket{{\sf y}_1}) \, .
\end{eqnarray} which yields
\begin{eqnarray} \label{devilencoding}
\ket{\Phi_1} &=& {\textstyle {1 \over 2}} ( \ket{\sf x_0y_0}+\ket{\sf
x_0y_1}+\ket{\sf x_1y_0}+\ket{\sf x_1y_1}) \, , \nonumber \\
\ket{\Phi_2} &=& {\textstyle {1 \over 2}} ( \ket{\sf x_0y_0}-\ket{\sf
x_0y_1}-\ket{\sf x_1y_0}+\ket{\sf x_1y_1}) \, , \nonumber \\
\ket{\Phi_3} &=& {\textstyle {1 \over \sqrt{2}}} ( \ket{\sf x_0y_0}-\ket{\sf
x_1y_1}) \, , \nonumber \\
\ket{\Phi_4} &=& -{\textstyle {1 \over \sqrt{2}}} ( \ket{\sf x_0y_1}-\ket{\sf
x_1y_0}) \, .
\end{eqnarray}
To see this, we again decompose the input state (\ref{theencoding}) again into a
state of the form (\ref{bla}).
Proceedings as in the previous subsection, we
find
\begin{eqnarray}
\ket{\psi_1} &=& \ket{\psi_{\rm in}} \, , \,
\ket{\psi_2} = Z_2 \big( \pi \big) \, Z_2 \big( \pi \big) \, \ket{\psi_{\rm in}}
\, , \nonumber \\
\ket{\psi_3}&=& P^{1-}_{\rm filter} \ket{\psi_{\rm in}} \, , \,
\ket{\psi_4} = -P^{2-}_{\rm filter} \ket{\psi_{\rm in}} \, .
\end{eqnarray}
One application of the quantum parity filter (\ref{filter}) is {\em
teleportation with insurance}, which now
requires less resources than previously proposed schemes \cite{Bose99}. Suppose,
a given state $\alpha \,
|0\rangle + \beta \, |1 \rangle$ of source A is to be teleported to another
target source B prepared in ${1
\over \sqrt{2}} (|0\rangle + |1 \rangle)$. Application of the quantum filter to
the combined state of the two
sources, then ultimately transfers this state into $\alpha \, |00 \rangle + \beta \, |11
\rangle$ or $\alpha \, |01
\rangle + \beta \, |10 \rangle$. In order to complete the teleportation, the
state of B should be disentangled
from the state of source A without revealing the coefficients $\alpha$ and
$\beta$. This can be
achieved by measuring source A on the basis given by $|\pm \rangle \equiv {1
\over \sqrt{2}} (|0\rangle \pm |1 \rangle)$. Depending on
the outcome of this measurement, a further local operation on the state of B
might be required.
We proceed next with the description of possible ways to realise the photon encoding.
\section{Entangled Atom-Photon generation from Atom-Cavity Systems}
\label{photonencoder}
\subsection{Introduction}
The purpose of this subsection is to show how a highly efficient encoder of a
stationary qubit to the flying qubit can be performed in
the context of an atom-cavity system. We
use an
atom-cavity photon gun \cite{Law97,Kuhn99,Duan03,Saavedra00, Gheri98,Ciaramicoli01,Maurer04,Keller04} in which either Raman transfer with
detuning or the
so-called stimulated Raman adabiatic passage (STIRAP) \cite{Bergmann98,Oreg84}
is employed. Gheri {\em et al.} \cite{Gheri98} were the first to exploit a limited
version of encoding with this technique. Here, we review by
following closely
the formalism of Duan {\em et al.} \cite{Duan03}, how a
deterministic single photon encoder with a single or double
$\Lambda$-type level configuration trapped in a cavity can be implemented in principle. We
illustrate this using the STIRAP method as an example although the Raman
transfer method using large detuning resulting in adiabatic elimination of the
excited
state will yield the same general conclusion. As in
\cite{Gheri98,Saavedra00}, we make use of the superposition
principle and the fact that no-cross couplings can occur between different
subspaces as described by the system Hamiltonian to show how efficient encoding to the
photon
is possible.
\subsection{Photon gun encoder} \label{encodertrick}
We consider the situation of a degenerate double $\Lambda$-type atom trapped in an optical
cavity. Specifically, each $\Lambda$ system $i$ where
$i=0,1$ consists of two ground states $\ket{u_i}$ and $\ket{v_i}$ as well as an
excited state $\ket{e_i}$ as shown in Fig.~\ref{polencoder}. We assume that the
atom
is initially prepared in the state
\begin{equation}
\ket{\psi(t=0)}=(\alpha_0\ket{u_0}+\alpha_1\ket{u_1})\ket{0}_{\rm {\rm
cav}}\ket{{\rm vac}},
\end{equation} where $\ket{0}_{\rm {\rm cav}}$ and $\ket{{\rm vac}}$
denotes the cavity vacuum field and the
external photon vacuum field respectively.
\begin{figure}
\begin{center}
\psfrag{O}{$\Omega$}
\psfrag{v0}{$v_0$}
\psfrag{v1}{$v_1$}
\psfrag{e0}{$e_0$}
\psfrag{e1}{$e_1$}
\psfrag{u0}{$u_0$}
\psfrag{u1}{$u_1$}
\psfrag{g}{$g$}
\includegraphics{photongun1}
\end{center}
\caption{Schematic view of a single photon polarisation encoder and level
configuration of the atomic
structure containing the qubit.} \label{polencoder}
\end{figure}
Lasers with time-dependent Rabi
frequency
$\Omega_i(t)$ with frequency
$\omega_L$ resonantly couple levels $u_i$ to $e_i$. The cavity resonantly couples levels $v_i$ to $e_i$ with coupling
strength
$g_i$ with the cavity resonant frequency $\omega_{{\rm cav}}$.
In addition, the cavity field associated with creation(destruction) operator
$a_i^{\dagger}(a_i)$ corresponding to both $\Lambda$ systems $i$
is required to be orthogonal (in this case, in polarisation) with respective decay rates $\kappa_i$. The
external photon fields can be described by
creation(destruction) operators $b_i(\omega)^{\dagger}(b_i(\omega))$ with
mode $i$ and frequency $\omega$. They satisfy the commutation rules
$[b_i(\omega_1),b^{\dagger}_j(\omega_2)]=\delta_{ij}\delta_{\omega_1\omega_2}$.
To solve the evolution of the system, it is convenient to borrow the concept of
quantum jump formalism \cite{Hegerfeldt93,Dalibard92,Carmichael93} with a
non-event (i.e. no spontaneously emitted photon emitted outside the cavity mode) evolution
described by a non-Hermitian Hamiltonian.
The conditional non-Hermitian Hamiltonian ($\hbar=1$), in the case where no
photon
is spontaneously emitted outside the cavity mode, in the interaction picture
with respect to the free evolution is given by
\cite{Tregenna02,Duan03}
\begin{equation} \label{Hencoder}
H(t)=H_{\rm atom-cavity}(t)+H_{\rm cavity-env}(t)
\end{equation} where
\begin{equation}
H_{\rm atom-cavity}(t)=\sum_i
\Omega_i(t)\ket{e_i}\bra{u_i}+g_ia_i\ket{e_i}\bra{v_i}+{\rm H.c.}-{\rm
i}\frac{\Gamma_i}{2}\ket{e_i}\bra{e_i} \, ,
\end{equation}
\begin{equation}
H_{\rm cavity-env}(t)=\sum_i {\rm i}\sqrt{\frac{\kappa_i}{2
\pi}}\int_{-\omega_b}^{\omega_b} d\omega
a_i^{\dagger}b_i(\bar{\omega}){\rm e}^{-{\rm i}\omega
t}+{\rm H.c.} \, ,
\end{equation} and $\bar{\omega}=\omega_{{\rm cav}}+\omega$ and $\omega_b$ is
the bandwidth. We assume $\omega_b$ to be relatively large but it has to be
much smaller than $\omega_{\rm cav}$. Within this bandwidth, the coupling
between the free field and the cavity mode is approximately
constant and given by $\sqrt{\frac{\kappa_i}{2 \pi}}$. The cavity
coupling $g_i$, which derives from the quantisation of the vacuum field in the
cavity, is a function of the cavity spatial mode function and the relevant
dipole transition element \cite{Chen04}. For the purpose of illustration, they
can be set to be real and equal without loss of generality. We also see
the effect of the damping factor $\Gamma_i$ on the excited
state $\ket{e_i}$ due to the possibility of spontaneous emission.
We now define the states $\ket{D_i(t)}$ and
$\ket{B_i(t)}$ such that
\begin{eqnarray}\label{dark}
\ket{D_i(t)}&=&\cos\vartheta_i(t) \ket{u_i}\ket{0}_{\rm {\rm
cav}}-\sin\vartheta_i(t)\ket{v_i}a_i^{\dagger}\ket{0}_{\rm {\rm
cav}} \nonumber \, , \\
\ket{B_i(t)}&=&\sin\vartheta_i(t) \ket{u_i}\ket{0}_{\rm {\rm
cav}}+\cos\vartheta_i(t)\ket{v_i}a_i^{\dagger}\ket{0}_{\rm {\rm
cav}}\, ,
\end{eqnarray} where
\begin{equation}
\cos\vartheta_i(t)=\frac{g_i}{\sqrt{g_i^2+\Omega_i^2(t)}}.
\end{equation} One can easily see that $\ket{D_i(t)}$ and $\ket{B_i(t)}$ are
orthogonal to each other. An important point, as
pointed out by Duan {\em et. al.} is that the spatial dependence of
$\cos\vartheta_i(t)$ can be made to vanish provided that
$\Omega_i(t)$ and $g_i$ share the same cavity spatial mode structure. This can
be accomplished by
collinear pumping where the external pumping laser couples to a similar spatial
cavity mode of a different polarisation relative to the
one used in generating the cavity photon that subsequently leaks out and is
encoded to the atomic state. This suggests that the atom need
not really be cooled to the
Lamb-Dicke limit for operation which removes a huge experimental challenge.
However, the same authors point out that cooling is still important to maintain
a long trap lifetime of the atom in the cavity.
When $\Gamma_i$ and $\kappa_i$ vanish, $\ket{D_i(t)}$ is an exact eigenstate
with
zero eigenvalue of the Hamiltonian $H(t)$ and hence,
known as a dark state. A system initially in the dark state
always stays in the dark state provided that
the adiabatic following condition is fulfilled. This is the essence of the
STIRAP process, which allows for robust coherent state transfer by remaining always in a
dark state. In the subspace defined by the
Hamiltonian $H(t)$ and making the assumption that
only one photon excitation can be put to the external field, the general state
after time $t$ is given by
\begin{eqnarray}\label{conditionalstate}
\ket{\psi(t)}&=&\sum_i \alpha_i ((c_{D_i}(t)\ket{D_i(t)}+c_{B_i}(t)\ket{B_i(t)}+
\nonumber \\
&&c_{e_i}(t)\ket{e_i}\ket{0}_{\rm cav})\ket{{\rm vac}}+ \ket{v_i}\ket{0}_{\rm cav}
\int_{-\omega_b}^{\omega_b} d\omega s(\omega,t)_i\ket{\bar{\omega}}_i)\, ,
\end{eqnarray} where $\ket{\bar{\omega}}_i=b_i^\dagger(\bar{\omega})\ket{\rm
vac}$. Note that the last term with the external photon excitation is
associated with the state $\ket{v_i}\ket{0}_{\rm
cav}$. This can be inferred by looking at the Hamiltonian given in Eq.
(\ref{Hencoder}). An external photon
excitation of mode $i$ comes only through the
annihilation of a cavity photon $a_i^{\dagger}\ket{0}_{\rm cav}$ from $H_{\rm
cavity-env}(t)$ which was created accompanying the
projection of the atomic state to $\ket{v_i}$ from
$H_{\rm atom-cavity}(t)$.
Up to now, we have not made any adiabatic approximations. We can calculate the
probability $P_{\rm cond}(t)=|| \,\ket{\psi(t)}||^2$ of the system
evolving according to $H(t)$. It is this evolution that yields the photon in
the cavity mode which leaks out subsequently. Otherwise, spontaneous emission
occurs and this takes place
with the probability $P_{\rm spon}(t)=1-P_{\rm
cond}(t)$.
Explictly, this is given by
\begin{equation}
P_{\rm spon}(t)=1-\sum_i |\alpha_i|^2(|c_{D_i}(t)|^2+|c_{B_i}(t)|^2+|c_{e_i}(t))|^2+\int_{-\omega_b}^{\omega_b} d\omega
|s(\omega,t)_i|^2)
\end{equation}
When $P_{\rm spon}(t)$ is small or close to $0$, the
system can yield an effective photon source on demand.
Now, the adiabatic condition, which we take as an ansatz (see further discussion by Duan {\em et al.} \cite{Duan03}), results in a very slow change of
$\cos\vartheta_i(t)$ which implies that the time derivatives of
$\ket{D_i(t)}$ and $\ket{B_i(t)}$ vanish. This condition also implies that the
population of
$\ket{B_i(t)}$ and $\ket{e_i}$ is virtually zero and can be effectively
neglected. This allows us to
simplify the calculation of the evolution of $\ket{\psi(t)}$ with the Schrodinger's equation by calculating the
two time-dependent
coefficients given by
\begin{eqnarray}
\dot{c}_{D_i}(t)&=&-\sqrt{\frac{\kappa_i}{2\pi}}\sin\vartheta_i(t)\int_{-\omega_
b}^{\omega_b} d\omega
s(\omega,t)_i {\rm e}^{-{\rm i}\omega t} \, , \nonumber \\
\dot{s}(\omega,t)_i&=& \sqrt{\frac{\kappa_i}{2\pi}}
c_{D_i}(t) \sin\vartheta_i(t) {\rm e}^{{\rm i}\omega t} \, .
\end{eqnarray}
The solutions of the two coefficients are given approximately by
\begin{eqnarray}\label{cgsolution}
c_{D_i}(t)&=& \exp(-\frac{\kappa_i}{2} \int_0^{t} dt' \sin^2 \vartheta_i(t'))
\, ,
\nonumber \\
s(\omega,t)_i&=& \sqrt{\frac{\kappa_i}{2\pi}}\int_0^t dt' {\rm e}^{{\rm
i}\omega t'} c_{D_i}(t') \sin\vartheta_i(t') \, .
\end{eqnarray} We have used the Markovian approximation in the process where the
limits of integration of $\omega$ is artificially
extended to $-\infty$ and $\infty$ due to the large bandwidth $\omega_b$ to
yield a delta function.
To have generated an external photon with unit probability by time $t=\tau$, we
should start with
$c_{D_i}(0)=1$ at time $t=0$ and end up with $c_{D_i}(\tau)=0$ by looking
at Eq. (\ref{conditionalstate}) where $\tau$ is
chosen to be a
characteristic time in the tail end of the pumping laser pulse when the
amplitude is near zero.
This is fulfilled by choosing a large $\tau$ and(or) increasing the laser Rabi
frequency $\Omega_i(t)$ by looking at Eq. (\ref{cgsolution}). Note that $\kappa_i^{-1}$
must be smaller than $\tau$. Otherwise, adiabatic following will imply coherent
return to the initial state with $c_{D_i}(\tau)=1$ with no external photon
generated\cite{Kuhn99} if $\kappa_i^{-1} \gg \tau$.
We can define the pulse shape by the Fourier transform of the spectral evelope
\begin{equation}
f(t,\tau)_i=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} d\omega
s(\omega,\tau)_i{\rm e}^{-{\rm
i}\omega t}.
\end{equation}
From Eq. (\ref{cgsolution}), we see that the pulse shape is given by
\begin{equation}\label{magicpulse}
f(t,\tau)_i=\sqrt{\kappa_i} \sin\vartheta_i(t)c_{D_i}(t)
\end{equation}
For simplicity, we assume all parameters related to different
$i$ to be the same i.e. $(\Omega_i(t)=\Omega(t),
g_i=g,\Gamma_i=\Gamma)$. We also assume that after a photon is generated, we
recycle the state of the system from $\ket{v_i}$ to
$\ket{u_i}$. In the ideal limit of strong cavity
coupling $g^2\gg \Gamma \kappa$ and adiabatic following, we are able to perform
a deterministic
mapping of the form
\begin{equation} \label{effectiveencoding}
\sum_i \alpha_i\ket{u_i}\ket{0}_{\rm {\rm cav}}\ket{{\rm vac}}
\to \sum_i
\alpha_i\ket{u_i}\ket{0}_{\rm {\rm cav}}\int_{-\omega_b}^{\omega_b} d\omega
s(\omega,t)_i b^{\dagger}_i(\bar{\omega})\ket{\rm vac}
\end{equation}
This implies that we can encode the state of the atoms to the
externally generated photons with a STIRAP process and was first demonstrated by
Gheri {\em et. al.} \cite{Gheri98} in the regime of Raman transfer with a large detuning.
Now, all these above calculations invoke the crucial assumption of adiabatic
following. The adiabatic following condition is
well defined in the limit where $\Gamma_i$ and $\kappa_i$ vanish.
Specifically, the evolution time $\tau$ must be longer than
the inverse of the frequency splitting gap between the dark state $\ket{D_i(t)}$
and the rest of
the eigenstates \cite{Kuhn99}. In our case, the gap $\delta$ is
$\sqrt{g_i^2+\Omega_i^2(t)}$ and the adiabatic condition is given by
\begin{equation}
(g_i^2+\Omega_i^2(t))\tau^2 \gg 1.
\end{equation} Note that this condition does not imply adiabatic following when
$\Gamma_i$ and $\kappa_i$ is non-zero due to the
atom-cavity system coupling to an infinite continum of modes. Therefore, a
fuller description necessitates numerical simulation.
Duan {\em et. al.} \cite{Duan03} have already performed such simulations and showed that
spontaneous emission loss is negligible in the limit of
strong coupling where $g^2/\kappa \Gamma \gg 1$. Empirically, $P_{\rm
spon}$ scales as $\kappa \Gamma /4 g^2$. Furthermore, it was found
that for strong coupling, the analytically calculated pulse shape $f(t,\tau)_i$
based on the adiabatic following ansatz agrees very well with
numerical simulations. If we restrict ourselves to the less general case of a
single $\Lambda$ subsystem characterised for example by $\alpha_i=0$, we recover
exactly the usual single photon gun \cite{Law97,Kuhn99}.
Finally, we consider the case where the same laser driving pulse is offset by a
time $T_i$ and we operate in the single $\Lambda$ subsystem, dropping all
subscripts corresponding to subsystem for readability. For convenience, we can
define the effective photon creation operator as $B^\dagger (t_i, \tau)$
\begin{equation}
B^\dagger (t, \tau)=\int_{-\omega_b}^{\omega_b} d\omega {\rm e}^{{\rm i} \omega
t}s(\omega,\tau)b^\dagger(\bar{\omega}).
\end{equation}
We find that when $|t_i-t_j|\gg \tau$ \cite{Gheri98,Saavedra00},
\begin{equation}
\left[B^\dagger (t_i, \tau),B^\dagger (t_j, \tau) \right]\to 0.
\end{equation}
This implies that photons created by laser driving pulse offset by a time
separation much larger than the driving pulse duration $\tau$ can be considered
to be in different modes and hence orthogonal. In fact,$ \left[B^\dagger (t_i,
\tau),B^\dagger (t_j, \tau)
\right] $ which depends essentially on the temporal overlap of two identical
pulses offset by $|t_i-t_j|$
approximately decay exponentially with the ratio $\frac{|t_i-t_j|}{\tau}$ and
thus typically, a
time-separation $|t_i-t_j|$ of the order of $\tau$ might already be sufficient
to achieve the photon
orthogonality condition \cite{Saavedra00}. This leads us to the possibility
of time-bin encoding in which photons are created by an early or late driving pulse
with the appropriate time
separation
dependent on the initial ground state of the atom.
\begin{figure}
\psfrag{O}{$\Omega$}
\psfrag{v1}{$v_1$}
\psfrag{e1}{$e_1$}
\psfrag{u0}{$u_0$}
\psfrag{u1}{$u_1$}
\psfrag{g}{$g$}
\begin{center}
\includegraphics{photongun2}
\end{center}
\caption{Schematic view of a single photon time-bin encoder and level
configuration of the atomic structure
containing the qubit.} \label{binencoder}
\label{photongun2}
\end{figure}
One way to implement such an encoding is to
first swap the atomic states $\ket{u_0}$ and $\ket{u_1}$. Then a laser pulse
with
increasing Rabi frequency should excite the $u_1$-$e_1$ transition (see
Fig.~\ref{binencoder}) at time $t_0$ for
example. This transfers the atom into the state $\ket{v_1}$
and places one excitation into the field of the strongly coupled optical cavity,
if the atom was initially
prepared in $|u_0 \rangle$. The
photon then leaks out through the outcoupling mirror of the resonator. The
encoding operation, which is
feasible with present technology \cite{Duan03}, is completed by transfering
$|v_1 \rangle$ back into $|u_1
\rangle$, swapping again the states
$|u_0 \rangle$ and $|u_1 \rangle$ and repeating the above described photon
generation process at a later time
$t_1$. The above process therefore describes the mapping of a form
\begin{equation}
\sum_i \alpha_i\ket{u_i}\ket{0}_{\rm cav}\ket{{\rm vac}} \to \sum_i
\alpha_i\ket{u_i} \ket{0}_{\rm cav}
B^\dagger(t_i,\tau)\ket{{\rm vac}}
\end{equation} In general, time-bin encoding which requires a simpler energy level structure compared to polarisation encoding may find
realisations in systems such as quantum dots and NV color centers where a double $\Lambda$-type configuration may not be easily found.
We now proceed
to the description of photon pair measurement.
\section{Measurements on photon pairs}\label{photonmeasurement}
We give 2 examples of measurements on the photon pairs based on the concrete
choice given in
(\ref{angelencoding}). The first method is suitable for polarisation encoded
photon pairs. The second is
suitable for dual-rail encoded photon pairs. In general, depending on the
initial choice of encoding,
conversion between encodings may be required. For example, one might need to
convert time-bin encoding to
either polarisation or dual-rail encoding and method 1 or 2 can be used
respectively for photon pair
measurements.
It is worth mentioning that our scheme has the same robustness from slow and
unknown phase fluctutations along
the photon paths due to the same reason outlined for example in Ref.
\cite{Simon03}. Due to the fact that we use
coincidence measurement for our Bell-state detection, any slow phase error on
the photons contributes only to
a global phase factor in the stationary qubits. This also implies that our
scheme does not require
interferometric stability as is the case of most schemes requiring coincidence
detection. We first describe a canonical Bell-state
measurement in polarisation encoding.
\subsection{Canonical Bell-state measurement}
Bell-state measurement on a photon pair is an important tool used widely in
quantum information processing with photons. It is crucial for quantum
teleportation \cite{Bennett93}, quantum
dense coding \cite{Bennett92} as well as entanglement swapping
\cite{Zukowski93}. Recently, Browne and Ruldoph
\cite{Browne05} have exploited Bell-state measurement for the efficient
construction of a photonic cluster
state. However, a complete Bell
measurement cannot be realised with unit success probability in a purely linear
optics based setup \cite{Lutkenhaus99}. This can be thought as a main limitation
to purely linear optics
based quantum computation. We show here a canonical example of how a partial
Bell measurement can be realised with the aid of a beam splitter
\cite{Braunstein95}. We recall the basis (\ref{completeBellintro})
states of a complete Bell basis as
\begin{eqnarray}
&&\ket{\Phi^\pm}=\frac{1}{\sqrt{2}}(a_{1,h}^\dagger a_{2,v}^\dagger \pm
a_{1,v}^\dagger a_{2,h}^\dagger ) \ket{0}_{\rm vac} \, , \nonumber \\
&&\ket{\Psi^\pm}=\frac{1}{\sqrt{2}}(a_{1,h}^\dagger a_{2,h}^\dagger \pm
a_{1,v}^\dagger a_{2,v}^\dagger ) \ket{0}_{\rm vac} \, .
\end{eqnarray} Here, $a_{i,\lambda}^\dagger$ refers to a photon creation
operator for spatial mode $i$ with polarisation $\lambda$. These 2-photon Bell
states are sent, one in each input arm, into a 50:50 beam splitter which is
described by the matrix $B(\frac{1}{2},1)$ in Chapter \ref{firework}. We fix the
convention that the spatial modes defined by the two
input ports are defined as spatial modes 1 and 2 and that defined by the two
output ports are
defined as spatial mode 3 and 4. It can be shown that the basis Bell states at
the input will transform to the output ports as
\begin{eqnarray}
&&\ket{\Phi^+} \to \frac{1}{\sqrt{2}}(a_{4,v}^\dagger
a_{4,h}^\dagger-a_{3,v}^\dagger a_{3,h}^\dagger)\ket{0}_{\rm vac} \, , \nonumber
\\
&&\ket{\Phi^-} \to \frac{1}{\sqrt{2}}(a_{3,v}^\dagger
a_{4,h}^\dagger-a_{4,v}^\dagger a_{3,h}^\dagger)\ket{0}_{\rm vac} \, , \nonumber
\\
&&\ket{\Psi^+} \to \frac{1}{2\sqrt{2}}((a_{4,v}^\dagger)^2+
(a_{4,h}^\dagger)^2-(a_{3,v}^\dagger)^2- (a_{3,h}^\dagger)^2)\ket{0}_{\rm vac}
\, ,
\nonumber \\
&&\ket{\Psi^-} \to \frac{1}{2\sqrt{2}} ((a_{4,v}^\dagger)^2-
(a_{4,h}^\dagger)^2-(a_{3,v}^\dagger)^2+ (a_{3,h}^\dagger)^2)\ket{0}_{\rm vac}
\, .
\end{eqnarray} We see that $\ket{\Phi^\pm}$ is indicated by detecting photons of
different polarisations in the same and different output ports respectively.
Unfortunately, $\ket{\Psi^\pm}$ cannot be distinguished by simple photon
detection and hence, the simple beam splitter cannot implement a complete Bell
measurement with unit efficiency. However, the following product states,
$\frac{1}{\sqrt{2}}(\ket{\Psi^+}+\ket{\Psi^-})=a_{1,h}^\dagger a_{2,h}^\dagger
\ket{0}_{\rm vac}$ and
$\frac{1}{\sqrt{2}}(\ket{\Psi^+}-\ket{\Psi^-})=a_{1,v}^\dagger a_{2,v}^\dagger
\ket{0}_{\rm vac}$ transform as
$\frac{1}{2}((a_{3,h}^\dagger)^2-(a_{4,h}^\dagger)^2)\ket{0}_{\rm vac}$ and
$\frac{1}{2}((a_{3,v}^\dagger)^2-(a_{4,v}^\dagger)^2)\ket{0}_{\rm vac}$
respectively. The output states in this case are distinguishable. Therefore, the
measurement performed in this
case is a partial
Bell measurement with two Bell states and two product states constituting the
measurement basis.
\subsection{Measurement for polarisation encoded photon pair}
We have shown that sending two polarisation encoded photons through the
different input ports of a 50:50 beam
splitter together with
polarisation sensitive measurements in the $\ket{\sf h}/\ket{\sf v}$-basis in
the output ports would result in
a measurement of the
states ${\textstyle {1 \over {\sqrt 2}}}(\ket{\sf hv}\pm\ket{\sf vh})$,
$\ket{\sf hh}$ and $\ket{\sf vv }$. To
measure the states $\ket{\Phi_i}$ defined in Section \ref{insurance} or \ref{teleportation}, we
therefore proceed as shown in Fig.~\ref{moon2}(a) and
perform the mapping
$U_1 = |{\sf h} \rangle
\langle {\sf a_1}| + |{\sf v} \rangle \langle {\sf a_2}|$ and $U_2 = |{\sf h}
\rangle
\langle {\sf b_1}| + |{\sf v} \rangle \langle {\sf b_2}|$ on the photon coming
from source $i$. For the states defined Section \ref{insurance}, using Eq.~(\ref{angelencoding}), we see
that this
corresponds to the single qubit rotations
\begin{eqnarray}
U_1 &=& {\textstyle {1 \over \sqrt{2}}} \, \big[ \, |{\sf h} \rangle \big(
\langle {\sf h}| + \langle {\sf
v}| \big) + |{\sf v} \rangle
\big( \langle {\sf h}| - \langle {\sf v}| \big) \, \big] \, , \nonumber \\
U_2 &=& {\textstyle {1 \over \sqrt{2}}} \, \big[ \, |{\sf h} \rangle \big(
\langle {\sf h}| + \langle {\sf
v}| \big) - {\rm i} \, |{\sf
v} \rangle \big( \langle {\sf h}| - \langle {\sf v}| \big) \, \big] \, .
\end{eqnarray}
After leaving the beam splitter, the photons should be detected in the $\ket{\sf
h}/\ket{\sf v}$-basis. A
detection of two ${\sf h}$ (${\sf v}$) polarised photons indicates a measurement
of $|\Phi_1 \rangle$
($|\Phi_2 \rangle$). Finding two photons of different polarisation in the same
(different) detectors
corresponds to a detection of $|\Phi_3 \rangle$ ($|\Phi_4 \rangle$).
\subsection{Measurement for dual-rail encoded photon pair}
Alternatively, one can redirect the generated photons ( for example, if the photons are time-bin encoded) to the different input
ports of a $4 \times 4$ symmetric
multiport beam splitter as shown in Fig.~\ref{moon2}(b). A symmetric multiport redirects
each incoming photon with
equal probability to any of the possible output ports and can therefore be used
to erase the which-way
information of the incoming photons as we have mentioned in Chapter \ref{firework} and \ref{fusion}. If $a_n^\dagger$ ($b_n^\dagger$)
denotes
the creation operator for a
photon in input (output) port $n$,
the effect of the multiport can be summarised as
\begin{equation} \label{scatter}
a_n^\dagger \to \sum_m U_{mn} b_m^\dagger,
\end{equation}where $U_{mn}$ is the probability amplitude to redirect a photon
from the $n$th input port to the $m$th
output port. For the implementation of either
a two qubit universal phase gate or
a parity filter, one should direct the input
$\ket{\sf x_0}$ ($\ket{\sf x_1}$) photon from source 1 to input port 1 (3) and
to direct a $\ket{\sf y_0}$
($\ket{\sf y_1}$) photon from source 2 to input port 2 (4). If
$\ket{\rm vac}$ denotes the state with no photons in the setup, this results in
the conversion $|{\sf x_0y_0}
\rangle \to a_1^\dagger
a_2^\dagger \, \ket{{\rm vac}}$, $|{\sf x_0y_1} \rangle \to a_1^\dagger
a_4^\dagger \, \ket{{\rm vac}}$,
$|{\sf x_1y_0} \rangle \to a_2^\dagger
a_3^\dagger \, \ket{{\rm vac}}$ and $|{\sf x_1y_1} \rangle \to a_3^\dagger
a_4^\dagger \, \ket{{\rm vac}}$.
This conversion should be
realised such that the photons enter the multiport at the same time.
For two-qubit universal gate implementation, $U_{mn}$ is given by
\begin{equation}
U_{mn} = {\textstyle {1 \over 2}} \, {\rm i}^{(m-1)(n-1)}.
\end{equation} In such a case, the multiport is also known as a Bell multiport
which was introduced in Chapter \ref{firework}.
Using Eq.~(\ref{scatter}), one can show that the network transfers
the basis states $\ket{\Phi_i}$, with the choice Eq. (\ref{angelencoding}) as
\begin{eqnarray}
|\Phi_1 \rangle & \to & {\textstyle {1 \over 2}} \, \big( b_1^{\dagger \, 2} -
b_3^{\dagger \, 2} \big) \,
\ket{\rm vac} \, , \,
|\Phi_2 \rangle \to - {\textstyle {1 \over 2}} \, \big( b_2^{\dagger \, 2} -
b_4^{\dagger \, 2} \big) \,
\ket{\rm vac} \, , \nonumber \\
|\Phi_3 \rangle & \to & {\textstyle {1 \over {\sqrt 2}}} \, \big( b_1^\dagger
b_4^\dagger - b_2^\dagger
b_3^\dagger \big) \, \ket{\rm vac} \, , \,
|\Phi_4 \rangle \to - {\textstyle {1 \over {\sqrt 2}}} \, \big( b_1^\dagger
b_2^\dagger - b_3^\dagger
b_4^\dagger \big) \, \ket{\rm vac} \, .
\end{eqnarray}
Finally, detectors measure the presence of photons in each of the possible
output ports. The detection of two
photons in the same output port, namely in 1 or 3 and in 2 or 4, corresponds to
a measurement of the
state $|\Phi_1 \rangle$ and $|\Phi_2 \rangle$, respectively.
The detection of a photon in ports 1 and 4 or in
2 and 3 indicates a measurement of the state $\ket{\Phi_3}$, while a photon in
the ports 1 and 2 or in 3 and 4
indicates the state
$\ket{\Phi_4}$.
\begin{figure}
\begin{minipage}{\columnwidth}
\begin{center}
\resizebox{\columnwidth}{!}{\rotatebox{0}{\includegraphics{moon3.eps}}}
\end{center}
\vspace*{-0.5cm}
\caption{Linear optics networks for the realisation of a measurement of the
basis states $\ket{\Phi_i}$ defined in Sections
(\ref{insurance},\ref{teleportation}) after
encoding the photonic qubits
in the polarisation degrees of two photons (a) or into four different spatial
photon modes (b) involving
either a beam splitter (BS) or a
$4 \times 4$ Bell multiport beam splitter.} \label{moon2}
\end{minipage}
\end{figure}
On the other hand, to implement a parity filter and hence teleportation with
insurance, another symmetric
multiport with $U_{mn}$ given by
\begin{eqnarray*} \label{matrix4}
U &=& {\textstyle {1 \over 2}} \left( \begin{array}{rrrr}
1 & 1 & 1 & 1\\
1 & 1& -1 & -1\\
1 & -1 & 1 & -1\\
1 & -1 & -1 & 1 \end{array} \right)
\end{eqnarray*} should be used. One can again show that the network transfers
the appropriate basis states $\ket{\Phi_i}$ defined in Eq. (\ref{devilencoding}) as
\begin{eqnarray}
|\Phi_1 \rangle & \to & {\textstyle {1 \over 2}} \, \big( b_1^{\dagger \, 2}
- b_3^{\dagger \, 2} \big) \, \ket{\rm vac} \, , \,
|\Phi_2 \rangle \to {\textstyle {1 \over 2}} \, \big( b_2^{\dagger \, 2}
- b_4^{\dagger \, 2} \big) \, \ket{\rm vac} \, , \nonumber \\
|\Phi_3 \rangle & \to & {\textstyle {1 \over {\sqrt 2}}} \, \big( b_1^\dagger
b_2^\dagger - b_3^\dagger
b_4^\dagger \big) \, \ket{\rm vac} \, , \,
|\Phi_4 \rangle \to {\textstyle {1 \over {\sqrt 2}}} \, \big( b_1^\dagger
b_4^\dagger - b_2^\dagger
b_3^\dagger \big) \, \ket{\rm vac} \, .
\end{eqnarray}
The detection of two photons in the same output port, namely in 1 or 3 and in 2
or 4, corresponds to a
measurement of the state $|\Phi_1 \rangle$ and $|\Phi_2 \rangle$, respectively
as in the case for two-qubit phase
gate implementation.
However now, the detection of a photon in ports 1 and 4 or in
2 and 3 indicates a measurement of the state $\ket{\Phi_4}$, while a photon in
the ports 1 and 2 or in 3 and 4
indicates the state $\ket{\Phi_3}$.
We now discuss an interesting feature of 2-photon coincidence measurement which we can exploit.
\subsection{Time-resolved detection for non-identical photon \\ sources}
So far, we have assumed identical atom-cavity systems and therefore, photons
generated from the photon sources
only differ in their encoded degree of freedom (for example, polarisation,
time-bin etc.). Non-identical
atom-cavity systems in general yield different temporal photon pulse shapes as
well as spatial modes and thus introduce an additional
degree
of freedom which allow for the origin of the photon pulse to be determined.
Generally, this leads to errors
in gate implementation as will be seen in this section. In principle, this can
be fixed by
pulse shape engineering if one has full knowledge of the cavity parameters.
Otherwise, a time-resolved
detection technique combined with spatial filters allows us to remove the
which-way information due to non-identical pulse shapes.
Recently, it has been shown how the Hong-Ou-Mandel dip, in which
total photon indistinguishability is normally a necessary requirement, can still
be observed even
with two distinguishable photons, provided one performs time-resolved
postselection \cite{Legero03, Legero04}. Time-resolved postselective detection
is
the essential mechanism that wipes away the which-way information as first
suggested by {\. Z}ukowski {\em et al.} \cite{Zukowski93,Zukowski95}.
It is convenient to start our discussion using the formalism of Legero {\em et
al.} \cite{Legero03,Legero04} with the simple situation
where two
distant atoms, labelled 1 and 2, are entangled with the aid of a
beam splitter. Now, we consider a beam splitter where ports 1 and 2 are the input
ports and ports 3 and 4 define the output ports, each containing an ideal photon detector. Correspondingly, the photon annihilation and
creation operator for port
$j$ and polarisation $i$ in frequency mode $\omega$ is denoted as $b_{ij}(\omega)$
and $b_{ij}^\dagger(\omega)$ respectively. By convention, the $j$th atom is
placed
at the $j$th input port. For each $j$th atom, we assume that its initial state is given by
$c_{0j}\ket{u_{0j}}+c_{1j}\ket{u_{1j}}$. The $j$th atomic state is subsequently
encoded (see Section \ref{encodertrick}) in the photon state as
\begin{equation}
\ket{\psi}_j=\sum_{i,k=0,1}c_{ij}\ket{u_{ij}} \int_{-\omega_b}^{\omega_b}
d\omega
s_j(\omega)\gamma_{ikj}b_{kj}^\dagger(\bar{\omega})\ket{\rm vac}.
\end{equation} such that the orthogonality condition $\sum_{k=0,1}
\gamma_{ikj}\gamma^*_{lkj}=\delta_{il}$ is fulfilled. The coefficients
$\gamma_{ikj}$ are introduced to allow the generated photons to be
transformed by arbitrary single qubit rotation. In addition, we have dropped the
index for photon generation time $\tau$ since it is
inconsequential to our discussion here. We have assumed perfect redundant
encoding, where we have set without
loss of generality, all cavity and laser driving parameters to be independent
of polarisation. In principle, photon generation need not be perfect and there
will generally be terms that do not
contribute to any photon in the total atom-photon state vector. These terms can
be neglected for our discussion as we use 2-photon detection to herald
entanglement, thereby
allowing us to disregard non-photon contributing terms. In contrast, this is not
possible in 1-photon detection protocols. We then define
the total
input state as $\ket{\Psi}_{\rm in}=\ket{\psi}_1\otimes\ket{\psi}_2$. We
further define the unitary transformed total output state, before photon
detection by $\ket{\Psi}_{\rm out}=S\ket{\Psi}_{\rm in}$ where $S$ is a unitary
operator that defines the beam splitter transformation. Specifically, we can set
\begin{eqnarray}\label{transforma}
S^{\dagger}b_{i3}(\omega)S&=&U_{31,i}b_{i1}(\omega)+U_{32,i}b_{i2}(\omega) \, ,
\nonumber \\
S^{\dagger}b_{i4}(\omega)S&=&U_{41,i}b_{i1}(\omega)+U_{42,i}b_{i2}(\omega) \, ,
\end{eqnarray} where $U_{mn,i}$ refers to the probability amplitude of
redirecting a photon of polarisation $i$ from input port $n$ to output port $m$.
Since we are dealing with time-resolved detection, it is convenient to use the
definition of the time-dependent electric field amplitude operator $E_{\lambda
j}^+(t)$ for the $j$th port and polarisation $'\lambda'$ where $b_{\lambda
j}(\omega)=\alpha_{0} b_{0j}(\omega)+\alpha_{1} b_{1j}(\omega)$ and
$|\alpha_{0}|^2+|\alpha_{1}|^2=1$ given by \cite{Ou99a,Legero03}
\begin{equation}
E_{\lambda j}^+(t)=\frac{1}{\sqrt{2 \pi}} \int_{0}^{\infty} d\omega K(\omega)
{\rm e}^{-{\rm i}\omega
t}b_{\lambda j}(\omega).
\end{equation} As in Legero {\em et al.} \cite{Legero03} and Gardiner and Zoller
\cite{Gardiner04}, we choose $K(\omega) \approx 1$ for reasons of
normalisation and the fact that $s_j(\omega)$ is
strongly peaked around $\omega_{\rm cav}$. Accordingly, using Eq.
(\ref{transforma}), the transformation of
the electric field operator is thus given by
\begin{eqnarray}\label{transformE}
S^{\dagger}E_{\lambda 3}^+(t)S&=&U_{31,\lambda}E_{\lambda
1}^+(t)+U_{32,\lambda}E_{\lambda 2}^+(t) \, , \nonumber \\
S^{\dagger}E_{\lambda 4}^+(t)S&=&U_{41,\lambda}E_{\lambda
1}^+(t)+U_{42,\lambda}E_{\lambda 2}^+(t) \, .
\end{eqnarray}
It is convenient to note that for the input ports 1 and 2,
\begin{eqnarray} \label{pulseshape}
E_{\lambda j}^+(t)\ket{\psi_j}&=&\sum_{i,k,l=0,1}\frac{c_{ij} }{\sqrt{2
\pi}}\ket{u_{ij}}\int_{-\omega_b}^{\omega_b} d\omega \int_{0}^{\infty} d\tilde
\omega {\rm e}^{-{\rm i}\tilde \omega t}
\alpha_{l} b_{lj}(\tilde \omega)s_j(\omega) \gamma_{ikj} b_{kj}^{\dagger}(\bar{
\omega})\ket{\rm vac} \nonumber \\
&=& \sum_{i,k=0,1}\frac{c_{ij} \gamma_{ikj} \alpha_{k}}{\sqrt{2 \pi}}
\ket{u_{ij}} {\rm e}^{-{\rm i}\omega_{\rm cav} t}
\int_{-\omega_b}^{\omega_b} d\omega {\rm e}^{-{\rm i}\omega t} s_j(\omega)
\ket{\rm vac}
\nonumber \\
&=& \sum_{i,k=0,1} c_{ij} \gamma_{ikj} \alpha_{k}{\rm e}^{-{\rm i}\omega_{\rm
cav} t} f_j(t)\ket{u_{ij}}\ket{\rm vac}
\end{eqnarray} where
\begin{equation}
f_j(t)=\frac{1}{\sqrt{2 \pi}}\int_{-\omega_b}^{\omega_b} d\omega {\rm e}^{-{\rm
i}\omega t} s_j(\omega)
\end{equation} is the pulse shape of the photon \cite{Duan03} in the $j$th input
given by the Fourier transform of its frequency spectrum
$s_j(\omega)$.
We suppose that a photon is detected at port 3 at time $t_3$ with the
polarisation $'a'$ and port 4 at time $t_4$ with the polarisation
$'b'$. The unnormalised conditional state of the system, as in the formalism by
Legero {\em et al.} \cite{Legero03} and also Gardiner and
Zoller \cite{Gardiner04} is therefore given by
$\ket{\Psi}_{\rm cond}$,
\begin{eqnarray}
\ket{\Psi}_{\rm cond}&=&E_{a3}^+(t_3)E_{b4}^+(t_4)\ket{\Psi}_{\rm out} \nonumber
\\
&=& SS^{\dagger}E_{a3}^+(t_3)SS^{\dagger}E_{b4}^+(t_4)S\ket{\Psi}_{\rm in}
\nonumber \\
&=&
S(U_{31,a}U_{42,b}E_{a1}^+(t_3)E_{b2}^+(t_4)+U_{32,a}U_{41,b}E_{a2}^+(t_3)E_{b1}
^+(t_4))\ket{\Psi}_{\rm in} \, . \nonumber \\
\end{eqnarray}
For the sake of concreteness, we assume a 50:50 beam splitter so that
$U_{mn,a}=U_{mn,b}$ is polarisation insensitive and
$U_{31,k}=U_{41,k}=U_{32,k}=-U_{42,k}=\frac{1}{\sqrt{2}}$. We also set
$\gamma_{ijk}=\delta_{ij}$ for simplicity. We further set $a='0'$ and $b='1'$
and require $c_{ij}=\frac{1}{\sqrt{2}}$.
Using the relations given by Eq. (\ref{pulseshape}), $\ket{\Psi}_{\rm
cond}$ can then be simplified to
\begin{equation}
\ket{\Psi}_{\rm
cond}=\frac{1}{4}(-f_1(t_3)f_2(t_4)\ket{u_{01}}\ket{u_{12}}+f_1(t_4)f_2(t
_3)\ket{u_{11}}\ket{u_{02}})
\end{equation} where we have conveniently dropped all the vacuum terms as well
as the inconsequential global phase factor for
readibility. For $\ket{\Psi}_{\rm cond}$ to describe a maximally entangled singlet state (which we would like to prepare), it is
necessary that the condition,
\begin{equation}\label{condition}
f_1(t_3)f_2(t_4)=f_1(t_4)f_2(t_3)
\end{equation} holds. This condition can be fulfilled unconditionally if the
photon pulse shapes originating from both atoms are similar, i.e.
$f_1(t)={\rm k}f_2(t)$ for some complex constant k. This means the fidelity of
the entangled state
is guaranteed to be unity as long as a photon is detected in both output
ports 3 and 4 irrespective of the time of detection. This is easily explained as
the time of detection does not reveal the origin of the photon given that the
photon pulse shapes are identical. In the case of distinguishable pulse shapes,
one can still fulfil the condition given by Eq. (\ref{condition}) by setting
$t_3=t_4$, thus requiring perfect time-resolved coincidence detection. Note that
this does
not require any preknowledge of the pulse shape in either cavity to achieve
arbitrary high fidelity. This is an illustration of the power of
measurement-based
approach to quantum computation in contrast to a fully coherent-based approach.
Furthermore, since k is an unspecified constant that is
complex, this implies that the introduction of any unknown slowly varying phase
factor (with respect to the photon generation and
detection time) in the path of the photons produces no observable effect on the
fidelity of operation.
So far, we have assumed that the photon pulses from the 2 cavities have the same
central frequency $\omega_1=\omega_2=\omega_{\rm cav}$. Further, suppose
that the 2 cavities
are detuned relative to each other by $\Delta \omega=\omega_1-\omega_2$, with
everything else being identical. This is then equivalent to introducing a
time-dependent phase factor to the pulse shape such that
\begin{equation}
f_1(t)=f_2(t)\exp(-{\rm i \Delta \omega t}).
\end{equation} In this case, pulse similarity can still be obtained for $\Delta
\omega (t_1-t_2)=2n\pi$. This condition was also previously predicted and used
in the context of observing Hong-Ou-Mandel dip for photons of 2 different
central frequencies but identical pulse shape \cite{Legero03,Legero04}.
It is useful to determine analytically, within the
approximation of adiabatic theorem, how the fidelity of entangled state
preparation is degraded for 2 different cavity parameters. Firstly, it is
convenient to recall formula for the pulse shape from Eq. (\ref{magicpulse})
\begin{equation}
f_j(t)=\sqrt{\kappa_j}\sin\theta_j(t)\exp(-\frac{\kappa_j}{2}\int_0^t d\tau
\sin^2 \theta_j(\tau))
\end{equation} where
\begin{equation}
\sin\theta_j(t)=\frac{\Omega_j(t)}{\sqrt{\Omega^2_{j}(t)+g_j^2}}
\end{equation} and $\kappa_j,\Omega^2_j(t),g_j$ are the cavity decay, Rabi
frequency of the driving laser and cavity coupling of the $j$th atom-cavity
respectively. It is then obvious that the below condition
with all other parameters being equal, guarantees pulse shape similarity, namely
\begin{equation}
\left|\frac{\Omega_{1}(t)}{g_{1}}\right|=\left|\frac{\Omega_{2}(t)}{g_{2}}\right
|.
\end{equation} Assuming that this condition is fulfilled, we now consider
$\kappa_1\ne\kappa_2$. The time-dependent fidelity $F(t_3,t_4)=|\bra{\Psi^-}\hat
\Psi \rangle_{\rm cond}|^2$
given in terms
of pulse shape functions is then given by\footnote{$\ket{\Psi^-}=\frac{1}{\sqrt{2}}(\ket{u_{01}}\ket{u_{12}}-\ket{u_{11}}
\ket{u_{02}})$ and $\ket{\hat \Psi \rangle_{\rm
cond}}=\ket{\Psi \rangle_{\rm cond}}/||\,\ket{\cdot}||$}
\begin{equation}
F(t_3,t_4))=\frac{(f_1(t_3)f_2(t_4)+f_1(t_4)f_2(t_3))^2}{2(f_1(t_3)^2f_2(t_4)^2+
f_1(t_4)^2f_2(t_3)^2)} \, .
\end{equation} Following this, it is straightforward to show that the average
fidelity $F_{\rm av}(b,a)$ with detectors integration time
from $t=a$ to $t=b$ is given by
\begin{equation}
F_{\rm av}(b,a)=\frac{1}{2}\left(1+\frac{(\int_a^b dt f_1(t)f_2(t))^2}{\int_a^b
dt f_1(t)^2 \int_a^b dt f_2(t)^2}\right) \, ,
\end{equation} and is related to the overlap between the two pulse shape functions.
It is interesting to also consider the case where the detected
polarisation is the same. In this case, $\ket{\Psi}_{\rm cond}$ is given by
\begin{equation}
\ket{\Psi}_{\rm
cond}=(-f_1(t_3)f_2(t_4)+f_1(t_4)f_2(t_3))\sum_{i=0,1}\alpha_i\ket{u_{i1}}\otimes\sum_{i=0,1}\alpha_i\ket{u_{i1}}.
\end{equation} For pulse similarity condition, this implies that
$||\,\ket{\Psi}_{\rm cond}||^2$ vanishes, which is essentially the Hong-Ou-Mandel
effect.
As an example, we calculate the fidelity and joint probability density of
2-photon coincidence detection for the case of a Gaussian driving pulse with
pulse width $\tau=40/\kappa_2$, being centered in
$20/\kappa_2$, with width $\sqrt{2}\tau/10$,
${\rm max}(\Omega_j^2(t)/g_j^2)=9$ and $t_3$ and $t_4$ normalised to
$\kappa_2^{-1}$.
\begin{figure}
\begin{center}
\psfrag{t3}{$t_3$} \psfrag{t4}{$t_4$}
\includegraphics[scale=0.3]{timeselect5}
\includegraphics[scale=0.3]{timeselect5p}
\end{center}
\caption{(a). Fidelity of entangled state and (b). Joint probability density of
2-photon coincidence detection for $\kappa_1=0.5 \kappa_2$ for a guassian
driving pulse with pulse width $\tau=40/\kappa_2$,being centered in
$20/\kappa_2$, with width $\sqrt{2}\tau/10$ and fixing
$max(\Omega_j^2(t)/g_j^2)=9$.}\label{minsk:timeselect5}
\end{figure}
\begin{figure}
\begin{center}
\psfrag{t3}{$t_3$} \psfrag{t4}{$t_4$}
\includegraphics[scale=0.3]{timeselect7}
\includegraphics[scale=0.3]{timeselect7p}
\end{center}
\caption{(a). Fidelity of entangled state and (b). Joint probability density of
2-photon coincidence detection for $\kappa_1=0.7 \kappa_2$ for the same driving
condition as above.} \label{minsk:timeselect7}
\end{figure}
\begin{figure}
\begin{center}
\psfrag{t3}{$t_3$} \psfrag{t4}{$t_4$}
\includegraphics[scale=0.3]{timeselect9}
\includegraphics[scale=0.3]{timeselect9p}
\end{center}
\caption{(a). Fidelity of entangled state and (b). Joint probability density of
2-photon coincidence detection for $\kappa_1=0.9 \kappa_2$ for the same driving
condition as above} \label{minsk:timeselect9}
\end{figure}
From Fig.~\ref{minsk:timeselect5}, Fig.~\ref{minsk:timeselect7} and
\ref{minsk:timeselect9} it can
be seen that the fidelity is always unity for perfect coincidence. The more
similar the cavities, the greater the tolerance of the
fidelity. In the limit of identical cavities, coincidence photon pairs can be
detected at any time interval with no effect on the fidelity. For comparision, the joint probability density
for 2-photon detection is also calculated and the more different the
cavities are, the lower the probability density of obtaining perfect coincidence generally. One
observes the interesting trend of almost perfect fidelity at photon detection
times at the leading and tail-end of the pulse even at non-perfect coincidence.
This is attributed to the fact that the photon pulse shape at these times is
relatively flat even for 2 non-identical cavities and in no way contributes to
any which-way information. Mathematically, this corresponds to the condition
where $f_1(t) \approx {\rm k}f_2(t)$ at the times $t$ during the leading or
tail-end
of the photon pulses. Of course, the probability density for such coincidence is
low as can be seen from the calculation. Without time-resolved detection, the
average fidelity $F_{\rm av}(\infty,0)$ corresponding to
Fig.~\ref{minsk:timeselect5}, Fig.~\ref{minsk:timeselect7} and
\ref{minsk:timeselect9} is 0.94, 0.98 and 0.99 respectively, consistent
with the fact that dissimilarities of cavities generally yield lower
fidelity.
We have given an example of how a time postselective 2-photon coincidence
detection can yield high fidelity of entangled state preparation. It is
straightforward to extend this to gate operation as described in this chapter.
For
example, one can assume the same setup as before except that now the following
holds
\begin{equation}
\gamma_{001}=\gamma_{011}=\gamma_{101}=-\gamma_{111}=\gamma_{002}=\gamma_{102}={\rm
i}\gamma_{012}=-{\rm i}\gamma_{112}.
\end{equation}
Repeating the same procedure as before and assuming the same detection syndrome with $a='0'$ and $b='1'$,
one can show that the fidelity
of gate operation is maximal when the pulse similarity condition is fulfilled.
\section{Conclusions and discussions }
We have shown that despite the fact that linear optics based Bell-state measurements on
photons are incomplete, that does not
prevent the deterministic implementation of a gate between distant qubits.
Therefore, rather surprising, we
show that it is not necessary to demonstrate deterministic entanglement
generation in order to achieve a
universal two-qubit gate with unit efficiency\footnote{The insurance aspect of this
scheme allows processing of a single copy of unknown
input state in principle. In contrast to cluster state implementation of
two-qubit gates, the non-deterministic teleportation of an
unknown unencoded input state into the
cluster without insurance can destroy the input qubit if the teleportation fails.}.
Furthermore, this scheme is intrinsically
interferometrically stable if one uses polarisation
encoding and measurement due to the same reason as mentioned in
Chapter~\ref{firework}.
In the real world, nonideal situations of photon loss, inefficient photon detectors and photon
generation
would all lower the gate
probability of success. Indeed, given photon number-resolving detectors (with no
dark counts) of quantum efficiency $p_d$ each, the
probability of single-shot gate failure
$p_F$, when two photons are not detected, is given by $1-p_d^2$. One can also
account for photon loss in the factor $p_d$.
The solution towards fault-tolerant quantum computation in this scheme is to
make a cluster
state of high fidelity. High fidelity can be achieved because our scheme is a
2-photon heralded scheme. In the case
where only one or no photon is detected, this is equivalent to tracing out the
photon degrees of freedom.
This effect can be removed by simply
destroying the two qubits by measuring them in the computational basis in which an
attempted cluster bond is
required. This does not decrease
the fidelity of the cluster state. Therefore, in the presence of imperfections
such as
photon loss and inefficient photon detectors, distributed quantum computation
can still be performed with high
fidelity by building a cluster state of distant qubits. Further discussions on
cluster state buildup can be found in the work by Barrett
and Kok \cite{Barrett04} and Lim {\em et al.}\footnote{Details of efficient cluster state buildup based on the insurance scenario described in this chapter can be found here.} \cite{Lim05}, and they lie out of the scope of this thesis. In
addition, Benjamin {\em et al.} \cite{Benjamin05} have also
recently argued how the insurance scenario in our
scheme can lead to a higher efficiency in building cluster states. We now proceed to the
next chapter where we highlight an important application of the results of this chapter on generating photon
entanglement on demand.
\chapter{Summary and Outlook}\label{outlook}
The work of this thesis
demonstrates various closely related aspects of quantum information processing
with
single photons as motivated in Chapter \ref{overview}. Hopefully, it adds new
perspectives to the relationship between single
photons and their sources and the implication to quantum information processing
in general. The summary
of the main work is as follows.
In Chapter \ref{firework}, we showed that a wide range of highly entangled
multiphoton states, including {\em W}-states, can be prepared
by interfering {\em single} photons inside a Bell multiport beam splitter and
using postselection. The described setup, being photon
encoding independent can be used to generate polarisation, time-bin and
frequency encoded multiphoton entanglement, even when using only a single
photon source. The success probability has a surprisingly non-monotonic
decreasing trend as the number of photons increases.
In Chapter \ref{fusion}, we demonstrated how the HOM dip can be generalised
to multiphoton coincidence detection in multiport
beamsplitters. We considered the canonical symmetric Bell multiport and show
that the HOM dip can be observed for all $N \times N$ Bell
multiports where $N$ is even but not necessarily when $N$ is odd. Note that this observation
applies generally to all bosons, of which photons
are an example, thus having wide applicability. For the sake of completeness, we
also discussed multifermionic scattering through a
multiport and showed that identical fermions always leave the output
ports of the multiport separately.
In Chapter \ref{hummingbird}, we proposed a scheme for implementing a
multipartite quantum filter that uses entangled photons as a
resource. It is shown that the success probability for the 2-photon parity
filter can be as high as ${1 \over 2}$, which is the highest
that has so far been predicted without the help of universal two-qubit quantum gates.
Furthermore, the required number of ancilla photons is the
least of all current parity filter proposals. Remarkably, the quantum filter
operates with probability ${1 \over 2}$ even in the
$N$-photon case, regardless of the number of photons in the input state.
In Chapter \ref{minsk} we described the efficient implementation
of eventually deterministic two-qubit gate operations between single photon
sources, despite the restriction of the no-go theorem
on deterministic Bell measurement with linear optics. No entangled ancilla
photons and photon-feed into cavities are needed.
The key principle for our approach is based on source encoding to the photon
that is generated as well as measurements in a mutually
unbiased basis with respect to the computational basis. The described approach
is highly general and lends wide implementation to various
types of single photon sources. Furthermore, the scheme is still robust even in
the case of dissimilarities of the photon sources, a
testament to the unique character of a measurement-based approach to quantum
computing. Our approach also gives fresh perspectives on the
use of mutually unbiased basis in quantum computation, besides existing
applications in quantum cryptography and for solving the Mean King's problem.
In Chapter \ref{demand}, we used ideas from Chapters \ref{firework} and
\ref{minsk} to show how multiphoton entanglement on demand can be
realised. Generally speaking, any multiphoton qubit state can be generated on
demand in a distributed manner. At the same time, we also
relate a duality relation between preparing photon entanglement and atom
entanglement which may lead to new perspectives in multiport designs for
entanglement generation.
In Chapter \ref{photon}, we showed, using a setup closely resembling a Young
double-slit experiment, that dipole-dipole as well as
2-photon entanglement can be generated with photons emitted from two distant
dipole sources in free space (i.e. without the aid of
linear optics setup). The scheme is highly robust to the dipole excitation
imperfections. In the case of two sources, the
entanglement arises under the condition of two emissions in certain spatial
directions and leaves the dipoles in a maximally
entangled state. This work adds new perspectives to current views on the entanglement
generation using measurements.
The outlook and possible extensions to the work in this thesis are manifold. For
example, the work in Chapter \ref{firework} and \ref{fusion} is mainly
restricted to Bell multiports due to a cyclic symmetry which
we exploited. It is interesting to examine a greater variety of multiports
defined by redirection or transfer matrices of various symmetries and their
implications on multiphoton scattering and entanglement generation. Here, we have
confined ourselves to analysing pure states and have not considered
photon mixed-states for simplicity. This might lead to applications like the
characterisation of photons or multiports. We also did not consider EPR photon
pairs as possible inputs to multiports which may yield
exciting possibilities in multiphoton entanglement generation with higher
probability of success. One could also extend this
work to the investigation of POVMs with detectors and multiports. An extension
of Chapter \ref{fusion} may find application in
experiments with particles with exotic statistics such as anyons\footnote{I thank Vlatko Vedral for stimulating discussions.}. Finally, multiports with weak
nonlinearities may yield interesting prospects in enhanced multiphoton state
preparation due to cooperative enhancement\footnote{I thank Jim Franson
for stimulating discussions.}.
In Chapter \ref{hummingbird}, we have dealt with a simple setup for a
multiphoton filter. It is interesting to see how this can be
extended to arbitrary multiphoton gates and if the probability of success could be
increased by combining approaches using an $N \times N$
multiport. With the aid of an arbitrary photon ancilla, generated perhaps by a
multiphoton source on demand, one might be able to
implement a programmable multiphoton gate. Further extensions to this work,
hinted by the duality relation obtained in Chapter
\ref{demand}, might lead to a multiatom filter implementation.
Chapter \ref{minsk} presents possible extensions to higher dimensional quNit
operations or
direct multiatom gate implementation. This might be
implemented with the aid of a linear optics multiport. For example in this
chapter, we have
already used multiports for measurements leading to useful
gate implementations. These techniques could be extended to new and
interesting results. An intriguing observation of the choice of
mutually unbiased basis used in this chapter yields, on suitable rearragement of
the coefficents of the computational basis, a $4 \times
4$ Fourier transform matrix, which defines a Bell multiport\footnote{I thank Thomas Durt for
bringing this to my attention.}. The relationship
between mutually unbiased basis and
multiports may be worth investigating.
The work of Chapter \ref{photon} may in
principle be extended to multiple dipoles or various energy level structures
and
by taking into account of dipole-dipole
interaction or various means of dipole excitation. So far, we have restricted
ourselves to the simplest case of two initially excited
dipoles which is experimentally reasonable. Spatial polarisation correlations or
intensity correlations may, for example be exploited
for certain search tasks as demonstrated by Agarwal {\em et al.}
\cite{Agarwal04}.
In this thesis, we have confined ourselves to discrete quantum information
processing with photons and have said nothing about the
equally rich field of continuous variable processing with photons or even a
hybrid field of discrete-continuous variables. It may be that
the work here can be extended to such domains and might lead to analogous applications.
Finally, although we have focused our attention to single photons, many parts of
this thesis may find analogous applications in other
flying qubits such as electrons, which in contrast to photons, have
fermionic statistics. It is noteworthy that many linear optical
operations on photons can also be implemented on electrons. At the same
time, there exist an intriguing prospect of setups, such as
doped fibers, that modify the quantum statistics of the photons. One might
envision new capabilities of quantum information processing
with single photons using multiport setups consisting of such doped fibers in
the future. It is hoped that the work in this thesis adds to the overall
development as well as inspiring new research in quantum information processing.
\chapter{Introduction}\label{overview}
Photons are natural carriers of quantum information owing to their long
lifetimes and ease of distribution and this constitutes the main motivation
for this thesis. In quantum information processing, entanglement plays its role in
diverse applications such as quantum cryptography, implementation of universal
quantum gates, tests of non-locality, and is prevalent in
all known quantum algorithms that provide an exponential speedup compared to
classical algorithms. Entanglement, a still elusive concept, is strictly defined as the situation where a quantum state cannot be
decomposed into a convex sum of tensor product density matrix states. The ability to generate or manipulate entanglement is thus a
key ingredient to
quantum information processing. In this thesis, we focus primarily on
various aspects of entangled state generation, detection, manipulation and
exploitation for quantum information processing using single
photons. We
consider novel means of how single photon sources can be manipulated through the
photons they emit and vice-versa the way photons can be
manipulated with the aid of single photon sources. Furthermore, as will be
seen, these two apparently different tasks can often be
exploited for each other. Therefore, an alternative title to this thesis could well be ``Photon
assisted quantum computation". We start here by giving a
brief overview to quantum information processing bringing single photons into
a general context. A more detailed survey of the
research done in this thesis can be found in the relevant chapters.
\section{Brief Introduction to quantum information processing and single
photons}
Quantum information processing is a remarkably diverse and interdisciplinary
field. In the words of Knill and Nielsen \cite{Knill02}, it is ``The science
of the theoretical, experimental and technological areas covering the use of
quantum mechanics for communication and computation." It includes quantum
information theory, quantum communication, quantum computation, quantum
algorithms and their complexity and quantum control. In general, these
fields are not mutually exclusive and often have substantial overlaps. It is therefore somewhat artificial to attempt to classify them as
separate subfields.
Early ideas of quantum information processing began with Feynman, who considered
the question of efficient
simulation of a quantum
system \cite{Feynman82}. He speculated that the only efficient
simulation of a quantum system that
could be achieved would come from another quantum
system. Following that, Deutsch and Jozsa \cite{Deutsch85,Deutsch92}
demonstrated
the existence of quantum algorithms that are more efficient than classical
algorithms. Later, Shor \cite{Shor94}, building on the work of Simon
\cite{Simon94} as well as Deutsch and Jozsa, demonstrated a quantum
algorithm for prime factorisation that is exponentially faster than any known
classical algorithm. Both the Shor and Deutsch-Jozsa algorithm, as well as the
Simon algorithm are actually special cases of the more general algorithm for the
problem known as the Hidden Subgroup Problem(HSP). In fact, all known algorithms
belonging to HSP class, at least for the case of finite Abelian groups, are exponentially more efficient than the best known
corresponding classical algorithms. Finally, Grover \cite{Grover96} demonstrated a
fundamentally different algorithm that is $\sqrt{N}$ faster than the best known
classical algorithm for an $N$ element database search without any partial
information. Entanglement appears to be necessary for
quantum algorithms that yield exponential speedups compared to classical
algorithms \cite{Linden01,Harrow03}.
The basic logical unit in each of these algorithms are so-called qubits,
which hold the quantum information. Each of these qubits can be in any
superposition between two orthogonal logical states, denoted $\ket{0}$ and
$\ket{1}$, constituting the computation basis states,
in analogy to classical bits `0' and `1' in classical computing. For example, a qubit $\ket{\psi}$ can be written as a state vector in a 2-dimensional
Hilbert space given by $\ket{\psi}=a\ket{0}+b\ket{1}$ where $a$ and $b$ are arbitrary normalised complex coefficients. In contrast, a
classical bit can only be in the state `0' or `1'. In addition, there are many possible physical realisations of a qubit. For
example, the computational basis states can be the Zeeman ground states of atoms, the direction of the spin of electrons or polarisation
states of photons. The coherent evolution of many qubits which can be in an
arbitrary superposition\footnote{Note that with $N$ qubits, the Hilbert space spans an exponentially large dimension given by $2^N$.} can be
thought as a mechanism that enables massive
parallelism in the computation, hence leading to a possible exponential speedup
compared to classical
computation. At the same time, any $N$-qubit unitary
operation can be decomposed to two-qubit unitary operations \cite{Barenco95}. It is important to
note that there exist two-qubit gates, which together with arbitrary single qubit
operations, can simulate any two-qubit unitary operation
\cite{DiVincenzo95}.
Any two-qubit gate fufilling the above universality criterion is known as a universal two-qubit gate. Notable examples of such gates are
CNOT
and CZ gates, which perform a controlled non-trivial single qubit unitary
operation on a target qubit
dependent on the state of the control qubit. Specifically, given two input qubits, a control and target one, the CNOT operation flips
the target qubit only if the control qubit is in the logical state $\ket{1}_{\rm c}$. Similarly, the CZ gate yields a negative sign to
the target qubit $\ket{1}_{\rm t}$ only if the control qubit is also in the state $\ket{1}_{\rm c}$.
\begin{figure}
\begin{minipage}{\columnwidth}
\begin{center}
\vspace*{0cm}
\psfrag{Psi}{$\ket{\psi}$}
\psfrag{Esi}{$\ket{\Psi^+}$}
\psfrag{HPsi}{$H\ket{\psi}$}
\psfrag{H}{$H$}
\psfrag{HPH}{$HPH$}
\resizebox{\columnwidth}{!}{\rotatebox{0}{\includegraphics{teleportation}}}
\end{center}
\vspace*{0cm}
\caption{Teleportation of a unitary operation such as a Hadamard gate $H$ over the input state $\ket{\psi}$. The action of a Hadamard
gate is defined by $H\ket{0}=\ket{+}=\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$ and $H\ket{1}=\ket{-}=\frac{1}{\sqrt{2}}(\ket{0}-\ket{1})$. The entangled ancilla $\frac{1}{\sqrt{2}}(\ket{0+}+\ket{1-})$ is given by
acting $H$ on $\ket{\Psi^+}=\frac{1}{\sqrt{2}}(\ket{00}+\ket{11})$. A Bell measurement between one of the qubits of the entangled
ancilla and the input state $\ket{\psi}$ is performed. $P$ consists of a local unitary operation that depends on the measurement syndrome
deriving from the Bell measurement. After measurement and the operation $HPH$, the teleported state becomes $H\ket{\psi}$. Note that
ordinary teleportation is given by replacing the gate $H$ with an identity operator.} \label{tele}
\end{minipage}
\end{figure}
In addition, application of these two-qubit gates to a suitable product state can entangle
the input state and we shall now make a short detour to illustrate some basic properties of an entangled state. For example, the state
$\frac{1}{\sqrt{2}}(\ket{10}-\ket{11})$ is not entangled because it can be written as a tensor product of two qubit states given by
$\frac{1}{\sqrt{2}}(\ket{0}-\ket{1})\otimes\ket{1}$. We now treat this as an input state where the left qubit is treated as the control
and the right qubit is treated as the target. The application of a CNOT gate to this separable state yields a
particularly interesting maximally entangled bipartite state, known as the singlet state $\ket{\Phi^-}$ given by
$\ket{\Phi^-}=\frac{1}{\sqrt{2}}(\ket{01}-\ket{10})$ where the control and target subscripts have been dropped for clarity. This state is
invariant under any qubit rotation applied equally to the two qubits. This means that in the basis given by $\ket{0'}$ and $\ket{1'}$,
the state $\ket{\Phi^-}$ is again given by $\frac{1}{\sqrt{2}}(\ket{0'1'}-\ket{1'0'})$. The two parties each holding a qubit will
always measure different basis states, no matter what common basis states they share. We now try to construct a two party non-entangled
state that might yield a similar measurement syndrome. For example, the non-entangled mixed-state $\rho_{\rm no
ent}=\frac{1}{2}(\ket{01}\bra{01}+\ket{10}\bra{10})$ will no doubt yield positive correlations of different states in the measurement basis
$\ket{0}$ and $\ket{1}$. Unfortunately, this will no longer be true in another measurement basis, say
$\ket{\pm}=\frac{1}{\sqrt{2}}(\ket{0}\pm\ket{1})$. This shows that an entangled state can have stronger correlations than is possible
compared to a non-entangled state. Due to the existence of these special correlations, a bipartite entangled state for example, cannot be thought of
as two separate parties. We give briefly an example of the exploitation of this correlation. Suppose we have an input state given by
$\ket{\psi}=\alpha\ket{0}+\beta\ket{1}$ together with an ancilla $\ket{\Phi^-}$. We can then write the tripartite state
$\ket{\psi}\ket{\Phi^-}$ as $\alpha\ket{001}-\alpha\ket{010}+\beta\ket{101}-\beta\ket{110}$ omitting the normalisation factor. If a projective
measurement is performed on the first two parties such that a maximally entangled state say
$\ket{\Psi^+}=\frac{1}{\sqrt{2}}(\ket{00}+\ket{11})$ is detected (an example of a Bell measurement), this immediately projects the third
qubit to the state $\alpha\ket{1}-\beta\ket{0}$, which is local unitary equivalent to the original input qubit $\ket{\psi}$. We have
therefore transferred the input state by measurement to one of the qubits in the entangled ancilla using the correlation found in $\ket{\Phi^-}$ as well in $\ket{\Psi^+}$.
This is also known as teleportation \cite{Bennett93}.
We have earlier defined what entanglement is by saying what it is not! Although
we have already some limited success on entanglement measures (see Ref. \cite{Plenio05} and references therein) and criteria
\cite{Horodecki96,Peres96} to help us establish whether a state is entangled or not, the full understanding of what entanglement really is remains
elusive\footnote{This is the author's personal perception.}.
Returning to the discussion on universal gates, one might assume that such gate
operations should be accomplished by coherent means, for example, with a controlled evolution of the Hamiltonian governing interactions
between qubits possibly with an external control agent, such as a laser beam, with the Cirac-Zoller gate for trapped ions
\cite{Cirac95} as a famous example.
This is however too restrictive and it is worth commenting briefly on
approaches which use entangled
resources to simulate universal gates with a measurement-based approach instead
of using purely coherent
evolutions. These approaches may be important for a future scalable quantum
computing implementation. Notable examples
are teleportation-based \cite{Gottesman99}, and cluster
state \cite{Briegel01,Raussendorf01,Raussendorf03} approaches. Both approaches require the preparation of a highly entangled ancilla
which
subsequently acts as a useful resource for quantum computation. The basic philosophy of the measurement-based approach is to bury all the
``difficult" quantum operations in the offline preparation of the entangled ancilla. Quantum computation then proceeds by measurement,
which is hopefully an easier operation. Generally, to simulate any $N$-qubit operation by these approaches, we require at least two-qubit
interactions or gates for the preparation of the entangled ancilla. In particular, cluster state approaches allow for
universal
quantum computation without the need of coherent qubit to qubit interaction once the cluster state\footnote{A cluster state is prepared
for example by first initialising a lattice arrangement of qubits in the state$\frac{1}{\sqrt{2}}(\ket{0}+\ket{1})$. A CZ gate is then
performed between each nearest neighbour to form the cluster state.} has been
prepared. Appropriate single qubit measurements in a cluster state allows for any quantum algorithm to be simulated. This was proven by
Raussendorf {\em et al.} \cite{Raussendorf03} by exploiting the correlations found in the cluster state, which is a highly entangled one. These measurements destroy the entanglement of the
cluster state and hence, the cluster state is not reusable. Therefore, the term ``one-way quantum computing" is used interchangeably with
cluster state quantum computing. In the
so-called teleportation based approach, the desired unitary operation is ``teleported" onto an output state with the help of a suitably
prepared entangled resource and Bell measurements (see (\ref{completeBellintro})). Refer to Fig.~\ref{tele} for an example of the teleportation of a single
qubit unitary operation. Note that this can be extended to any multiqubit unitary operation. For a general discussion of the measurement-based
approach, see Ref. \cite{Aliferis04,Childs05}.
There is however another kind of approach that seems to share properties of both the coherent and measurement-based approaches. Examples
are given in \cite{Beige00a,Franson04} where a Zeno-type measurement induces a coherent evolution. A Zeno effect can
be understood as the process of halting an evolution based on continuous strong measurements. This is a very useful tool to freeze
undesired evolution. Applied to cavity QED \cite{Beige00a}, an environment induced Zeno-type effect suppresses the cavity decay, that would usually decohere the system. Applied to photons, \cite{Franson04} the Zeno effect can prevent the undesired 2-photon occupation, associated
with a failure event, in a doped fiber with a very large 2-photon absorption cross-section and with negligible 1-photon absorption cross-section. Therefore, it
is a special kind of ``deterministic" postselection.
In parallel to these developments, came the invention of quantum error correction codes by Calderbank, Shor and Steane (CSS)
\cite{Shor95,Calderbank96, Steane96}. It was initially thought that this was impossible due to the notion that quantum states are
fragile, characterised with a continuous degree of freedom and generally subjected to noise of continuous nature which leads to
decoherence.
Furthermore, the quantum no-cloning theorem \cite{Wootters82} ruled out the naive method of state copy to combat against noise, as often
used in classical communication and computation. CSS however showed that quantum error correction was possible with the help of encoding
operations and the measurement of error syndromes. This important result led to the concept of fault-tolerant quantum computation where
one
can asymptopically approach error-free computation with suitable encodings and error corrections provided that the error probability of
gates do not exceed a certain threshold \cite{Gottesman98}.
Therefore, a lot of effort both experimentally and theoretically, has been
focussed on the physical implementation of universal two-qubit gates.
General criteria
for a scalable quantum computing system were formulated by DiVincenzo
\cite{DiVincenzo00}. Note that this criteria, based on the conventional gate model for quantum computation, have been formulated before
the recent development of new paradigms of quantum computation, such as measurement-based approaches to quantum computation or even
hybrid models. A relook of this criteria may be timely. To date,
gate implementation has been implemented using NMR techniques on a molecule (perfluorobutadienyl iron complex) \cite{Vandersypen01} where
a seven-qubit
Shor's algorithm for the prime factorisation of the number 15 was demonstrated.
In
trapped ions, the Cirac-Zoller gate \cite{Schmidt-Kaler03}, a geometric two-ion phase gate \cite{Leibfried03} the Deutsch-Jozsa
algorithm \cite{Gulde03}, determinstic teleportation of ions \cite{Barrett04a,Riebe04}, quantum error correction \cite{Chiaverini04} as
well as a semi-classical quantum Fourier transform \cite{Chiaverini05} has been demonstrated. These systems consist of qubits which are
stationary with a possibly long decoherence time which makes them suitable as quantum memories. On
the other hand, disadvantages of using stationary qubits alone include
the requirement for precise coherent control. Furthermore, interaction with
remote
stationary qubits is difficult.
Alternatively, single photons, generally loosely thought of as a single excitation in the electromagnetic field, are natural flying
qubits with
long decoherence time (compared to gate operations) and are useful for the distribution of quantum
information. At optical frequencies, the background photon count rate is virtually zero. Furthermore, photons are bosons and they obey
the following
commutation rules,
\begin{equation}
\left[a_i, a_j^\dagger \right]=\delta_{ij}\, \, , \left[a_i, a_j
\right]=\left[a_i^\dagger, a_j^\dagger \right]=0
\end{equation}where $a_i$ $(a_i^\dagger)$ is the photon annihilation (creation)
operator for a certain mode $i$,
$\left[\hat{a},\hat{b}\right]=\hat{a}\hat{b}-\hat{b}\hat{a}$ and $\delta_{ij}=1$
for
$i=j$ or $0$ otherwise. Photons can in general be described in various encodings
or degree of freedom,
such as polarisation, spatial
or frequency, or even angular momentum. For example, in polarisation
encoding, one can assign the logical
qubit $\ket{0}_L$ and $\ket{1}_L$ to any two orthogonal polarisations, such as
the horizontal and vertical
polarisations.
Single qubit operations for photons
are extremely easy \cite{James01,Englert01} to implement with waveplates, polarisation rotators etc. However, there exists practically no coupling
between photons
in vacuum and hence
a two-qubit gate implementation between photons is difficult, which is one of the
reasons why photons are so stable. Indeed, an early proposal \cite{Milburn89} of a photonic universal three-qubit conditional SWAP gate, known as the Fredkin gate, requires
Kerr nonlinearity to produce intensity-dependent phase shifts. The Kerr nonlinearity required is extremely huge\footnote{See Ref. \cite{Turchette95} for a proof-of-principle demonstration with cavity QED.} if operation at the single photon level is required, which pose a severe experimental challenge. One of the early explorations of how quantum logic can be simulated (inefficiently and requiring exponential
resources) with linear optical elements alone is found in the paper by
Cerf {\em et al.} \cite{Cerf98}. The word ``linear optics''\footnote{This definition would certainly include squeezing which is not part
of the standard linear optical quantum information processing toolbox. We do not have to include squeezing in this thesis, although weak
squeezing with photon detectors can result in a heralded single photon source. The linear optical quantum information processing toolbox
we consider consist only of photon sources, detectors, beam splitters and phase plates. } is defined in the sense in
which the
Hamiltonian that describe the photon transformation has only at most quadratic terms in photon creation or destruction operators. In this
way, the resulting Heisenberg equations of motion are linear in terms of photon creation or destruction operators. Cerf {\em et al.}'s
scheme is however not generally applicable to quantum computation with different photons as it operates on a Hilbert space of two degrees
of freedom(polarisation and momentum) on the same photon instead of different photons.
Following that, a very important no-go theorem by L{\"u}tkenhaus {\em et al.} \cite{Lutkenhaus99} showed that complete Bell state
measurement with unit efficiency is impossible with linear optics resource alone, despite having ancillas and conditional measurements as
resources. Note that their work covers the case where the Bell state is defined with two photons regardless of the type of encoding,
which applies generally to quantum
computation with different photons. Further work
\cite{Calsamiglia01} (see also related work by Vaidman and Yoran \cite{Vaidman99}) in this direction led to the result that given no
ancillas
as resources, linear optics-based Bell measurement yields a success probability of at most $50 \%$ (see Chapter \ref{minsk} where such an
example is given).
We define, without loss of generality, the basis
states of a complete Bell measurement as
\begin{eqnarray} \label{completeBellintro}
&&\ket{\Phi^\pm}=\frac{1}{\sqrt{2}}(a_{1,h}^\dagger a_{2,v}^\dagger \pm
a_{1,v}^\dagger a_{2,h}^\dagger ) \ket{0}_{\rm vac} \, , \nonumber \\
&&\ket{\Psi^\pm}=\frac{1}{\sqrt{2}}(a_{1,h}^\dagger a_{2,h}^\dagger \pm
a_{1,v}^\dagger a_{2,v}^\dagger ) \ket{0}_{\rm vac} \, .
\end{eqnarray} Here, $a_{i,\lambda}^\dagger$ refers to a photon creation
operator for spatial mode $i$ with polarisation mode $\lambda$. Bell states, which are maximally entangled two-qubit states, provide
quantum correlations which feature as a crucial ingredient in many aspects of quantum information processing such as teleportation,
entanglement swapping etc..
Later, the seminal
paper by Knill {\em et al.} \cite{Knill01} demonstrated that quantum
computing can be implemented efficiently (i.e. with polynomial resource) with
photons and linear optics elements
if one has deterministic single photon sources with perfect photon-resolving
detectors. They proposed a photon nonlinear gate operation based on photon interference in a linear optics setup together with
postselection. Their scheme also makes use of a teleportation based approach \cite{Gottesman98} with the help of Bell state measurements.
They managed to
approach near $100 \%$ efficient Bell measurement with the aid of asymtopically
large number of highly entangled photons without contradicting the no-go theorem of L{\"u}tkenhaus {\em et al.} Franson {\em et al.}
\cite{Franson02} have subsequently improved this scaling tremendously, with feedforward corrections, from the failure rate of $1/n$ to $1/n^2$ where $n$ is the number
of ancilla photons. Probabilistic gate operations, based on Ref. \cite{Knill01} with some clever improvements,
between photons have since been
demonstrated
experimentally \cite{Brien03,Pittman03,Gasparoni04,Zhao05} and serve as a testbed for
quantum computation.
Unfortunately, approaches using purely photon and
linear optics alone seem to require huge practical resources for scaling even if
they are polynomial \cite{Scheel03,Scheel04a,Eisert05}. In principle, this can be alleviated through a photonic
cluster state computation model in which the cluster state
can be built in an efficient manner \cite{Browne05,Nielsen04}. The cluster state than serves as a universal palette for any quantum
computation that should proceed by measurement with unit efficiency in
principle. A recent working demonstration of a postselected 4-photon cluster state quantum computation is found in Ref. \cite{Walther05}.
It is however still
necessary to implement photon memory and this is currently still a great experimental challenge.
Going by a different thread from the usual linear optics quantum computation, it has been recently shown that
relatively weak, but non-zero Kerr nonlinearity
\cite{Munro05,Nemoto04} is sufficient for implementing universal gates between photons with unit efficiency. The surprising thing is that
one does not really need strong Kerr nonlinearity for this. The trick is to use a homodyne measurement with an intense coherent state
source to compensate for the weak nonlinearity. This promising approach has many applications useful to photonic based quantum
computation. Besides implementing photonic gates with unit efficiency, it can be used as a photon counting non-demolition
measurement or to turn a weak coherent pulse into a heralded single photon source.
One might envision a hybrid approach using the best properties of both
stationary and flying qubits (photons)
which is a key feature in this thesis. Motivations of such hybrid
approaches have been first considered by Van Enk {\em et al.} \cite{Enk97,Enk98} in quantum
networking, where information can be
sent to distant nodes via flying qubits between stationary nodes consisting of stationary qubits. The stationary qubit (for example, atoms
or ions) function as a qubit with long decoherence time as well as acting as a quantum memory. Such an approach opens the possibility
of
distributed quantum computing.
The basic
component of such a network requires stationary qubit to flying qubit interfaces
which is commonly found in cavity QED and atomic ensemble implementations.
Parallel developments in the field of quantum communication which essentially
involves the exchange of classical or quantum information through classical and
quantum channels includes quantum cryptography, teleportation, distributed quantum computation
etc. In the field of quantum cryptography, also widely known as quantum key
distribution, protocols such as BB84, Ekert \cite{Bennett84,Ekert91} show the
possibility of two parties establishing a secret key with no possibility of an
eavesdropper being able to share any part of the secret key. The main principles
used are the quantum no-cloning theorem and the fact that a
measurement of a state
generally disturbs the original state. The eavesdropper attempting to learn anything of the secret key necessarily reduces the measured
correlation observed between the two rightful parties, Alice and Bob. Such an observation signals the presence of a possible eavesdropper
if the correlation is below a certain bound. Again, due to their nature of being
flying qubits, all experiments to-date
implementing quantum key distributions
involve photons \cite{Peng05,Kurtsiefer02,Gisin02}. Particularly, Ekert's
protocol requires the preparation of an
entangled pair of photons. Related to Ekert's protocol is the so-called Bell's
inequality violation test \cite{Bell65,Clauser69}. This is a deep test for
ruling out a local hidden variable theory that can make predictions similar to
quantum mechanics. Such a test involves the repeated preparation of an entangled pair of particles followed by independent measurements
on each of the qubits to obtain a statistical correlation function. All local hidden-variable theories will yield a bound in the
correlation function. According to quantum mechanics, this bound can be violated.
The violation has been widely demonstrated\footnote{There exist two loopholes applying to experiments demonstrating the violation
of the Bell's inequality. One is the lightcone loophole that would still allow a possible local realistic interpretation. The other is
the detection loophole where the whole ensemble may not violate the Bell's inequality although the detected subensemble is perceived to
violate it. To date, there has been no experiments that closes both loopholes.} for the case where at least one of the particles is a
photon. For that of two photons, the violation has been observed from atomic cascade emission \cite{Aspect82} as well from spontaneous
parametric downconversion \cite{Ou88}. Particularly interesting,
the
experiment performed by Blinov {\em et al.} \cite{Blinov04,Moehring04} demonstrated
entanglement between an ion and a photon or in other words, a stationary and a
flying qubit. They also demonstrated for the first time, a Bell inequality violation between particles of different species, namely an
atom and a photon. This provides a building block to
distributed quantum computation between distant ions assisted by photons.
Teleportation also plays an especially important role in quantum communication. Augmented with quantum repeaters \cite{Briegel98} based
on entanglement purification
\cite{Bennett96a}, states can be transferred with high fidelity through teleportation with a robustly created perfect entangled ancilla.
Experiments
with photons over long distances have also been performed
\cite{Ursin04,Riedmatten04} further illustrating the use of photons as an
information
carrier.
As attractive as it is to use single photons in quantum information processing, five major sources of decoherence and errors are
relevant.
They are interferometric stability, mode matching (spatial and temporal), photon loss as well as detector accuracy and efficiency.
Various aspects of these issues will be addressed in this thesis although we do not claim to fully resolve all these issues.
We have also seen in this section how important entanglement generation and manipulation of single
photons is to the field of quantum information processing.
This short introduction, in which we have not discussed those aspects of quantum information theory which are out of the scope of this
thesis, obviously
does not do justice
to the wide field of quantum
information processing. The interested reader is invited to refer to the
book by Nielsen and Chuang
\cite{Nielsen00} for an excellent exposition.
\section{Thesis Overview}
The central theme of this thesis
is the manipulation and preparation of qubits (be it stationary or flying qubits) with single photons. The bulk of the
research work based on this theme is
described from Chapters \ref{firework} to
\ref{photon} and a brief overview is given as follows.
In Chapter \ref{firework}, we show that a wide range of highly entangled
multiphoton states, including {\em W}-states, can be prepared
by interfering {\em single} photons inside a Bell multiport beam splitter and
using postselection. Multiphoton entanglement being an
important resource for linear optics quantum computing motivates the work in
this chapter. The results that we obtain is photon encoding independent and thus
have wide applicability. We perform further studies on the multiport in the next
chapter for a different application.
In Chapter \ref{fusion}, we study an important aspect of multiphoton
interference, namely, the generalised Hong-Ou-Mandel(HOM) effect
that plays a crucial role to many aspects of linear optics based quantum
computation with photons. The famous HOM dip for two photons,
where two identical photons entering separate input arms of a 50:50
beamsplitter never exit in separate output arms, plays an important role
in quantum information processing such as the characterisation of single photon
sources, Bell measurements etc. Here, we present a new generalisation of
the HOM dip for multiparticle scattering through a multiport.
In Chapter \ref{hummingbird}, we propose a scheme for implementing a
multipartite quantum filter that uses entangled photons as a
resource. Such filters have applications in the building of cluster states and
are shown
to be universal. The scheme that we propose is highly efficient and uses the
least resources of all comparable current schemes.
In Chapter \ref{minsk}, we describe an architecture of distributed quantum
computing that can be realised with single photon sources
without the need of highly entangled ancilla states. The ability to perform
gate operations between arbitrary qubits, and not only
between next neighbours, yields a significant improvement of the scalability of
quantum computing architectures. This can be achieved
with the help of distributed quantum computing, where the information of
stationary qubits is encoded in the states of flying
qubits (i.e. single photons), which then allow to establish a communication
between distant sources. We describe the implementation
of an eventually deterministic universal two-qubit gate operation between single photon
sources, despite the restriction of the no-go theorem
on deterministic Bell measurements with linear optics. This is a novel demonstration of an efficient repeat-until-success architecture to
quantum computation.
In Chapters
\ref{firework} and \ref{photon}, the entangled photons are shown to be generated
postselectively or at best preselectively. Ideally, one would like to generate these entangled photons on demand. Interestingly,
by combining ideas of photon interaction with their sources together with
measurements from Chapters \ref{firework} and \ref{minsk}, we
show in Chapter \ref{demand} that distributed photon entanglement can be
generated on demand. This can then serve as a useful tool for
the diverse applications already mentioned.
So far, linear optics has played a crucial component in the preceding chapters.
Penultimately, in Chapter \ref{photon}, we do not
consider any linear optics manipulation of light at all. Indeed, we recall the Young's
double-slit experiment in the
context of two distant dipole sources in free space without cavities.
Experiments have shown that interference fringes can be observed by coherent
light scattered by the dipole sources. Taking a step
further, we show that polarisation entanglement can also be produced by
initially unentangled {\em distant} single photon sources in
free space which at the same time also results in entanglement between the
sources. This adds new perspectives to common notions where it
is widely
thought that photon polarisation entanglement can only be obtained via pair
creation within the {\em same} source or via postselective measurements on
photons that overlapped within their coherence time inside a linear optics
setup.
Finally, we close in Chapter \ref{outlook} with a summary and give
limitations and an outlook of the work of
this thesis.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 8,696
|
\section{Introduction and statement of results}
Let $\Gamma$ be a finite group and let $L/K$ be a finite extension of
fields with $\Gal(L/K) \cong \Gamma$ (for brevity, we say: $L$ is a
$\Gamma$-extension of $K$). Then $L$ is a module over the group algebra
$K[\Gamma]$, and $K[\Gamma]$ carries the structure of a $K$-Hopf algebra. This
makes $L$ into a $K[\Gamma]$-Hopf-Galois extension of $K$. There may be
other $K$-Hopf algebras $H$ which act on $L$ so that $L$ is an
$H$-Hopf-Galois extension. Such Hopf-Galois structures were investigated by
Greither and Pareigis \cite{GP}, who showed how the determination of
all Hopf-Galois structures on a given separable field extension $L/K$
could be reduced to a question in group theory. In particular, any
Hopf algebra $H$ which gives a Hopf-Galois structure on $L$ has the
property that $L \otimes_K H = L[G]$ as $L$-Hopf algebras, where $G$
is some regular group of permutations of $\Gamma$. Thus $G$ and
$\Gamma$ have the same order, but in general they need not be
isomorphic. We will refer to the isomorphism class of $G$ as the {\em type} of
the Hopf-Galois structure, and will say that the Hopf-Galois
structure is {\em abelian} (resp.~{\em nilpotent}) if $G$ is abelian
(resp.~nilpotent).
For some groups $\Gamma$ it is known that every Hopf-Galois structure
on a $\Gamma$-extension must have type
$\Gamma$. This holds for cyclic groups of order $p^n$ with $p>2$
prime and $n \geq 1$ \cite{K}, for elementary abelian groups of
order $p^2$ with $p>2$ \cite{unique}, for cyclic groups of order $n$
with $(n, \varphi(n))=1$ (where $\varphi$ is Euler's totient function)
\cite{unique}, and for non-abelian simple groups \cite{simple}. On the
other hand, there are many groups $\Gamma$ for which there are
Hopf-Galois structures whose type is different from $\Gamma$, the smallest
cases being the two groups of order 4 \cite{unique}. Indeed, if
$\Gamma$ is abelian then there may be Hopf-Galois structures which are
not abelian, or even nilpotent. For example, if $\Gamma$ is cyclic of order
$pq$, where $p$, $q$ are primes such that $q|(p-1)$, then $L/K$ admits
$2(q-1)$ Hopf-Galois structures which are not nilpotent, in addition
to the unique (classical) one of type $\Gamma$ \cite{pq}.
This phenomenon was investigated in some detail in \cite{NYJM}, where
it was shown that any abelian extension $L/K$ of even degree $n>4$
admits a non-abelian Hopf-Galois structure, and that the same holds
for many abelian groups of odd order. On the other hand, some new
groups $\Gamma$ were given in \cite{NYJM} for which all
Hopf-Galois structures are of type $\Gamma$ (cf.~Remark \ref{not-best}
below).
In this paper, we supplement the results of \cite{NYJM} by considering
the situation where $\Gamma$ and $G$ are both abelian or, more
generally, both nilpotent. We will show that the enumeration of such
Hopf-Galois structures can be reduced to the case of groups of prime
power order.
Let $e(\Gamma,G)$ denote the number of Hopf-Galois structures of type
$G$ on a $\Gamma$-extension $L/K$. Thus the total number of
Hopf-Galois structures on $L/K$ is given by
$$ e(\Gamma) = \sum_{G} e(\Gamma,G), $$
where the sum is over all isomorphism classes of groups $G$ of order
$|\Gamma|$. We also write
$$ \eab(\Gamma) = \sum_{G\ \small{\mathrm{abelian}}} e(\Gamma,G), \qquad
\enil(\Gamma) = \sum_{G\ \small{\mathrm{nilpotent}}}
e(\Gamma,G), $$
where the sum is over all isomorphism types of abelian
(resp.~nilpotent) groups $G$ of order $|\Gamma|$. Thus $\eab(\Gamma)$
(resp.~$\enil(\Gamma)$) is the number of abelian (resp.~nilpotent)
Hopf-Galois structures on $L/K$. Recall that a finite group $\Delta$
is nilpotent if it is the
direct product of its Sylow subgroups \cite[(5.2.4)]{Rob}. In
particular, if $\Delta$ is abelian, or if $\Delta$ is a $p$-group for some prime
number $p$, then $\Delta$ is nilpotent.
Let $n$ be the degree of the extension $L/K$. We write the prime
factorisation of $n$ as
$$ n = \prod_{p | n} p^{v_p}, $$
where the product is over the distinct prime factors $p$ of $n$. If
$\Gamma$ is nilpotent, we can correspondingly write $\Gamma$ as a direct
product of groups
\begin{equation} \label{nilp-decomp}
\Gamma = \prod_{p | n} \Gamma_p,
\end{equation}
where $\Gamma_p$ is the (unique) Sylow $p$-subgroup of $\Gamma$ and
has order $p^{v_p}$. By Galois theory, we can then decompose $L$ as
$$ L = \bigotimes_{p | n } L_p, $$ (tensor product over $K$) where
$L_p$ is a $\Gamma_p$-extension of $K$. If, for each $p$, we
take a Hopf-Galois structure on
$L_p/K$, say of type $G_p$ and with corresponding $K$-Hopf algebra
$H_p$, then the Hopf algebra $H=\bigotimes_{p | n} H_p$ acts in the
obvious way on $L$,
giving $L/K$ a Hopf-Galois structure of type $G= \prod_{p | n}
G_p$. This Hopf-Galois structure is necessarily nilpotent,
and is abelian if and only if each $G_p$ is abelian.
We will see that if $\Gamma$ is nilpotent then {\em every} nilpotent
Hopf-Galois structure on $L/K$ arises in this way. This is the key
observation in the proof of our first main result:
\begin{theorem} \label{nilp-thm}
Let $\Gamma$ be a nilpotent group of order $n$. Then for each
nilpotent group $G$ of order $n$ we have $e(\Gamma,G)= \prod_{p | n}
e(\Gamma_p,G_p)$.
\end{theorem}
Taking the sum over all isomorphism types of nilpotent (resp.~abelian)
groups $G$ of order $n$, we immediately obtain:
\begin{corollary} \label{nilp-ab}
For a finite nilpotent group $\Gamma$, we have
$$ \enil(\Gamma)=\prod_{p | n} e(\Gamma_p) \mbox{ and }
\eab(\Gamma)=\prod_{p | n} \eab(\Gamma_p). $$
\end{corollary}
As an application of Theorem \ref{nilp-thm}, we will determine
the number of nilpotent (resp.~abelian) Hopf-Galois structures
on a cyclic extension of arbitrary finite degree. Before stating the
result, we fix some notation. For $m \geq 1$, let $C_m$
denote the cyclic group of order $m$, and, for $v \geq 3$, let $D_{2^v}$
(resp.~$Q_{2^v}$) denote the dihedral (resp.~generalized
quaternion) group of order $2^v$. Also, for $n \geq 1$, let
$r(n)$ be the radical of $n$:
$$ r(n) = \prod_{p | n} p. $$
\begin{theorem} \label{cyc-thm}
Let $\Gamma$ be a cyclic group of order $n$.
\begin{itemize}
\item[(i)]
If $n$ is not divisible by $4$, then
$$ \enil(\Gamma)=\eab(\Gamma)=e(\Gamma,\Gamma)= \frac{n}{r(n)}. $$
Thus every nilpotent Hopf-Galois structure on a cyclic extension of
degree $n$ is cyclic, and hence abelian.
\item[(ii)] If $n \equiv 4 \pmod{8}$, then again
$$ \enil(\Gamma)=\eab(\Gamma)= \frac{n}{r(n)}, $$
but
$$ e(\Gamma,\Gamma)=e(\Gamma, C_2 \times C_{n/2}) =
\frac{n}{2r(n)}. $$
Thus every nilpotent Hopf-Galois structure on a cyclic extension of
degree $n$ is abelian, but only half of them are cyclic.
\item[(iii)] If $n$ is divisible by 8, so $n=2^v n'$ with $v \geq 3$
and $n'$ odd, then
$$ \enil(\Gamma)= \frac{3n}{2r(n)} \mbox{ and }
\eab(\Gamma)=e(\Gamma,\Gamma)= \frac{n}{2r(n)}, $$
with
$$ e(\Gamma,D_{2^v} \times C_{n'}) = e(\Gamma,Q_{2^v} \times C_{n'}) =
\frac{n}{2r(n)}, $$
Thus every abelian Hopf-Galois structure on a cyclic extension of
degree $n$ is cyclic, although there are also Hopf-Galois structures which
are nilpotent but not abelian.
\end{itemize}
\end{theorem}
For a finite abelian $p$-group $\Gamma$, Featherstonhaugh, Caranti and
Childs \cite{FCC} have given conditions under which every abelian
Hopf-Galois structure on a $\Gamma$-extension must have type
$\Gamma$. Combining this with Theorem \ref{nilp-thm}, we will obtain
the following result in the abelian case.
\begin{theorem} \label{ab-thm}
Let $\Gamma$ be a finite group of order $n = \prod_p p^{v_p}$, and
suppose that, for each prime factor $p$ of $n$, either $v_p <
p-1$ or $p \leq3$, $v_p < p$. Then every abelian Hopf-Galois structure
on a $\Gamma$-extension has type $\Gamma=\Gal(L/K)$. Equivalently,
$\eab(\Gamma)=e(\Gamma,\Gamma)$.
\end{theorem}
Combining Theorems \ref{cyc-thm} and \ref{ab-thm} with a result of
L.~E.~Dickson \cite{Dickson} dating from 1905, we obtain some new
cyclic groups $\Gamma$ for which {\em every} Hopf-Galois structure has
type $\Gamma$:
\begin{theorem} \label{all-Gamma-thm}
Suppose that $n= \prod_p p^{v_p}$ satisfies the following conditions:
\begin{itemize}
\item[(i)] $v_p \leq 2$ for all primes $p$ dividing $n$;
\item[(ii)] $p \nmid (q^{v_q}-1)$ for all primes $p$, $q$ dividing
$n$;
\item[(iii)] $ 4 \nmid n$.
\end{itemize}
Then a cyclic extension of degree $n$ admits precisely
$n/r(n)$ Hopf-Galois structures, all of which are of cyclic type.
\end{theorem}
\medskip
\noindent {\sc Acknowledgment:} The author thanks Lindsay Childs and Tim Kohl
for email correspondence about this work, which led to a
simplification of some of the arguments.
\section{Nilpotent Hopf-Galois Structures}
In this section we prove Theorem \ref{nilp-thm}.
We first recall the method of counting Hopf-Galois
structures on a $\Gamma$-extension for an arbitrary finite group $\Gamma$.
It was shown in \cite{GP} that these Hopf-Galois structures
correspond to regular permutation groups on $\Gamma$ which are
normalized by the group $\lambda(\Gamma)$ of left multiplications by
elements of $\Gamma$. (Recall that a permutation group $H$ on a set $X$
is regular if, given $x$, $y \in X$, there is a unique $h \in H$ with
$hx=y$.) Thus finding all Hopf-Galois structures with a given type $G$
amounts to finding all regular subgroups in the group
$\Perm(\Gamma)$ of permutations of $\Gamma$ which are isomorphic
to $G$ and are normalized by $\lambda(\Gamma)$. It was shown in
\cite{unique} that this problem can be reframed as a calculation
inside $\Hol(G) = \rho(G) \cdot \Aut(G)$, the holomorph of $G$,
which is usually a much smaller group than $\Perm(\Gamma)$. Here $\rho\;
\colon G \lra \Perm(G)$ is the right regular representation
$\rho(g)(x)=xg^{-1}$ for $g$, $x \in G$. As further reformulated by
Childs (see e.g.~\cite[\S7]{Ch00}), this gives the following method
of counting Hopf-Galois structures. A homomorphism
$\beta \colon \Gamma \lra \Hol(G)$ will be called a regular embedding if it is
injective and its image is a regular group of permutations on
$G$. Two such embeddings will be called equivalent if they are conjugate by
an element of $\Aut(G)$. Then the number $e(\Gamma,G)$ of Hopf-Galois
structures of type $G$ on a $\Gamma$-extension
is the number of equivalence classes of regular
embeddings of $\Gamma$ into $\Hol(G)$.
We will need the following general result.
\begin{proposition} \label{cent-prop}
Let $N$ be a regular subgroup of $\Hol(G)$. Then the centralizer of $N$
in $\Hol(G)$ has order dividing $|G|$.
\end{proposition}
\begin{pf}
We can regard $\Hol(G)$ as a subgroup of the group $B=\Perm(G)$ of all
permutations of $G$. By \cite[Lemma 2.4.2]{GP}, the centralizer of $N$
in $B$ is canonically identified with the opposite group of $N$, so in
particular has order $|N|=|G|$. The centralizer of $N$ in $\Hol(G)$ is a
subgroup of this, so has order dividing $|G|$.
\end{pf}
If $G$ is a nilpotent group, its Sylow subgroups $G_p$ are
characteristic subgroups. We therefore have direct product decompositions
\begin{equation} \label{dec-Aut}
\Aut(G) = \prod_{p | n} \Aut(G_p),
\end{equation}
and hence
\begin{equation} \label{dec-Hol}
\Hol(G) = \prod_{p | n} \Hol(G_p).
\end{equation}
Now suppose that $\Gamma$ and $G$ are nilpotent groups of order $n$,
and that we are given a homomorphism $\beta_p \colon \Gamma_p \lra
\Hol(G_p)$ for each $p|n$. Using (\ref{nilp-decomp}) and (\ref{dec-Hol}), we
can define a homomorphism
\begin{equation} \label{dp-beta}
\beta = \left(\prod_{p|n} \beta_p\right) \colon \Gamma \lra \Hol(G).
\end{equation}
It is clear that if each $\beta_p$ is a regular embedding then so is
$\beta$. This construction corresponds to taking tensor products of
Hopf-Galois structures on field extensions of prime-power degrees, as
described in \S1.
Not every homomorphism $\beta \colon \Gamma \lra \Hol(G)$ arises as
such a product. For any primes $p$, $q$ dividing $n$, let
$\iota_p \colon \Gamma_p \lra \Gamma$ be the inclusion induced by the
direct product decomposition (\ref{nilp-decomp}) of $\Gamma$, and let
$\pi_q \colon \Hol(G) \lra \Hol(G_q)$ be the projection induced by
(\ref{dec-Hol}). Given a homomorphism $\beta \colon \Gamma \lra
\Hol(G)$, let $\beta_{pq}$ be the composite homomorphism $\beta_{pq}=\pi_q \circ
\beta \circ \iota_p \colon \Gamma_p \lra \Hol(G_q)$. Then $\beta$ is
determined by its matrix of components $(\beta_{pq})$. For each $q$,
the images of the $\beta_{pq}$ must centralize each other in
$\Hol(G_q)$, since the $\Gamma_p$ centralize each other in $\Gamma$.
Conversely, a matrix of homomorphisms
$(\beta_{pq})$, $\beta_{pq} \colon \Gamma_p \lra \Hol(G_q)$,
determines a homomorphism $\beta \colon \Gamma \lra \Hol(G)$, provided
only that, for each $q$, the images of the $\beta_{pq}$ centralize
each other in $\Hol(G_q)$.
We can determine from the matrix $(\beta_{pq})$ whether $\beta$
is a regular embedding:
\begin{lemma} \label{diag}
Let $\Gamma$ and $G$ be nilpotent, and let $\beta \colon \Gamma \lra
G$ correspond to the matrix of homomorphisms $(\beta_{pq})$ as above.
Then $\beta$ is a regular embedding if and only if
$\beta_{pp} \colon \Gamma_p \lra \Hol(G_p)$ is a regular embedding for each $p$.
\end{lemma}
\begin{pf}
First observe that $\beta_{pp}(\Gamma_p)$ is the unique Sylow
$p$-subgroup in the subgroup $\pi_p \circ \beta(\Gamma)$ of $\Hol(G_p)$,
and hence is normal in $\pi_p \circ \beta(\Gamma)$.
If $\beta$ is regular then $\pi_p \circ \beta(\Gamma)$ is transitive
on $G_p$. Then, by Proposition \ref{perm-prop} below, the number of
orbits of $\beta_{pp}(\Gamma_p)$ on $G_p$ divides both $|G_p|=p^{v_p}$
and $|\pi_p \circ \beta(\Gamma)/\beta_{pp}(\Gamma)|$ (which is coprime
to $p$). Thus $\beta_{pp}$ is transitive, and hence regular, on
$G_p$.
Conversely, suppose that each $\beta_{pp}$ is a regular embedding. We write
$e_G$ for the identity element of $G$. Consider the subsets
$X=\beta(\Gamma) e_G$ and $Y=\beta(\Gamma_p) e_G$ of $G$. Clearly $|Y|\leq
|\Gamma_p|$, and the regularity of $\beta_{pp}$ ensures that $|Y| \geq
|G_p|=|\Gamma_p|$. Hence $|Y|=|\Gamma_p|$. As $\beta(\Gamma_p)$ is
normal in $\beta(\Gamma)$, Proposition \ref{perm-prop} shows that all
orbits of $\beta(\Gamma_p)$ on $X$ have the same size. One such orbit
is $Y$, so $|X|$ is divisible by $|\Gamma_p|$. This holds for all $p$,
so $X=G$ and $\beta$ is a regular embedding.
\end{pf}
In the above proof, we used the following simple fact about
permutation groups:
\begin{proposition} \label{perm-prop}
Let $H$ be a finite group acting transitively on a set $X$, and let
$N$ be a normal subgroup of $H$. Then the orbits of $N$ on $X$ all
have the same size, and the number of these orbits divides both
$|X|$ and $|H/N|$.
\end{proposition}
\begin{pf}
Let $N$ have $m$ orbits on $X$, and let $Nx$ and $Ny$ be two such
orbits. Then $y=hx$ for some $h\in H$, and $Ny=Nhx=hNx$. This shows
that the quotient group $H/N$ acts on the set $\{Nx\}$ of orbits of
$N$, and that this action is transitive. It follows firstly that these
orbits have the same size, so that $m$ divides $|X|$, and secondly that
$m$ divides $|H/N|$.
\end{pf}
\begin{pf-nilp-thm}
Let $\beta \colon \Gamma \lra \Hol(G)$ be a regular embedding, and let
$(\beta_{pq})$ be the corresponding matrix of homomorphisms. By Lemma
\ref{diag}, each $\beta_{pp}$ is a regular embedding of $\Gamma_p$
into $\Hol(G_p)$. For $p \neq q$,
the image of the homomorphism $\beta_{pq} \colon \Gamma_p \lra
\Hol(G_q)$ must centralize the regular subgroup $\beta_{qq}(\Gamma_q)$
of $\Hol(G_q)$, and so must be a $q$-group by Proposition
\ref{cent-prop}. But $\beta_{pq}(\Gamma_p)$ is a $p$-group since
$\Gamma_p$ is. Thus $\beta_{pq}$ is the trivial homomorphism
whenever $p \neq q$. This means that the matrix $(\beta_{pq})$ is
``diagonal'' and $\beta$ is just the product $\beta= (\prod_p
\beta_{pp})$ as in (\ref{dp-beta}). Conversely, given a regular
embedding $\beta_p \colon \Gamma_p \lra \Hol(G_p)$ for each $p$, the
homomorphism $(\prod_p \beta_p) \colon \Gamma \lra G$ is a regular
embedding. It is immediate that these two constructions are mutually
inverse.
We have just established a bijection between regular embeddings
$\beta \colon \Gamma \lra \Hol(G)$ and families of regular embeddings
$\beta_p \colon \Gamma_p \lra \Hol(G_p)$ for each $p | n$. It follows
from (\ref{dec-Aut}) that two regular embeddings $\beta$, $\beta'$
are conjugate by an element of $\Aut(G)$ if and only if, for each $p$, their
components $\beta_p$, $\beta'_p$ are conjugate by an element of
$\Aut(G_p)$. Hence the equivalence classes of
regular embeddings $\beta \colon \Gamma \lra \Hol(G)$ correspond
bijectively to families of equivalence classes of regular embeddings
$\beta_p \colon \Gamma_p \lra \Hol(G_p)$. This shows that
$e(\Gamma,G)=\prod_p e(\Gamma_p,G_p)$.
\end{pf-nilp-thm}
\section{Hopf-Galois structures on cyclic extensions}
For cyclic extensions whose degree is a power of a prime $p$, all the
Hopf-Galois structures are already known. We recall the results.
\begin{lemma} \label{kohl-etc}
\begin{itemize}
\item[(i)] For $n=p^v$ with $p>2$ and $v\geq 1$, we have
$e(C_n)=e(C_n,C_n)=p^{v-1}$.
\item[(ii)] For $n=2$, we have $e(C_2)=e(C_2,C_2)=1$; for $n=4$, we
have $e(C_4)=2$ with $e(C_4,C_4)=e(C_4,C_2\times C_2)=1$.
\item[(iii)] For $n=2^v$ with $v \geq 3$, we have $e(C_n)=3 \cdot
2^{v-2}$ with $e(C_n,C_n)=e(C_n,D_n)=e(C_n,Q_n)=2^{v-2}$.
\end{itemize}
Thus, for a prime power $n=p^v$, we have $e(C_n)=n/r(n)$ except in the
case $p=2$, $v \geq 3$, when $e(C_n)=3n/(2r(n))$.
\end{lemma}
\begin{pf}
(i) is equivalent to Kohl's result \cite{K} that, for an odd prime $p$, a cyclic Galois
extension of degree $p^r$ admits $p^{r-1}$ Hopf-Galois
structures, all of cyclic type. Similarly, (ii) follows from
\cite{unique} and (iii) from \cite{2power}.
\end{pf}
Theorem \ref{cyc-thm} follows directly from Lemma
\ref{kohl-etc} and Theorem \ref{nilp-thm}.
\section{Abelian Hopf-Galois Structures}
In this section, we prove Theorems \ref{ab-thm} and
\ref{all-Gamma-thm}.
From \cite[Theorem 1]{FCC} we have the following result:
\begin{lemma} \label{FCC-lemma}
Let $\Gamma$ be an abelian $p$-group of $p$-rank $m$, with $p>m+1$. Then
$\eab(\Gamma)=e(\Gamma,\Gamma)$.
\end{lemma}
\begin{pf-ab-thm}
Let $G$ be an abelian group of order $n$, and let $\Gamma_p$,
$G_p$ be the Sylow $p$-subgroups of $\Gamma$, $G$ as usual. If
$v_p<p-1$ then certainly $p>m+1$ where $m$ is the $p$-rank of $G_p$,
so, by Lemma \ref{FCC-lemma}, $e(\Gamma_p,G_p)=0$ unless
$G_p=\Gamma_p$. If $p=3$ and $v_3=2$ then either $\Gamma_3 =
C_9$, when by Lemma \ref{kohl-etc}(i) we have $e(\Gamma_3,G_3)=0$
unless $G_3=\Gamma_3$, or $\Gamma_3 = C_3 \times C_3$, when the
same conclusion holds by \cite{unique}. If $p=2$ and $v_2=1$ then
$\Gamma_2 = C_2$ and $G_2 = C_2$. Thus the hypotheses of
Theorem \ref{ab-thm} ensure that
$\eab(\Gamma_p)=e(\Gamma_p,\Gamma_p)$ for all $p$. By Corollary
\ref{nilp-ab} we then have
$$ \eab(\Gamma) = \prod_{p | n} e(\Gamma_p, \Gamma_p) =
e(\Gamma,\Gamma), $$
and every abelian Hopf-Galois structure on $L/K$ has type
$\Gamma$.
\end{pf-ab-thm}
To prove Theorem \ref{all-Gamma-thm}, we need the following old result of
L.~E.~Dickson \cite{Dickson} (see also \cite[\S5.5, Exercise 24,
p.~189]{DF}):
\begin{lemma} \label{dickson}
Let $n$ have prime factorisation $\prod_p p^{v_p}$. Then every group
of order $n$ is abelian if and only if $v_p \leq 2$ for each prime $p$
dividing $n$, and $p \nmid (q^{v_q}-1)$ for all primes $p$, $q$
dividing $n$.
\end{lemma}
\begin{pf-all-Gamma-thm}
Let $\Gamma$ be a cyclic group of order $n$.
The conditions of Theorem \ref{all-Gamma-thm} imply those of Theorem
\ref{ab-thm}, so that every abelian Hopf-Galois structure on a
$\Gamma$-extension has cyclic type. On the other hand,
the hypotheses of Lemma \ref{dickson} are also satisfied. Thus every
group of order $n$ is abelian, and therefore every Hopf-Galois structure is
abelian. It follows that all the Hopf-Galois structures are cyclic. By
Theorem \ref{cyc-thm}(i), the number of Hopf-Galois structures is
therefore $n/r(n)$.
\end{pf-all-Gamma-thm}
\begin{remark} \label{not-best}
In Theorem \ref{all-Gamma-thm}, there are no non-abelian Hopf-Galois
structures for the rather trivial reason that there are no non-abelian
groups of the appropriate order. This result is certainly not best
possible, since if $n=p^2 q^2$ for primes $2<q<p$ with $(q,p+1)>1$
(e.g.~$q=3$, $p=11$), or if $n=p^3 q$ for distinct primes $p$, $q$ with
$(p,q-1)=(q,p^2-1)=1$ but $(q,p^3-1)>1$ (e.g.~$p=7$, $q=19$), then a
cyclic extension of degree $n$ admits only cyclic Hopf-Galois
structures \cite[Theorems 24, 25]{NYJM}. In both cases, non-abelian groups of
order $n$ exist, but a partial analysis of their holomorphs shows
that they cannot arise as the type of a Hopf-Galois structure on a
cyclic extension.
\end{remark}
\section{Abelian Hopf-Galois structures on abelian extensions}
In this final section we describe an alternative approach to Theorem
\ref{nilp-thm} in the case that $\Gamma$ and $G$ are both abelian
(restated as Theorem \ref{ab-case} below). This avoids the use of
Proposition \ref{cent-prop}, and instead is based upon a
result of Caranti, Dalla Volta and Sala \cite{CDVS} which underlies Lemma
\ref{FCC-lemma}. It therefore shows how the ideas in \cite{FCC} extend
to a finite abelian group $\Gamma$ which is not of prime-power order.
An important ingredient in the proof of Lemma \ref{FCC-lemma} (though
not of the original weaker version in Featherstonhaugh's thesis \cite{F}) is a
correspondence between regular subgroups of $\Hol(G)$ for an abelian
group $G$ and certain multiplication operations $\cdot$ on $G$.
This correspondence was first observed in \cite[Theorem
1]{CDVS} for vector spaces over a field $F$. The case
$F=\F_p$ (the field of $p$ elements) covers elementary
abelian $p$-groups $G$. It was noted in \cite{FCC} that the same
argument works for any finite $p$-group; indeed, this is what is
required to prove Lemma \ref{FCC-lemma}. It is easily verified that the
argument of \cite{CDVS} is still valid for arbitrary abelian
groups. Here is the result in that setting.
\begin{lemma} \label{CDVS-corr}
Let $(G,+)$ be an abelian group with identity element $0$. Then there
is a one-to-one correspondence between regular abelian subgroups $T$
of $\Hol(G)$ and binary operations $\cdot$ on $G$ which make
$(G,+,\cdot)$ into a commutative, associative (non-unital) ring with
the property that every element of $G$ has an inverse under the circle
operation $x \circ y = x + y + x \cdot y$ (so $(G,\circ)$ is an
abelian group, whose identity element is again $0$). Under this
correspondence, the subgroup $T$ of $\Hol(G)$ corresponding to $\cdot$
is $\{ \tau_g \; : \; g\in G\}$, where $\tau_g(x)=g \circ x$ for all
$x \in G$.
\end{lemma}
We next investigate the Sylow subgroups of (the additive group of)
such a ring.
\begin{proposition} \label{fin-ring}
Let $(R,+, \cdot)$ be a finite associative non-unital ring, and for each prime
$p$ dividing its order, let $R_p$ be the Sylow $p$-subgroup of
$(R,+)$. Then $R_p$ is an ideal (and hence a subring) of $R$, and
$R$ is the direct product of its subrings
$R_p$. Moreover, every element of $R$ has an inverse under
$\circ$ if and only if the same is true in each $R_p$.
\end{proposition}
\begin{pf}
Let $r \in R_p$, and let $s \in R$ be arbitrary. If $p^e$ is the
exponent of $R_p$ then, by associativity, $p^e(r \cdot s)=(p^e r)\cdot s=0
\cdot s =0$, so that $r \cdot s \in R_p$. Similarly $s \cdot r \in
R_p$. In particular, if $r \in R_p$ and $s \in R_p$ then $r \cdot s
\in R_p$, and if $r \in R_p$
and $s \in R_q$ with $p \neq q$ then $r \cdot s \in R_p \cap R_q$ so
$r \cdot s = 0$. Hence
$R_p$ is both an ideal and a subring of $R$, and $R$ is the direct
product of its subrings $R_p$.
Suppose now that every $r \in R$ has a $\circ$-inverse. If $r \in
R_p$ has $\circ$-inverse $s$ in $R$ then
$s=-r-r\cdot s \in R_p$, so $r$ has $\circ$-inverse $s$ in
$R_p$. Conversely, suppose that $\circ$-inverses exist in each $R_p$. Let $r\in R$. We can
write $r= \sum_p r_p$ with $r_p \in R_p$ for each $p$. If $s_p$ is the
$\circ$-inverse of $r_p$ in $R_p$ then $s= \sum_p s_p$ is the
$\circ$-inverse of $r$ in $R$.
\end{pf}
\begin{corollary} \label{T-syl}
In Lemma \ref{CDVS-corr}, the Sylow $p$-subgroup $T_p$ of $T$ is $\{
\tau_g \; : \; g \in G_p\}$.
\end{corollary}
\begin{pf}
If $g$, $h \in G_p$ then $g \circ h =g+h +g \cdot h\in G_p$ by Proposition
\ref{fin-ring}. But $\tau_g (\tau_h(x))=g \circ (h \circ x)=(g \circ
h) \circ x= \tau_{g \circ h}(x)$. The non-empty subset $\{\tau_g \; : \; g
\in G_p\}$ of the finite abelian group $T$ is therefore closed under composition, and
hence is a subgroup. Since its cardinality is $|G_p|$ and
$|G|=|T|$, it is the Sylow $p$-subgroup $T_p$.
\end{pf}
\begin{theorem} \label{ab-case}
Let $\Gamma$ and $G$ be abelian groups of order $n$. Then
$$ e(\Gamma,G)= \prod_{p | n} e(\Gamma_p,G_p). $$
\end{theorem}
\begin{pf}
Let $\beta \colon \Gamma \lra \Hol(G)$ be a regular embedding. Then
$T=\beta(\Gamma) \cong \Gamma$ is a regular subgroup of $\Hol(G)$
which by Lemma \ref{CDVS-corr} gives a multiplication $\cdot$ on $G$
making $G$ into a ring. Then $T = \{ \tau_g \; : g \in G\}$, where
the $\tau_g$ are defined using the $\circ$-operation obtained from
$\cdot$. By Proposition \ref{fin-ring}, $G$ is the direct product of
its subrings $G_p$. Since $\circ$-inverses exist in $G$, they exist in
$G_p$, so that the multiplication on $G_p$ corresponds via Lemma
\ref{CDVS-corr} to a regular subgroup $T'_p$ of $\Hol(G_p)$. Writing
elements of $G= \prod_p G_p$ as tuples $g=(g_p)_p$ with $g_p \in G_p$,
we have
$$ \tau_g(x) = g + x + g \cdot x = (g_p + x_p + g_p \cdot x_p )_p $$
for any $x=(x_p)_p \in G$. It follows that $T'_p$ consists of the
restrictions to $G_p$ of the $\tau_{g_p}$ for $g_p \in G_p$. By
Corollary \ref{T-syl}, the $\tau_{g_p}$ are precisely the elements of
the Sylow $p$-subgroup $T_p=\beta(\Gamma_p)$ of $T$. Thus $\beta$
induces a regular embedding $\beta_p \colon \Gamma_p \lra \Hol(G_p)$
for each $p$, where $\beta_p(h)$ for $h \in G_p$ is merely the restriction of
$\beta(h)$ to $G_p$. If we form the product $\beta^* = \left(
\prod_p \beta_p \right) \colon \Gamma \lra \Hol(G)$ as in
(\ref{dp-beta}), then $T^*=\beta^*(\Gamma)$ is a regular subgroup of
$\Hol(G)$ which induces the operation $\cdot$ on each $G_p$. By Lemma
\ref{CDVS-corr} and Proposition \ref{fin-ring} we then have $T^*=T$
and so $\beta^*=\beta$. Thus every regular embedding $\beta$ comes
from a family of regular embeddings $\beta_p$. As in the proof of
Theorem \ref{nilp-thm}, it follows that $e(\Gamma,G)=\prod_p
e(\Gamma_p,G_p)$.
\end{pf}
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{"url":"https:\/\/www-fourier.ujf-grenoble.fr\/ratio2019\/details.php?type=expose_1&orateur=Poonen","text":"# Expos\u00e9\n\nChoix de l'expos\u00e9\u00a0:\n\n# The local-global principle for stacky curves\n\n## Bjorn Poonen\n\nFor smooth projective curves of genus $$g$$ over a number field, the local-global principle holds when $$g=0$$ and can fail for $$g=1$$, as has been known since the 1940s. Stacky curves, however, can have fractional genus. We construct stacky curves of genus $$1\/2$$ that violate the local-global principle, and show that $$1\/2$$ cannot be reduced. This is joint work with Manjul Bhargava.","date":"2021-07-28 03:16:48","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7888109683990479, \"perplexity\": 419.0750168440805}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-31\/segments\/1627046153521.1\/warc\/CC-MAIN-20210728025548-20210728055548-00715.warc.gz\"}"}
| null | null |
Q: Why does Java not put the filename in args? In C and C++, the main method holds the filename in the first position of the array at argv[0]. In Java, however, the filename is not included in the args string array.
Is there a practical reason for this? I understand that this makes iterating through command line arguments 0-based instead of 1-based, but is there a benefit? Was the filename just deemed to be useless?
A: In some cases a program can be run in different ways and exhibit different behavior on how it is called. If you call vim as vi, it runs in a compatibility mode. Sometimes it is to try to maintain one version of several related programs - for example mailq and newaliases on many unix systems are a link to sendmail so that these programs stay in sync)
Java programs are typically invoked as:
% java -jar foo.jar args
% java Foo args
The first version is where you have a Manifest file that indicates the main class, the second version runs the main method in the class Foo found in the class path.
The information presented for Java is either a path to the jar or the name of the class being invoked.
The location of the jar isn't important enough to be something to code from (and was actually not part of the original spec). A Jar can be named anything really, and often includes version numbers. Whats more, there's no guarantee that the class was even stored in a .jar (it could have been extracted).
Invoking a Java application with -jar has only one way to enter it - the class defined in the Manifest. There's no renaming that can be done.
The other option, of invoking it with the class name points directly to the execution unit. Furthemore, it can't be named multiply - you can't have Bar.class be the code for class Foo it just doesn't work that way.
This should show that there's really no point to passing the information of argv[0] in the C sense to a Java application - its either going to be java, meaningless and arbitrary, or the name of the class that is being invoked (that you are already executing code out of (you could do something like getClass().getEnclosingClass().getName() if you were desperate...)).
There is a point here, you can define multiple Main methods in classes in a .jar or on the class path. And you could have them behave differently just as if there was a series of if statements based on what argv[0] was.
I have in the past had code akin to java -cp Foo.jar com.me.foo.Test which invoked the Test class's Main method rather than the one defined in the one defined in the Manifest.
A: actually there's no benefit with it, it really depends on the syntax of the programming language you are using if it is 0-based or 1-based. the variable (you refer to as filename) also depends on the language, it can be different in other languages just follow the correct syntax of the language you are using.
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\section{Our Algorithm}
In this section, we describe our topological algorithm for
finding a best conformal matching between
two non-degenerate road networks, $G_1$ and $G_2$:
\begin{enumerate}
\item
Create quasi-unique labels for each vertex, $v$,
in $G_1$ and $G_2$ based on the degrees of the nodes
at distance at most $k$ from $v$, for an input parameter, $k$.
(We show in our experimental section that choosing $k$
between 5 and 8 tends to give the best results.)
\item
Choose a good
pair of starting nodes, $s_1 \in G_1$ and $s_2\in G_2$, with the same
quasi-unique label, $L$, and, for each such pair having label $L$,
perform the following:
\begin{enumerate}
\item
Perform a breadth-first search (BFS) matching of the corresponding portions
in $G_1$ and $G_2$ that are respectively reachable from $s_1$ and $s_2$
according to a greedy conformal matching that emanates out from these starting
nodes.
\item
Save this conformal matching that starts from $s_1$ and $s_2$ if it is the best
(highest cardinality) such matching found so far for this quasi-unique label.
\end{enumerate}
\item
Commit the conformal matching that began with $s_1$ and $s_2$, removing
all matched nodes as candidates for starting nodes.
\item
Repeat the above process for another good pair of starting nodes, if such
a pair of nodes still remains.
\end{enumerate}
We describe these steps in more detail below.
\subsection{Labeling Vertices}
The first step of our algorithm is
to give each vertex, $v$, in $G_1$ and $G_2$ a quasi-unique label, based on
the degrees of the nodes at distance at most $k$ from $v$, for a given
parameter, $k$.
This approach is similar to a labeling method used in
the (exact) graph isomorphism
algorithm by Weisfeiler and Leman (WL)~\cite{grohe2000isomorphism}.
Specifically, we begin by determining the degree, deg$(v)$, of each
vertex, $v$.
Then we create a list for each vertex, $v$,
which contains its degree, followed by the degrees of nodes at distance $1$
from $v$, nodes at distance $2$ from $v$, and so on, up to a distance $k$,
where $k$ is an input parameter for this step.
So as to make sure that these labels are quasi-unique, we add the degrees of
these nodes at distance at most $k$ from $v$ according to a canonical
ordering, which in our case is a lexicographically minimum breadth-first
search (BFS) ordering.
This BFS ordering sorts the immediate neighbors of $v$ according
to a lexicographically minimum cyclic ordering of $v$'s neighbors based on
their degrees, and then it performs a
BFS from this queue, adding nodes
to the queue based on the cyclic ordering of edges around each vertex so long
as they are at distance at most $k$ from $v$.
We return a dictionary for $G_i$ (for $i=1,2$),
which we call {masterTable}$({G_i})$, such
that each entry in this dictionary is a list of vertices having the same
quasi-unique label.
That is, the keys we use to index the (list) entries in masterTable$({G_i})$
are the label$[v]$ lists produced by our quasi-labeling method.
The pseudocode for this step is given in Algorithm \ref{alg:find_seed}.
\input{find_seed}
With respect to the efficiency for performing this step, note that
the time needed for this step is dominated by our doing
a BFS from each node, $v$, to explore those other nodes at distance $k$
from $v$.
Since the vertices of a road network have degree bounded by some parameter,
$d$, this step takes worst-case time $O(d^kn)$, for a road network of $n$ nodes.
In practice, $k$ is a constant, $d$ is usually $3$ or $4$,
and the graph is rather sparse; hence,
this step runs in $O(n)$ time in practice.
\subsection{Choosing Pairs of Starting Nodes}
After we have labeled each vertex of $G_1$ and $G_2$ with quasi-unique
labels, we need to choose a pair
of starting nodes in $G_1$ and $G_2$ with the same
label to start the matching process.
If we are able to
find a unique pair of nodes having the same label, then we can take them
as starting nodes and start our matching.
But it may happen that
we don't have any such unique pair of nodes; that is,
it might be the case that
there are at least $3$ nodes from $G_1\cup G_2$
for each quasi-unique label of vertices in the master table.
For each distinct label, $L$,
let $n_1(L)$ denote the number of vertices in $G_1$ with label $L$
and let $n_2(L)$ denote the number of vertices in $G_2$ with label $L$.
As mentioned above, if we have a label, $L$, such that
$n_1(L)=n_2(L)=1$, then we choose the unique pair of vertices,
$s_1\in G_1$ and $s_2\in G_2$, with label $L$
as a good pair of starting vertices.
Otherwise, we would like to choose a pair,
$s_1\in G_1$ and $s_2\in G_2$, that maximizes the probability that
there is a large conformal matching of the connected components
of $G_1$ and $G_2$ respectively containing $s_1$ and $s_2$, such that
$s_1$ and $s_2$ have the same quasi-unique label, $L$.
For any such label, $L$, the number of such candidate pairs is
$n_1(L)\cdot n_2(L)$; hence, to maximize the probability of finding a good
pair of starting nodes, we choose a pair, $s_1$ and $s_2$, that minimizes
the product, $n_1(L)\cdot n_2(L)$,
since the probability such
a pair actually correspond to corresponding nodes in $G_1$ and $G_2$,
conditioned on their having the same label, $L$,
is at least $1/(n_1(L)\cdot n_2(L))$.
We then perform a flooding-based search from each such $s_1$ and $s_2$ with
label $L$, committing to the pairing that results in the largest matched
components in $G_1$ and $G_2$.
Then, we remove all the matched vertices
in $G_1$ and $G_2$ from consideration (since they are now matched),
and we repeat our search for another good pair of starting seed vertices.
In order to perform such searches and updates quickly, we use an auxiliary
priority queue
data structure that stores each quasi-unique label, $L$, according
to its priority, $n_1(L)\cdot n_2(L)$.
Such
products can be found by taking the product
of lengths of both lists for each label used as a key
in \emph{masterTable}.
As we are performing our greedy matching processes,
we also need to
update these lists by removing each matched pair of nodes.
Of course,
this will also change the product for each label, so we have to update
labels in our priority queue to now have new priorities.
Since these products are always integers in the range $[1,C]$,
for some parameter, $C\le n^2$, let us use
a van~Emde~Boas tree~\cite{VANEMDEBOAS197780,vanEmdeBoas1976}
(vebTree) for storing non-zero products,
$n_1(L)\cdot n_2(L)$, for each label, $L$, as well as a
hash table, \emph{productTable}, that gives us the product for any existing
label, $L$.
This allows us to perform searches, updates, and finding of labels with
minimal product values in $O(\log\log C)$ time.
Every time the algorithm needs a pair of starting nodes, it finds
a label, $L$, with minimum
product, $n_1(L)\cdot n_2(L)$,
from \emph{vebTree}.
If there are multiple labels having that product, we randomly choose
any one of them.
After finding the required label, we take a pair
of nodes having the same label from the \emph{masterTable}. After
finding the starting pair of nodes, we update these data structures,
and the productTable,
so that we don't consider this pair of nodes again.
Note
this approach works even when we have unique pair of nodes having
the same label. In that scenario, the product will be $1$ and that
will be minimum product in \emph{vebTree}.
With respect to efficiency, we can do all the setup for this step
in $O(n\log\log C)$ time.
Moreover, we can determine already at this point what is the maximum
product, $n_1(L)\cdot n_2(L)$, over all labels, $L$, for a given
value of $k$. Since $k$ is a constant for real-world road networks
and there is an inverse relationship between $k$ and the size of
these products, we can perform a (binary) search to choose $k$
so that the maximum product size is bounded by some constant, $C$.
The running time of this search would be $O(n)$ for constants $k$ and
$C$.
There is a tradeoff, however, between using a large value for $k$ and getting
good matches, since two starting nodes are paired only if their
quasi-unique labels are the same, that is, if the
respective portions of the road
network at distance $k$ from these nodes is the same.
Since we are considering road networks that are evolving, we
therefore don't want to set too high a value for $k$.
Thus, we would like to choose $k$ as small as possible so that the
products, $n_1(L)\cdot n_2(L)$, are bounded by a constant, $C$.
(Say, $C\le 24$.)
As we note in the experimental section of this paper, choosing $k$
between 5 and 8 seems to work well in practice for this purpose.
\subsection{Flood-based Conformal Matching}
After finding a starting pair of nodes, we start our
greedy BFS matching process.
We begin by marking the starting nodes as
matched and we add them to our current tentative matching.
As we perform our BFS matching process, we will tentatively be matching up
additional pairs of nodes from $G_1$ and $G_2$, updating our
supporting data structures as we go, e.g., to tentatively remove each
such pair from consideration in vebTree.
Moreover,
if a starting node has more
than one lexicographically minimum ordering of the degrees of its
neighbors, then we also consider each such ordering of the edges,
performing our BFS matching process for each.
Tentative matchings are compared on the basis of number of matched nodes
and the matching with maximum number of matched nodes is taken as
best matching.
This raises an important implementation detail, which we should
probably discuss before going on to other details. Our matching algorithm
considers different pairs of starting vertices (and even
possibly different starting orientations of their incident edges), looking for
the pair that produces the largest portions of matching subgraphs.
Thus, we may have tentative matches that need to be undone
so that other tentative matches can be considered.
There are at least two possible ways to deal with this
branch-and-bound element in our conformal matching
algorithm.
One way is to checkpoint our supporting data structures, like
vebTree, masterTable, and productTable, saving the
version that produced the best tentative match so far.
This is the method we use, for example, in the version
of our algorithm that we implemented
for our experiments, since it is easy to implement.
Another way is to perform a two-phase commit, where we perform updates
to global copies of these data structures, but keep a history of the
updates we have performed during a tentative matching, so that we can
then roll back these updates if we do not commit to that tentative
matching (because there is another one that gave a larger number of
matched vertices).
This is the version of our algorithm
that we analyze for our theoretical analysis.
Given that there is some method that allows us to roll back to an
earlier state of our supporting data structures, vebTree,
masterTable,
and productTable, let us discuss in more detail how our conformal BFS
proceeds.
Once we map the neighbors around a pair of starting nodes, as discussed
above, we flood-search both graphs using a conformal-matching BFS.
When we reach any other
node except a starting node in the flooding, we know the edge we
are coming from and as we are following clocking ordering around
any node, there will be exactly one ordering around that node in
which we can traverse and map the neighbors with another graph, so as
to be forming a conformal matching.
Figure \ref{fig:order} shows an example.
\input{ordering_fig}
For matching any two nodes, $v_1\in G_1$ and $v_2\in G_2$,
that are not starting nodes, they should satisfy following properties:
\begin{itemize}
\item both $v_1$ and $v_2$ should be unmatched.
\item degree of $v_1$ should be same as $v_2$.
\end{itemize}
If any of these two conditions fail, we don't match $v_1$ and $v_2$
and we terminate that branch of the BFS. If both the conditions are satisfied,
then we mark $v_1$ and $v_2$ as matched, add them to current matching
and the queue for the BFS.
Then we remove them from \emph{masterTable}, \emph{vebTree} and
\emph{productTable}, so that they are not considered again in
the matching
process. The pseudocode for this step in our algorithm is given as Algorithm
\ref{alg:process_nodes}.
\input{processNodes}
When there is no further branch that can be matched, our BFS search terminates.
If this is the best tentative matching for the given quasi-unique
label, $L$,
then we tentatively save the matching corresponding to this BFS to the total
matching.
Then we check if there is still any remaining pair of seed nodes
having this same label.
If so, then we perform another conformal BFS for this next pair of
seed vertices.
Once we have completed performing a tentative matching for each pair
of seed nodes having the same quasi-unique label, $L$, we commit to
the matching for this label that produced the largest match.
Then we check if
\emph{vebTree} is empty or not.
If \emph{vebTree} is empty, we terminate the
algorithm and return the total matching. If not, we repeat our search
for a quasi-unique label, $L$, having the smallest product,
$n_1(L)\cdot n_2(L)$, and repeat the above conformal BFS for that
label.
The pseudocode for this step in
our algorithm is given as Algorithm \ref{alg:matching}.
\input{find_matching}
Each time
we explore subgraphs of $G_1$ and $G_2$ for a particular starting
pair, $s_1$ and $s_2$, that are in the starting label set of pairs
for some quasi-unique label, $L$, and one of the deg$(s_1)$ possible
orientations of edges, we traverse subgraphs of some size at most,
$n(L)\le n$, where $n(L)$ is the size of the largest match for the
label $L$.
Thus, the running time of this part of our algorithm is
at most
$O(n(L)\log\log C)$, where $C$ is the maximum value of a
product, $n_1(L')\cdot n_2(L')$, for some label, $L'$.
If $d$ is the maximum degree in a road network (e.g., $d\le 8$),
then the total worst-case running time of our BFS matching algorithm is
therefore
\[
O\left(dC\sum_L n(L)\log\log C\right)
=
O(dCn\log\log C),
\]
since $\sum_L n(L) \le n$, because the maximum amount of nodes we can
ultimately match in a pair of non-degenerate road networks is $n$.
Combining this with the theoretical analysis of the other steps
in our matching algorithm implies that the total running time
of our entire algorithm is
$O(d^kn+dCn\log\log C)$, where $d$ is the maximum degree of
a road network, $k$ is the distance we choose for producing
quasi-unique labels, and $C$ is the maximum value of a product,
$n_1(L)\cdot n_2(L)$, for any label, $L$.
Thus, in the practical case when $d$, $k$, and $C$ are constants, our
matching algorithm runs in $O(n)$ time.
\section{Experiments}
In this section, we provide an empirical evaluation of our topological flood-based matching. All of our experiments were ran on data from the U.S. TIGER/Line road network database~\cite{tiger-line-database}.
\subsection{Preprocessing the Data}
The TIGER/Line database provides the road networks in two different file formats: shapefile and TIGER/Line ASCII format. The data the shapefile format provides allows a graph to be created that not only has a node for every intersection of two roads, but also nodes to indicate the curvature of a road. That is, the format allows for curved roads to be represented as a sequence of many two-degree vertices. Therefore, for the preprocessing of files in the shapefile format, we simply take the first and the last vertex for each road to avoid introducing unnecessary two-degree vertices. With this approach to processing files in the shapefile format and fact that the TIGER/Line ASCII format lends itself to easy conversion to the definition of a road network given in the introduction, our algorithm performs well on both file formats.
\subsection{Tuning the Seed-labeling Parameter}
Let us consider the choice of the value for $k$, the parameter that
is input to Algorithm~\ref{alg:find_seed}
that defines distance to which to perform a lexicographic BFS so
as to improve the uniqueness
of vertex labels.
To characterize this uniqueness factor, let us
define the \emph{approximation ratio} of a
labeling as $a/b$, where $a$ is the number of pairs of nodes
with the same label
and $b$ is the minimum of the number of nodes in the two graphs.
Intuitively,
if $k$ is small, there will likely be many pairs of nodes $(u,v)$ with $u$
in $G_1$ and $v$ in $G_2$ that both have label $L$ where
$n_1(L) \cdot n_2(L)$ is large.
For example, labels
like ``44444", which indicates a four-way intersection that leads
to four other four-way intersections, are likely to
be common, and many other
examples like this are likely from real-world. As many of these
products are expected to be large,
we would expect the approximation ratio to be larger for
smaller value for $k$, because we could be possibly finding many
pairs of vertices with the same label
that should not actually be matched. For instance, we might find
two vertices labeled ``44444" even though they are not similar
beyond their immediate neighbors. We expect to run into this situation
only when the product is large since our algorithm matches the pair
of vertices for a given label that maximizes the number of nodes
matched.
As we increase the value of $k$, we would expect that the
approximation ratio to decrease.
That is,
if $k$ is large, we expect there to be more labels $L'$ such that $n_1(L')
\cdot n_2(L')$ is small or even 1, as the labels should become more
distinct as $k$ increases. Because the labels are expected
to be more distinct in this case,
it should be less likely to find pairs of vertices with those labels,
causing the approximation ratio to decrease.
The histograms in Figures~\ref{fig:amador-hists} and
\ref{fig:alameda-hists} exemplify the preceding interpretation of
the parameter $k$. The $x$-axis indicates the physical distance between
every node and its pair partner(s) with the same quasi-unique label,
$L$, using the longitude and latitude values given
from the database. The distance is determined using the haversine
formula, which yields the shortest distance between two points on
a sphere~\cite{shumaker1984astronomical}.
(Although our algorithm doesn't use geometric information to determine
matching pairs, we used geometric information in this experiment to
empirically validate our approach.)
Ideally, all pairs should be at distance 0 from each other.
As we expected, larger
$k$ values minimize the physical distances between pairs of nodes
with the same label, which
gives us a more accurate matching; hence, it reduces the number
of false pairs that our algorithm needs to consider. A histogram that is highly
skewed is desirable, as that implies that the number of
incorrect nodes being falsely matched is small. Note that the Amador County
data from 2000 and 2006 in Figure~\ref{fig:amador-hists} included
6,970 and 6,784 nodes, respectively, and the Alameda County data
from 2000 and 2006 in Figure~\ref{fig:alameda-hists} included 52,566
and 51,054 nodes, respectively.
Figure~\ref{fig:approx-plot} shows the change of the approximation
ratio with respect to the change in $k$ for Amador County. The
decrease in the approximation ratio with the increase in $k$ again
matched our intuition. The plot with the same $x$-axis and $y$-axis
values for Alameda County started at a similar approximation ratio
and decreased at a similar rate, so it was omitted.
\begin{figure}[htb]
\begin{center}
{
\includegraphics[width=0.7\columnwidth]{approx_ratio_amador_county}
}
\end{center}
\vspace*{-12pt}
\caption{Change in approximation ratio for Amador County, CA from 2000 to 2006}
\label{fig:approx-plot}
\end{figure}
We also plot the change in the maximum product with respect to $k$ in Figure~\ref{fig:max-product}. As described in Section 2, the maximum product is the value $\max\{n_1(L)\cdot n_2(L) : L \text{ is a label generated by Algorithm~\ref{alg:find_seed}}\}$. As expected, the maximum product decreases as $k$ increases. Note that only for San Francisco County does the maximum product reach 1. This is due to the fact that for the other counties, there are labels that do not change as $k$ increases as the nodes the labels correspond to are in small connected components e.g.\ ``121" is the cause of this in San Mateo County.
\begin{figure}
\begin{center}
{
\includegraphics[width=0.8\columnwidth]{max_prod_against_k}
}
\end{center}
\vspace*{-12pt}
\caption{Change in maximum product for Napa, San Francisco, and San Mateo Counties with road networks from 2000 and 2006}
\label{fig:max-product}
\end{figure}
\subsection{Example Output of Our Algorithm}
In this subsection, we provide a visualization of the matching our
algorithm created for Del Norte County, CA. We performed the
matching with $k = 3$ and then took four snapshots of the matching
to enlarge the details. For Figures~\ref{fig:ex-matching-1},
\ref{fig:ex-matching-2}, \ref{fig:ex-matching-3}, and
\ref{fig:ex-matching-4}, a node is colored blue if it was matched
and red otherwise. Furthermore, a node with a white box above it
from the first image containing number $i$ matches the node with a
white box above it containing the number $i$ from the second image.
First, consider Figure~\ref{fig:ex-matching-1}. Solely based off of geographic location, it is clear that the nodes are being matched to the correct area. After further inspection, it can be seen that the graph has remained nearly the same around the white boxes containing ``1", ``3", ``7", and ``8". Near each of these white boxes, our matching algorithm has matched the correct nodes, indicated by all of the blue nodes surrounding said boxes. Figure~\ref{fig:ex-matching-1} also demonstrates the issue of using a small value for $k$. The yellow boxes in Figure~\ref{fig:ex-match-new-1} indicate nodes that have been matched to other nodes in the graph from Figure~\ref{fig:ex-match-old-1} that are not included in the image. This incorrect matching is due to the fact that when $k$ is small, as mentioned earlier, it is likely that many nodes will end up with the same label, yielding a higher likelihood of incorrectly matching two nodes that should not be matched.
Second, consider Figure~\ref{fig:ex-matching-2}. The white boxes in these figures are here to indicate that the matching algorithm is performing properly in many parts of the graph. As we are just using topological features, we also get some unexpected matching as shown in Figure~\ref{fig:ex-matching-2}. The two yellow boxes in Figure~\ref{fig:ex-match-old-2} are matched to the two yellows boxes in Figure~\ref{fig:ex-match-new-2}. A new vertex was added in the 2006 graph that caused the matching of the vertices under the yellow boxes to occur in the wrong place. Because we are only using topological features, our matching algorithm cannot distinguish between the new vertex and the old one that it should be matching to.
Last, consider Figures~\ref{fig:ex-matching-3} and \ref{fig:ex-matching-4}. Many more white boxes were included to show the success of our matching algorithm in these portions of the graph.
\begin{figure*}[hbtp]
\begin{center}
\subfloat[{$2000$}]
{
\label{fig:ex-match-old-1}
\includegraphics[width=0.3\textwidth]{matching_example_old_1}
}
\subfloat[{$2006$}]
{
\label{fig:ex-match-new-1}
\includegraphics[width=0.3\textwidth]{matching_example_new_1}
}
\end{center}
\vspace*{-12pt}
\caption{First example of a portion of a matching for Del Norte
County, CA where $k = 3$.}
\label{fig:ex-matching-1}
\end{figure*}
\begin{figure*}[hbtp]
\vspace*{-12pt}
\begin{center}
\subfloat[{$2000$}]
{
\label{fig:ex-match-old-2}
\includegraphics[width=0.25\textwidth]{matching_example_old_2}
}
\subfloat[{$2006$}]
{
\label{fig:ex-match-new-2}
\includegraphics[width=0.25\textwidth]{matching_example_new_2}
}
\end{center}
\vspace*{-12pt}
\caption{Second example of a portion of a matching for Del Norte County, CA where $k = 3$}
\label{fig:ex-matching-2}
\end{figure*}
\begin{figure*}[hbtp]
\vspace*{-12pt}
\begin{center}
\subfloat[{$2000$}]
{
\includegraphics[width=0.2\textwidth]{matching_example_old_3}
}
\subfloat[{$2006$}]
{
\includegraphics[width=0.2\textwidth]{matching_example_new_3}
}
\end{center}
\vspace*{-12pt}
\caption{Third example of a portion of a matching for Del Norte
County, CA where $k = 3$.}
\label{fig:ex-matching-3}
\end{figure*}
\begin{figure*}[hbtp]
\vspace*{-12pt}
\begin{center}
\subfloat[{$2000$}]
{
\includegraphics[width=0.2\textwidth]{matching_example_old_4}
}
\subfloat[{$2006$}]
{
\includegraphics[width=0.2\textwidth]{matching_example_new_4}
}
\end{center}
\vspace*{-12pt}
\caption{Fourth example of a portion of a matching for Del Norte
County, CA where $k = 3$.}
\label{fig:ex-matching-4}
\end{figure*}
\subsection{Detailed Analysis}
We ran our algorithm on $40$ different counties in California ranging
from small counties to big counties. The results for our
experiments are shown in Table~\ref{tab:overall-results}. Each row
gives analysis about one particular county where $G_1$ is obtained
from TIGER/Line ASCII format from the year $2000$ and $G_2$ is
obtained from TIGER/Line ASCII format from the year $2006$. The
column titled ``seed time" indicates the time taken to find the
seed vertices for the given value of $k$ and the column titled
``match time" indicates the amount of time taken for the topological
flood-based matching algorithm. We ran the experiments on a machine
with $3.1$ GHz Intel Core i7 CPU and $16$ GB of RAM and report the
timings in seconds. The last column titled ``thresh. ratio" is the
ratio of number of pairs of matched vertices within $5$ miles of
each other to the total number of pairs of matched vertices which
gives up the quality of matching. We can see from the table that
thresh. ratio is always greater than $0.9$ which tells us that our
algorithm performs well on all kinds of inputs.
\include{table_results}
Figure~\ref{fig:runtime-against-size} plots the experiments in Table~\ref{tab:overall-results} with the total running time (seed time plus match time) as the $y$-axis and the size of the smaller graph as the $x$-axis. The red line represents the function $0.003n\log\log n$. Therefore, it seems that the variable $C$ defined in Section 2 is much less than $n$, which is good for the running time of our algorithm.
\begin{figure}
\begin{center}
{
\includegraphics[width=\columnwidth]{runtime_against_size}
}
\end{center}
\caption{Running times for our algorithm on the graphs given in Table~\ref{tab:overall-results}}
\label{fig:runtime-against-size}
\end{figure}
\section{Introduction}
Road network algorithms
are an important topic of study in Geographic Information
Systems (GIS), in that road networks facilitate transportation
and are the products of social, geographic, economic, and political forces.
In addition, road networks are interesting data types, in that they combine
both geometric information and graph-theoretic information.
\ifFull
(E.g., see~\cite{Abraham2011,Eppstein:2008,eppstein2009going,Yang98}.)
\else
(E.g., see~\cite{Eppstein:2008}.)
\fi
Formally, we view a road networks as a graph,
where we create a vertex for every
road intersection or major jog,
and we create an edge for every pair of such vertices that have a road
segment that joins them.
In addition, some road networks are annotated with geometric/geographic
information, such as the GPS coordinates of some vertices or
labels identifying road names.
Nevertheless, because road networks may contain
many vertices and edges without such geometric/geographic
information, we are interested in this paper in studying road networks from
strictly a topological viewpoint, that is, as embedded graphs.
Specifically,
we are interested in the problem of determining how road
networks evolve over time,
e.g., highlighting places
where new roads and bridges are added and where old roads and bridges are
removed.
(See Figure~\ref{fig:sf}.)
\begin{figure}[hbt!]
\centering
\includegraphics[width=.8\columnwidth, trim= 1in 3.1in 1in 2.5in, clip]{SF1915.pdf}
\\[0.1in]
\includegraphics[width=.8\columnwidth, trim= 1in 2.5in 1in 2.5in, clip]{SF2016.pdf}
\caption{A map of San Francisco from 1915 and one from 2016 (taken
from OpenStreetMap).
The top image is in the public domain; the bottom image is licensed under
the Open Database License, CC BY-SA.
Note that most of the roads in both maps are not labeled.
}
\label{fig:sf}
\end{figure}
\pagebreak
\subsection{Problem Definition}
Viewed topologically in terms of their graph properties,
road networks are embedded graphs,
that is,
the edges incident on each vertex are
given in a particular order (i.e., clockwise or counterclockwise),
which defines a topological structure
for the graph
known as a \emph{rotation system} (e.g., see~\cite{JGT:JGT3190020402}).
Road networks are not typically planar graphs (e.g., see~\cite{Eppstein:2008}),
however, since there are edge crossings, for example, at overpasses.
Thus, we cannot in general apply algorithms for planar graphs to road networks.
Nevertheless, the vertices in road networks have bounded degrees
(since the number of roads that meet at a single junction cannot
be arbitrarily large); hence, a road network with $n$ vertices has
$O(n)$ edges.
Given two undirected graphs, $G_1$ and $G_2$,
an \emph{isomorphism} of $G_1$ and $G_2$ is a bijection, $f$, from the vertices
of $G_1$ to the vertices of $G_2$ such that $(u,v)$ is an edge in
$G_1$ if and only
if $(f(u),f(w))$ is an edge in $G_2$ (e.g., see~\cite{McKay201494}).
In the \emph{subgraph isomorphism} problem, we are given two graphs,
$G_1$
and $G_2$, and asked to determine whether there is
a subgraph of $G_1$ isomorphic to $G_2$.
This problem is NP-complete, even if $G_1$ is an embedded planar graph,
by a reduction from
the planar Hamiltonian circuit problem~\cite{doi:10.1137/0205049}.
\ifFull
By the same argument, finding a
maximum-cardinality subgraph that is isomorphic to both $G_1$ and $G_2$ is
NP-hard.
\fi
Thus, defining the best matching
between two road networks simply in terms of a maximum common subgraph
is unlikely to lead to a polynomial-time algorithm.
So let us restrict the types of matchings we consider.
Suppose we are given a subgraph, $G_1'$, of a graph, $G_1$, and
a subgraph, $G_2'$ of a graph, $G_2$, such that
$G_1$ and $G_2$ are embedded graphs, i.e., having specified rotation systems.
Suppose further that $f$ is an isomorphism from $G_1'$ to $G_2'$.
We say that $f$ is \emph{conformal} if it satisfies
the following conditions:
\begin{enumerate}
\item
For every vertex $v$ in $G_1'$,
$v$ has the same degree in $G_1$ as $f(v)$ has in $G_2$.
That is, we only match vertices having the same degree.
\item
For every pair of incident edges, $(v,u)$ and $(v,w)$, in $G_1'$,
$(v,u)$ precedes $(v,w)$ in the clockwise order of edges around
$v$ in $G_1$ if and only if
$(f(v),f(u))$ precedes $(f(v),f(w))$ in
the clockwise order of edges around $f(v)$ in $G_2$.
That is, we
match vertices consistently with the edge orderings around each vertex.
\end{enumerate}
Since road evolution tends to involve adding or removing whole roads or
neighborhoods, we restrict our notion of road network evolution in this paper
to be defined in terms conformal matchings.
There is still one more restriction that we need to add, however, which deals
with degeneracies that are unlikely to occur in real-world road networks.
Suppose we are
given two road networks, $G_1$ and $G_2$, and
a maximum-cardinality subgraph, $G_1'$, having
a conformal matching, $f$, to a subgraph of $G_2$ (which is how we determine
the parts of $G_1$ that are the same in $G_2$).
We say that $G_1$ and $G_2$ are \emph{degenerate}
if, for any vertex $v$ in $G_1'$ and edge $(v,w)$ in $G_1$,
we can change the assignment, $f(w)$, for $w$
and still have $f$ be a conformal matching
(even allowing for $f(w)$ to be undefined).
Since our intended applications involve the second road network being
a newer copy of the first, a maximum-cardinality subgraph with a conformal
matching identifies
the portions of the road network that have not changed over time; hence,
the portions outside of this maximum-cardinality subgraph identify the
portions that have changed.
Thus, we argue that such applications involve non-degenerate graphs, since
it is unlikely, for example, for us to encounter
an $8\times 16$ grid that evolves into a grid-like annulus of 64 nodes
with radius 8, which would be degenerate.
Given that such configurations are likely to be rare in the real world,
we are interested in this paper only in
finding maximum conformal matchings in
non-degenerate pairs of road networks, which is a problem
we refer to as the \emph{map evolution} problem.
Incidentally,
the map evolution problem should not be confused with the \emph{map matching}
\ifFull
problem (e.g., see~\cite{Liu:2012,Lou:2009,Newson:2009}),
\else
problem (e.g., see~\cite{Liu:2012,Lou:2009}),
\fi
which is the unrelated problem of matching a trajectory of
(possibly noisy) GPS
coordinates, as might be produced by a moving vehicle, to the
geometry of the road network in which the trajectory is traveling.
\subsection{Prior Related Work}
As noted above, the map evolution
problem is related to the graph isomorphism problem,
which has a rich history (e.g., see~\cite{grohe2000isomorphism,McKay201494}),
due to the fact that it is not known
to be NP-complete, and the best known worst-case algorithm
runs in quasipolynomial time~\cite{babai2015graph},
but the problem tends to be feasible
in practice (e.g., see~\cite{McKay201494}).
For the generalized approximate graph isomorphism problem,
which is NP-hard, Arvind {\it et al.}~\cite{Arvind2012} give a
quasipolynomial approximation algorithm.
Such algorithms are necessarily not taking advantage of any efficiencies,
however, that could come from topological considerations like our restrictions
to embedded graphs and conformal matchings.
The map evolution problem is also related to
the \emph{map alignment} problem,
which is also known as \emph{GIS conflation}
\ifFull
(e.g., see~\cite{brown1995automated,rosen1985match,saalfeld1988conflation}).
\else
(e.g., see~\cite{rosen1985match,saalfeld1988conflation}).
\fi
In this problem, one is given two road networks, including both topological
information (such as vertex-edge-face relationships)
and geometric information (such as vertex coordinates and edge directions
and lengths),
and one is interested in computing a ``most likely'' matching between
the two networks.
Rosen and Saalfeld~\cite{rosen1985match,saalfeld1988conflation}
develop an iterative process involving a human operator
based on matchings that use
topology/geometry classifications of the vertices,
edges, and faces of the maps.
Xiong~\cite{xiong2000three} extends these topological/geometric
approaches using more sophisticated classifications.
Savary and Zeitouni~\cite{savary2005automated}
and
Zhang~\cite{zhang2009methods} extends these approaches further by including
additional properties, such as geographic data, including road names and shapes.
Their use of geometry, however, implies that all of these conflation
methods are not strictly topological algorithms and their performance degrades
when roads or vertices lack geometric or geographic information.
Detecting changes in road networks
and geographic regions has also been studied from the perspective
of image processing, e.g., using satellite images
\ifFull
(e.g., see~\cite{fonseca1996registration,%
shapiro1980structural,shapiro1981structural,ventura1990image}).
\else
(e.g., see~\cite{ventura1990image}).
\fi
For example,
Zhang and Couloigner~\cite{qiaoping2004automatic} use image analysis
to extract polylines defining roads and match them between two images
of the same geographic region taken at different times.
Such image-analysis approaches are inherently geometric, however; hence these
are also not strictly topological algorithms and do not apply when image
data is not available.
Our topological approach is more closely aligned with the work
of Eppstein {\it et al.}~\cite{eppstein2009approximate}, which uses
a topological approach for approximately
matching for quadrilateral meshes used in
computer-generated animations.
Our approach differs from their methods, however,
in that we do not consider faces in our matching algorithm (since road
network faces can be large and complex), whereas their method crucially depends
on matching faces (which in their application are always
quadrilaterals or triangles).
\subsection{Our Results}
In this paper, we study the map evolution problem,
for matching two road network
graph of same area but from different time, by using only topological
properties.
The primary motivation for this approach is to show that the map
evolution problem
problem can be solved effectively using only topological information.
Thus, this gives GIS practitioners a tool that can be applied for solving
the map evolution problem even for problem instances where geometric and
geographic information is missing, such as in older hand-drawn maps, pairs of
maps where only one of them is derived from an image, pairs of
maps annotated in different languages, or maps missing geographic and
geometric annotations due to scaling resolution.
We develop an algorithm for the map evolution
problem that runs in polynomial time for finding conformal matchings
between non-degenerate embedded graphs, such as real-world road networks.
Our algorithm uses a breadth-first flooding technique that begins each
flooding phase by finding potentially-matching ``seed'' vertices using
a labeling technique similar to that used in the
the Weisfeiler-Leman (WL) graph isomorphism
algorithm (e.g., see~\cite{grohe2000isomorphism}).
So as to limit the amount of flooding done in subgraphs that ultimately are
determined not to match,
our algorithm is probabilistic in nature---when we don't have any
pair of unique starting nodes, we take the pair which minimizes
an estimate of the probability of a wrong match.
We provide verification of our algorithm
in experiments
and case studies
that show empirically that our algorithm produces good matches
in practice.
\section{Conclusion}
We have given a purely topological algorithm for determining the
changes that occur between two road networks, and we have provided
both theoretical and experimental analysis to show that our algorithm
is effective and efficient.
We therefore feel that this algorithm provides a good tool for
solving the map evolution problem when geometric or geographic
features are missing from one or both of the road networks being
considered.
\subsection*{Acknowledgments}
This article reports on work supported by the Defense Advanced
Research Projects Agency under agreement no.~AFRL FA8750-15-2-0092.
The views expressed are those of the authors and do not reflect the
official policy or position of the Department of Defense or the
U.S.~Government.
This work was also supported in part by the U.S.~National Science Foundation
under grants 1228639 and 1526631.
We would like to thank David Eppstein for several helpful
discussions related to the topics of this paper.
\bibliographystyle{abbrv}
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
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|
The Illinois Department of Transportation (IDOT) is currently engaged in preliminary engineering and environmental studies (Phase I) to improve Illinois Route 62 (Algonquin Road) from Illinois Route 25 to Illinois Route 68. The project is located in the Village of Barrington Hills in Cook and Kane Counties. Other nearby communities that have the potential to be affected by this improvement are the Villages of Algonquin, Carpentersville, and South Barrington and Kane and Cook Counties. The purpose of the study is to evaluate alternatives to address safety concerns and accommodate existing and projected year 2040 travel demands. This improvement is not currently included in the Department's FY 2019-2024 Proposed Highway Improvement Program, however it will be included for future funding consideration.
This website has been created to provide two-way communication between the public and the project team. Educational information, materials, and updates about the Illinois Route 62 Phase I study will be posted here. You can provide comments through the website, which will be reviewed and considered by the project team. The goal is to provide the public with the opportunity to express their needs and concerns regarding the project and to keep them informed about the project development process.
To see upcoming events related to this project, please click on the Project Updates tab.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 4,856
|
#include "graphd/graphd.h"
/**
* @brief To sleep, perchance to dream.
*
* This calback is invoked once every second on all the head
* requests of all non-CPU-queued sessions, to take care of
* time outs while the session is off doing something else
* or waiting for something.
*
* If the session times out, it must be unsuspended by this
* callback.
*
* @param data opaque application data; in our case, the graphd handle
* @param srv service module pointer
* @param now current time in microseconds
* @param session_data opaque application per-session data, gses
* @param request_data opaque application per-request data, greq
* @return 0 on success, nonzero error codes on unexpected
* system errors.
*/
int graphd_sleep(void* data, srv_handle* srv, unsigned long long now,
void* session_data, void* request_data) {
graphd_handle* g = data;
graphd_session* gses = session_data;
graphd_request* greq = request_data;
cl_handle* cl = gses->gses_cl;
graphd_runtime_statistics report;
graphd_runtime_statistics my_use;
cl_assert(cl, gses != NULL);
cl_assert(cl, greq != NULL);
cl_assert(cl, g != NULL);
/* Only if we don't have an error yet and haven't been served yet...
*/
if (greq->greq_error_message != NULL ||
greq->greq_req.req_done & (1 << SRV_RUN))
return 0;
/* Haven't started processing this yet?
*/
if (!greq->greq_runtime_statistics_started) return 0;
cl_enter(cl, CL_LEVEL_SPEW, "req_id=%llu",
(unsigned long long)greq->greq_req.req_id);
/* cumulative - saved starting point = my use */
greq->greq_runtime_statistics_accumulated.grts_endtoend_micros =
now -
greq->greq_runtime_statistics_accumulated.grts_endtoend_micros_start;
greq->greq_runtime_statistics_accumulated.grts_endtoend_millis =
greq->greq_runtime_statistics_accumulated.grts_endtoend_micros / 1000;
my_use = greq->greq_runtime_statistics_accumulated;
if (greq->greq_soft_timeout) graphd_runtime_statistics_max(&report);
if (graphd_runtime_statistics_exceeds(
&my_use, &greq->greq_runtime_statistics_allowance, &report)) {
if (greq->greq_soft_timeout) {
char buf[200];
char const* ex;
ex = graphd_cost_limit_to_string(&report, buf, sizeof buf);
greq->greq_soft_timeout_triggered =
cm_strmalcpy(greq->greq_req.req_cm, ex);
if (greq->greq_soft_timeout_triggered == NULL)
graphd_request_error(greq, "SYSTEM out of memory");
} else {
/* Fail the request with a
"took too long" error.
*/
cl_assert(cl, !graphd_replica_protocol_session(gses));
graphd_request_error(greq, "COST allowance exceeded");
}
/* Returning an error does not require exclusive access
* to the database. Or shared access, for that matter!
*/
graphd_request_xstate_set(greq, GRAPHD_XSTATE_NONE);
}
cl_leave(cl, CL_LEVEL_SPEW, "leave");
return 0;
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 1,731
|
NEW YORK — A stock drop is never reassuring — except when it could have been worse.
"We'll take that trading pattern any time," said Arthur Hogan, chief market analyst at Jefferies & Co. He said he came into work anticipating the Dow to drop between 1 percent and 2 percent Tuesday after the index jumped 6.8 percent Monday — its biggest gain since late October.
The bulk of Tuesday's market retreat was in financial stocks — the companies that have been rising the most recently.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 416
|
Q: how to overwrite the inherited property values of derived class with that of base class instance I have a Human class with simple two properties Age and Name. Then, I have another class Woman derived from the Human class and it has one extra properties FavCream. Now, I have two different instances of these classes say h1(of base class Human) and w1 (of derived class Woman) and I simply want to overwrite all the derived properties of w1 with the corresponding values from h1.
However, I did not found this built-in. I do not like to create a public method on Woman class which manually does the assignment job. I think, I can create a utility function using reflection feature but looking for proven way to deal with this.
Is this impossible? Can not we have any advantage of inheritance in this scenario? or Is this illogical?
Please have patience with me. Thanks!
class Program
{
private static Woman _aWoman;
static void Main(string[] args)
{
var aHuman = new Human() { Age = 25, Name = "Ram" };
Program._aWoman = new Woman() { Age = 22, FavCream = "Ponds" , Name = "Sita"};
Console.WriteLine("Woman is " + Program._aWoman.ToString()); // prints Age = 22 Name = Sita FavCream = Ponds
Console.WriteLine("Human is " + aHuman.ToString()); // prints Age = 25 Name = Ram
**// Do something here so woman derived property has auto over written with property of human instance.**
Console.WriteLine("after overwriting, woman is " + _aWoman.ToString());
// The result I m looking for here is "Age = 25 Name = Ram FavCream = Ponds"
}
}
internal class Human
{
public int Age { get; set; }
public string Name { get; set; }
public override string ToString()
{
return "Age = " + this.Age + " Name = " + this.Name;
}
}
internal class Woman: Human
{
public string FavCream { get; set; }
public override string ToString()
{
return base.ToString() + " FavCream = " + this.FavCream;
}
}
UPDATE:
I ended on creating a method and copied the base class properties using reflection. Also, I tried to enhance this with generic restriction but not succeeded yet.
Thanks everyone for such quick reply.
// Trying to make it generic... but could not enforce contraint that T must be base class of 'this'
public void Overwrite<T>(T baseInstance) where T : Human
{
Type baseType = baseInstance.GetType();
Type derivedType = this.GetType();
PropertyInfo[] propInfos = null;
propInfos = baseType.GetProperties();
foreach (PropertyInfo eachProp in propInfos)
{
PropertyInfo baseProp = (baseType).GetProperty(eachProp.Name);
object baseVal = baseProp.GetValue(baseInstance, null);
eachProp.SetValue(this, baseVal, null);
}
}
A: Yes you can do it using reflection
To fetch all the properties on Human
// get all public static properties of MyClass type
PropertyInfo[] propertyInfos;
propertyInfos = typeof(Human).GetProperties(BindingFlags.Public |
BindingFlags.Static);
You can use PropertyInfo GetValue() and SetValue() Methods for getting from Human and Setting it to Woman.
Look at MSDN : SetValue : http://msdn.microsoft.com/en-us/library/axt1ctd9.aspx
GetValue : http://msdn.microsoft.com/en-us/library/b05d59ty.aspx
Instead of assigning values at runtime, I would suggest you overload implicit or explicit operator and simply do a casting from Human to Woman.
Woman newWoman = (Woman)H1;
All the logic for converting Human to Woman would be dumped in the overrloaded method.
A: The Human class instance that you create is totally separate to the Woman class instance.
It's difficult to see what you're trying to achieve, perhaps this?
w1.Age = h1.Age;
w1.Name = h1.Name;
A: What you are asking for is illogical. Even if you had two instances of Woman, w1 and w2, there is no builtin mechanism for copying all of the properties from w1 to w2.
You will need to implement this using reflection, but that is a fairly straightforward operation.
A: Apart from other answers you can simply write some function in Human class which will copy properties you're interested in:
class Human
{
...
public void CopyFrom(Human h)
{
this.Age = g.Age;
this.Name = g.Name;
}
}
Other solution can be using automapper
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 9,142
|
package assembler
import "math"
func Sqrt(x float32) float32 {
return float32(math.Sqrt(float64(x)))
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 8,260
|
#include <Cocoa/Cocoa.h>
#include <ApplicationServices/ApplicationServices.h>
#include <Foundation/Foundation.h>
#include <IOKit/IOKitLib.h>
#include <IOKit/Graphics/IOGraphicsLib.h>
#include "ga-osx.h"
static CGDirectDisplayID displayID;
static CGContextRef cgctx;
static CGRect rect;
static int imagesize = 0;
static void *imgData = NULL;
static struct gaImage gaimage;
static CGContextRef
CreateARGBBitmapContext(CGImageRef inImage) {
CGContextRef context = NULL;
CGColorSpaceRef colorSpace;
void * bitmapData;
int bitmapByteCount;
int bitmapBytesPerRow;
size_t pixelsWide = CGImageGetWidth(inImage);
size_t pixelsHigh = CGImageGetHeight(inImage);
bitmapBytesPerRow = (pixelsWide * 4);
bitmapByteCount = (bitmapBytesPerRow * pixelsHigh);
colorSpace = CGColorSpaceCreateWithName(kCGColorSpaceGenericRGB);
if (colorSpace == NULL) {
fprintf(stderr, "Error allocating color space\n");
return NULL;
}
bitmapData = malloc( bitmapByteCount );
if (bitmapData == NULL) {
fprintf (stderr, "Memory not allocated!");
CGColorSpaceRelease( colorSpace );
return NULL;
}
context = CGBitmapContextCreate (bitmapData,
pixelsWide,
pixelsHigh,
8, // bits per component
bitmapBytesPerRow,
colorSpace,
kCGImageAlphaPremultipliedLast);
if (context == NULL) {
free (bitmapData);
fprintf (stderr, "Context not created!");
}
CGColorSpaceRelease( colorSpace );
return context;
}
int
ga_osx_init(struct gaImage *image) {
CGImageRef screen;
//
displayID = CGMainDisplayID();
screen = CGDisplayCreateImage(displayID);
if((cgctx = CreateARGBBitmapContext(screen)) == NULL) {
CGImageRelease(screen);
return -1;
}
image->width = CGImageGetWidth(screen);
image->height = CGImageGetHeight(screen);
image->bytes_per_line =
(CGImageGetBitsPerPixel(screen)>>3) * image->width;
bcopy(image, &gaimage, sizeof(gaimage));
//
CGRect myrect = { {0, 0}, {image->width, image->height} };
rect = myrect;
imagesize = image->height * image->bytes_per_line;
//
CGImageRelease(screen);
//
return 0;
}
void
ga_osx_deinit() {
CGContextRelease(cgctx);
if(imgData) {
free(imgData);
}
return;
}
int
ga_osx_capture(char *buf, int buflen, struct gaRect *grect) {
CGImageRef screen;
void *data;
//
if(grect == NULL && buflen < imagesize)
return -1;
if(grect != NULL && buflen < grect->size)
return -1;
//
screen = CGDisplayCreateImage(displayID);
CGContextDrawImage(cgctx, rect, screen);
data = CGBitmapContextGetData(cgctx);
if(data != NULL) {
if(grect == NULL) {
bcopy(data, buf, imagesize);
} else {
int i;
char *src, *dst;
src = (char *) data;
src += gaimage.bytes_per_line * grect->top;
src += RGBA_SIZE * grect->left;
dst = (char*) buf;
//
for(i = 0; i < grect->height; i++) {
bcopy(src, dst, grect->linesize);
src += gaimage.bytes_per_line;
dst += grect->linesize;
}
}
}
CGImageRelease(screen);
//
return grect == NULL ? imagesize : grect->size;
}
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 330
|
Zac Efron Reunites With Ex Girlfriend Sami Miro In Japan After Nasty Split?
Jenny Cox
Zac Efron and his ex-girlfriend Sami Miro are both in Japan. While their simultaneous trip to Japan just may be a coincidence, the fact that Zac and Sami took a trip out to Japan right before their split adds a level of mystery to this happenstance.
Zac and Sami's latest Instagram updates reveal that they are in Japan. While they are not photographed nor referenced together, the fact that they are in the same country thousands of miles away from Hollywood is piquing their fans' curiosity. Are they ready to be friends?
Check out his ex-girlfriend's Instagram post in Japan.
A post shared by SAMI MIRÓ (@samimiro)
After releasing a lot of bro comedies this summer, the 28-year-old actor also got himself to Japan, treating himself to multiple ice cream cones.
A post shared by Zac Efron (@zacefron)
In another picture she took in Tokyo, Sami wrote "Tokyo is Motherland," showing her deep love for the country.
There are a lot of speculation revolving around these to posts. Are they patching things up after their unpleasant split in Japan? Are they getting back together? Or is this a pure coincidence and they aren't even aware of each other's presence in east Asia?
The last time that Zac Efron was in Osaka, he was with his ex-girlfriend, Sami Miro. They traveled also to southeast Asia, taking in the sights and sounds of Vietnam.
Recently, Zac Efron confirmed the fact that he can't seem to bounce back after his split with girlfriend Sami Miro. At first it seemed like their parting of ways was amicable, but time revealed that it was much nastier than anyone suspected.
The 28-year-old star is having a hard time being single while famous. Not only is he having a hard time weeding out girls who are only interested in him because he is famous, but he can't seem to get a break on Tinder, either!
"Dating is something I'll never be able to do," Zac said according to People Magazine. "As in the dictionary definition of dating, because one way or another I've impacted that person's life and they'll soon realize it."
"Amazingly, when I signed up for Tinder, nobody swiped me," he added. "They thought it was fake… That never happened."
Sami Miro recently went to Twitter to reveal that he has been super clingy after the split and has been annoying her. She deleted a lot of tweets she posted, but the followers have documented them.
One of them was, "Oh, when he texts you saying he's trying so hard to move on but you already moved on months ago," which is pretty obvious in pointing out Zac Efron as the clingy boyfriend that will not stop contacting her.
She also allegedly tweeted, "when you've never been happier and he's still a hot mess."
The one that still is up on her account alludes to the drama that has been swirling around her and Zac Efron.
Missed your chance bye bye #iaintsorry
— Sami Miró (@SamiMiro) July 19, 2016
Sounds like there still is a lot of bad blood between them!
Unlike Zac, Sami has not had a hard time moving on from her ex-boyfriend. In fact, just days after they split, she was seen with a new guy in Los Angeles.
"Sami Miro was spotted letting her hair down on Tuesday night, enjoying a dinner date in the company of handsome DJ Alex Andre at ROKU Sunset in West Hollywood," reported the Daily Mail in early May. "The newly-single model was dressed to impress for her night on the town, donning a black bodycon minidress."
Do you think there is ever a chance that Zac Efron and Sami Miro will end up back together? Let us know in the comments below!
[Photo by Jordan Strauss/AP Images]
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 4,029
|
void cx18_process_vbi_data(struct cx18 *cx, struct cx18_mdl *mdl,
int streamtype);
int cx18_used_line(struct cx18 *cx, int line, int field);
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 6,364
|
Q: how to render or display data in react useeffect hook
I am able to get data in react JSON object, now I want to convert the JSON to string data and display into the page, Please find the below screenshot.
export const QvcPollsTab = () => {
const [spData, setSpData] = useState<string>();
-------------------------------------------------------
return (
<Provider theme={theme}>
<Flex.Item>
<div>
<div>
<Text content={`Hello test ${spData}`} />
------------------------------------------------------------
I can see the JSON data as is attached in image, but I am not able to iterate the data and display individual field like fields values. Please help.
A: You cannot directly bind the JSON object to DOM element. You have to loop through the list items & then bind individual property to DOM.
Check below article for detailed information: SharePoint Framework: Retrieve and Display SharePoint List Items using REST API and React JS
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 4,081
|
Q: How can I get multiple data, from a php file, with ajax? I want to get data from submit.php page and put them in some tags in my page. in the example below I can put just on data in tag that has id: "shortcutTitle".
I get more than one data from my php file(from database) and I want them to sit on various tags in my page.
function submit(shortcutid,shortcuttitle,shortcutlink,icon)
{
var xmlhttp;
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("shortcutTitle").value=xmlhttp.responseText;
}
}
xmlhttp.open("GET","submit.php?shid="+shortcutid+"&shtitle="+shortcuttitle+"&shlink="+shortcutlink+"&shicon="+icon,true);
xmlhttp.send();
}
A: You would want to use json_encode to send a javascript object back to the XML request
<?php
echo json_encode( array('name' => 'tehlulz', 'id' => '1') );
?>
So from the results, you can then call the output by:
var ajaxResuts = {};
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
ajaxResuts = JSON.parse( xmlhttp.responseText );
alert("Returned: " + ajaxResults.name);
}
|
{
"redpajama_set_name": "RedPajamaStackExchange"
}
| 3,118
|
{"url":"https:\/\/socratic.org\/questions\/how-do-you-graph-y-in-x-1","text":"0\n\n# How do you graph y=In(-x)?\n\npreview \u00d7\n\u2022 Edit wording or topic\nUpdate the question or its topic\nThen teach the underlying concepts\nDon't copy without citing sources\npreview\n?\n\n#### Explanation\n\nExplain in detail...\n\n#### Explanation:\n\nI want someone to double check my answer\n\u2022 Show activity\n\u2022 @jmicheli JMicheli Could you clarify your question? It sounds like you might be asking how to graph ln(-x) but I'm not sure.\n\nIf I and n are separate variables then we can't make a graph without knowing the values.\n1 year ago\n\u2022 @jimh Jim H @JMicheli The student has almost certainly made the common error of using a capital I rather than a lower case l to abbreviate the natural logarithmic function.\n1 year ago\n\u2022 16 minutes ago\n\u2022 16 minutes ago\n\u2022 23 minutes ago\n\u2022 26 minutes ago\n\u2022 2 minutes ago\n\u2022 3 minutes ago\n\u2022 10 minutes ago\n\u2022 10 minutes ago\n\u2022 13 minutes ago\n\u2022 15 minutes ago\n\u2022 16 minutes ago\n\u2022 16 minutes ago\n\u2022 23 minutes ago\n\u2022 26 minutes ago","date":"2018-06-19 17:38:30","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.5570111870765686, \"perplexity\": 5723.793569361702}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 20, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-26\/segments\/1529267863109.60\/warc\/CC-MAIN-20180619173519-20180619193519-00487.warc.gz\"}"}
| null | null |
Submit site search.
Join the Federal City Council
District Strong
Washington Housing Initiative
Metro Reform
Langston Initiative
Monumental Sports & Entertainment: Feeding the Frontline with Small Business and Nonprofit Partners
In response to COVID-19, Monumental Sports & Entertainment (MSE), headed by FC2 Trustee Ted Leonsis, has paused gatherings celebrating competition, athleticism, musical and visual talent and kinship. But they have also created a program to summon the same community spirit that is needed now more than ever.
As part of its Feeding the Frontlines initiative, the company's foundation is partnering with two small businesses that normally cater food to the Wizards and Capitals teams to provide meals to first responders and health care workers at DC area hospitals and COVID-19 testing sites. They are also partnering with the nonprofit DC Central Kitchen, where food-industry trainees prepare donated meals.
Since the end of March, donations from Wizards, Mystics, Go-Go and Caps players and coaches, along with MSE staff, have totaled over $100,000 to fund nearly 5,000 meals to five District-area hospitals. A $100 donation to the MSE Foundation will provide between 10-12 meals. Their goal is to provide more than 1,000 meals per week.
They welcome fans and community members to join in donating towards providing healthy meals. You can make a contribution here (please note "Feeding the Frontlines" in the memo field).
Federal City Council
©2021 Federal City Council | Privacy Policy | Accessibility Policy
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 9,332
|
\section{Introduction}
Many efforts in the literature, over several decades, have been devoted to the
construction of gauge theories of gravity, mainly motivated \cite{Neeman1998}
by: \vskip0.3cm \noindent(i) conceptual simplicity within the geometric
framework of fibre-bundle picture of modern field theories; \vskip0.3cm
\noindent(ii) the search for quantum gravity (see \cite{Espo2011} and the
references therein), that should unify quantum theory and general relativity.
Such an unification is logically compelling if one thinks that gravity couples
to the energy-momentum tensor of matter, but matter fields are defined and
treated via quantum theory, and the semiclassical analysis where gravity
remains classical is a hybrid scheme that cannot hold at all energy scales.
\vskip0.3cm \noindent(iii) attempts of building unified theories of all
fundamental interactions. The various models can be classified according to
the relevant structure group in the fibre \cite{Neeman1998}. Within the
Poincar\'{e} group, this has been either the translations, or the Lorentz
group, or the whole Poincar\'{e} group. The enlargement of the group has
involved using the conformal, or special affine or affine groups. Moreover,
the so-called supergroups have included the graded-Poincar\'{e} or the
superconformal group. Interestingly, re-expressing the action of supergravity
in Yang--Mills form has made it possible to simplify enormously the proof of
its invariance under local supersymmetry transformations \cite{van2005}.
An alternative view is also available, which is convenient, if not natural, if
one is interested in the geometry of the space of histories and its
application to functional integrals in quantum field theory \cite{dewitt2003}.
Within this framework, gauge theories are all field theories whose action
functional $S$ is invariant under an infinite-dimensional Lie group. Gauge
invariance is then expressed by the existence of vector fields $Q_{\alpha}$
($\alpha$ being a Lie-algebra index) on the space of histories that leave
invariant $S$, i.e.
\begin{equation}
Q_{\alpha}S=0.
\end{equation}
All known gauge theories belong to one of the following families: \vskip0.3cm
\noindent(i) Type I: The Lie bracket $[\;,\;]$ of such vector fields is a
linear combination of the $Q_{\alpha}$ themselves, i.e.
\begin{equation}
\lbrack Q_{\alpha},Q_{\beta}]=C_{\;\alpha\beta}^{\gamma}\;Q_{\gamma
},\;C_{\alpha\beta}^{\gamma}=-C_{\beta\alpha}^{\gamma},
\end{equation}
the coefficients $C_{\;\alpha\beta}^{\gamma}$ being structure constants, hence
independent of the field variables $\varphi^{i}$.
Interestingly, Maxwell's electrodynamics, Yang--Mills theory and Einstein's
gravity share the property of being type-I theories, which was overlooked for
a long time and even nowadays, since Yang--Mills theory was first formulated
in Minkowski space-time, and only there can its perturbative renormalizability
be proved, this property being spoiled by space-time curvature
\cite{dewitt2003}. \vskip0.3cm \noindent Type II: The structure constants of
the previous equation are replaced by structure functions ${\widetilde{C
}_{\;\alpha\beta}^{\gamma}$, i.e. dependent on field variables, so that
\begin{equation}
{\frac{\delta}{\delta\varphi^{i}}}{\widetilde{C}}_{\;\alpha\beta}^{\gamma
}\not =0.
\end{equation}
An example is given by simple supergravity in $4$ space-time dimensions,
endowed with auxiliary fields. \vskip0.3cm \noindent Type III: The right-hand
side of the defining equation acquires also an inhomogeneous term, so that
\begin{equation}
\lbrack Q_{\alpha},Q_{\beta}]={\widetilde{C}}_{\;\alpha\beta}^{\gamma
}Q_{\gamma}+T_{\alpha\beta}^{i}{\frac{\delta S}{\delta\varphi^{i}}},
\end{equation}
where $T$ is skew-symmetric in the Lie-algebra indices. An example is provided
by simple and extended supergravity in any number of space-time dimensions,
without auxiliary fields.
Now we can revert to a topic more immediately related to the calculations we
are interested in, i.e. the formulation of Einstein's theory as $SL(2,C)$
gauge theory, following \cite{carmeli}. We study in detail the equations for
nonlinear gravito-electromagnetic waves and their solutions.
\section{Electrodynamics and general nonlinear constitutive equations}
To set the context and specify notation, we review established results in this
section \cite{dup/gol/sht}. We consider only flat spacetime, so that the
metric is given by the Minkowski tensor $\eta_{\mu\nu}=\left(
1,-1,-1,-1\right) $, $x^{\mu}=\left( ct,x^{i}\right) $, $\mu,\nu
,\dots\,=\,0,1,2,3$; $i,j,\dots\,=\,1,2,3$; with $\partial_{\mu
=\partial\diagup\partial x^{\mu}\,=\,\left[ c^{-1}\partial\diagup\partial
t,\nabla\right] $. The antisymmetric Levi-Civita tensor is written
$\varepsilon^{\mu\nu\rho\sigma}$, with $\varepsilon^{0123}=1$.
With tensor notation, we denote by $F_{\mu \nu}$ the electromagnetic field
tensor and by $G_{\mu \nu}$ the tensor expressing the constitutive
equations (see below), while their Hodge dual tensors are
$\tilde{F}^{\mu\nu}=\frac{1}{2}\varepsilon^{\mu\nu\rho\sigma}
F_{\rho\sigma}$ and $\tilde{G}
^{\mu\nu}=\frac{1}{2}\varepsilon^{\mu\nu\rho\sigma}G_{\rho\sigma}$,
and hence Maxwell's equations become
\begin{equation}
\partial_{\mu}\tilde{F}^{\mu\nu}=0,\ \ \ \ \ \ \partial_{\mu}G^{\mu\nu}
=j^{\nu},\label{max4}
\end{equation}
where $j^{\mu}=\left( c\rho,\mathbf{j}\right) $ is the 4-current. The first
equations in (\ref{max4}) imply $F_{\mu\nu}=\partial_{\mu}A_{\nu}
-\partial_{\nu}A_{\mu}$, where $A_{\mu}$ is an Abelian gauge field; but in
general there is no similar representation for $G_{\mu\nu}$. With covariant
notation we have $2$ independent invariants
\begin{equation}
X=\dfrac{1}{4}F_{\mu\nu}F^{\mu\nu},\ \ \ \ \ \ \ \ Y=\dfrac{1}
{4}F_{\mu\nu}\tilde{F}^{\mu\nu}.\label{xy}
\end{equation}
Then the nonlinear constitutive equation of \cite{gol/sht1,gol/sht2} is
\begin{equation}
G_{\mu\nu}=N\left( X,Y\right) F_{\mu\nu}+cM\left( X,Y\right) \tilde
{F}_{\mu\nu}\,.\label{con2
\end{equation}
Although the constitutive equations (\ref{con2}) are fairly general, they do
not take account of a variety of possibilities, such as anisotropic media
\cite{dmi1}, chiral materials where derivative terms enter \cite{vin1},
piroelectric and ferromagnetic materials, and so forth. Therefore it was
proposed \cite{dup/gol/sht} to generalize (\ref{con2}) by introducing the
\textit{constitutive tensors\/} $S_{\mu\nu}$, $R_{\mu\nu}^{\rho\sigma}$, and
$Q_{\mu\nu}^{\rho\sigma\lambda_{1}\ldots\lambda_{n}}$, $n=1,2,3,\,\dots$ and
write a general nonlinear constitutive equation
\begin{align}
G_{\mu\nu} & =S_{\mu\nu}\,+\,R_{\mu\nu}^{\rho\sigma}F_{\rho\sigma
\,+\,Q_{\mu\nu}^{\rho\sigma\lambda_{1}}D_{\lambda_{1}}F_{\rho\sigma
}\label{con}\\
& +\,Q_{\mu\nu}^{\rho\sigma\lambda_{1}\lambda_{2}}D_{\lambda_{1}
D_{\lambda_{2}}F_{\rho\sigma}\,+\ \ldots\,+\,Q_{\mu\nu}^{\rho\sigma\lambda
_{1}\ldots\lambda_{n}}D_{\lambda_{1}}D_{\lambda_{2}}\ldots D_{\lambda_{n
}F_{\rho\sigma}\ .
\end{align}
Evidently the formula (\ref{con}), taken together with Maxwell's equations,
includes all possibilities discussed so far, as well as new ones,
defining the general nonlinear electromagnetic theory. We impose Lorentz
covariance on the constitutive equations, the constitutive tensors will depend
on the fields through the invariants $X$ and $Y$, as follows: $S_{\mu\nu}$ is
a constant independent of $X$ and $Y$, while
\begin{equation}
R_{\mu\nu}^{\rho\sigma}=R_{\mu\nu}^{\rho\sigma}\left( X,Y\right)
,\ \ \ \ \ Q_{\mu\nu}^{\rho\sigma\lambda_{1}\ldots\lambda_{n}}=Q_{\mu\nu
}^{\rho\sigma\lambda_{1}\ldots\lambda_{n}}\left( X,Y,\ldots\right) ,
\end{equation}
where \textquotedblleft\ldots\textquotedblright\ denotes covariant derivatives
of the invariants $X$, $Y$ up to $n$-th order. Obviously $S_{\mu\nu}$ is
antisymmetric, $R_{\mu\nu}^{\rho\sigma}$ is antisymmetric in its upper and
lower indices separately, and the $Q_{\mu\nu}^{\rho\sigma\lambda_{1
\ldots\lambda_{n}}$ are antisymmetric in their upper and first two lower
indices; with respect to the $\lambda_{i}$, they are symmetric Lorentz tensors.
If the equations of motion for the nonlinear theory derive from a Lagrangian
$L\left( X,Y\right) $ which is a scalar function of the invariants $X$ and
$Y$ but does not depend on their derivatives, then from the usual definitions
together with Eq. (\ref{con2}) we have
\begin{equation}
G_{\mu\nu}=-\dfrac{\partial L\left( X,Y\right) }{\partial X}F_{\mu\nu
}-\dfrac{\partial L\left( X,Y\right) }{\partial Y}\tilde{F}_{\mu\nu
}.\label{gf
\end{equation}
Comparing (\ref{con}) and (\ref{gf}) gives us the constitutive tensors,
\begin{align}
& S_{\mu\nu}=0,\nonumber\\
& R_{\mu\nu}^{\rho\sigma}=-\dfrac{\partial L\left( X,Y\right) }{\partial
X}\delta_{\lbrack\mu}^{\rho}\delta_{\nu]}^{\sigma}-\dfrac{\partial L\left(
X,Y\right) }{\partial Y}\varepsilon_{\lbrack\mu\nu]\lambda\delta
\eta^{\lambda\rho}\eta^{\delta\sigma},\label{rl}\\
& Q_{\mu\nu}^{\rho\sigma\lambda_{1}\ldots\lambda_{n}}=0.\nonumber
\end{align}
Now we can relate this formulation of nonlinear electrodynamics with gravity.
\section{Gravity as a gauge theory}
Here we consider the formulation of the Einstein gravity as $SL\left(
2,C\right) $ gauge theory \cite{carmeli}. The gauge potentials are
interpreted in terms of an affine connection on a complex vector bundle. In
the $SL\left( 2,C\right) $ gauge theory they are taken to be the dyad
components of the spinor connection, and are therefore the Newman--Penrose
spin coefficients.
Let $\Gamma_{\mu\ \ B}^{\ A}$ be the complex components of the spinor
connection $\nabla$, i.e.
\begin{equation}
\nabla_{\mu}\xi_{a}^{\ A}=\Gamma_{\mu\ \ B}^{\ \ A} \; \xi_{a}^{\ B},
\end{equation}
where $\xi_{a}^{A}$ is a spinor basis of the complex vector bundle. The dyad
components of the spinor connection are defined by
\begin{equation}
B_{\mu\ a}^{\ \ \ \ b} \equiv\Gamma_{\mu\ A}^{\ \ \ \ \ B} \;
\xi_{a}^{\ A} \; \xi_{\ B}^{b}.
\end{equation}
The field strength of the gauge theory (spinor curvature tensor) can be
expressed through $\Gamma_{\mu}^{\ \ AB}$ as
\begin{equation}
\hat{F}_{\mu\nu\ A}^{\ \ \ \ \ B}=\partial_{\nu}\Gamma_{\mu\ A} ^{\ \ \ \ \ B
-\partial_{\mu}\Gamma_{\nu\ A}^{\ \ \ \ \ B} +\Gamma_{\mu\ A}^{\ \ \ \ \ C} \;
\Gamma_{\nu\ C}^{\ \ \ \ \ B}-\Gamma_{\nu\ A}^{\ \ \ \ \ C} \; \Gamma_{\mu
\ C}^{\ \ \ \ \ B}.
\end{equation}
Thus, we can rewrite $\hat{F}_{\mu\nu\ A}^{\ \ \ \ \ B}$ directly through the
Riemann curvature tensor $R_{\alpha\beta\mu\nu}$ as follows:
\begin{equation}
\hat{F}_{\mu\nu\ a}^{\ \ \ \ \ b}=\dfrac{1}{2} R_{\ \ \beta\mu\nu}^{\alpha}
\sigma_{\alpha ac^{\prime}}\sigma^{\beta bc^{\prime}}, \label{fr
\end{equation}
where $\sigma_{\alpha ab^{\prime}} \equiv\sigma_{\alpha AB^{\prime}} \xi
_{a}^{\; A} {\overline\xi}_{b^{\prime}}^{\; B^{\prime}}$, the $\sigma_{\alpha
AB^{\prime}}$ being the Infeld--van der Waerden symbols which turn tensors
into spinors and express the isomorphism between the tangent space at a point
of space-time and the tensor product of unprimed spin-space and primed
spin-space. In Minkowski space-time they reduce to the Pauli spin matrices.
The field strength obeys the Bianchi identity
\begin{equation}
\nabla_{\rho}\hat{F}_{\mu\nu\ A}^{\ \ \ \ \ B} +\mathrm{permutation}=0,
\label{bian
\end{equation}
which are indeed the second set of the Maxwell equations. The Riemann tensor
$R_{\alpha\beta\mu\nu}$ can be decomposed into the Weyl tensor $W_{\alpha
\beta\mu\nu}$, the Ricci tensor $R_{\mu\nu}$ and the Ricci scalar $R$ as
\begin{equation}
R_{\mu\nu\rho\sigma}=W_{\mu\nu\rho\sigma}+g_{\mu\lbrack\rho}R_{\sigma]\nu
}+g_{\nu\lbrack\rho}R_{\sigma]\mu}+\dfrac{1}{2}Rg_{\mu\lbrack\sigma}
g_{\rho]\nu}. \label{r
\end{equation}
The Einstein equations are
\begin{equation}
R_{\mu\nu}=k\left( T_{\mu\nu}-\dfrac{1}{2}g_{\mu\nu}T\right) .
\end{equation}
Thus, the decomposition (\ref{r}) becomes
\begin{equation}
R_{\mu\nu\rho\sigma}=W_{\mu\nu\rho\sigma}+2kg_{[\rho\lbrack\mu}R_{\nu
]\sigma]]}-\dfrac{1}{2}kg_{\mu\lbrack\sigma}g_{\rho]\nu}T. \label{rc
\end{equation}
A corresponding expression can be written for the field-strength tensor
\begin{equation}
\hat{F}_{\mu\nu}=\hat{F}_{\mu\nu}^{W}+k\tau_{\mu\nu}, \label{ff
\end{equation}
\begin{equation}
F_{\mu\nu}=G_{\mu\nu}-S_{\mu\nu
\end{equation}
\begin{align}
G_{\mu\nu} & =S_{\mu\nu}\,+\,R_{\mu\nu}^{\rho\sigma}F_{\rho\sigma
}\,+0\Longrightarrow\\
\text{Einstein gravity }R_{\mu\nu}^{\rho\sigma} & =\delta_{\mu}^{\rho}
\delta_{\nu}^{\sigma},S_{\mu\nu}=-k\tau_{\mu\nu},Q=0.
\end{align}
This suggests the general structure
\begin{align}
S_{\mu\nu} & =C\left( x\right) \delta_{\lbrack\mu}^{\rho}\delta_{\nu
]}^{\sigma} \varepsilon_{\lbrack\mu\nu]\lambda\delta},\\
R_{\mu\nu}^{\rho\sigma} & =A\left( X,Y\right) \delta_{\lbrack\mu}^{\rho
}\delta_{\nu]}^{\sigma}+B\left( X,Y\right) \varepsilon_{\lbrack\mu
\nu]\lambda\delta}\eta^{\lambda\rho}\eta^{\delta\sigma},
\end{align}
\medskip while $Q$ is an as yet unspecified tensor $Q_{\mu\nu}^{\rho
\sigma\lambda}$.
Let us introduce the covariant derivative
\begin{equation}
D_{\mu}^{\left( B\right) }X=\partial_{\mu}X-\left[ B_{\mu},X\right] ,
\end{equation}
then we can write the Bianchi identity as
\begin{equation}
\varepsilon^{\mu\nu\rho\sigma}D_{\nu}^{\left( B\right) }\hat{F}_{\rho\sigma
}=0. \label{e0
\end{equation}
By using the formula (\ref{ff}) we obtain
\begin{equation}
\varepsilon^{\mu\nu\rho\sigma}D_{\nu}^{\left( B\right) }\hat{F}_{\rho\sigma
}^{W}=kJ^{\mu}, \label{ej
\end{equation}
where $J^{\mu}$ is the current corresponding to the matter
\begin{equation}
J^{\mu}=-\varepsilon^{\mu\nu\rho\sigma}D_{\nu}^{\left( B\right) }\tau
_{\rho\sigma}\sim T.
\end{equation}
Now we are ready to compare the above formulas with the standard nonlinear
electrodynamics: the Maxwell equations (\ref{max4}) and the constitutive
equations (\ref{con}). We can identify (\ref{con}) with (\ref{ff}) and obtain
\begin{align}
G_{\mu\nu} & =\hat{F}_{\mu\nu}^{W},\\
F_{\mu\nu} & =\hat{F}_{\mu\nu},\\
S_{\mu\nu} & =-k\tau_{\mu\nu},\\
j^{\mu} & =kJ^{\mu}.
\end{align}
It follows from (\ref{ej}) that
\begin{equation}
\varepsilon^{\mu\nu\rho\sigma}D_{\nu}^{\left( B\right) } G_{\rho\sigma
}=j^{\mu}.
\end{equation}
\medskip
To obtain the analogy with the first set of the standard Maxwell equations
$\partial_{\mu}G^{\mu\nu}=j^{\nu}$, we introduce the Hodge dual tensor
\begin{equation}
\tilde{G}^{\mu\nu}=\dfrac{1}{2}\varepsilon^{\mu\nu\rho\sigma}G_{\rho\sigma}.
\end{equation}
Then we have
\begin{equation}
D_{\nu}^{\left( B\right) }\tilde{G}^{\mu\nu}=\dfrac{1}{2}j^{\mu}.
\end{equation}
\section{Nonlinear gravito-electromagnetism}
Here we will turn to 3-dimensional notation for convenience and examples. Let
us consider the gravito-electromagnetism in the weak-field approximation
(following, e.g., \cite{clark}). Recall the standard Maxwell equations in SI
units \cite{jackson
\begin{align}
\operatorname{curl}\mathbf{E} & =-\dfrac{\partial\mathbf{B}}{\partial
t},\ \ \ \ \ \ \operatorname{div}\mathbf{B}=0,\nonumber\\
\operatorname{curl}\mathbf{H} & =\dfrac{\partial\mathbf{D}}{\partial
t}+\mathbf{j},\ \ \ \operatorname{div}\mathbf{D}=\rho, \label{max3a
\end{align}
where $\mathbf{E}$ is the electric field, $\mathbf{B}$ is the magnetic field,
$\rho$ is charge density, $\mathbf{j}$ is electric current density. In the
linear case
\begin{equation}
\mathbf{B}=\mu_{0}\mathbf{H},\ \ \ \mathbf{D}=\varepsilon_{0}\mathbf{E,}
\label{be
\end{equation}
In the nonlinear case these equations can be presented in the form
\cite{fus/sht/ser}
\begin{align}
\mathbf{D} & =M\left( I_{1},I_{2}\right) \mathbf{B}+\dfrac{1}{c^{2
}N\left( I_{1},I_{2}\right) \mathbf{E},\nonumber\\
\mathbf{H} & =N\left( I_{1},I_{2}\right) \mathbf{B}-M\left( I_{1
,I_{2}\right) \mathbf{E}, \label{con1
\end{align}
where the invariants are
\begin{equation}
I_{1}=\mathbf{B}^{2}-\dfrac{1}{c^{2}}\mathbf{E}^{2},\ I_{2}=\mathbf{B
\cdot\mathbf{E.\
\end{equation}
Their gravitational analogues in SI are
\begin{align}
\operatorname{curl}\mathbf{E}_{g} & =-\dfrac{\partial\mathbf{B}_{g
}{\partial t},\ \ \ \ \ \ \operatorname{div}\mathbf{B}_{g}=0,\label{maxg0}\\
\operatorname{curl}\mathbf{B}_{g} & =\dfrac{1}{c^{2}}\dfrac{\partial
\mathbf{E}_{g}}{\partial t}+\dfrac{1}{\varepsilon_{g}c^{2}}\mathbf{j
_{g},\ \ \ \operatorname{div}\mathbf{E}_{g}=\dfrac{1}{\varepsilon_{g}}\rho
_{g}, \label{maxg
\end{align}
where $\mathbf{E}_{g}$ is the static gravitational field (conventional
gravity, also called gravitoelectric for the sake of analogy), $\mathbf{B
_{g}$ is the gravitomagnetic field, $\rho_{g}$ is mass density, $\mathbf{j
_{g}$ is mass current density, $G$ is the gravitational constant,
$\varepsilon_{g}$ is the gravity permittivity (analog of $\varepsilon_{0})$.
Here
\begin{equation}
\varepsilon_{g}=-\dfrac{1}{4\pi G},\ \ \ \ \ \mu_{g}=-\dfrac{4\pi G}{c^{2}},
\label{em
\end{equation}
are the gravitational permittivity and permeability, respectively.
The main idea is to introduce analogues of $\mathbf{H}$ and $\mathbf{D}$ to
write (\ref{maxg}) in the Maxwell form for 4 fields in SI as
\begin{align}
\operatorname{curl}\mathbf{E}_{g} & =-\dfrac{\partial\mathbf{B}_{g
}{\partial t},\ \ \ \ \ \ \operatorname{div}\mathbf{B}_{g}=0,\label{maxgg0}\\
\operatorname{curl}\mathbf{H}_{g} & =\dfrac{\partial\mathbf{D}_{g}}{\partial
t}+\mathbf{j}_{g},\ \ \ \operatorname{div}\mathbf{D}_{g}=\rho_{g}.
\label{maxgg}
\end{align}
In the linear-gravity case
\begin{align}
\mathbf{D}_{g} & =\varepsilon_{g}\mathbf{E}_{g},\label{de}\\
\mathbf{B}_{g} & =\mu_{g}\mathbf{H}_{g},\label{bh}\\
\varepsilon_{g}\mu_{g} & =\dfrac{1}{c^{2}}. \label{mc
\end{align}
The main idea is the following: the linear-gravity case
(\ref{de})--(\ref{mc}) corresponds to weak approximation
and some special case of
gravitational field configuration. We generalize it
to nonlinear case which can
explain other configurations and non-weak fields, as in (\ref{con1}), by
\begin{align}
\mathbf{D}_{g} & =M_{g}\left( I_{g1},I_{g2}\right) \mathbf{B}_{g
+\dfrac{1}{c^{2}}N_{g}\left( I_{g1},I_{g2}\right) \mathbf{E}_{g
,\label{dg}\\
\mathbf{H}_{g} & =N_{g}\left( I_{g1},I_{g2}\right) \mathbf{B}_{g
-M_{g}\left( I_{g1},I_{g2}\right) \mathbf{E}_{g}, \label{hg
\end{align}
where the invariants are
\begin{equation}
I_{g1}=\mathbf{B}_{g}^{2}-\dfrac{1}{c^{2}}\mathbf{E}_{g}^{2},\ I_{g2
=\mathbf{B}_{g}\cdot\mathbf{E}_{g}\mathbf{.\ } \label{ig
\end{equation}
The gravity-Maxwell equations (\ref{maxgg0})--(\ref{maxgg}) together with the
nonlinear gravity-constitutive equations (\ref{dg})-(\ref{hg}) can give
a nonlinear electrodynamics formulation of gravity
(at least its particular cases).
\section{Linear gravito-electromagnetic waves}
The gravity-Maxwell equations for gravito-electromagnetic waves (far from
sources) are
\begin{align}
\operatorname{curl}\mathbf{E}_{g} & =-\dfrac{\partial\mathbf{B}_{g}}{\partial
t},\operatorname{div}\mathbf{B}_{g}=0,\\
\operatorname{curl}\mathbf{H}_{g} & =\dfrac{\partial\mathbf{D}_{g}}{\partial
t},\operatorname{div}\mathbf{D}_{g}=0
\end{align}
with generic values of permittivity and permeability (\ref{em}). Then
\begin{align}
\operatorname{curl}\mathbf{E}_{g} & =-\mu_{g}\dfrac{\partial\mathbf{H}_{g
}{\partial t},\operatorname{div}\mathbf{H}_{g}=0,\\
\operatorname{curl}\mathbf{H}_{g} & =\varepsilon_{g}\dfrac{\partial
\mathbf{E}_{g}}{\partial t},\operatorname{div}\mathbf{E}_{g}=0.
\end{align}
We differentiate the first equation with respect to time: $\operatorname{curl
\dfrac{\partial}{\partial t}\mathbf{E}_{g}=-\mu_{g}\dfrac{\partial
^{2}\mathbf{H}_{g}}{\partial t^{2}}$ $\Rightarrow\dfrac{1}{\varepsilon_{g
}\operatorname{curl}\left( \operatorname{curl}\mathbf{H}_{g}\right)
=-\mu_{g}\dfrac{\partial^{2}\mathbf{H}_{g}}{\partial t^{2}}.$ Since
$\operatorname{curl}\left( \operatorname{curl}\mathbf{H}_{g}\right)
=\operatorname{grad}(\operatorname{div}\mathbf{H}_{g})-\Delta\mathbf{H
_{g}=-\Delta\mathbf{H}_{g}$, then
\begin{equation}
\Delta\mathbf{H}_{g}=\varepsilon_{g}\mu_{g}\dfrac{\partial^{2}\mathbf{H}_{g
}{\partial t^{2}}.
\end{equation}
By analogy, from the second equation $\operatorname{curl}\dfrac{\partial
}{\partial t}\mathbf{H}_{g}=\varepsilon_{g}\dfrac{\partial^{2}\mathbf{E}_{g
}{\partial t^{2}}$ $\Rightarrow$ $\dfrac{-1}{\mu_{g}}\operatorname{curl
\left( \operatorname{curl}\mathbf{E}_{g}\right) =\varepsilon_{g
\dfrac{\partial^{2}\mathbf{E}_{g}}{\partial t^{2}}$. Hence we get the wave
equation for $\mathbf{E}_{g}$,
\begin{equation}
\Delta\mathbf{E}_{g}=\varepsilon_{g}\mu_{g}\dfrac{\partial^{2}\mathbf{E}_{g
}{\partial t^{2}}.
\end{equation}
\section{Nonlinear gravito-electromagnetic waves}
The differences begin with the constitutive equations (\ref{dg})--(\ref{hg}).
For simplicity put $M_{g}=0$. Then
\begin{align}
\mathbf{D}_{g} & =\dfrac{N}{c^{2}}\mathbf{E}_{g},\\
\mathbf{B}_{g} & =\dfrac{1}{N}\mathbf{H}_{g
\end{align}
where $N\equiv N_{g}\left( I_{g1},I_{g2}\right) $. The Maxwell equations
become (hereafter the dots denote time derivatives)
\begin{align}
\operatorname{curl}\mathbf{E}_{g} & =-\left( \dfrac{1}{N}\right) ^{\bullet
}\mathbf{H}_{g}-\dfrac{1}{N}\dfrac{\partial\mathbf{H}_{g}}{\partial
t},\label{1}\\
\operatorname{div}\left( \dfrac{1}{N}\mathbf{H}_{g}\right) &
=\mathbf{H}_{g}\operatorname{grad}\left( \dfrac{1}{N}\right) +\dfrac{1
{N}\operatorname{div}\left( \mathbf{H}_{g}\right) =0,\label{2}\\
\operatorname{curl}\mathbf{H}_{g} & =\dfrac{\dot{N}}{c^{2}}\mathbf{E
_{g}+\dfrac{N}{c^{2}}\dfrac{\partial\mathbf{E}_{g}}{\partial t},\label{3}\\
\operatorname{div}\left( \dfrac{N}{c^{2}}\mathbf{E}_{g}\right) &
=\mathbf{E}_{g}\operatorname{grad}\left( \dfrac{N}{c^{2}}\right) +\dfrac
{N}{c^{2}}\operatorname{div}\left( \mathbf{E}_{g}\right) =0.\label{4
\end{align}
Take derivative of (\ref{1}) with respect to time and get
\begin{equation}
\operatorname{curl}\dfrac{\partial}{\partial t}\mathbf{E}_{g}=-\left(
\dfrac{1}{N}\right) ^{\bullet\bullet}\mathbf{H}_{g}-2\left( \dfrac{1
{N}\right) ^{\bullet}\dfrac{\partial\mathbf{H}_{g}}{\partial t}-\dfrac{1
{N}\dfrac{\partial^{2}\mathbf{H}_{g}}{\partial t^{2}}.
\end{equation}
From (\ref{3}) it follows $\dfrac{\partial\mathbf{E}_{g}}{\partial t
=\dfrac{c^{2}}{N}\operatorname{curl}\mathbf{H}_{g}-\dfrac{\dot{N}
{N}\mathbf{E}_{g}$. Then we get
\begin{equation}
\operatorname{curl}\left( \dfrac{c^{2}}{N}\operatorname{curl}\mathbf{H
_{g}-\dfrac{\dot{N}}{N}\mathbf{E}_{g}\right) =-\left( \dfrac{1}{N}\right)
^{\bullet\bullet}\mathbf{H}_{g}-2\left( \dfrac{1}{N}\right) ^{\bullet
\dfrac{\partial\mathbf{H}_{g}}{\partial t}-\dfrac{1}{N}\dfrac{\partial
^{2}\mathbf{H}_{g}}{\partial t^{2}}.
\end{equation}
The left-hand side here is
\begin{align*}
& \operatorname{curl}\left( \dfrac{c^{2}}{N}\operatorname{curl}\mathbf{H
_{g}-\dfrac{\dot{N}}{N}\mathbf{E}_{g}\right) \\
& =\operatorname{grad}\dfrac{c^{2}}{N}\times\operatorname{curl}\mathbf{H
_{g}+\dfrac{c^{2}}{N}\operatorname{grad}\operatorname{div}\mathbf{H
_{g}-\dfrac{c^{2}}{N}\Delta\mathbf{H}_{g}-\dfrac{\dot{N}}{N
\operatorname{curl}\mathbf{E}_{g}-\operatorname{grad}\dfrac{\dot{N}}{N
\times\mathbf{E}_{g}.
\end{align*}
From (\ref{2}) we get $\operatorname{div}\left( \mathbf{H}_{g}\right)
=-N\mathbf{H}_{g}\operatorname{grad}\left( \dfrac{1}{N}\right) \neq0$. Thus,
the nonlinear analogue of the wave equation is
\begin{align}
& \operatorname{grad}\dfrac{c^{2}}{N}\times\operatorname{curl}\mathbf{H
_{g}+\dfrac{c^{2}}{N}\operatorname{grad}\operatorname{div}\mathbf{H
_{g}-\dfrac{c^{2}}{N}\Delta\mathbf{H}_{g}-\dfrac{\dot{N}}{N
\operatorname{curl}\mathbf{E}_{g}-\operatorname{grad}\dfrac{\dot{N}}{N
\times\mathbf{E}_{g}\\
& =-\left( \dfrac{1}{N}\right) ^{\bullet\bullet}\mathbf{H}_{g}-2\left(
\dfrac{1}{N}\right) ^{\bullet}\dfrac{\partial\mathbf{H}_{g}}{\partial
t}-\dfrac{1}{N}\dfrac{\partial^{2}\mathbf{H}_{g}}{\partial t^{2}}.
\end{align}
Note that if $N=const$, then we obtain the usual wave equation
\begin{equation}
\Delta\mathbf{H}_{g}=\dfrac{1}{c^{2}}\dfrac{\partial^{2}\mathbf{H}_{g
}{\partial t^{2}}.
\end{equation}
Take now the constitutive equations in the form
\begin{align}
\mathbf{D}_{g} & =M\mathbf{B}_{g}+\dfrac{N}{c^{2}}\mathbf{E}_{g},\\
\mathbf{H}_{g} & =N\mathbf{B}_{g}-M\mathbf{E}_{g},
\end{align}
where $N,M$ are constants. In absence of sources, the Maxwell equations
become
\begin{align}
\operatorname{curl}\mathbf{E}_{g} & =-\dfrac{\partial\mathbf{B}_{g}}{\partial
t},\ \ \ \operatorname{div}\mathbf{B}_{g}=0,\\
\operatorname{curl}\mathbf{H}_{g} & =\dfrac{\partial\mathbf{D}_{g}}{\partial
t},\ \ \ \operatorname{div}\mathbf{D}_{g}=0.
\end{align}
If we express the Maxwell equations through $\mathbf{E}_{g}$ and
$\mathbf{B}_{g}$, then second pair of equations become
\begin{align}
\operatorname{curl}\mathbf{H}_{g} & =\dfrac{\partial\mathbf{D}_{g}}{\partial
t},\\
N\operatorname{curl}\mathbf{B}_{g}-M\operatorname{curl}\mathbf{E}_{g} &
=M\dfrac{\partial\mathbf{B}_{g}}{\partial t}+\dfrac{N}{c^{2}}\dfrac
{\partial\mathbf{E}_{g}}{\partial t}.
\end{align}
Since $\operatorname{curl}\mathbf{E}_{g}=-\dfrac{\partial\mathbf{B}_{g
}{\partial t}$, from the last equation we get
\begin{equation}
\operatorname{curl}\mathbf{B}_{g}=\dfrac{1}{c^{2}}\dfrac{\partial
\mathbf{E}_{g}}{\partial t}.
\end{equation}
The second equation, $\operatorname{div}\mathbf{D}_{g}=0$, reduces to
$M\operatorname{div}\mathbf{B}_{g}+\dfrac{N}{c^{2}}\operatorname{div
\mathbf{E}_{g}=0$. Since $\operatorname{div}\mathbf{B}_{g}=0,$ we get
\begin{equation}
\operatorname{div}\mathbf{E}_{g}=0.
\end{equation}
Thus, using constitutive equations with constant $M$ and $N$ we have Maxwell
equations in terms of $\mathbf{B}_{g}$ and $\mathbf{E}_{g}$, i.e.
\begin{align}
\operatorname{curl}\mathbf{E}_{g} & =-\dfrac{\partial\mathbf{B}_{g}}{\partial
t},\ \ \ \operatorname{div}\mathbf{B}_{g}=0,\\
\operatorname{curl}\mathbf{B}_{g} & =\dfrac{1}{c^{2}}\dfrac{\partial
\mathbf{E}_{g}}{\partial t},\ \ \ \operatorname{div}\mathbf{E}_{g}=0.
\end{align}
At this stage, we get the wave equations in the standard way. The time
derivative of the first equation yields $\operatorname{curl}\dfrac{\partial
}{\partial t}\mathbf{E}_{g}=-\dfrac{\partial^{2}\mathbf{B}_{g}}{\partial
t^{2}}$ $\Rightarrow c^{2}\operatorname{curl}\left( \operatorname{curl
\mathbf{B}_{g}\right) =-\dfrac{\partial^{2}\mathbf{B}_{g}}{\partial t^{2}}$.
Since $\operatorname{curl}\left( \operatorname{curl}\mathbf{B}_{g}\right)
=\operatorname{grad}(\operatorname{div}\mathbf{B}_{g})-\Delta\mathbf{B
_{g}=-\Delta\mathbf{B}_{g}$, then
\begin{equation}
\Delta\mathbf{B}_{g}=\dfrac{1}{c^{2}}\dfrac{\partial^{2}\mathbf{B}_{g
}{\partial t^{2}}.
\end{equation}
By analogy $\operatorname{curl}\dfrac{\partial}{\partial t}\mathbf{B
_{g}=\dfrac{1}{c^{2}}\dfrac{\partial^{2}\mathbf{E}_{g}}{\partial t^{2}}$
$\Rightarrow$ $-\operatorname{curl}\left( \operatorname{curl}\mathbf{E
_{g}\right) =\dfrac{1}{c^{2}}\dfrac{\partial^{2}\mathbf{E}_{g}}{\partial
t^{2}}$, and we get the wave equation for $\mathbf{E}_{g}$,
\begin{equation}
\Delta\mathbf{E}_{g}=\dfrac{1}{c^{2}}\dfrac{\partial^{2}\mathbf{E}_{g
}{\partial t^{2}}.
\end{equation}
Thus, the gravi-electromagnetic waves $\mathbf{E}_{g}$ and $\mathbf{B}_{g}$
have speed $c$ and do not depend on the constants $M$ and $N$.
\section{Waves and constitutive equations for linear constitutive functions}
Let us consider the constitutive equations (\ref{dg})--(\ref{hg}) as linear
functions of invariants, i.e.
\begin{align}
M & =M_{g}\left( I_{g1},I_{g2}\right) =a_{m}I_{g1}+b_{m}I_{g2},\\
N & =N_{g}\left( I_{g1},I_{g2}\right) =c^{2}\varepsilon_{g}+a_{n
I_{g1}+b_{n}I_{g2},
\end{align}
$a_{m},b_{m},a_{n},b_{n}$ being some constants. From all the Maxwell equations
in material media, and in the absence of sources one finds
$\operatorname{curl}\mathbf{H}_{g}=\dfrac{\partial\mathbf{D}_{g}}{\partial t
$, $\operatorname{curl}\left( N\mathbf{B}_{g}-M\mathbf{E}_{g}\right)
=\dfrac{\partial}{\partial t}\left( M\mathbf{B}_{g}+\dfrac{N}{c^{2
}\mathbf{E}_{g}\right) $, and $N\operatorname{curl}\mathbf{B}_{g
-M\operatorname{curl}\mathbf{E}_{g}=M\dfrac{\partial\mathbf{B}_{g}}{\partial
t}+\dfrac{N}{c^{2}}\dfrac{\partial\mathbf{E}_{g}}{\partial t}$. Since
$\operatorname{curl}\mathbf{E}_{g}=-\dfrac{\partial\mathbf{B}_{g}}{\partial
t}$, from the last equation one gets
\begin{equation}
\operatorname{curl}\mathbf{B}_{g}=\dfrac{1}{c^{2}}\dfrac{\partial
\mathbf{E}_{g}}{\partial t}.
\end{equation}
The second equation, $\operatorname{div}\mathbf{D}_{g}=0$, reduces to
$\operatorname{div}\left( M\mathbf{B}_{g}+\dfrac{N}{c^{2}}\mathbf{E
_{g}\right) =0$, or $M\operatorname{div}\mathbf{B}_{g}+\dfrac{N}{c^{2
}\operatorname{div}\mathbf{E}_{g}=0$. Since $\operatorname{div}\mathbf{B
_{g}=0$, one gets
\begin{equation}
\operatorname{div}\mathbf{E}_{g}=0.
\end{equation}
\section{Inverse problem of nonlinear gravito-electromagnetism}
In electrodynamics the direct solution of the Maxwell equations together with
the nonlinear constitutive equations is a nontrivial and complicated task even
for simple systems \cite{gol/sht1,gol/sht2}. In previous sections we presented
some very special cases of the nonlinear functions $N$ and $M$. Here we
formulate the following inverse problem: if we have some particular solution
of the gravity-Maxwell equations (\ref{maxgg0})--(\ref{maxgg}), can we then
find the exact form of the corresponding nonlinear gravity-constitutive
equations (\ref{dg})-(\ref{hg})?
It is natural to consider the case of plane gravitational waves, when the
fields have only one space coordinate. We will show that even in this case one
can have a nontrivial nonlinearity. Let us choose $\mathbf{E}_{g}$ and
$\mathbf{B}_{g}$ mutually orthogonal and perpendicular to the direction of
motion
\begin{equation}
\mathbf{E}_{g}=\left(
\begin{array}
[c]{c
E\\
0\\
0
\end{array}
\right) ,\ \ \ \mathbf{B}_{g}=\left(
\begin{array}
[c]{c
0\\
0\\
B
\end{array}
\right) ,\label{e
\end{equation}
where $E\equiv E\left( t,y\right)$, $B\equiv B\left( t,y\right) $.
Now the invariants (\ref{ig}) become
\begin{align}
I_{g1} & =B^{2}-\dfrac{1}{c^{2}}E^{2}\equiv I,\label{i}\\
I_{g2} & =0\mathbf{.\ }
\end{align}
The use of the nonlinear gravity-constitutive equations (\ref{dg})-(\ref{hg})
gives for other fields
\begin{equation}
\mathbf{D}_{g}=\left(
\begin{array}
[c]{c}
\dfrac{1}{c^{2}}NE\\
0\\
MB
\end{array}
\right) ,\ \ \ \mathbf{H}_{g}=\left(
\begin{array}
[c]{c}
-ME\\
0\\
NB
\end{array}
\right) ,
\label{dh}
\end{equation}
where $N\equiv N\left(I\right)$, $M\equiv M\left(I\right)$ are the
sought for gravity-constitutive functions. They depend on $I$ only, because of
Lorentz invariance (see \cite{gol/sht1,gol/sht2}). Inserting the fields
(\ref{e}) and (\ref{dh}) into the gravity-Maxwell equations
(\ref{maxgg0})--(\ref{maxgg}) without sources gives us $3$ equations
(hereafter, a prime with the corresponding subscript denotes
the first partial derivative with respect to the variable
in the subscript, while dot denotes time derivative)
\begin{align}
E_{y}' & =\dot{B},\label{eb}\\
\left(NB\right)_{y}' & =\dfrac{1}{c^{2}}\left(NE\right)^{\cdot},
\label{nb}\\
\left(ME\right)_{y}' & =\left(MB\right) ^{\cdot}.
\label{me}
\end{align}
Now we take into account
that the gravity-constitutive functions $N$, $M$
depend only on the invariant $I$ and present (\ref{nb})--(\ref{me})
as the differential equations for them
\begin{align}
N_{I}'\left(BI_{y}'-\dfrac{1}{c^{2}}E\dot{I}\right) +N\left(
B_{y}'-\dfrac{1}{c^{2}}\dot{E}\right) & =0,
\label{ni}\\
M_{I}'\left(EI_{y}'-B\dot{I}\right) & =0,
\label{mi}
\end{align}
where we have exploited the identities
\begin{equation}
N_{y}'=N_{I}' I_{y}', \; M_{y}'=M_{I}' I_{y}',
\end{equation}
\begin{equation}
{\dot N}=N_{I}'{\dot I}, \; {\dot M}=M_{I}'{\dot I}.
\end{equation}
The second equation (\ref{mi}) can be immediately solved by
\begin{equation}
M\left( I\right) =\left\{
\begin{array}
[c]{c}
M_{0}={\rm const},\ \ \ \text{if }EI_{y}'\neq B\dot{I},\\
{\rm arbitrary},\ \ \ \text{if }EI_{y}'=B\dot{I}.
\end{array} \right.
\label{m1}
\end{equation}
The first equation (\ref{ni}) can be solved if
\begin{equation}
\lambda \equiv {\left(B_{y}'-{{\dot E}\over c^{2}}\right) \over
\left(BI_{y}'-{E {\dot I}\over c^{2}}\right)}
\end{equation}
depends only on $I$, which is a very special case.
One then has the differential equation
\begin{equation}
N_{I}'+\lambda\left(I\right)N=0,
\end{equation}
and its solution is
\begin{equation}
N\left( I\right) =N_{0}{\rm e}^{-\int\lambda
\left(I\right) {\rm d}I}.
\end{equation}
Otherwise, by using the expressions for $I_{y}'$ and
$\dot{I}$ from (\ref{i}), i.e.
\begin{equation}
I_{y}'=2BB_{y}'-{2EE_{y}'\over c^{2}}, \;
{\dot I}=2B{\dot B}-{2E{\dot E}\over c^{2}},
\end{equation}
we obtain
\begin{equation}
2N_{I}'\left(B^{2}B_{y}'+\dfrac{1}{c^{4}}E^{2}\dot{E}-\dfrac
{2}{c^{2}}EBE_{y}'\right) +N\left(B_{y}'-\dfrac{1}{c^{2}}\dot
{E}\right)=0,
\label{2ni}
\end{equation}
where the sum of terms in brackets is not a function of $I$ in general.
Usually, in the wave solutions the dependence of fields on frequency $\omega$
and wave number $k$ is the same, and therefore we can consider the
concrete choice
\begin{equation}
E\left(t,y\right) =f\left(\varepsilon\omega t+ky\right)\equiv f(X(t,y)) ,
\ \ \ B\left(t,y\right)
=g\left(\varepsilon\omega t+ky\right)\equiv g(X(t,y)) ,
\end{equation}
where $\varepsilon \equiv \pm 1$, with $f$ and $g$ arbitrary smooth
nonvanishing functions. Bearing in mind that
$$
E_{y}'=f_{X}'X_{y}'=k f_{X}', \;
B_{y}'=g_{X}'X_{y}'=k g_{X}',
$$
$$
{\dot E}=f_{X}'{\dot X}=\varepsilon \omega f_{X}', \;
{\dot B}=g_{X}'{\dot X}=\varepsilon \omega g_{X}',
$$
our Eq. (\ref{eb}) yields
\begin{equation}
kf_{X}'=\varepsilon \omega g_{X}'.
\end{equation}
Therefore
\begin{equation}
g\left(X\right) =\dfrac{k}{\varepsilon \omega}f
\left(X\right) +\alpha,
\end{equation}
where $\alpha$ is a constant, so that both $E$ and $B$ can be
expressed through one function only, i.e. $f$, and the invariant $I$
reads eventually as
\begin{equation}
I=\dfrac{1}{\omega^{2}}\left( k^{2}-\dfrac{\omega^{2}}{c^{2}}\right)
f^{2}+2\dfrac{k}{\varepsilon \omega}\alpha f+\alpha^{2}.
\end{equation}
The equations for the gravity-constitutive functions
take therefore the form
\begin{align}
N_{I}'\left[2I\left(k^{2}-{\omega^{2}\over c^{2}}\right)
+2 {\omega^{2}\over c^{2}} \alpha^{2} \right]
+N \left(k^{2}-{\omega^{2} \over c^{2}}\right) & =0,
\label{w1}\\
M_{I}'f_{X}'\left[{2f \over \varepsilon \omega}
\left(k^{2}-{\omega^{2}\over c^{2}}\right)+2k \alpha \right]
\alpha & =0,
\label{w2}
\end{align}
where we have exploited the identities
\begin{equation}
gI_{y}'-{f {\dot I}\over c^{2}}=\left(gf_{y}'
-{f {\dot f}\over c^{2}}\right)
\left[{2f \over \omega^{2}}\left(k^{2}-{\omega^{2}\over c^{2}}\right)
+2{k \over \varepsilon \omega}\alpha \right]
\end{equation}
\begin{equation}
gf_{y}'-{f {\dot f}\over c^{2}}={f_{X}'\over \varepsilon \omega}
\left[f \left(k^{2}-{\omega^{2}\over c^{2}}\right)
+k \varepsilon \omega \alpha \right],
\end{equation}
and, after some cancellations,
\begin{eqnarray}
\; & \; & \left[f \left(k^{2}-{\omega^{2}\over c^{2}}\right)
+k \varepsilon \omega \alpha \right]
\left[{2f \over \omega^{2}}\left(k^{2}-{\omega^{2}\over c^{2}}
\right)+2{k \over \varepsilon \omega}\alpha \right]
\nonumber \\
&=& 2 \left(k^{2}-{\omega^{2}\over c^{2}}\right)I
+2{\omega^{2}\over c^{2}}\alpha^{2},
\end{eqnarray}
while
\begin{equation}
fI_{y}'-g{\dot I}=(ff_{y}'-g{\dot f})
\left[{2f \over \omega^{2}}
\left(k^{2}-{\omega^{2}\over c^{2}}\right)
+2{k \over \varepsilon \omega}\alpha \right],
\end{equation}
\begin{equation}
ff_{y}'-g{\dot f}=f_{X}'(kf-\varepsilon \omega g)
=-\varepsilon \omega f_{X}' \alpha.
\end{equation}
The results of our analysis now depend on whether or not
$\alpha$ vanishes. Indeed, if $\alpha=0$, $M$ is arbitrary
and hence we obtain the equation
\begin{equation}
\left(k^{2}-{\omega^{2}\over c^{2}}\right)(2IN_{I}'+N)=0,
\end{equation}
which implies that either the dispersion relation
\begin{equation}
k^{2}-{\omega^{2}\over c^{2}}=0
\label{disp}
\end{equation}
holds, with $N$ kept arbitrary, or such a dispersion relation
is not fulfilled, while $N$ is found from the differential equation
\begin{equation}
2IN_{I}'+N=0,
\end{equation}
which is solved by
\begin{equation}
N(I)={N_{0}\over \sqrt{I}}.
\end{equation}
By contrast, if $\alpha$ does not vanish, $M$ equals a constant
$M_{0}$, while $N$ solves the more complicated equation
(\ref{w1}). At this stage, to be consistent with the dependence
of $N$ on $I$ only, we have to require again that the dispersion
relation (\ref{disp}) should hold, jointly with $N_{I}'=0$, which
implies the constancy of $N$: $N=N_{0}$.
\section{Concluding remarks}
We have proposed a way to take an extremely general approach to a nonlinear
formulation of gravity as classical electrodynamics. The framework formally
includes nonLagrangian as well as Lagrangian theories, and accommodates the
description of nonlocal effects. This is accomplished through generalized
constitutive equations (and constitutive tensors). We expect future research
directions to include the detailed development of new examples
within this framework.
\acknowledgments
S. Duplij thanks M. Bianchi, J. Gates, G. Goldin, A. Yu. Kirochkin,
M. Shifman, V. Shtelen, D. Sorokin, A. Schwarz, M. Tonin, A. Vainshtein,
A. Vilenkin for fruitful discussions. E. Di Grezia and G. Esposito are
grateful to the Dipartimento di Fisica of Federico II University, Naples, for
hospitality and support.
|
{
"redpajama_set_name": "RedPajamaArXiv"
}
| 5,955
|
Photobiomodulation – where it started and where is it going?
By Johnathan George
Juanita Anders Ph.D., Professor of Anatomy, Physiology and Genetics and Professor of Neuroscience at Uniformed Services University of the Health Sciences, presented the history of PBM and how the past has lead to the current obstacles to PBM's acceptance. The laser was invented in May of 1960 and by 1963 it was being studied for surgical applications. At the same time the laser was being evaluated as a replacement for the scalpel, its biomodulating effects on tissue were also being explored. In 1965 Endre Mester began experimenting with medical lasers as a treatment for cancer. While the cancer treatment failed, he noticed that the hair on the shaven mice treated with laser light regrew at an increased rate. Dr. Mester was the first to describe biostimulation and realize the therapeutic value of lasers.
While the use of laser as a scalpel replacement in surgery was quickly adopted, Anders discussed how the mechanisms of PBM were unknown and complex, and how it remains relatively unknown and little used today. Possibly due to the complexity of selecting frequency, dose, and modulation when the molecular mechanisms were unknown, many early experiments in PBM reported negative results. Anders also talked about how a major hurdle in the acceptance of PBM was the early failure to unify terminology. Dr. Mester began by describing the therapy as low power laser rays in 1968 and subsequently switched to biostimulation. Industry has used over eight different names for PBM from Cold Laser Therapy to Low Level Laser Therapy (LLLT). Only within the last few years have researchers and clinicians migrated to the common term "photobiomodulation" to describe the therapy.
Next, Michael Hamblin Ph.D., Principal Investigator for at the Wellman Center for Photomedicine at Massachusetts General Hospital, and the leading researcher on the mechanisms of PBM, presented the latest research on the mechanisms of PBM. The research he presented ranged from suppressing pain in mice with PBM to preconditioning mice with PBM for enhanced muscle performance.
In an open discussion later in the day attendees talked about the hurdles to PBM's adoption. The wide number and variety of potential applications for PBM seems to have been an unexpected hindrance to the acceptance of PBM. In each application of PBM there are different types of tissues, all with their own optical and biological parameters. In translating from biological mechanism to insurance-approved clinical application these parameters must be studied and proven individually. This presents a challenge to regulatory agencies and device manufacturers who must focus on specific applications for a device rather than the overly broad set of potential applications.
More to come from this afternoon's session on overcoming these hurtles as well as a closer look at several clinical applications of Photobiomodulation.
Host Donald Patthoff from the Academy Laser Dentistry kicked things off providing an overview of the goals of the Incubator.
Praveen Arany, current president of the North American Association for Photobiomodulation Therapy discussing the future of the technology – and for the community.
Biomedical Optics
OSA Incubator
Posted: 31 August 2015 by Johnathan George | with 0 comments
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
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"I have been skiing for over 50 years, 40 of them at PCMR. For the last 20 years I have been a part of a gang that skies every Tuesday morning before we go to Rotary lunch. I am an expert skier, but after passing age 60, I noticed that I was slowing down; over the years with the same gang, I was slowly moving further back in the pack on arrival at the lift. Part of this was that some younger folks had joined in over the years.
Jim did a demo of SlackBow for our Rotary club and then invited us in for a private session. I had a one-hour visit where Jim put me through the entire spectrum of activities he had devised. The following week, skiing with the same bunch in same conditions, I found myself standing at the lift at the bottom of the hill waiting for the gang to catch up. One of the gang is the father of the world champion ski racer and a very fine skier—my age—himself.
I may not be 25 anymore, but I am skiing better than I have in years.
one of Jim Klopman's SlackBow courses, and it was everything I hoped it would be. I felt much more comfortable on my skis, especially at speed or in difficult conditions, and overall much more balanced and sure of myself. I found it also helped my cycling, both road and mountain.
Jim Klopman, through his research and innovative thinking in and out of the box, has brought this critical life issue into a broader and more complete understanding.
"Quite a few years after my career as a professional ballet dancer and a few after a shorter stint as a Certified Personal Trainer, I thought I knew a thing or two about balance from my training in these two areas. Especially when I began training older adults, including yours truly. I observed firsthand how older people (me included) struggled with balance, sometimes with serious injuries. With younger ballet dancers, you are constantly challenging your balance with an endless variety of movement.
|
{
"redpajama_set_name": "RedPajamaC4"
}
| 5,727
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{"url":"https:\/\/gmatclub.com\/forum\/if-y-y-0-which-of-the-following-must-be-true-131099.html","text":"GMAT Question of the Day - Daily to your Mailbox; hard ones only\n\n It is currently 14 Dec 2018, 20:37\n\n### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we\u2019ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n## Events & Promotions\n\n###### Events & Promotions in December\nPrevNext\nSuMoTuWeThFrSa\n2526272829301\n2345678\n9101112131415\n16171819202122\n23242526272829\n303112345\nOpen Detailed Calendar\n\u2022 ### Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student\n\nDecember 14, 2018\n\nDecember 14, 2018\n\n10:00 PM PST\n\n11:00 PM PST\n\nCarolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.\n\u2022 ### Free GMAT Strategy Webinar\n\nDecember 15, 2018\n\nDecember 15, 2018\n\n07:00 AM PST\n\n09:00 AM PST\n\nAiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.\n\n# If y + | y | = 0, which of the following must be true?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nTAGS:\n\n### Hide Tags\n\nManager\nStatus: May The Force Be With Me (D-DAY 15 May 2012)\nJoined: 06 Jan 2012\nPosts: 220\nLocation: India\nConcentration: General Management, Entrepreneurship\nIf y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n21 Apr 2012, 20:15\n4\n10\n00:00\n\nDifficulty:\n\n15% (low)\n\nQuestion Stats:\n\n67% (00:36) correct 33% (00:42) wrong based on 867 sessions\n\n### HideShow timer Statistics\n\nIf y + | y | = 0, which of the following must be true?\n\nA. y > 0\nB. y \u2265 0\nC. y < 0\nD. y \u2264 0\nE. y = 0\n\nWhy is just E incorrect?\n\n_________________\n\nGiving +1 kudos is a better way of saying 'Thank You'.\n\nMath Expert\nJoined: 02 Sep 2009\nPosts: 51215\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n21 Apr 2012, 21:23\n4\n2\nboomtangboy wrote:\nIf y + | y | = 0, which of the following must be true?\n(A) y > 0\n(B) y\u22650\n(C) y < 0\n(D) y\u22640\n(E) y = 0\n\nWhy is just E incorrect?\n\nAbsolute value properties:\nWhen $$x\\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \\ expression\\leq{0}$$ then $$|some \\ expression|={-(some \\ expression)}$$. For example: $$|-5|=5=-(-5)$$;\n\nWhen $$x\\geq{0}$$ then $$|x|=x$$, or more generally when $$some \\ expression\\geq{0}$$ then $$|some \\ expression|={some \\ expression}$$. For example: $$|5|=5$$;\n\nSo, $$y+|y|=0$$ --> $$|y|=-y$$, which means that $$y\\leq{0}$$.\n\nAs for your doubt: question asks which of the following MUST be true, not COULD be true. Since all negative values of y satisfy $$|y|=-y$$ then it's not necessarily true that $$y=0$$.\n\nHope it's clear.\n_________________\n##### General Discussion\nIntern\nJoined: 28 Oct 2011\nPosts: 3\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n08 May 2012, 09:03\nBunuel wrote:\nboomtangboy wrote:\nIf y + | y | = 0, which of the following must be true?\n(A) y > 0\n(B) y\u22650\n(C) y < 0\n(D) y\u22640\n(E) y = 0\n\nWhy is just E incorrect?\n\nAbsolute value properties:\nWhen $$x\\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \\ expression\\leq{0}$$ then $$|some \\ expression|\\leq{-(some \\ expression)}$$. For example: $$|-5|=5=-(-5)$$;\n\nWhen $$x\\geq{0}$$ then $$|x|=x$$, or more generally when $$some \\ expression\\geq{0}$$ then $$|some \\ expression|\\leq{some \\ expression}$$. For example: $$|5|=5$$;\n\nSo, $$y+|y|=0$$ --> $$|y|=-y$$, which means that $$y\\leq{0}$$.\n\nAs for your doubt: question asks which of the following MUST be true, not COULD be true. Since all negative values of y satisfy $$|y|=-y$$ then it's not necessarily true that $$y=0$$.\n\nHope it's clear.\n\nHi ,\n\nThanks for the clear and concise explaination.\n\nJust wanted to clarify one thing.\n\nIn mods the two conditions I know are applied include; If x<0 or if x>=0. However in the above explaination you have used x<=0. Was that used for some particular reason or my concepts of absolute values are incorrect.\n\nThanks much\nSenior Manager\nJoined: 23 Oct 2010\nPosts: 350\nLocation: Azerbaijan\nConcentration: Finance\nSchools: HEC '15 (A)\nGMAT 1: 690 Q47 V38\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n08 May 2012, 09:15\nIf y + | y | = 0, which of the following must be true?\n\nplease note that | y | is always POSITIVE. so , in order to make this equation equal to zero, u need either y=0, or y <0.\n\nif y=0 , then you get 0+| 0|=0\n\nif y<0, then you get (-)+| -| =0 or (-)+(+)=0\ni\nthat is why y< or = 0\n_________________\n\nHappy are those who dream dreams and are ready to pay the price to make them come true\n\nI am still on all gmat forums. msg me if you want to ask me smth\n\nManager\nJoined: 21 Feb 2012\nPosts: 75\nLocation: India\nConcentration: Finance, General Management\nGMAT 1: 600 Q49 V23\nGPA: 3.8\nWE: Information Technology (Computer Software)\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n09 May 2012, 07:10\nBunuel wrote:\nboomtangboy wrote:\nIf y + | y | = 0, which of the following must be true?\n(A) y > 0\n(B) y\u22650\n(C) y < 0\n(D) y\u22640\n(E) y = 0\n\nWhy is just E incorrect?\n\nAbsolute value properties:\nWhen $$x\\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \\ expression\\leq{0}$$ then $$|some \\ expression|\\leq{-(some \\ expression)}$$. For example: $$|-5|=5=-(-5)$$;\n\nWhen $$x\\geq{0}$$ then $$|x|=x$$, or more generally when $$some \\ expression\\geq{0}$$ then $$|some \\ expression|\\leq{some \\ expression}$$. For example: $$|5|=5$$;\n\nSo, $$y+|y|=0$$ --> $$|y|=-y$$, which means that $$y\\leq{0}$$.\n\nAs for your doubt: question asks which of the following MUST be true, not COULD be true. Since all negative values of y satisfy $$|y|=-y$$ then it's not necessarily true that $$y=0$$.\n\nHope it's clear.\n\nHi Bunuel, why are we considering the case of y=0, as if y=0, then the expression\n|y|=-y makes no sense, because |0|=0. and there is no +0 or -0.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 51215\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n09 May 2012, 07:26\npiyushksharma wrote:\nBunuel wrote:\nboomtangboy wrote:\nIf y + | y | = 0, which of the following must be true?\n(A) y > 0\n(B) y\u22650\n(C) y < 0\n(D) y\u22640\n(E) y = 0\n\nWhy is just E incorrect?\n\nAbsolute value properties:\nWhen $$x\\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \\ expression\\leq{0}$$ then $$|some \\ expression|\\leq{-(some \\ expression)}$$. For example: $$|-5|=5=-(-5)$$;\n\nWhen $$x\\geq{0}$$ then $$|x|=x$$, or more generally when $$some \\ expression\\geq{0}$$ then $$|some \\ expression|\\leq{some \\ expression}$$. For example: $$|5|=5$$;\n\nSo, $$y+|y|=0$$ --> $$|y|=-y$$, which means that $$y\\leq{0}$$.\n\nAs for your doubt: question asks which of the following MUST be true, not COULD be true. Since all negative values of y satisfy $$|y|=-y$$ then it's not necessarily true that $$y=0$$.\n\nHope it's clear.\n\nHi Bunuel, why are we considering the case of y=0, as if y=0, then the expression\n|y|=-y makes no sense, because |0|=0. and there is no +0 or -0.\n\nNot, so. You can write |0|=-0 and there is nothing wrong in that.\n_________________\nSenior Manager\nJoined: 23 Oct 2010\nPosts: 350\nLocation: Azerbaijan\nConcentration: Finance\nSchools: HEC '15 (A)\nGMAT 1: 690 Q47 V38\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n09 May 2012, 07:33\nBunuel wrote:\n\nNot, so. You can write |0|=-0 and there is nothing wrong in that.\n\nhm, absolute value of an integer means how far this integer is from zero.\nso, absolute value of zero iz zero, since zero is zero far from zero (sounds like a quote of Alice from Wonderland hehe)\n-0 looks weird to me, since zero is neither positive, nor negative, and has no sigh. But still, I wont claim that my way of thinking is right. I will believe to Bunuel )) amazing life, every day is a new discovery )\n_________________\n\nHappy are those who dream dreams and are ready to pay the price to make them come true\n\nI am still on all gmat forums. msg me if you want to ask me smth\n\nManager\nJoined: 26 Dec 2011\nPosts: 93\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n15 May 2012, 03:32\nI have a question here.. if the the question was y + |y| = 2y , then can we say y>=0? given then 0+0 = 2(0). Please let me know in case I am doing something wrong. Thanks in advance.\nMath Expert\nJoined: 02 Sep 2009\nPosts: 51215\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n15 May 2012, 03:45\npavanpuneet wrote:\nI have a question here.. if the the question was y + |y| = 2y , then can we say y>=0? given then 0+0 = 2(0). Please let me know in case I am doing something wrong. Thanks in advance.\n\n$$y+|y|=2y$$ --> $$|y|=y$$ --> $$y\\geq{0}$$.\n_________________\nIntern\nJoined: 11 Jul 2012\nPosts: 40\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n03 Oct 2012, 02:31\nFor me the answer is C. (since the question asks MUST be true?)\nCan you anyone tell me why C is not correct\nMath Expert\nJoined: 02 Sep 2009\nPosts: 51215\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n03 Oct 2012, 02:36\nOusmane wrote:\nFor me the answer is C. (since the question asks MUST be true?)\nCan you anyone tell me why C is not correct\n\nCheck the solution here: if-y-y-0-which-of-the-following-must-be-true-131099.html#p1076758\n\ny<0 (C) is not correct because y+|y|=0 hods true when y=0 too, so y<0 is not necessarily true (not a must true statement).\n\nHope it's clear.\n_________________\nSenior Manager\nJoined: 13 Aug 2012\nPosts: 429\nConcentration: Marketing, Finance\nGPA: 3.23\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n04 Dec 2012, 01:03\nManipulate the equations:\ny + | y | = 0\n|y| = -y\n\n-y > 0 OR -y = 0\n\nThis means y could be 0 or y is less than 0.\n\nD. y\u22640\nE. y=0\n\n_________________\n\nImpossible is nothing to God.\n\nSenior SC Moderator\nJoined: 22 May 2016\nPosts: 2210\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n03 Oct 2017, 16:27\nboomtangboy wrote:\nIf y + | y | = 0, which of the following must be true?\n\nA. y > 0\nB. y\u22650\nC. y < 0\nD. y\u22640\nE. y = 0\n\nWhy is just E incorrect?\n\n$$y + |y| = 0$$\n$$|y| = 0 - y$$\n$$|y| = -y$$\n\nThe last expression means that $$y\\leq{0}$$. That rule can seem odd or counterintuitive.\n\nThe variable has a \"hidden\" negative sign. With the variable, it's hard to remember that there ARE two negative signs on RHS. We just do not (cannot) write the minus sign twice with the variable. These equations are equivalent, where y = -2:\n\n|-2| = -(-2) = 2\n|y| = -(y) = -y\n\nSo if $$y + |y| = 0$$, then $$|y| = -y$$ and\n\n$$y\\leq{0}$$\n\nIf none of the above occurs to you or if it makes no sense, pick and list three numbers: negative, 0, and positive.\n\nUse them to try to DISPROVE the answers. Even one example that defies the rule being tested makes \"must be true\" false.\n\n-2, 0, and 2\n\nA. y > 0\n$$y + |y| = 0$$. Try y = 0\n$$0 + |0| = 0$$. That works. $$y$$ does not have to be positive. REJECT\n\nB. y\u22650. Use -2\n$$y + |y| = 0$$\n$$-2 + |-2| = 0$$. That works. $$y$$ can be negative. REJECT\n\nC. y < 0. We know from (A) that $$y$$ CAN equal 0. REJECT\n\nD. y\u22640. Try 2\n$$y + |y| = 0$$\n$$2 + |2| \\neq{0}$$\n\nWe know from (A) that $$y$$ can equal 0.\nWe know from (B) that $$y$$ can be negative.\n\nAnd having tested +2, we know that $$y$$ CANNOT be positive.\n\nThis expression MUST be true. KEEP\n\nE. y = 0\nWe know from (B) that $$y$$ can be negative. Yes, $$y$$ can also be 0. But it does not have to be 0 -- it can be negative, e.g. -2. REJECT\n\nVP\nStatus: It's near - I can see.\nJoined: 13 Apr 2013\nPosts: 1354\nLocation: India\nConcentration: International Business, Operations\nGMAT 1: 480 Q38 V22\nGPA: 3.01\nWE: Engineering (Consulting)\nRe: If y + | y | = 0, which of the following must be true?\u00a0 [#permalink]\n\n### Show Tags\n\n09 Mar 2018, 08:20\nboomtangboy wrote:\nIf y + | y | = 0, which of the following must be true?\n\nA. y > 0\nB. y\u22650\nC. y < 0\nD. y\u22640\nE. y = 0\n\nWhy is just E incorrect?\n\nQuestion : If y + | y | = 0, which of the following must be true?\n\nmeans | y | = -y, which is possible when y <= 0\n\nHence (D)\n_________________\n\n\"Do not watch clock; Do what it does. KEEP GOING.\"\n\nRe: If y + | y | = 0, which of the following must be true? &nbs [#permalink] 09 Mar 2018, 08:20\nDisplay posts from previous: Sort by\n\n# If y + | y | = 0, which of the following must be true?\n\n new topic post reply Question banks Downloads My Bookmarks Reviews Important topics\n\n Powered by phpBB \u00a9 phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT\u00ae test is a registered trademark of the Graduate Management Admission Council\u00ae, and this site has neither been reviewed nor endorsed by GMAC\u00ae.","date":"2018-12-15 04:37:12","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 1, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.7581757307052612, \"perplexity\": 2182.519239000541}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.3, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2018-51\/segments\/1544376826715.45\/warc\/CC-MAIN-20181215035757-20181215061757-00371.warc.gz\"}"}
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Omar Al-Mukhtar Street, West Bay, Doha, Qatar
About QBA
QBA Members
QBA Membership
Qatar National Vision 2030
Qatar Bids
How To Establish a Company
Overview About Qatar
The State of Qatar is a sovereign and independent state in the Middle East, occupying a peninsula that juts into the Arabian Gulf. Since its complete independence from Britain in 1971, Qatar has emerged as one of the world's most important producers of oil and gas. It is an Islamic State whose laws and customs follow the Islamic tradition. Since 2013, the country has been governed by HH Sheikh Tamim bin Hamad bin Khalifa Al-Thani.
Location and geography
The State of Qatar is a peninsula located amid the western coast of the Arabian Gulf. The peninsular is approximately 100km across and extends 200km into the Gulf. Qatar includes several islands the largest of which are, Halul, Shraouh and Al-Asshat.. shares its southern border with Saudi Arabia and a maritime border Bahrain, the United Arab Emirates and Iran.
The land mainly consists of a flat rocky plain, covered with a range of low limestone outcroppings in Jebel Dukhan in the west and Jebel Fuyart in the north. This plain is mostly marked by its many inland seas (Khors), bays and basins called (al-Riyadh) over the northern and middle areas that are considered to be the most fertile lands housing different natural plants.
Qatar occupies an area of 11,521 square kilometers.
Qatar has a population of approximately 2,288,927 million.
Qatar's capital city is Doha (in Arabic, ad-Dawḥa, which means 'the big tree').
Doha (capital), Al-Wakrah, Al-Khor, Dukhan, Al-Shamal, Msaieed, Ras Lafan and others.
Islam is the official religion of the State of Qatar, and the Islamic Law (Sharia) shall be the principal source of its legislation.
Since the mid-1800s, Qatar has grown from a poor British protectorate known for pearling into one of the world's most important oil and gas producing countries. While there is increasing investment in non-energy sectors, oil and gas still account for more than half of the Gross Domestic Product. Due to its substantial reserves of oil and gas, the country has one of the highest incomes per capita in the world.
Arabic is the official language of the country, though English is widely spoken.
Qatar has a desert climate with hot summers, warm winters and scarce rainfall.
Qatari Riyal (1 Riyal = 100 Dirhams).
The Riyal is pegged to the US Dollar ($US 1 = QAR 3.65).
Qatar National Day is held on 18th December every year in celebration and commemoration of Sheikh Jassim Bin Mohammad Al-Thani, the founder of the State of Qatar.
GMT +3 hours.
Enter your e-mail and subscribe to our newsletter.
Address: Omar Al-Mukhtar Street, West Bay, Doha, Qatar
Phone: (+974) 4435-3120
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"redpajama_set_name": "RedPajamaCommonCrawl"
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AAAAAAAAAAAAA
BBBBBBBBBBBBB
CCCCCCCCCCCCC
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"redpajama_set_name": "RedPajamaGithub"
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Horst Meyer (født 20. juni 1941 i Hamburg, død 24. januar 2020 på Lanzarote) var en tysk roer, olympisk guldvinder, dobbelt verdensmester og firedobbelt europamester.
Meyer var en del otteren fra Ratzeburger RC, der vandt det vesttyske mesterskab i hvert af årene 1962-1968. Båden vandt i 1962 også VM-titlen, og det blev desuden til EM-titlen i hvert af årene 1963-1965. Han var også med i båden for det forenede Tyskland ved OL 1964 i Tokyo. Her vandt tyskerne deres indledende heat i ny olympisk rekord foran USA. Skønt amerikanerne dermed måtte gennem opsamlingsheat, kom de i finalen, som de efter den første kilometer dominerede og sikrede sig guldmedaljen, mens tyskerne blev nummer to foran Tjekkoslovakiet.
Efter 1965-sæsonen skete der en stor udskiftning i Ratzebuger-otteren, men Meyer fortsatte i båden og var til at vinde VM-guld i 1966 samt EM-guld i 1967. Han var igen med ved OL 1968 i Mexico City, hvor han var eneste genganger i båden fra legene fire år forinden. Denne gang repræsenterende var det Vesttyskland, han repræsenterede, og otteren vandt planmæssigt sit indledede heat. I finalen kom de lidt langsomt fra start, men ved 1500 meter-mærket var de i spidsen og holdt hjem til guldmedaljen, som blev vundet med næsten et sekund foran Australien på andenpladsen og Sovjetunionen på tredjepladsen, yderligere godt et sekund bagude. Ud over Meyer bestod besætningen hos guldvinderne af Wolfgang Hottenrott, Dirk Schreyer, Egbert Hirschfelder, Rüdiger Henning, Jörg Siebert, Lutz Ulbricht, Roland Böse (i finalen erstattet af Niko Ott) og styrmand Gunther Tiersch.
Guldbesætningen fra 1968 bar det olympiske flag ind ved åbningsceremonien til OL 1972 i München. Efter afslutningen af sin aktive karriere sad Meyer i Tysklands Olympiske Komité i flere omgange, og han sad i bestyrelsen for Sportshilfe. Senere involverede han sig i fredsbevægelsen og var med til at danne "Athletes for Peace against Nuclear Warriors", der var aktiv til kort efter Tysklands genforening. Han var uddannet i virksomhedsøkonomi og direktør for et konsulentfirma.
OL-medaljer
1968: Guld i otter
1964: Sølv i otter
Referencer
Eksterne henvisninger
Roere fra Tyskland
Personer fra Hamborg
Deltagere for Tyskland ved sommer-OL 1964
Deltagere for Vesttyskland ved sommer-OL 1968
Olympiske mestre fra Vesttyskland
Olympiske sølvmedaljevindere fra Tyskland
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package io.github.jeddict.relation.mapper.persistence.internal.jpa.metadata;
import java.io.IOException;
import java.net.URISyntaxException;
import java.net.URL;
import java.util.HashMap;
import java.util.Map;
import org.eclipse.persistence.internal.jpa.deployment.PersistenceUnitProcessor;
import org.eclipse.persistence.internal.jpa.metadata.MetadataHelper;
import org.eclipse.persistence.internal.jpa.metadata.MetadataProcessor;
import org.eclipse.persistence.internal.jpa.metadata.accessors.classes.EmbeddableAccessor;
import org.eclipse.persistence.internal.jpa.metadata.accessors.classes.EntityAccessor;
import org.eclipse.persistence.internal.jpa.metadata.xml.XMLEntityMappings;
import org.eclipse.persistence.internal.jpa.metadata.xml.XMLEntityMappingsReader;
import org.eclipse.persistence.internal.sessions.AbstractSession;
import org.eclipse.persistence.jpa.Archive;
public class JPAMMetadataProcessor extends MetadataProcessor {
public JPAMMetadataProcessor(AbstractSession session, ClassLoader loader, boolean weaveLazy, boolean weaveEager, boolean weaveFetchGroups, boolean multitenantSharedEmf, boolean multitenantSharedCache, Map predeployProperties, MetadataProcessor compositeProcessor) {
m_loader = loader;
m_session = session;
m_project = new JPAMMetadataProject(session, weaveLazy, weaveEager, weaveFetchGroups, multitenantSharedEmf, multitenantSharedCache);
m_predeployProperties = predeployProperties;
m_compositeProcessor = compositeProcessor;
if (m_compositeProcessor != null) {
m_compositeProcessor.addCompositeMemberProcessor(this);
m_project.setCompositeProcessor(m_compositeProcessor);
}
}
@Override
protected void initPersistenceUnitClasses() {
// 1 - Iterate through the classes that are defined in the <mapping>
// files and add them to the map. This will merge the accessors where
// necessary.
HashMap<String, EntityAccessor> entities = new HashMap<>();
HashMap<String, EmbeddableAccessor> embeddables = new HashMap<>();
for (XMLEntityMappings entityMappings : m_project.getEntityMappings()) {
entityMappings.initPersistenceUnitClasses(entities, embeddables);
}
// 2 - Iterate through all the XML entities and add them to the project
// and apply any persistence unit defaults.
for (EntityAccessor entity : entities.values()) {
// This will apply global persistence unit defaults.
m_project.addEntityAccessor(entity);
// This will override any global settings.
entity.getEntityMappings().processEntityMappingsDefaults(entity);
}
// 3 - Iterate though all the XML embeddables and add them to the
// project and apply any persistence unit defaults.
for (EmbeddableAccessor embeddable : embeddables.values()) {
// This will apply global persistence unit defaults.
m_project.addEmbeddableAccessor(embeddable);
// This will override any global settings.
embeddable.getEntityMappings().processEntityMappingsDefaults(embeddable);
}
}
@Override
protected void loadStandardMappingFiles(String ormXMLFile) {
URL rootURL = m_loader.getResource("");
URL ormURL;
Archive par = null;
try {
par = PersistenceUnitProcessor.getArchiveFactory(m_loader).createArchive(rootURL, null);
if (par != null) {
ormURL = par.getEntryAsURL(ormXMLFile);
if (ormURL != null) {
// Read the document through OX and add it to the project., pass persistence unit properties for any orm properties set there
XMLEntityMappings entityMappings = XMLEntityMappingsReader.read(ormURL, m_loader, m_project.getPersistenceUnitInfo().getProperties());
entityMappings.setIsEclipseLinkORMFile(ormXMLFile.equals(MetadataHelper.ECLIPSELINK_ORM_FILE));
m_project.addEntityMappings(entityMappings);
}
}
} catch (IOException | URISyntaxException e) {
throw new RuntimeException(e);
} finally {
if (par != null) {
par.close();
}
}
}
}
|
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Heads up for transporting animals!
Our ASPCA Relo team did some investigating on how to comply with last week's memo from the Washington State Vet Office regarding their updated health certificate requirements surrounding small animal interstate TO Washington state. In that memo, the WA State Vet Office relayed that the downloadable APHIS Form 7001 will no longer be valid as an acceptable health cert for small animal transport.
Speaking at length with both the California and Washington State Veterinary Offices, the WA State Vet office has recognized how difficult it will be for transporting organizations & vets (especially in CA) to comply with their updated health certificate policy. They've recognized that, currently, there isn't a viable alternative for small animal vets in CA to create electronic health certs for small animal transport. The WA State Vet Office has reassured us that they are looking into a solution that will eventually allow small animal transporters & vets to comply with their updated health cert policy. Until the WA State Vet Office makes an announcement regarding feasible alternatives, they've reassured us that the downloadable APHIS Form 7001 is still OKAY to use.
Summary: For the time being, CA vets and transporting organizations are okay to send small animal transports using the downloadable APHIS 7001 Form until we hear of any updates from the WA State Vet Office.
Please read on if you want the additional details on this situation.
The free mCVI is not available for CA vets yet.
The free WA eCVI (as well as any paper copies of this form) is only for use by WA vets.
The free APHIS platform, VSPS, currently does not have "small animal" option, so it cannot be used at this time.
Large animal (livestock) paper health certs that the CA state vet office issues and distributes for free is a viable alternative (under species to be transported, you would select "other" and write in "canine" or "feline); however these large animal health certs can only be ordered by a USDA accredited category 2 vet (large animal accredited). Most of our shelter partners do not have a USDA accredited category 2 vet, so this is not a feasible alternative for most of us either.
Please let me know if I can help you with any additional questions. We will be sure to update you of any additional information we learn.
In an effort to remain compliant with individual state requirements for small animal interstate transport, we are alerting out shelter partners about an updated requirement when participating in interstate transport TO the state of Washington.
Hard copy APHIS 7001 forms can still be used but are not easily accessible. The WA State Vet Office is instead recommending the use of electronic forms that will generate unique certificate identifiers when created and will continue to require a USDA accredited veterinarian's signature and ID #. Here is information about acceptable alternatives forms: https://agr.wa.gov/FoodAnimal/AnimalHealth/Docs/MemoToWAAccreditedVetsNoLongerApproving7001.pdf and here is detailed information on these acceptable alternatives: https://agr.wa.gov/FoodAnimal/AnimalID/ElectronicCertificates.aspx.
When speaking with the WA State Vet Office, they relayed they will give transporters a window of time to become compliant with this policy change as they understand it will take time for organizations to adapt.
|
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{"url":"https:\/\/openarchive.usn.no\/usn-xmlui\/handle\/11250\/218417\/browse?type=subject&value=Quark+gluon+plasma","text":"Now showing items 1-2 of 2\n\n\u2022 #### Measurement of Ds + production and nuclear modification factor in Pb-Pb collisions at \u221asNN = 2.76 TeV \ufeff\n\n(Journal article; Peer reviewed, 2016)\nThe production of prompt D s + mesons was measured for the first time in collisions of heavy nuclei with the ALICE detector at the LHC. The analysis was performed on a data sample of Pb-Pb collisions at a centre-of-mass ...\n\u2022 #### Transverse momentum dependence of D-meson production in Pb-Pb collisions at \u221asNN = 2.76 TeV \ufeff\n\n(Journal article; Peer reviewed, 2016)\nThe production of prompt charmed mesons D0, D+ and D\u2217+, and their antiparticles, was measured with the ALICE detector in Pb-Pb collisions at the centre-of-mass energy per nucleon pair, sNN\u2212\u2212\u2212\u221a, of 2.76 TeV. The production ...","date":"2021-03-03 11:12:31","metadata":"{\"extraction_info\": {\"found_math\": false, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 0, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.8735403418540955, \"perplexity\": 4211.457976441608}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-10\/segments\/1614178366959.54\/warc\/CC-MAIN-20210303104028-20210303134028-00119.warc.gz\"}"}
| null | null |
Nemaha Central USD 115 is a public unified school district headquartered in Seneca, Kansas, United States. The district includes the communities of Baileyville, Kelly, Oneida, Seneca, St. Benedict and nearby rural areas of Nemaha County.
Schools
The school district operates the following schools:
Nemaha Central High School
Nemaha Central Elementary and Middle School
History
In 1859, the first organized classes in Seneca were taught in the parlor of the Smith hotel. As the class sizes increased, they were moved to the Nemaha County Courthouse. In 1865, the community build a brick schoolhouse at the site of the current Saints Peter and Paul Catholic Church now in Seneca. The school employed two teachers, and it marked the beginning of a graded school system in the village.
In 1868, the school became known as District No. 11. To accommodate a growing population, the district built a four-room stone schoolhouse, adding two additional rooms a few years later. This building served the town for 13 years before the district demolished it in 1889.
In 1881, the district used money from a bequest to construct the Van Loan Memorial School Building. It was designated for children in the lower grades who lived south of the railroad tracks in Seneca. There were five graduates of this school in its first year. Classes were held there until 1890. In 2003, the Van Loan Building was demolished by a fire.
In 1889, the district started building the Seneca Public High School; it opened on April 28, 1890. The high school was made of pressed brick, and it was heated by steam. It was large enough to accommodate all the pupils in the Nemaha Valley District as well as pupils from other parts of the county.
In 1937, crowded conditions at the high school prompted plans for a new Seneca Grade School. The new grade school was finished in 1938, and it contained eight classrooms, an office, a stage, an auditorium / gymnasium, and a vocational shop and classroom. The building was planned so the district could add wings to accommodate future expansion. In February 1939, a cornerstone containing a copper box time capsule was placed on the southeast corner of the building. The grade school opened for grades 1–6 in August 1939.
In 1965, the State of Kansas started unifying small county school districts into larger school districts. The Seneca School District was combined with towns of Kelly, Kansas, and Corning, Kansas, into Unified School District No. 442. In 1966, the new school board selected the name Nemaha Valley for the new school district. According to Earl McGee, the superintendent of school, "My suggestion would be 'Nemaha Valley Schools' since the valley starts at Corning and runs past Kelly and Seneca."
On December 14, 1966, the Seneca Public High School was condemned. From 1967 to 1969, high school classes were held at the City Hall building, the Masonic Temple, and at the Seneca Grade School. Several bond issues to build a new school failed to pass election. Much of the disagreement centered around the location of the new high school. Some supported a central location for the three towns, while others favored building it in Seneca. In 1968, this disagreement eventually led to Corning transferring to the Centralia, Kansas school district. After Corning left the Nemaha Valley District, the high school bond issue specifying a new Nemaha Valley High School at Seneca passed. The old Seneca Public High School was demolished on July 1, 1968.
In 1969 the new Nemaha Valley High School opened for classes. During this same year, Kelly High School was closed, and its students attended NVHS. In March 1970, Nemaha Valley High School was dedicated.
In 1974, Saints Peter and Paul High School closed. In 1979, Kelly Junior High closed, and in 1981, Kelly grade school closed. During this time, a new Onida grade school and high school and Bern High school opened in the district.
This addition of students required expansion of the buildings. In 1983, the north wing on the grade school was added, and in 1993, the west wing of the grade school was added. The north wing is now the junior high. The west wing primarily houses elementary grade classes.
On March 10, 2007, the old section of Seneca Grade School was destroyed in a fire. For the remainder of the school year, classes were held in the public library annex in the basement of Bowman Law Office and Dental Practice, at the former restaurant Bob's Sirloin, and at the high school. A bond issue to build a new grade school and junior high school passed in November 2007. The new school was to be located south of the high school.
Nemaha Central USD 115 formed in 2011 by the consolidation of Nemaha Valley USD 442 and B&B USD 451.
In May 2014, state budget cuts resulted in closing of the neighboring B and B High School in Baileyville, Kansas. Those students were transferred to Nemaha Valley High School. This prompted a school renaming the preceding year, with students in grades 7 through 11 at Baileyville, students in grades K through 11 in Nemaha Valley, and students in Saints Peter and Paul Catholic School voting to change the name to Nemaha Central. The mascot was renamed the Thunder, and the new school colors were purple and Columbia blue. In August 2014, the school opened its doors as Nemaha Central. This also prompted changing the district number from 442 to 115.
See also
Kansas State Department of Education
Kansas State High School Activities Association
List of high schools in Kansas
List of unified school districts in Kansas
References
External links
School districts in Kansas
Education in Nemaha County, Kansas
2011 establishments in Kansas
School districts established in 2011
|
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| 7,237
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If you are Jacquelyn, please click here now to create your login!
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|
{
"redpajama_set_name": "RedPajamaC4"
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| 9,351
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The Uncharted Course
Parallels associate general counsel Nancy Norton's days are unpredictable, but she knows the international nature of her work will always be rewarding
By Jessica Montoya Coggins
Nancy Norton always knew she was destined for an international career. Her love of travel and interest in other cultures started early when her father took her to Ireland to visit relatives. "I became enamored with other countries when I was quite young," she says, "and I knew I wanted to be in business." She sought out opportunities to travel, learned another language, and pursued degrees that would allow her to build a career in international law. Now, Norton is the vice president and associate general counsel at Parallels, and she spends much of her time focusing on a wide variety of international business issues.
Parallels and its Odin brand are global leaders in cross-platform solutions and cloud services enabling. As in-house counsel, Norton enjoys that her position entails something different each day. "My schedule is never predictable, every day I am working on something completely different."
Embracing unpredictability is practically natural for Norton. She has recently found herself juggling negotiating an enterprise licensing deal in Brazil, managing a trademark issue in the Philippines, drafting a strategic partnership agreement in China, and resolving a finance matter in Greece.
The most practical issue for Norton is simply the time of day. She admits that it can be trying to align the business needs with multiple time zones. Based in Seattle, Norton recently negotiated a deal where the Parallels sales team was in Singapore and the customer was in Mumbai. Since she always defers to the customer's schedule and because the sales team wanted to close the deal by year-end, Norton spent six weeks on negotiation calls starting at 2 a.m.
She is online at 4 a.m. every day in order to reply to e-mails that have come in overnight and need a response before the day ends in one of the Parallels offices around the world. Norton also takes part in a weekly call at 6 a.m. with her team in Germany and Russia, and she often responds to urgent calls for pressing matters late at night in order to accommodate her colleagues in distant time zones. Though it might be a minor inconvenience for her (and it recently interrupted a concert she was attending in Seattle), Norton considers it just part of her job. "I don't sleep very much, so I guess I'm well-suited for the demands of the job," she notes with a laugh.
With a busy and ever-changing schedule, Norton has one activity that keeps her focused: running. Almost every day at lunchtime Norton can be found running along a river near Parallels' office. She says running helps to clear her head, relieve stress, and allows her to bond with colleagues since she often recruits employees from Parallels to run with her.
In addition, when she travels to a new country, the first thing Norton does upon arriving is go for a run. "Running helps me combat jet lag and allows me to explore the city before I spend the next few days sitting in a conference room." Norton recently ran in Munich and London and plans to run a half marathon in Sydney.
"My interest in international law and business have merged together to get me where I am today," says Norton. "I'm very fortunate because I love what I do."
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 637
|
Carley Reavill
– in the House of Commons at 4:31 pm on 20th February 1992.
All Commons debates on 20 Feb 1992
Dr Jack Cunningham Shadow Leader of the House of Commons 4:31 pm, 20th February 1992
On a point of order, Mr. Speaker. During Prime Minister's Questions today my hon. Friend the Member for Leeds, Central (Mr. Fatchett) put to the Prime Minister a question concerning the appalling tragedy that had befallen Carley Reavill. May we have it on the record that the Prime Minister seems to have been misadvised about this tragic case? Dr. Alison Shurtz has confirmed that Carley was diagnosed as having meningitis shortly after arrival at Queen Elizabeth II hospital, and she would have been transferred to another hospital if an intensive paediatric care bed had been available. Dr. Duncan Matthews, of the Great Ormond Street hospital, has said that critically ill children are being denied intensive care treatment because there is not any money to make beds available. It was the shortage of beds, caused by the shortage of funding in the national health service, that led to this appalling tragedy.
(Citation: HC Deb, 20 February 1992, c481)
Mr Geoffrey Dickens , Littleborough and Saddleworth
Further to that point of order, Mr. Speaker.
Mr Bernard Weatherill , Croydon North East
This is not a point of order for me. It seems to be a continuation of Question Time.
Try to make it a point of order for me.
Do you. Mr. Speaker, think that it would be helpful to you and to the House to know where that information came from so that a judgment might be made?
I do not think that it would be helpful to me. This is not a point of order for me.
|
{
"redpajama_set_name": "RedPajamaCommonCrawl"
}
| 1,090
|
StreamBaseKit is a Swift UI toolkit for [Firebase](https://www.firebase.com). It surfaces Firebase queries as streams that are synched with Firebase in real time, fetched incrementally, and can be merged or split into multiple sections. These streams can be easily plugged into UI elements like table views.
This kit also includes a [persistence layer](#persistence-layer) that makes it easy to persist objects in Firebase.
## Installing StreamBaseKit
Using Cocoapods, add to your `Podfile`:
```
pod 'StreamBaseKit', '~> 0.1'
```
And, if you don't have it already:
```
platform :ios, '8.0'
use_frameworks!
```
## Getting Started with StreamBaseKit:
If you don't have one already, sign up for a [Firebase account](https://www.firebase.com/signup/).
To use streams, you'll want to know about these classes:
Class | Description
-------|-------------
StreamBase | This is the main class that exposes a Firebase query as a Stream.
StreamBaseItem | Base class for objects that appear in streams.
StreamBaseDelegate | Delegate that is notified of stream changes.
StreamTableViewAdapter | Adapter from streams to UITableViews.
PartitionedStream | Split a stream into multiple sections.
TransientStream | Stream that's not connected to Firebase.
UnionStream | Stream for merging multiple streams.
QueryBuilder | Helper for composing Firebase queries.
To get started, you'll need to build from StreamBaseItem and StreamBase. Additionally, StreamTableViewAdapter provides some convenient functionality to connect streams with tables. Here's the basic outline:
```swift
MyItem.swift
class MyItem : StreamBaseItem {
var name: String?
func update(dict: [String: AnyObject) {
super.update(dict)
name = dict["name"] as? String
}
var dict: [String: AnyObject] {
var d = super.dict
d["name"] = name
return d
}
}
MyViewController.swift
class MyViewController : UIViewController {
var stream: StreamBase!
var adapter: StreamTableViewAdapter!
// etc...
override func viewDidLoad() {
// etc...
let firebaseRef = Firebase(url:"https://<YOUR-FIREBASE-APP>.firebaseio.com/")
stream = StreamBase(type: MyItem.self, ref: firebaseRef)
adapter = StreamTableViewAdapter(tableView: tableView)
stream.delegate = adapter
}
}
extension MyViewController : UITableViewDataSource {
func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
return stream.count
}
func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
let cell = tableView.dequeueReusableCellWithIdentifier("MyCell", forIndexPath: indexPath)
let item = stream[indexPath.row] as! MyItem
cell.titleLabel?.text = item.name
return cell
}
}
```
## Multiple Sections with PartitionedStream
PartitionedStream allows you to define a partition function for splitting a stream into mulitple sections. This is useful, eg, in splitting a roster of users into organizers and participants. Building off of the previous example:
```swift
User.swift
class User : StreamBaseItem {
var isOrganizer: Bool
// etc...
}
MyViewController.swift
class MyViewController : UIViewController {
var pstream: PartitionedStream!
// etc...
override func viewDidLoad() {
// etc...
pstream = PartitionedStream(stream: stream, sectionTitles: ["organizers", "participants"]) { ($0 as! User).isOrganizer ? 0 : 1 }
pstream.delegate = adapter
}
}
extension MyViewController : UITableViewDataSource {
func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
return pstream[section].count
}
func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
let cell = tableView.dequeueReusableCellWithIdentifier("MyCell", forIndexPath: indexPath)
let item = pstream[indexPath] as! User // Index partitioned stream with whole NSIndexPath
cell.titleLabel?.text = item.name
return cell
}
func tableView(tableView: UITableView, titleForHeaderInSection section: Int) -> String? {
return pstream.sectionTitles[section]
}
}
```
## Multiple Sections with Multiple Streams
PartitionedStream is convenient to use, but if the underlying data has hundreds or more elements you'll need to provide limits per section. You can do this by constructing multiple sections like this:
```swift
class MyViewController : UIViewController {
var organizerStream: StreamBase!
var participantStream: StreamBase!
var organizerAdapter: StreamTableViewAdapter!
var participantAdapter: StreamTableViewAdapter!
// etc...
override func viewDidLoad() {
// etc...
let organizerQuery = QueryBuilder(ref)
organizerQuery.limit = 100
organizerQuery.ordering = .Child("is_organizer")
organizerQuery.start = true
organizerStream = StreamBase(type: User.self, queryBuilder: organizerQuery)
organizerAdapter = StreamTableViewAdapter(tableView: tableView, section: 0)
let participantQuery = QueryBuilder(ref)
participantQuery.limit = 100
participantQuery.ordering = .Child("is_organizer")
participantQuery.end = false
participantStream = StreamBase(type: User.self, queryBuilder: participantQuery)
participantAdapter = StreamTableViewAdapter(tableView: tableView, section: 1)
}
}
extension MyViewController : UITableViewDataSource {
func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
return (section == 0) ? organizerStream.count : participantStream.count
}
func tableView(tableView: UITableView, cellForRowAtIndexPath indexPath: NSIndexPath) -> UITableViewCell {
let cell = tableView.dequeueReusableCellWithIdentifier("MyCell", forIndexPath: indexPath)
let user: User
if indexPath.section == 0 {
user = organizerStream[indexPath.row] as! User
} else {
user = participantStream[indexPath.row] as! User
}
// etc...
return cell
}
}
```
## Placeholders and Incremental Fetching with TransientStream and UnionStream
Chat histories can grow long, so it's important to enable them to be incrementally fetched. To do this, we need to be able to insert temporary placeholders into the table to provide a ui for the fetch, and to actually perform the additional fetch. There are further details (such as knowing the actual size of the stream!), but this is a rough sketch of how this would work:
```swift
class MyViewController : UIViewController {
let maxToFetch = 100
var stream: StreamBase!
var unionStream: UnionStream!
var transientStream: TransientStream!
var fetchmore: FetchMoreItem?
override func viewDidLoad() {
// etc...
stream = StreamBase(type: MyItem.self, ref: firebaseRef, ascending: false, limit: maxToFetch)
transient = TransientStream()
unionStream = UnionStream(sources: stream, transient)
unionStream.delegate = self
}
func fetchMoreTapped(sender: UIButton) {
stream.fetchMore(maxToFetch, start: stream[0].key!) {
transient.remove(fetchmore)
fetchmore = nil
}
}
}
```
Here we extend the StreamBaseDelegate rather than use the StreamTableViewAdapter because we need to manipulate state in the controller.
```swift
extension MyViewController : StreamBaseDelegate {
override func streamDidFinishInitialLoad(error: NSError?) {
super.streamDidFinishInitialLoad(error)
if stream.count > maxToFetch {
fetchmore = FetchMoreItem(key: dropLast(stream[0].key))
transient.add(fetchmore)
}
}
}
```
# Persistence Layer
StreamBaseKit also includes a persistence layer that uses a declarative approach: you state where something is stored, and the layer takes care of the rest. For example,
```swift
registry.resource(Group.self, "/group/@")
registry.resource(GroupMessage.self, "/group_message/$group/@")
```
This states that groups are stored under "/group", and messages, which are logically contained in groups, under "/group_message". (Recall that it's not a good practice to store them under "/group", say in "/group/$group/message/@", because fetches of the group would also fetch all of the messages.)
The "@" means that an auto-id is generated for create operations, and the object's key is used for updates and destroys. The "$" means that the value must be looked up using a ResourceContext... more on that below.
To use the persitence layer, you'll want to know about these classes:
Class | Description
-------|-------------
ResourceBase | Core of persistence layer.
ResourceContext | Helper for managing context in persistence layer.
ResourceRegistry | Protocol for registering resources.
## Registering Resources with Persistence Layer
You register resources with an the ResourceBase using the ResourceRegistry protocol. One approach is to put the ResourceBase in a singleton. For example:
```swift
Environment.swift
class Environment {
var resourceBase: ResourceBase!
static let sharedEnv: Environment = {
let env = Environment()
let firebase = Firebase(url: "https://<YOUR-FIREBASE-APP>.firebaseio.com")
env.resourceBase = ResourceBase(firebase: firebase)
let registry: ResourceRegistry = env.resourceBase
registry.resource(Group.self, "/group/@")
registry.resource(GroupMessage.self, "/group_message/$group/@")
// ...
return env
}()
}
```
## Using the ResourceContext Stack
In the initial view controller, use the Environment singleton to create the root resource context:
```swift
InitialViewController.swift
class InitialViewController : UIViewController {
var rootResourceContext: ResourceContext!
// ...
override func viewDidLoad() {
super.viewDidLoad()
rootResourceContext = ResourceContext(base: Environment.sharedEnv.resourceBase, resources: nil)
// ...
```
Now, this view controller can now create, update or delete Groups using the root resource context. For example:
```swift
let group = Group()
group.name = "group name"
rootResourceContext.create(group)
```
Recall that "$group" in "/group_message/$group/@" indicates a context key which must be filled in in order to persist a GroupMessage. The ResourceContext is responsible for doing this. Say you had a GroupViewController which allows users to message groups. In your initial view controller, before pushing the group view controller onto your navigation controller, you'd do something like this:
```swift
override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
switch(segue.destinationViewController) {
case let groupVC as GroupViewController:
groupVC.resourceContext = resourceContext.push(["group": sender as! Group])
// ...
```
Now, when you call ```resourceContext.create(GroupMessage())``` in GroupViewController it will know how to resolve the key "$group". Similarly, if you went deeper and could "like" messages in groups, you could do that by pushing yet another ResourceContext onto the stack transiently. That might look like:
```swift
func messageLikeTouched(sender: MessageLikeControl) {
resourceContext.push(["message": sender.message]).create(MessageLike())
}
```
## Counters
It's also possible to specify counters which get incremented or decremented when objects are created or destroyed. Say you wanted to keep track of how many messages were in your group. You'd register a counter like:
```swift
registry.counter(Group.self, "message_count", GroupMessage.self)
```
The ResourceBase will take care of incrementing and decrementing the "message_count" when messages are created and destroyed. Since groups are registered under "/group/@", this counter will appear here: "/group/$group_key/message_count".
Note that this counter is maintained client-side, and so can become inconsistent over time. For example, Firebase [transactions are not persisted across app restarts](https://www.firebase.com/docs/ios/guide/offline-capabilities.html), so if the user makes changes while offline, and then closes the app, counters may not be updated.
## Extending ResourceBase
ResourceBase has a number of hooks for subclasses to use when extending it. There are hooks for create, update and destroy that are invoked before, after commit to local storage, and after commit to remote storage. There is also a hook for [logging so a server can handle side effects](https://medium.com/@spf2/action-logs-for-firebase-30a699200660).
# Building the Example
A simple example project is included. To build it:
```
$ git clone https://github.com/movem3nt/StreamBaseKit.git
$ cd StreamBaseKit/StreamBaseExample
$ pod install
$ open StreamBaseExample.xcworkspace
```
Set the active scheme to StreamBaseExample, and then hit command-R.
# Comparison with FirebaseUI-iOS
An alternative library to consider is [FirebaseUI-iOS](https://github.com/firebase/FirebaseUI-iOS). It is the official client library for Firebase. It's written in Objective-C instead of Swift, and is simpler.
StreamBaseKit grew out of building Movem3nt, a complex social application, and addresses a variety of problems encountered in doing so. For example, iOS table views will auto-scroll if content is inserted on top, but Firebase appends new data to the bottom. To make these work well together for messaging-type apps, one needs to invert both the firebase collection and the table view.
StreamBaseKit also makes it easy to add more advanced functionality like splitting a collection into multiple table sections, and inserting transient content into the table like a "fetch more" control for incremental fetching.
The resource layer makes it much to keep your database persistence logic and ui view controller logic separate. It also provides the convenient counter feature.
Another consideration is that Firebase also has a Android UI library. StreamBaseKit does not (yet).
|
{
"redpajama_set_name": "RedPajamaGithub"
}
| 9,097
|
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| null | null |
/*
* The asset info dictionary, used to find and load assets. The
* dictionary is produced by the build system and loaded by a separate
* script.
*/
declare module Assets {
interface FontInfo {
name: string;
bold: boolean;
italic: boolean;
size: number;
ascender: number;
descender: number;
height: number;
char: string;
glyphcount: number;
glyph: Int16Array; // initially a string
kern: Int8Array; // initially a string
}
interface ImageSetInfo {
[name: string]: string;
}
interface SpriteMap {
[name: string]: number[];
}
interface AssetInfo {
fonts: FontInfo[];
images: ImageSetInfo;
sprites: SpriteMap;
}
}
declare var AssetInfo: Assets.AssetInfo;
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"redpajama_set_name": "RedPajamaGithub"
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{"url":"https:\/\/faculty.math.illinois.edu\/Macaulay2\/doc\/Macaulay2-1.18\/share\/doc\/Macaulay2\/DGAlgebras\/html\/_is__Homology__Algebra__Trivial.html","text":"# isHomologyAlgebraTrivial -- Determines if the homology algebra of a DGAlgebra is trivial\n\n## Synopsis\n\n\u2022 Usage:\nisTriv = isHomologyAlgebraTrivial(A)\n\u2022 Inputs:\n\u2022 Optional inputs:\n\u2022 GenDegreeLimit => ..., default value infinity, Option to specify the maximum degree to look for generators\n\u2022 Outputs:\n\u2022 isTriv, ,\n\n## Description\n\nThis function computes the homology algebra of the DGAlgebra A and determines if the multiplication on H(A) is trivial.\n\n i1 : R = ZZ\/101[a,b,c,d]\/ideal{a^4,b^4,c^4,d^4} o1 = R o1 : QuotientRing i2 : S = R\/ideal{a^3*b^3*c^3*d^3} o2 = S o2 : QuotientRing i3 : A = acyclicClosure(R,EndDegree=>3) o3 = {Ring => R } Underlying algebra => R[T ..T ] 1 8 3 3 3 3 Differential => {a, b, c, d, a T , b T , c T , d T } 1 2 3 4 o3 : DGAlgebra i4 : B = A ** S o4 = {Ring => S } Underlying algebra => S[T ..T ] 1 8 3 3 3 3 Differential => {a, b, c, d, a T , b T , c T , d T } 1 2 3 4 o4 : DGAlgebra i5 : isHomologyAlgebraTrivial(B,GenDegreeLimit=>6) o5 = true\n\nThe command returns true since R --> S is Golod. Notice we also used the option GenDegreeLimit here.\n\n i6 : R = ZZ\/101[a,b,c,d]\/ideal{a^4,b^4,c^4,d^4} o6 = R o6 : QuotientRing i7 : A = koszulComplexDGA(R) o7 = {Ring => R } Underlying algebra => R[T ..T ] 1 4 Differential => {a, b, c, d} o7 : DGAlgebra i8 : isHomologyAlgebraTrivial(A) o8 = false\n\nThe command returns false, since R is Gorenstein, and so HA has Poincare Duality, hence the multiplication is far from trivial.\n\n## Ways to use isHomologyAlgebraTrivial :\n\n\u2022 \"isHomologyAlgebraTrivial(DGAlgebra)\"\n\n## For the programmer\n\nThe object isHomologyAlgebraTrivial is .","date":"2021-09-28 10:35:44","metadata":"{\"extraction_info\": {\"found_math\": true, \"script_math_tex\": 0, \"script_math_asciimath\": 0, \"math_annotations\": 0, \"math_alttext\": 0, \"mathml\": 0, \"mathjax_tag\": 0, \"mathjax_inline_tex\": 0, \"mathjax_display_tex\": 0, \"mathjax_asciimath\": 1, \"img_math\": 0, \"codecogs_latex\": 0, \"wp_latex\": 0, \"mimetex.cgi\": 0, \"\/images\/math\/codecogs\": 0, \"mathtex.cgi\": 0, \"katex\": 0, \"math-container\": 0, \"wp-katex-eq\": 0, \"align\": 0, \"equation\": 0, \"x-ck12\": 0, \"texerror\": 0, \"math_score\": 0.6279861330986023, \"perplexity\": 4165.307455349869}, \"config\": {\"markdown_headings\": true, \"markdown_code\": true, \"boilerplate_config\": {\"ratio_threshold\": 0.18, \"absolute_threshold\": 10, \"end_threshold\": 15, \"enable\": true}, \"remove_buttons\": true, \"remove_image_figures\": true, \"remove_link_clusters\": true, \"table_config\": {\"min_rows\": 2, \"min_cols\": 3, \"format\": \"plain\"}, \"remove_chinese\": true, \"remove_edit_buttons\": true, \"extract_latex\": true}, \"warc_path\": \"s3:\/\/commoncrawl\/crawl-data\/CC-MAIN-2021-39\/segments\/1631780060677.55\/warc\/CC-MAIN-20210928092646-20210928122646-00499.warc.gz\"}"}
| null | null |
[](https://travis-ci.org/christophercliff/mke-bus)
This project aims to provide an independent, open and open source interface to the [Milwaukee Country Transit System][mcts] real time [bus location data][bustime]. It's currently accessible at [https://mke-bus.herokuapp.com/][herokuapp].
## Features
- [x] Publicly accessible—no access key required
- [x] Web and human friendly [HAPI-REST][hapi] API
- [x] Cross-domain requests with [JSONP][jsonp] or [CORS][cors]
- [x] [Fully documented API][herokuapp]
- [x] [JavaScript client][client] (for the browser and node.js)
- [ ] Interface for spatial querying bus stops
- [ ] Streaming interface for bus arrival times
- [ ] User-facing web and mobile applications
## Contributing
See [CONTRIBUTING][contributing] for details.
## License
MIT, see [LICENSE][license] for details.
[mcts]: http://www.ridemcts.com/
[bustime]: http://realtime.ridemcts.com/bustime/home.jsp
[herokuapp]: https://mke-bus.herokuapp.com/
[client]: https://github.com/christophercliff/mke-bus-client
[hapi]: https://github.com/jheising/HAPI
[jsonp]: http://en.wikipedia.org/wiki/JSONP
[cors]: https://developer.mozilla.org/en-US/docs/Web/HTTP/Access_control_CORS
[contributing]: https://github.com/christophercliff/mke-bus/blob/master/CONTRIBUTING.md
[license]: https://github.com/christophercliff/mke-bus/blob/master/LICENSE.md
|
{
"redpajama_set_name": "RedPajamaGithub"
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| 4,279
|
Die Reichspräsident-Ebert-Kaserne ist eine Kaserne der Bundeswehr in Hamburg-Iserbrook. Sie wurde von 1935 bis 1937 erbaut. Ursprünglich hieß sie Iserbrook-Kaserne, seit 1965 ist sie nach Friedrich Ebert benannt, dem ersten Reichspräsidenten der Weimarer Republik.
Geschichte
Im Rahmen der Aufrüstung der Wehrmacht wurde ab Mitte der 1930er-Jahre ein ehemaliges Klinikgelände ("Landrat Scheiff Krankenhaus") in Hamburg-Iserbrook für militärische Nutzungen umgebaut, erweitert und zunächst durch Flugabwehreinheiten der Luftwaffe genutzt.
Während der Besatzung durch die Britische Rheinarmee nach dem Zweiten Weltkrieg wurde die Kaserne als Reading Barracks bezeichnet und durch verschiedene Einheiten des "Labour Service" genutzt.
Die Bundeswehr begann 1957 mit der Übernahme des Geländes und schloss diese am 1. April 1958 ab. Während der Nutzung durch die Bundeswehr waren eine Vielzahl verschiedener Einheiten und Dienststellen hier stationiert. Die bedeutenderen sind die seit 1958 dort stationierte Bundeswehrfachschule, das Landeskommando Hamburg sowie die zeitweise dort stationierte Führungsakademie der Bundeswehr.
Stationierte Einheiten der Bundeswehr
Denkmalschutz
Einige Teile der Kasernengebäude stehen seit der Neufassung des Hamburger Denkmalschutzgesetzes im Jahre 2013 unter Schutz. Sehr viele der heutigen Gebäude stammen noch aus der Bauzeit der Kaserne, einige Gebäude im südlichen Teil sind älter und wurden in den 1900er-Jahren ursprünglich als Klinikgebäude errichtet. Insgesamt gilt die Bausubstanz als wenig verändert und zeigt an vielen Stellen noch den für die Errichtungszeit typischen Bauschmuck.
Die in Verbindung mit der Kaserne errichteten Dienstwohnungen am Lachmannweg und an der Osdorfer Landstraße stehen ebenfalls unter Denkmalschutz.
Bildergalerie
Bundeswehr in Hamburg
In unmittelbarer Nachbarschaft zur Reichspräsident-Ebert-Kaserne befinden sich zwei weitere Kasernen der Bundeswehr. Die Generalleutnant-Graf-von-Baudissin-Kaserne liegt im benachbarten Stadtteil Hamburg-Osdorf und ist ca. 2,5 km entfernt. Die Clausewitz-Kaserne befindet sich in Hamburg-Nienstedten und ist Luftlinie ca. 1 km entfernt. Der mittlerweile geschlossene Mobilmachungsstützpunkt Sülldorf in Hamburg-Sülldorf befand sich ca. 3 km entfernt.
Weblinks
MGFA.de – Standortdatenbank des Militärgeschichtlichen Forschungsamtes Potsdam (stationierte Einheiten in der Reichspräsident-Ebert-Kaserne über Suche nach Postleitzahl 22589 erreichbar)
Einzelnachweise
Kaserne in Hamburg
Kulturdenkmal in Hamburg-Iserbrook
Friedrich Ebert als Namensgeber
Erbaut in den 1930er Jahren
Kaserne der Bundeswehr
Bauwerk im Bezirk Altona
|
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"redpajama_set_name": "RedPajamaWikipedia"
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| 8,791
|
Temperature sensor delivers multi-point sensing, high accuracy
Dallas, Tex. — Dallas Semiconductor, a wholly owned subsidiary of Maxim
Integrated Products, has introduced the DS28EA00, a 1-wire digital temperature
sensor with ±0.5°C accuracy and a new chain-mode signaling and protocol feature that automatically determines the physical location of individual sensors in an
environment where multiple sensors are connected to a common 1-wire line.
The DS28EA00 is suitable for applications that require accurate multi-point digital temperature measurements with minimal wiring complexity and cost. Typical applications include rack-based equipment such as wireless base stations, central office switches, enterprise servers, HVAC, and process monitoring.
Available in an 8-pin microSOP package, the DS28EA00 measures temperature over the -40°C to 85°C range and is factory calibrated in a liquid bath to provide a worst-case conversion error of ±0.5° within the ambient-measurement-temperature range of -10°C to 85°; it provides a ±2.0°C worst-case conversion error within the ambient range of -40° to -10°. Device temperature sensing and digitization is produced from the differential VBE output of a matched-NPN transistor pair and a 1st-order sigma-delta ADC. Conversion resolution is user programmable with 9-bit to 12-bit options.
The DS28EA00 operates over the Dallas Semiconductor 1-wire interface with a protocol
that delivers power as well as communication over a single connection. This unique interface enables multiple 1-wire devices, including the DS28EA00, to be multi-dropped on the line and operate independently. With power and communication over a single contact, the DS28EA00 minimizes wiring complexity and associated cost in a multi-device system, said the company.
Like all 1-wire devices, the DS28EA00 has a unique 64-bit internal serial
number that is used in a multi-drop environment to select a specific device for temperature results. Said to be the first of its kind to determine the physical location of multiple temperature sensors, the DS28EA00 includes a chain-mode command set and two signal pins for daisy-chaining multi-dropped devices. A 1-Wire host controller then uses the chain-mode command protocol to learn the 64-bit serial number of each part in the chain, starting with the first device on the line and proceeding sequentially.
This sequence knowledge is directly linked to the physical location of the DS28EA00 in the system. In addition, the two chain-mode daisy-chain signal pins perform a dual-purpose. They can be used to detect the physical location of a device in a network, or they can be used as a 5 V/4 mA GPIO to control LED indicators or other input/output signaling.
Prices: Starts at $2.27 (1,000-up, FOB USA)
Datasheet: www.maxim-ic.com/DS28EA00
Dallas Semiconductor Corp. , 1-972-371-4000, www.maxim-ic.com
0 comments on "Temperature sensor delivers multi-point sensing, high accuracy"
|
{
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