text stringlengths 16 3.88k | source stringlengths 60 201 |
|---|---|
a typical theoretical physicist does for a living: guess the
equation!
F µν ≈ J µ
No, derivatives not right.
∂F µν /∂xν ≈ J µ
No, constants not right.
F µν /∂xµ = J µ
1
c
Correct, amazingly! (even sign)
µ = 0:
∂F 0ν /∂xν = ρ
∂F 0i/∂xi = ρ
F0i = −Ei
F 0i = Ei
So � · E� = ρ verified!
Electromagnetism in a ... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
� = ρ in all dimensions.
Notation: Circle S� is a 1D manifold, the boundary of a ball B2
Sphere S2(R) : x2
1
Ball B3(R) : x2
2
+ x + x
1
2
2
+ x + x
2
2
3 = R2
3 ≤ R2
2
5
Lecture 3
8.251 Spring 2007
When talking about S2(R), call it S2 (R = 1 implied)
Vol(S1) = 2π
Vol(S2) = 4π
Vol(S3) = 2π2
Vol(Sd−1) =
d
22π... | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
E(r) =
Γ(d/2) q
2πd/2 rd−1
Electric field of a point charge in d dimensions.
If there are extra dimensions, then would see larger E at very small distances.
7 | https://ocw.mit.edu/courses/8-251-string-theory-for-undergraduates-spring-2007/184b1fe03c3d32c9ac0a860af62b877c_lec3.pdf |
Introduction to Algorithms: 6.006
Massachusetts Institute of Technology
Instructors: Erik Demaine, Jason Ku, and Justin Solomon
Recitation 3
Recitation 3
Recall that in Recitation 2 we reduced the Set interface to the Sequence Interface (we simulated
one with the other). This directly provides a Set data structur... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
return self.A.get_at(i + 1)
return None
def find_prev(self, k):
# O(log n)
return None
if len(self) == 0:
i = self._binary_search(k, 0, len(self) - 1)
x = self.A.get_at(i)
if x.key < k:
if i > 0:
else:
return x
return self.A.get_at(i - 1)
return None
def insert(self, x):
if len(self.A) == 0:
self.A.inse... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
algorithms for sorting
small numbers of items because they are easy to understand and implement. Both algorithms are
incremental in that they maintain and grow a sorted subset of the items until all items are sorted.
The difference between them is subtle:
• Selection sort maintains and grows a subset the largest i ... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
def insertion_sort(A):
2
for i in range(1, len(A)):
j = i
while j > 0 and A[j] < A[j - 1]:
# Insertion sort array A
# O(n) loop over array
# O(1) initialize pointer
# O(i) loop over prefix
A[j - 1], A[j] = A[j], A[j - 1] # O(1) swap
j = j - 1
# O(1) decrement j
In-place and Stability
Both insertion sort an... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
_sort(A, a = 0, b = None):
2
if b is None:
b = len(A)
if 1 < b - a:
c = (a + b + 1) // 2
merge_sort(A, a, c)
merge_sort(A, c, b)
L, R = A[a:c], A[c:b]
i, j = 0, 0
while a < b:
# Sort sub-array A[a:b]
# O(1) initialize
# O(1)
# O(1) size k = b - a
# O(1) compute center
# T(k/2) recursively sort left
# T... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
can be made stable with only a small modification. Can you
modify the implementation to make it stable? We’ve made CoffeeScript visualizers for the merge
step of this algorithm, as well as one showing the recursive call structure. You can find them here:
https://codepen.io/mit6006/pen/RYJdOG
https://codepen.io/mit600... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
tree. When f (n) grows asymptotically faster than n
, the work done at each level decreases
geometrically so the work at the root dominates; alternatively, when f (n) grows slower, the work
done at each level increases geometrically and the work at the leaves dominates. When their growth
rates are comparable, the w... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
and
show that your recurrence relation satisfies all conditions required by the relevant case. There are
even stronger (more general) formulas1 to solve recurrences, but we will not use them in this class.
Exercises
1. Write a recurrence for binary search and solve it.
Solution: T (n) = T (n/2) + O(1) so T (n) = O(... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
T (n) = 4T (n/2) + O(n)
Solution: T (n) = O(n2), height log2 n degree-4 tree, O(2k) work per node at height k.
1http://en.wikipedia.org/wiki/Akra-Bazzi_method
MIT OpenCourseWare
https://ocw.mit.edu
6.006 Introduction to Algorithms
Spring 2020
For information about citing these materials or our Terms of Use, vis... | https://ocw.mit.edu/courses/6-006-introduction-to-algorithms-spring-2020/1869dbf640ded6b31f1bd369d2001ef5_MIT6_006S20_r03.pdf |
8.701
0. Introduction
0.2 Course Organization
Introduction to Nuclear
and Particle Physics
Markus Klute - MIT
1
Course Setup
Inverted classroom in online setting (see additional video A.02)
Lectures are organized in short video discussing specific
concepts or methods. Each video is supplemented with slides,
text, ... | https://ocw.mit.edu/courses/8-701-introduction-to-nuclear-and-particle-physics-fall-2020/18834425462a7fb4f77dd5b94bedfd54_MIT8_701f20_lec0.2.pdf |
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-1
Lecture 5 - PN Junction and MOS
Electrostatics (II)
pn Junction in Thermal Equilibrium
September 22, 2005
Contents:
1. Introduction to pn junction
2. Electrostatics of pn junction in thermal equilibrium
3. The depletion approximation
4. Contact potent... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
cid:0)(cid:0)
(b)
metal contact to(cid:13)
n side
Doping distribution of abrupt p-n junction:
p-region
Na
N
0
n-region
Nd
x
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-5
What is the carrier concentration distribution in thermal
equilibrium?
First think of two sides separately:
p-region
n-region... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-8
3. The depletion approximation
• Assume QNR’s perfectly charge neutral
• assume SCR depleted of carriers (depletion region)
• transition between SCR and QNR’s sharp
(must calculate where to place −xpo and xno)
Na
po, no
depletion(cid:13)
approximatio... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
−qNadx
(x + xpo)
• 0 < x < xno
E(x) = qNd
(cid:15)s
(x − xno)
• xno < x
E(x) = 0
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-11
• Electrostatic potential
(with φ = 0 @ no = po = ni):
φ =
kT
q
ln
no
ni
φ = −
kT
q
ln
po
ni
In QNR’s, no, po known ⇒ can determine φ:
in p-QNR: po = Na ⇒ φp = −kT
in n... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
φ(x) = φp + qNa
2(cid:15)s
(x + xpo)2
• 0 < x < xno
φ(x) = φn − qNd
2(cid:15)s
(x − xno)2
• xno < x
φ(x) = φn
Almost done...
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-14
Still don’t know xno and xpo ⇒ need two more equations
1. Require overall charge neutrality:
qNaxpo = qNdxno
2. Require φ(x)... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
2(cid:15)sφB
qNd
∝
1
√
Nd
|Eo| '
v
u
u
u
u
u
t
2qφBNd
(cid:15)s
√
Nd
∝
The lowly-doped side controls the electrostatics of the pn
junction.
ρ
qNd
-qNa
ρ
qNd
ρ
qNd
x(cid:13)
x(cid:13)
x
-qNa
-qNa
6.012 - Microelectronic Devices and Circuits - Fall 2005
Lecture 5-17
4. Contact potentials
Potential distribution in therma... | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
⇒ from contact to contact, there is no potential build-
up across pn junction | https://ocw.mit.edu/courses/6-012-microelectronic-devices-and-circuits-fall-2005/188f6340d4d10b0e923d1c0166c8fba1_lec5.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.01 Single Variable Calculus
Fall 2006
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
Lecture 5
18.01 Fall 2006
Implicit
5
Lecture
Differentiation
and
Inverses
Implicit Differentiation
d
dx
(x a) = ax a−1 .
Example 1.... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/18d6a86a30a4bc046c5f7034d47587f1_lec5.pdf |
x2 .
So y = ±
1 − x2. Let us look at the positive case:
√
1
y = + 1 − x2 = (1 − x 2) 2
dy
dx
=
1
−
(1 − x 2)
2
(−2x) =
√
−x
1 − x2
=
−x
y
�
� �
1
2
1
Lecture 5
18.01 Fall 2006
Now, let’s do the same thing, using implicit differentiation.
d �
dx
(x 2) +
x 2 + y 2 = 1
2�
d
dx
x 2 ... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/18d6a86a30a4bc046c5f7034d47587f1_lec5.pdf |
= f (x)
f −1(y) = x
d
dx
(f −1(y)) =
d
dx
(x) = 1
By the chain rule:
= 1
d
dy
(f −1(y))
dy
dx
and
(f −1(y)) =
d
dy
1
dy
dx
2
Lecture 5
18.01 Fall 2006
So, implicit differentiation makes it possible to find the derivative of the inverse function.
Example. y = arctan(x)
tan y = x
d
dx
[tan(y)] =
d... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/18d6a86a30a4bc046c5f7034d47587f1_lec5.pdf |
) = f −1(y) = x. To graph g and f together we need to write g as a
function of the variable x. If g(x) = y, then x = f (y), and what we have done is to trade the
variables x and y. This is illustrated in Fig. 2
f −1(f (x)) = x f −1 f (x) = x
◦
f (f −1(x)) = x f
◦
f −1(x) = x
Figure 2: You can think about f −1 as t... | https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/18d6a86a30a4bc046c5f7034d47587f1_lec5.pdf |
§ 18. Lattice codes (by O. Ordentlich)
Consider the n-dimensional additive white Gaussian noise (AWGN) channel
Y X Z
+
=
where Z ∼ N (0, In×n) is statistically independen
t
reliably over this channel, under the power constraint
of the input X. Our goal is to communicate
∥ ∥2 ≤ SNR
X
1
n
where SNR is the signal-to-noise... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
1 Lattice Definitions
A lattice Λ is a discrete subgroup of
lattice Λ in R
R
n is spanned by some n n matrix G such that
×
n which is closed under reflection and real addition. Any
Λ = {t = Ga ∶ a ∈ Z
n}.
185
We will assume G is full-rank. Denote the nearest neighbor quantizer associated with the lattice Λ
by
where ties... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
ver
and therefore
∩
+ V) S] + t
Dt = [(−t
= A− +
t
t.
Vol(S) =
∑
Λ
t
∈
Vol
(A ) =
t
∑
∈Λ
t
Vol
(A− + ) =
t
t
Vol
(Dt) = Vol(V).
∑
∈Λ
t
S =
⊍
∈Λ
t
At =
⊍
∈Λ
t
A−t =
⊍
Λ
t
∈
Dt − t,
[S] mod Λ =
Dt = V.
⊍
∈Λ
t
186
Corollary
(
∣
det G
)∣. In Particular, Vol
(V
) = ∣
)∣
(
det G .
18.1. If S is a fundamental cell of a latti... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
reduction. Thus, a lattice decoder is indeed much more “structured” than ML decoder for a
random code.
R , where
the
Note that for an additive channel Y X Z, if X Λ we have that
=
+
∈
Pe = Pr
(QΛ(Y
)
≠ X) = Pr(Z ∉ V).
(18.4)
We therefore see that the resilience of a lattice
Since we know that Z will be inside a ball of... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
as a ball with the same volume.
that is
go
od for
coding is as resilient to semi
187
(a)
Figure 18.1: (a) shows a lattice in R
effective ball.
(b)
2, and (b) shows its Voronoi region and the corresponding
18.2 First Attempt at AWGN Capacity
√
n with reff(Λ) =
Assume we have a lattice Λ ⊂ R
n(1 + δ) that is good for codi... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
)
Y v
on the shifted
output. Since
Pe = Pr Q
( Λ(Y − v) ≠ t
) = Pr(Z ∉ V).
188
reff heme over
with n. Taking δ → 0 we see that any rate
log(SNR) can be achieved reliably. Note that for this coding scheme (encoder+decoder) the
∥
Since Λ is good for coding and
additive semi norm-ergodic noise channel will vanish
an
R 1
<
... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
�c Λf is defined as [CS83, For89, EZ04]
lattice code (sometimes also called “Voronoi constellation”)
based on the nested lattice
(18.8)
Proposition 18.2.
L ≜ Λf ∩ Vc.
∣L∣ =
Vol(Vc)
Vol(Vf )
.
Thus, the codebook L has rate R = 1 log ∣L∣ = log Γ
n
(Λf , Λc).
Proof. First note that
Let
and note that
Λf ≜
⊍
t
∈L
(t
+ Λc).
(... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
Uniform(B(reff(Λ)). By the isoperimetric inequality [Zam14]
The average transmission
∼
L
L
V
σ2(Λ) ≥
1
n
E∥W∥2 =
r2
eff(Λ)
+
n 2
.
exhibits
( )
(
B
reff Λ .
a good tradeoff between average power and volume if its second moment is close
A lattice Λ
to that of
Definition 18.5 (Goodness for MSE quantization). A sequence of lat... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
rate is
be a
nested
SNR − (cid:15) , whereas the fine lattice is good for coding and has r2
(
)
1
therefore
(
eff Λf
SNR
(
1
n
1
2
log
(
log (
log
)
ol(Vc)
V
Vol(Vf )
r2
eff(Λc)
r2
eff(Λf )
SNR(1 − (cid:15))
SNR
1+SNR (1 + (cid:15))
)
⎛
⎝
R =
=
→
→
1
2
1
2
log
(1 + SNR) ,
190
⎞
⎠
(18.9)
Figure 18.2: An example of a nested... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
Proof.
[
that v
or any v
F
V]
+
mod Λ
∈ R
n the set v
= V
and Vol
+ V
( +
v
fundamen
a
is
(V)
=
V)
Vol
tal cell of Λ. Th
us, by Proposition 18.1 we have
∈
. Thus, for any v R
n
X∣V = v ∼ [v + U] mod Λ ∼ Uniform(V).
The Crypto Lemma ensures that 1
1 (cid:15) SNR, but our power constraint was X 2
∥ ∥ ≤
n
nSNR. Since X is... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
2 α2
<
+ ( −
1 α 2SNR.
)
(18.11)
(18.12)
(18.13)
(18.14)
dic,
Since Z is semi norm-ergo
is good for MSE
α . Setting α SNR 1 SNR , such as to minimize the upper bound on σ2
variance σ2
( )
eff
ariance σ2
in effective v
er the Voronoi region of a lattice that
quantization, Theorem 18.1 implies that Zeff is semi norm-ergodic... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
fixed shift u that achieves the
same, or better, Pe,avg. However, for a fixed shift the error probability is no longer independent of
the chosen message.
18.4 Dirty Paper Coding
Assume now that the channel is
Y = X S
+ + Z,
192
Z is a unit variance semi norm-ergodic noise, X is subject to the same power constraint
where... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
construction is based
n to form
on starting with a linear code over a prime finite field, and embedding it periodically in R
a lattice.
k n
Definition 18.6 (p-ary Construction A). Let p be a prime number, and let F ∈ Z
×
p
matrix whose entries are all members of the finite field Zp. The matrix F generates a
code
×
be a k n
... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
if all
that for any x Λ F
minimum distance. Thus, Λ F is indeed
F are distinct, then Λ F has a finite
C( )
number of distinct codewords in F is
the
pk, so the
ery integer shift of the unit cube
p k.
−
Similarly, we can construct a nested lattice pair from
˜
a are some ve
and
∈
−
in
)
(V
≤
obtained by taking only the firs... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
matrix
]
). Also, in order to specify the codebook , only
L
we take the elements of F to be i.i.d. and
If
nested
lattice
p O n 1 (cid:15) 2
( + )/ )
= (
coarse lattice are good for both coding and for MSE quantization [OE15].
ble of
codes. It can be shown that if p grows fast enough with the dimension n (taking
suffices)... | https://ocw.mit.edu/courses/6-441-information-theory-spring-2016/18e413df88c6303554cc45a6b3b876b6_MIT6_441S16_chapter_18.pdf |
Massachusetts Institute of Technology
6.042J/18.062J, Fall ’05: Mathematics for Computer Science
Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld
September 16
revised September 20, 2005, 1273 minutes
Solutions to InClass Problems Week 2, Fri.
Problem 1. Subset takeaway1 is a two player game involving a fixed finit... | https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/194a6380ce2c26852fde38af8e2daa39_cp2fsol.pdf |
is
therefore a win for the 2nd player.
Case 3:[1st player chooses a size 2 subset {a, b}] Then 2nd player should choose the complemen
tary size 2 subset {c, d}. Here a, b, c, d are the numbers 1,2,3,4 in some order. So the possible
remaining moves are the four other subsets of size 2 and the four subsets of size 1.... | https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/194a6380ce2c26852fde38af8e2daa39_cp2fsol.pdf |
some other size 1 subset will also available as a
possible move, and Player 2 should choose any such size 1 subset. This either leaves
no more moves, or the situation of a game of size 2, which Player 2 will win in either
case.
We leave it to the reader to check that the moves prescribed in the subcases 3.1–3 will ... | https://ocw.mit.edu/courses/6-042j-mathematics-for-computer-science-fall-2005/194a6380ce2c26852fde38af8e2daa39_cp2fsol.pdf |
18.782 Introduction to Arithmetic Geometry
Lecture #15
Fall 2013
10/29/2013
As usual, k is a perfect field and k is a fixed algebraic closure of k. Recall that an affine
(resp. projective) variety is an irreducible alebraic set in An = An(k) (resp. Pn = Pn(k)).
¯
¯
¯
15.1 Rational maps of affine varieties
Before defining rati... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
advantage of this approach is that there is then a one-to-one correspondence between
morphisms and the functions they define. If φ = (φ1, . . . , φn) and ψ = (ψ1, . . . , ψn) define
the same function from X to Y then each of the polynomials φi−ψi in k[x1, . . . , xm] contains
X in its zero locus and therefore lies in the... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
− x3x4 in A (which is in fact a
Example 15.1. Consider the the zero locus X of x x
variety) and the rational function r = x1/x3 = x4/x2. At the point P = (0, 1, 0, 0) ∈ X the
value x1(P )/x3(P ) is not defined, but x4(P )/x2(P ) = 0 is defined, and the reverse occurs
for P = (0, 0, 1, 0) ∈ X. But we can assign a meaningf... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
3
¯
¯
¯
We now associate to r the function from dom(r) to k that maps P to
¯
r(P ) = (f /g)(P ) = f (P )/g(P ),
where g is chosen so that g(P ) = 0 and gr = f
k[X]. Now if r is actually an element of
¯k[X], then r is regular at every point in X and we have dom(r) = X. The following lemma
says that the converse holds.
L... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
if it is regular.
Proof. A morphism is clearly a regular rational map. For the converse, apply Lemma 15.3
to each component of φ = (φ1, . . . , φn).
We now want to generalize the categorical equivalence between affine varieties and their
function fields, analogous to what we proved in the last lecture for affine varieties a... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
want to restrict our attention to rational maps whose image is
dense in its codomain.
Definition 15.7. A rational map φ : X → Y is dominant if φ(dom(φ)) = Y .
If φ : X → Y and ψ : Y → Z are dominant rational maps then the intersection of
φ(dom(φ)) and dom(ψ) must be nonempty; the complement of dom(ψ) in Y is a proper
cl... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
, . . . , αn, but there may be algebraic relations between the αi that make
many of these expressions equivalent. In any case, R is isomorphic to the quotient of the
polynomial ring k[x1, . . . , xn] by an ideal I corresponding to all the algebraic relations that
exist among the αi. The ring R is an integral domain (si... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
¯
¯
Proof. The proofs of parts (i) and (iii) follow the proof of Theorem 14.8 verbatim, with
coordinate rings replaced by function fields, only part (ii) merits further discussion. As
discussed above, we can write K (cid:39) k(Y ) and L (cid:39) k(X) for some varieties X and Y , and
¯
¯
3
3 y morphism K
an
isomorphisms,... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
domain and the image of θ∗ is the zero element. This
is in turn equivalent to showing that if r ◦ θ∗ = θ(r) vanishes at every point in its domain
then r = 0. But the only element of k(X) that vanishes at every point in its domain is the
zero element, and θ is injective, so we are done.
¯
¯
¯
Corollary 15.9. The categor... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
have done for maps between affine varieties to
maps between projective varieties. This is completely straight-forward, we just need to
account for the equivalence relation on Pn.
Recall from Lecture 13 that although we defined the function field k(X) of a projective
variety X ⊆ Pn as the function field of any of its non-emp... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
φn) in its
class with each component regular at P and at least one nonzero at P . The open subset
of X at which φ is regular is denoted dom(φ).
Remark 15.15. We can alternatively represent the rational map φ : X → Y as a tuple
of homogeneous polynomials in k[x0, . . . , xm] that all have the same degree and not all of
... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
ive varieties. The proofs are exactly the same, modulo the equivalence
relations for projective points and rational maps between projective varieties. Alternatively,
one can simply note that any dominant rational map X → Y of projective varieties restricts
to a dominant rational map between any pair of the nonempty affin... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
regular (it is not defined at
−1), but it is
dominant (but not surjective). Thus X is birationally equivalent to A1, but not isomorphic
to A1
, as expected. The function fields of X and A1 are both isomorphic to k(t).
¯
√
Now let us consider the corresponding map of projective varieties. The projective closure
X of X in ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
: A2 → A2 defined by φ(x, y) = (x, xy) from
Lecture 14, where we noted that the image of φ is not closed (but it is dense in A2, so φ is
dominant). Let us now consider the corresponding rational map ϕ : P2 → P2 defined by
x xy
:
z2
z
but this is not
the case! It is not regular at (0 : 1 : 0).
This is not an accident. As ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
cid:79) 201(cid:22)
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/194e47df6e1f016ee3fb2719abe7e8d3_MIT18_782F13_lec15.pdf |
Topic 3 Notes
Jeremy Orloff
3 Line integrals and Cauchy’s theorem
3.1 Introduction
The basic theme here is that complex line integrals will mirror much of what we’ve seen for multi-
variable calculus line integrals. But, just like working with is easier than working with sine and
cosine, complex line integrals are ea... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
so the right hand side of this equation is
That is, it is exactly the same as the expression in Equation 1a.
(()) ′() .
∫
2 along the straight line from 0 to 1 + .
Example 3.1. Compute ∫
Solution: We parametrize the curve as () = (1 + ) with 0 ≤ ≤ 1. So ′() = 1 + . The line
integral is
1
∫
2 = ∫
0
2(1 + )2... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
curve in from
0 to
1 then
′() = (
1) − (
0).
∫
3 LINE INTEGRALS AND CAUCHY’S THEOREM
3
Proof. This is an application of the chain rule. We have
(())
= ′(()) ′().
So
′() = ∫
′(()) ′() = ∫
∫
(())
= (()) = (
1) − (
0).
(cid:243)
(cid:243)
(cid:243)
(cid:243)
Another equivalent way to state the fun... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
3.7.
Theorem 3.8. If () has an antiderivative in an open region , then the path integral ∫
path independent for all paths in .
Proof. Since () has an antiderivative of (), the fundamental theorem tells us that the integral
only depends on the endpoints of , i.e.
() is
() = (
1) − (
0)
∫
where
0 and
1 are th... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
same line integral. This shows that the line integrals
Figure (i)
Figure (ii)
3.5 Examples
using the fundamental theorem.
Example 3.10. Why can’t we compute ∫
Solution: Because doesn’t have an antiderivative. We can also see this by noting that if had an
antiderivative, then its integral around the unit circle ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
1 = ∫
∫ 2
′() = (endpoint) − (start point) = 0.
It equals 0 because the start and endpoints are the same.
(ii) As usual, we parametrize the unit circle as () = e with 0 ≤ ≤ 2. So, ′() = e and the
integral becomes
2
1
2
∫
= ∫
0
1
e2
e = ∫
0
2
e− = −e− = 0.
2
(cid:243)
(cid:243)
(cid:243)0
3.6 Cauchy... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
(− − ) = 0.
∫
∫
+ = ∫
We get 0 because the Cauchy-Riemann equations say = , so − = 0.
To see part (i′) you should draw a few curves that intersect themselves and convince yourself that
they can be broken into a sum of simple closed curves. Thus, (i′) follows from (i).1
( − ) = 0.
Part (ii) follows from (i) and ... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
to prove that the Cauchy-Riemann equations given in Equation 3 hold for (). The
0 to , which can be used to compute (). To compute
figure below shows an arbitrary path from
the partials of we’ll need the straight lines that continue to + ℎ or + ℎ.
To prepare the rest of the argument we remind you that the fundamenta... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
is analytic and = . □
′
3.7 Extensions of Cauchy’s theorem
8
(6)
Cauchy’s theorem requires that the function () be analytic on a simply connected region. In cases
where it is not, we can extend it in a useful way.
Suppose is the region between the two simple closed curves
oriented in a counterclockwise directio... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
at the origin. Thus,
Case (ii): we will show that
() = 0.
∫
1
() = 2.
∫
2
Let
3 be a small circle of radius centered at 0 and entirely inside
2.
Figure for part (ii)
xyxyC1RC2xyC2C3R 3 LINE INTEGRALS AND CAUCHY’S THEOREM
10
By the extended Cauchy theorem we have
() = ∫
() = ∫
0
3
∫
2
2
= 2.
Here,... | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
MIT OpenCourseWare
https://ocw.mit.edu
18.04 Complex Variables with Applications
Spring 2018
For information about citing these materials or our Terms of Use, visit: https://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-04-complex-variables-with-applications-spring-2018/1968cab6de7c96ead1c83001459a007b_MIT18_04S18_topic3.pdf |
18.782 Introduction to Arithmetic Geometry
Lecture #10
Fall 2013
10/8/2013
In this lecture we lay the groundwork needed to prove the Hasse-Minkowski theorem
for Q, which states that a quadratic form over Q represents 0 if and only if it represents 0
over every completion of Q (as proved by Minkowski). The statement sti... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
all of its elements lie in Zp and at least one lies in Z×
gives several equivalent definitions of the Hilbert symbol.
Lemma 10.2. For any a, b ∈ Q×
p , the following are equivalent:
(i) (a, b)p = 1.
(ii) The quadratic form z2 − ax2 − by2 represents 0.
(iii) The equation ax2 + by2 = z2 has a primitive solution.
(iv) a ∈ ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
1.
(iii) (a, c)p = 1 =⇒ (a, c)p(b, c)p = (ab, c)p.
(iv) (c, c)p = (−1, c)p.
c) to Qp. For (i) we have N (1) = 1. For (ii),
Proof. Let N denote the norm map from Qp(
−c = N (−c) for c ∈ Q×2 and −c = N (
c) otherwise. For (iii), If a and b are both norms
in Q(
c), then so is ab, by the multiplicativity of the norm map; c... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
find a non-trivial solution
to z2 − ux2 − vy2 = 0 modulo p. If we then fix two of x, y, z so that the third is nonzero,
Hensel’s lemma gives a solution over Zp.
Remark 10.6. Lemma 10.5 does not hold for p = 2; for example, (3, 3)2 = −1.
Theorem 10.7. Let p be an odd prime, and write a, b ∈ Q×
α, β ∈ Z and u, v ∈ Z×
p as ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
. Applying Corol-
(p, v)p
(cid:17)
v
p
(cid:16) u
p
lary 10.3 we have
(pu, pv)p = (pu, pv)p(−pv, pv)p = (−p2uv, pv)p = (−uv, pv)p = (pv, −uv)p
Applying the formula in the case α = 1, β = 0 already proved, we have
(pv, −uv)p =
(cid:19)
(cid:18) −uv
p
=
(cid:18) −
1
p
(cid:19) (cid:18) u
p
(cid:19) (cid:18) v
p
(cid:19)
... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
, z0)
modulo 8 must have y0 or z0 odd; if not, then z2
0, are divisible
by 4, but this means x0 is also divisible by 2, which contradicts the primitivity of (x0, y0, z0).
Lifting w0 to a root of f (w) over Z2 yields a solution to the original equation.
0, and therefore 2ux2
0 and vy2
Theorem 10.9. Write a, b ∈ Q×
Then
... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
)(cid:47)(cid:72)(cid:70)(cid:87)(cid:88)(cid:85)(cid:72)(cid:3)(cid:20)(cid:19)(cid:3)(cid:54)(cid:68)(cid:74)(cid:72)(cid:3)
2
2
We now note the following corollary to Theorems 10.4, 10.7, and 10.9.
Corollary 10.10. The Hilbert symbol (a, b)p is a nondegenerate bilinear map. This means
that for all a, b, c ∈ Q×
p we ... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
have (b, c)p = ( w ) =p
) =p
−1.
−
−
γ
p
p
3
3 For p = 2, we have
(a, c) (b, c) = (−1)(cid:15)(u)(cid:15)(w)+αω(w)+γω(u)(−1)(cid:15)(v)(cid:15)(w)+βω(w)+γω(v)
2
2
= (−1)((cid:15)(u)+(cid:15)(v))(cid:15)(w)+(α+β)ω(w)+γ(ω(u)+ω(v))
= (−1)(cid:15)(uv)(cid:15)(w)+(α+β)ω(w)+γω(uv)
= (ab, c)2,
where we have used the fact that... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
b)p = 1.
p
Proof. We can assume without loss of generality that a, b ∈ Z, since multiplying each of a
and b by the square of its denominator will not change (a, b)p for any p. The theorem holds
if either a or b is 1, and by the bilinearity of the Hilbert symbol, we can assume that
a, b ∈ {−1} ∪ {q ∈ Z>0 : q is prime}.
... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
to (−1)
(2, b)2 = (−1)
ω(b)
(cid:54)∈ {2, b}, while
.
2
p
Case 5: a and b are distinct odd primes. Then (a, b)p = 1 for all p (cid:54)∈ {2, a, b}, while
(a, b)
p =
(cid:0)
(cid:0)
(−1)(cid:15)(a)(cid:15)(b)
(cid:1)
a
b
b (cid:1)
a
if p = 2,
if p = b,
if p = a.
Since (cid:15)(x) = (x − 1)/2 mod 2, we have
(cid:8... | https://ocw.mit.edu/courses/18-782-introduction-to-arithmetic-geometry-fall-2013/196f7a627ae6d75ab009bbef96e7c8c9_MIT18_782F13_lec10.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.727 Topics in Algebraic Geometry: Algebraic Surfaces
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
ALGEBRAIC SURFACES, LECTURE 2
LECTURES: ABHINAV KUMAR
Remark. In the definition of (L, M ) we wrote M = OX (... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
D.
·
We say C is numerically equivalent to D, C ≡ D, if C E = D E for every
t1
t2
·
divisor E on X.
We have an intersection pairing Div X × Div X → Z which factors through
Pic X × Pic X → Z, which shows that linear equivalence = ⇒ num equivalence.
alg equivalence (map to P1 defined by (f )) and
In fact, lin equi... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
2g(C) − 2 = C.(C + K) (genus formula). Note: C 2 = deg (OX (C) ⊗ OC ) by
definition. I/I 2 is the conormal bundle, and is ∼ O(−C) ⊗ OC , while NC/X is
the normal bundle ∼= O(C) ⊗ OC .
Theorem 1 (Riemann-Roch). χ(L) = χ(OX ) + 1
2 (L2 − L ωX ).
=
·
Proof. L−1 · L ⊗ ω−1 = χ(OX ) − χ(L) − χ(ωX ⊗ L−1) + χ(ωX ). By Serre... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
C in embedding by nH) for larger n).
1.2. Hodge Index Theorem.
·
·
Lemma 1. Let D1, D2 be two divisors on X s.t. h0(X, D2) = 0
h0(X, D1 + D2).
. Then h0(X, D1) ≤
Proof. Let a = 0
H0(X, D1 + D2) is injective.
∈ H 0(X, D2). Then H 0(X, D1) → H 0(X, D1) ⊗k H 0(X, D2) →
�
a
Proposition 1. If D is a divisor on X with... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
so D · H > 0.
Corollary 1. If D is a divisor on X and H is a hyperplane section on X s.t.
(D · H) = 0 then D2 ≤ 0 and D2 = 0 ⇔ D ≡ 0.
Proof. Only the last statement is left to be proven. If D �≡ 0 but D2 = 0, then ∃E
E� = (H 2)E − (E · H)H, and get D · E� = (H 2)D · E = 0
on X s.t. D.E = 0. Let
and H E� = 0. Thus... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
and C.D equals C ∪ D under
(4)
H 2(X, Q�(1)) × H 2(X, Q�(1)) → H 4(X, Q�(2)) ∼= Q�
The map NumX � C → [C] ∈ H 2 is an injective map. The intersection numbers
define a symmetric bilinear nondegenerate form on M (= NumX ⊗Z R). Let h
be the class in M of a hyperplane section on X. We can complete to a basis for
j. By... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
L by Ln to assume L =
OX (D), D effective.
(5)
0 → Ln−1 → Ls0
n → Ln ⊗ OD → 0
Ln ⊗ OD = Ln|D is ample on D (since L · D = L2 > 0) so H 1(Ln|D) = 0 for
n >> 0.
(6)
H 0(Ln) → H 0(L |D) → H 1(Ln−1) → H 1(Ln) → 0
n
for n >> 0 = ⇒ h1(Ln) ≤ h1(Ln−1) so h1(Ln) stabilizes and the map H 1(Ln−1) →
H 1(Ln) is an isomorphi... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
s.t. X˜ � π−1(p)
X � {p} is an isomorphism and
π−1(p) is a curve ∼ P1 (called the exceptional curve). We explicitly construct
this as follows: take local coordinates at p, i.e x, y ∈ mpOX,p defined in some
neighborhood U of p. Shrink U if necessary so that p is the only point in U
where x, y both vanish. Let U˜ ⊂ U... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
X˜ = Proj
(7)
I˜/I˜2 = OX � (1) so NY �,X˜ = OY � (−1).
If C is an irreducible curve on X passing through P with multiplicity m, then
the closure of π−1(C � {p}) in X˜ is an irreducible curve C˜ called the strict
transform of C. π∗C defined in the obvious way: think of C as a Cartier divisor,
defined locally by som... | https://ocw.mit.edu/courses/18-727-topics-in-algebraic-geometry-algebraic-surfaces-spring-2008/198274c0c471d31fc05d600e28e403db_lect2.pdf |
8.04: Quantum Mechanics
Massachusetts Institute of Technology
Professor Allan Adams
2013 February 12
Lecture 3
The Wavefunction
Assigned Reading:
E&R 16,7, 21,2,3,4,5, 3all NOT 4all!!!
1all, 23,5,6 NOT 2-4!!!
Li.
12,3,4 NOT 1-5!!!
Ga.
3
Sh.
In classical mechanics, the configuration or state of a system i... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
��nd. Furthermore, ψ must be singly-valued and not
“stupid”; the latter point will be elaborated later.
Let us examine this set of examples in further detail. The first wavefunction ψ1 is sharply
peaked at a particular value of x, and the probability density, being its square, is likewise
peaked there as well. This ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
positions for these two wave-
functions are ill-defined, so they are not well-localized, and the uncertainty in the position
is large in each case.
8.04: Lecture 3
3
The fifth wavefunction is multiply-valued, so it is considered to be “stupid”. It does not
have a well-defined probability density.
Note the normaliza... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
hand, ψ1 and ψ2 do not look like sinusoidal waves, so it
is difficult to define a wavelength and therefore a momentum for each, and the respective
momentum uncertainties are large. These qualitatively satisfy the uncertainty relation.
In general, given a wavefunction, once the uncertainty in the position is determined,... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
that for a superposition state
the probability density
ψ(x) = αψa(x) + βψb(x),
p(x) = |αψa(x) + βψb(x)|2 = |αψa(x)|2 + |βψb(x)|2 + α�βψ�(x)ψb(x) + αβ�ψa(x)ψ�
a
b (x)
exhibits quantum interference aside from the usual addition of probability!
For example, let us consider ψ5 = ψ1 + ψ2 from our previous set of examp... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
made
even better with three states of different definite momenta. We could do this for arbitrarily
large countable n: as a state of definite momentum is
ψ(x; k) = e ikx
8.04: Lecture 3
5
except for normalization, a superposition of states of definite momentum
ψ =
ikj x
αj e
j
could have a very well-localized p... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
of states eikx with definite momenta
p = k as
ψ(x) =
√1
∞
2π −∞
˜
ψ(k)e
ikx dk.
(0.8)
Furthermore, ψ˜(k) gives the exact same information as ψ(x) about the
quantum state, so once one is known, the other can be found automat
ically as well.
What do the Fourier transforms of wavefunctions look like? Let us look... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
x has a Fourier transform that is well-localized around a given wavevector, and
that wavevector is the frequency of oscillation as a function of x.
So what then is p(k)? This is the probability density that the particle described by the
wavefunction ψ(x) has a momentum p = k. The expression turns out to be surprisin... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/19862abcdf8e4f7cf1321d060cd6dd7c_MIT8_04S13_Lec03.pdf |
3.15: Transistors in ‘forward active’ mode
Common base circuit
E
B
C
Input
IE,VEB
Output
IE,VCB
IC,mA
5
0
_
IE = 5mA
3mA
1 mA
IE = 0
0
+
_VcB
Figure by MIT OCW.
This is the easiest to visualize, though of limited usefulness. We use one power supply to put EB
into forward bias (p side positive) and another one to put... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1996e5617f952ee322c739d9e24f4ff9_lecture7b.pdf |
supply between E and C (this is chosen to make sure BC
is in reverse bias).
The base current IB is primarily composed of electrons that contribute to the forward current
through EB. In the EB junction, there is a relation between the forward currents of holes and of
electrons. These currents are in the ratios of the... | https://ocw.mit.edu/courses/3-15-electrical-optical-magnetic-materials-and-devices-fall-2006/1996e5617f952ee322c739d9e24f4ff9_lecture7b.pdf |
LECTURE 11
LECTURE OUTLINE
Review of convex progr. duality/counterexamples
Fenchel Duality
•
•
Conic Duality
•
Reading: Sections 5.3.1-5.3.6
Line of analysis so far:
Convex analysis (rel. int., dir. of recession, hy-
•
perplanes, conjugacy)
MC/MC - Three general theorems: Strong dual-
•
ity, existence of dual optimal s... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
⇤ = f ⇤ and there exists a dual optimal
solution.
⌘
⌘
(b) f ⇤ = q⇤, and (x⇤, µ⇤, ⌃⇤) are a primal and
dual optimal solution pair if and only if x⇤
is feasible, µ⇤ ≥
x⇤
arg min L(x, µ⇤, ⌃⇤), µ⇤j gj(x⇤) = 0,
0, and
⌘
x X
⌦
j
2 COUNTEREXAMPLE I
Strong Duality Counterexample: Consider
•
minimize f (x) = e−
subject to x1 ... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
⇤
e−
√x1x2 =
0
1
⇤
⇧
if u > 0,
if u = 0,
if u < 0,
1
0.8
0.6
0.4
0.2
0
0
√x1x2
e−
5
10
15
x1 = u
5
10
x2
20
0
20
15
Connection with counterexample for preserva-
•
tion of closedness under partial minimization.
4 COUNTEREXAMPLE II
Existence of Solutions Counterexample:
, f (x) = x, g(x) = x2. Then x⇤ = 0 is
•
Let X =
th... | https://ocw.mit.edu/courses/6-253-convex-analysis-and-optimization-spring-2012/19c72f4d87303765c3925112f5926282_MIT6_253S12_lec11.pdf |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.