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Lecture 7
8.321 Quantum Theory I, Fall 2017
39
Lecture 7 (Sep. 27, 2017)
7.1 Spin Precession In a Magnetic Field
Last time, we began discussing the classic example of precession of a spin- 1 particle in a magnetic
field. The Hamiltonian of this system is
2
With B = Bzˆ, this becomes
The energy levels are
giving a level ... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/42767c0b651a33268189c7ef1f87630e_MIT8_321F17_lec7.pdf |
. For example, if we initially have |ψ(cid:105) = |+(cid:105), then
which has
|ψ(t)(cid:105) = e−iωt/2|+(cid:105) ,
Prob(cid:0)Sz
(cid:126)=
(cid:1)
=2
1
(7.10)
(7.11)
for all times. This is why energy eigenstates are often called stationary states. Because time
evolution is generated by the Hamiltonian, energy eigenst... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/42767c0b651a33268189c7ef1f87630e_MIT8_321F17_lec7.pdf |
)
(cid:18) ωt
2
(cid:19)(cid:21)
(cid:18) ωt
2
(cid:126)
2
=
cos(ωt) .
(7.16)
We see that this oscillates as a function of time, with angular frequency ω, as we expect from our
classical intuition. What can we learn from this calculation? If we have a general initial state
represented by nˆ on the Bloch sphere, then nˆ... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/42767c0b651a33268189c7ef1f87630e_MIT8_321F17_lec7.pdf |
ˆ .
(7.22)
This is the same equation as for classical spins, but the interpretation is entirely different because
S is now an operator. If we take the expectation value of each side of this expression, we will get
the same answers that we found in the Schr¨odinger picture.
7.2 Particle in a Potential
We have said, on ge... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/42767c0b651a33268189c7ef1f87630e_MIT8_321F17_lec7.pdf |
, in practice this can often be very difficult.
In the Heisenberg picture, we simply care about the operators,
(cid:126)
(cid:126)
xH(t) = eiHt/ xe−iHt/ ,
pH(t) = eiHt/(cid:126)pe−iHt/(cid:126) .
(7.27)
Using the Heisenberg picture equation of motion, we see that the operator xH evolves according to
dxH
dt
=
1
i(cid:126)... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/42767c0b651a33268189c7ef1f87630e_MIT8_321F17_lec7.pdf |
we take the expectation values of Eqs. (7.29) and (7.30), we find
d(cid:104)x(cid:105)
dt
d(cid:104)p(cid:105)
dt
=
= −
(cid:104)p(cid:105)
,
m
(cid:28) dV
dx
(cid:29)
.
(7.31)
This result is called Ehrenfest’s theorem.
It is important, when using the second equation, to
remember to differentiate the potential first befor... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/42767c0b651a33268189c7ef1f87630e_MIT8_321F17_lec7.pdf |
x(0) +
p(0)t
m
+
q
2m
Et2 .
7.2.2 Example: Simple Harmonic Oscillator
Recall the simple harmonic oscillator Hamiltonian,
H =
p2
2m
+
1
2
mω2x2 .
(7.37)
(7.38)
We are likely all familiar with the approach in the Schr¨odinger picture. In the Heisenberg picture,
we have
dx
dt
=
p
m
,
dp
dt
=
−mω2x .
Solving these equation... | https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/42767c0b651a33268189c7ef1f87630e_MIT8_321F17_lec7.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
18.969 Topics in Geometry: Mirror Symmetry
Spring 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
MIRROR SYMMETRY: LECTURE 10
DENIS AUROUX
1. The Quintic (contd.)
Recall that we had a quintic mirror family Xˇ
ψ with L... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/42bc2e161167860681a541acc4cfc821_MIT18_969s09_lec10.pdf |
equation in the form
(4)
3
d4 �
ˇ
[Ω] +
dz4
k=0
ck(z)
[ ˇ
Ω] = 0
dk
dzk
1
2
DENIS AUROUX
�3
ckWk = 0. By Griffiths transversality ( dk Ω has no (0, 3)
ˇ
dzk
k=0
Then W4 +
component unless k ≥ 3), W0 = W1 = W2 = 0. Moreover,
�
d2Ωˇ
Xˇ dz2
�
�
dΩˇ
Xˇ dz
d2Ωˇ
dz2
d3Ωˇ
dz3
d2
dz2
W2 =
0 =
(5)
∧... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/42bc2e161167860681a541acc4cfc821_MIT18_969s09_lec10.pdf |
c
1
q −
2
c
1950750
c
2
1
q 2 −
2
2
c
10277490000
c
6
1
q 3 + · · ·
3
2
MIRROR SYMMETRY: LECTURE 10
3
Now we can describe the mirror symmetry: there exists a basis of H 2(X, Z) ∼= Z
(where X is the original quintic) given by the Poincar´e dual {e} of a hyperplane
s.t., writing [B + iω] = te, q = exp(2πit... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/42bc2e161167860681a541acc4cfc821_MIT18_969s09_lec10.pdf |
2 + 27 n3 +
8
d>0
= 5 + n1q + 8 n2 +
�
�
n1 q 3 + 64 n4 +
27
n2 +
8
�
n1 q 4 +
64
· · ·
Matching these gives
c1 = −5.
•
• n1 =
575 5
c2
· = c2
2875
: classical algebraic geometry tells us that 2875 is the
number of lines on a quintic, c2 = 1.
• n2 = 609250 (had been calculated by Sheldon Katz, 198... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/42bc2e161167860681a541acc4cfc821_MIT18_969s09_lec10.pdf |
on the symplectic side. For this, we will construct the Fukaya (A∞)-category,
4
DENIS AUROUX
which is roughly the category whose objects are Lagrangian submanifolds, whose
morphisms are intersections, and whose algebraic structures (differential, prod
uct, etc.) are governed by J-holomorphic disks. On the complex ... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/42bc2e161167860681a541acc4cfc821_MIT18_969s09_lec10.pdf |
�(CF, δ) is invariant under Hamiltonian isotopies. The motivation for this was
to understand Arnold’s conjecture on Lagrangian intersections. From that point
of view, HF is an obstruction to displacement of Lagrangians: in general, if we
have a topological isotopy between two Lagrangian submanfiolds, a pair of inter... | https://ocw.mit.edu/courses/18-969-topics-in-geometry-mirror-symmetry-spring-2009/42bc2e161167860681a541acc4cfc821_MIT18_969s09_lec10.pdf |
Lecture 7
Burke’s Theorem and Networks of Queues
Eytan Modiano
Massachusetts Institute of Technology
Eytan Modiano
Slide 1
�Burke’s Theorem
• An interesting property of an M/M/1 queue, which greatly
simplifies combining these queues into a network, is the
surprising fact that the output of an M/M/1 queue with arriva... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/42cb7759de031ae20aae04632773fef4_Lecture7.pdf |
(packets in system) left by a
(forward) departure is independent of the past departures
–
In backward process the state is independent of future arrivals
Eytan Modiano
Slide 4
NETWORKS OF QUEUES
Exponential
Exponential
Poisson
Poisson
M/M/1
λ
λ
M/M/1
?
Poisson
λ
• The output process from an M/M/1 queue... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/42cb7759de031ae20aae04632773fef4_Lecture7.pdf |
er packets catch up with the long packets
• Similar to phenomenon that we experience on the roads
– Slow car is followed by many faster cars because they
catch up with it
Eytan Modiano
Slide 8
Jackson Networks
• Independent external Poisson arrivals
• Independent Exponential service times
– Same job has indepen... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/42cb7759de031ae20aae04632773fef4_Lecture7.pdf |
time)
Independent M/M/1 queues
–
– Surprising result given that arrivals to each queue are neither
Poisson nor independent
– Similar to Kleinrock’s independence approximation
– Reversibility
Exogenous outputs are independent and Poisson
The state of the entire system is independent of past exogenous departures
... | https://ocw.mit.edu/courses/6-263j-data-communication-networks-fall-2002/42cb7759de031ae20aae04632773fef4_Lecture7.pdf |
Tutorial #2
Verilog
Simulation
Toolflow
Tutorial Notes Courtesy of Christopher Batten
% v c s m i p s . v
s i m v
/
% .
- R P P &
% v c s
6.884 – Spring 2005
02/16/05
T02 – Verilog 1
Figure by MIT OCW.
A Hodgepodge Of Information
− CVS source management system
− Browsing a CVS repository with viewcvs
− Makef... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
Checkout a working copy
Update working dir vs. repos
Add new files/dirs to repos
See how working copy differs
Set the $CVSEDITOR environment variable to change
which editor is used when writing log messages
6.884 – Spring 2005
02/16/05
T02 – Verilog 5
6.884 CVS Repository
Use the following commands to checkou... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
Spring 2005
02/16/05
T02 – Verilog 9
CVS – Multiple Users
F.2
commit
Conflict!
User
A
User
B
F.3
F.2
6.884 – Spring 2005
02/16/05
T02 – Verilog 10
CVS – Multiple Users
F.2
update
Merges F.2 changes
and denotes conflicts
With <<<< >>>>
User
A
User
B
F.2/3
F.2
6.884 – Spring 2005
02/16/05
T02... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
-test.dat
% cvs checkout examples
% cd examples/gcd
% mkdir build-gcd-rtl
% cd build-gcd-rtl
% ../configure.pl ../config/gcd_rtl.mk
% make simv
% ./simv
% vcs –RPP &
6.884 – Spring 2005
02/16/05
T02 – Verilog 15
Modifying the config/*.mk Files
Analagous to mkasic.pl .cfg files except these use
standard make cons... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
/cbatten/mips2stage
% cd 2005-spring/cbatten/mips2stage
% mkdir build
% cd build
% ../configure.pl ../config/mips2stage.mk
% make simv
% make run-tests
% <run simv with some other tests>
6.884 – Spring 2005
02/16/05
T02 – Verilog 18
Writing SMIPS Assembly
Our assembler takes as input SMIPS assembly code
with... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
, r21
loop: beq zero, zero, loop
.set at
TEST_CODEEND
Assembler directive which
tells the assembler not to
use r1 (at = assembler
temporary)
6.884 – Spring 2005
02/16/05
T02 – Verilog 22
Writing SMIPS Assembly (reorder)
By default the branch delay slot is not visible! The
assembler handles filling the bran... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
0: 8f5a1534 lw $k0,5428($k0)
1014: 409a6000 mtc0 $k0,$12
1018: 3c1a0000 lui $k0,0x0
101c: 275a1400 addiu $k0,$k0,5120
1020: 03400008 jr $k0
1024: 42000010 rfe
00001100 <__testexcep>:
1100: 401a6800 mfc0 $k0,$13
1104: 00000000 nop
<snip>
00001400 <__testcode>:
1400: 24010001 li $at,1
1404: 4081a800 mtc0 $at,$21
00... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
05800 mtc0 $zero,$11
100c: 3c1a0000 lui $k0,0x0
1010: 8f5a1534 lw $k0,5428($k0)
1014: 409a6000 mtc0 $k0,$12
1018: 3c1a0000 lui $k0,0x0
101c: 275a1400 addiu $k0,$k0,5120
1020: 03400008 jr $k0
1024: 42000010 rfe
00001100 <__testexcep>:
1100: 401a6800 mfc0 $k0,$13
1104: 00000000 nop
<snip>
00001400 <__testcode>:
140... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
] [ireg=%x] [rd1=%x] [rd2=%x] [wd=%x] tohost=%d",
cycle, mips.fetch_unit.pc, mips.exec_unit.ir,
mips.exec_unit.rd1, mips.exec_unit.rd2, mips.exec_unit.wd, tohost);
cycle = cycle + 1;
end
CYC: 0 [pc=00001000] [ireg=xxxxxxxx] [rd1=xxxxxxxx] [rd2=xxxxxxxx] [wd=00001004] tohost= 0
CYC: 1 [pc=00001004] [ireg=08000500] [r... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
waveforms
#include <smipstest.h>
TEST_SMIPS
TEST_CODEBEGIN
.set noat
addiu r1, zero, 1
mtc0 r1, r21
loop: beq zero, zero, loop
.set at
TEST_CODEEND
CYC: 0 [pc=00001000] [ireg=xxxxxxxx] [rd1=xxxxxxxx] [rd2=xxxxxxxx] [wd=00001004] tohost= 0
CYC: 1 [pc=00001004] [ireg=08000500] [rd1=00000000] [rd2=00000000] [wd=00001... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
toolchain working!)
– You must verify that the checkoff procedure works
– Your verilog will be checked out automatically Friday at 1pm
Lab Assignment 2
– Synthesize your two stage mips processor
– Assigned on Friday and due the following Friday, Feb 25
– I will work on a synthesis tutorial over the weekend and ema... | https://ocw.mit.edu/courses/6-884-complex-digital-systems-spring-2005/42da3ce96010a59f449da886e13685c0_t02_verilog_tut.pdf |
2.997 DecisionMaking in LargeScale Systems
MIT, Spring 2004
February 17
Handout #6
Lecture Note 4
1
Averagecost Problems
In the average cost problems, we aim at finding a policy u which minimizes
Ju(x) = lim sup
1
T →∞ T
E
�
T −1
�
�
�
�
x0 = 0 .
gu(xt) �
t=0
(1)
Since the state space is finite, it c... | https://ocw.mit.edu/courses/2-997-decision-making-in-large-scale-systems-spring-2004/42e1a355d7a74c79abb0bf9c959068e5_lec_4_v1.pdf |
, x∗, . . . . . . ,
� �� � � �� � � �� �
λ1
u
h(x)
λ2
u
Let Ti(x), i = 1, 2, . . . be the stages corresponding to the ith visit to state x∗, starting at state x. Let
⎡
u(x) = E ⎣
λi
t=Ti (x)
�Ti+1 (x)−1 gu(xt)
Ti+1(x) − Ti(x)
⎤
⎦
Intuitively, we must have λi
u(x), since we have the same
transition probab... | https://ocw.mit.edu/courses/2-997-decision-making-in-large-scale-systems-spring-2004/42e1a355d7a74c79abb0bf9c959068e5_lec_4_v1.pdf |
a version of Bellman’s equation for computing λ∗ and h∗. Approximating
J ∗(x, T ) as in (3, we have
J ∗(x, T + 1) = min ga(x) +
Pa(x, y)J ∗(y, T )
�
a
λ∗(T + 1) + h∗(x) + o(T ) = min ga(x) +
Pa(x, y) [λ∗T + h∗(y) + o(T )]
�
�
�
y
�
y
�
a
�
λ∗ + h∗(x) = mina ga(x) +
�
�
y Pa(x, y)h∗(y)
(4)
Therefor... | https://ocw.mit.edu/courses/2-997-decision-making-in-large-scale-systems-spring-2004/42e1a355d7a74c79abb0bf9c959068e5_lec_4_v1.pdf |
(x) = λ∗, ∀x,
J ∗
u∗ (x) ≤ Ju(x), ∀u.
Proof: Let u = (u1, u2, . . . ). Let N be arbitrary. Then
TuN −1 h∗ ≥ T h∗ = λ∗e + h∗
TuN −2 (h∗ + λ∗e)
TuN −2 TuN −1 h∗
≥
= TuN −2 h∗ + λ∗e
≥ T h∗ + λ∗e
= h∗ + 2λ∗e
2
Then
Thus,we have
T1T2 · · ·
TN −1h∗ ≥ N λ∗e + h∗
�
E
N −1
�
�
gu(xt) + h∗(xN ) ≥ (N − 1)λ∗e + h... | https://ocw.mit.edu/courses/2-997-decision-making-in-large-scale-systems-spring-2004/42e1a355d7a74c79abb0bf9c959068e5_lec_4_v1.pdf |
the
average cost Ju∗ (x) is the same for all initial states. It is easy to come up with examples where this is not
the case. For instance, consider the case when the transition probability is an identity matrix, i.e., the state
visits itself every time, and each state incurs different transition costs g(·). Then the ... | https://ocw.mit.edu/courses/2-997-decision-making-in-large-scale-systems-spring-2004/42e1a355d7a74c79abb0bf9c959068e5_lec_4_v1.pdf |
5
Properties of Linear,
Time-Invariant
Systems
In this lecture we continue the discussion of convolution and in particular ex-
plore some of its algebraic properties and their implications in terms of linear,
time-invariant (LTI) systems. The three basic properties of convolution as an
algebraic operation are that it i... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/431b597316940ea786c72a16b8cd6371_MITRES_6_007S11_lec05.pdf |
a parallel combination can be collapsed into a
single system whose impulse response is the sum of the two individual ones.
In looking at and understanding the algebraic properties of convolution, it is
worthwhile to recognize that convolution as an algebraic property relates to
addition in exactly the same way that mul... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/431b597316940ea786c72a16b8cd6371_MITRES_6_007S11_lec05.pdf |
is zero also.
In this lecture we illustrate the properties discussed above with some
systems. The problems associated with this lecture provide the opportunity to
explore these properties further.
In Lecture 3 in discussing the continuous-time impulse function, we indi-
cated some inherent difficulty with defining the ... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/431b597316940ea786c72a16b8cd6371_MITRES_6_007S11_lec05.pdf |
(t)
* h(t)
-00
Commutative:
x[n] * h [n] = h[n] *x[n]
x(t) * h(t)
= h(t) *x(t)
TRANSPARENCY
5.2
The commutativity
property of
convolution.
u[n] -> au[n]
=anu[n] * u[n]
u(t) * eatu(t)
= e-atu(t) *u(t)
Signals and Systems
5-4
TRANSPARENCY
5.3
Three algebraic
properties of
convolution.
Commutative:
x * h
h * x
Associat... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/431b597316940ea786c72a16b8cd6371_MITRES_6_007S11_lec05.pdf |
-~Or
U at
-46~
MARKERBOARD
5.2
-ov- LII
~ievs
COIIASOL I +
tV(*)O
t<0o
L\T ea'
'-t4ety
,
vx
,c \
0
0
-T
0
0
Q74' 0
+9A 6
S0
Pcc
do.tov-:
~
-00
,j\n
,kT-Y3
00o
nIIt\
YJi~V t 0 ::::: cz
>~ v~oI
Ce
~z ~-~~*'
= JL"N-13 + Nr-v'J
I
Properties of Linear, Time-Invariant Systems
XCO3
a(cL=8t
At+2*Sc*' u2 t
tt
SCn3*8t... | https://ocw.mit.edu/courses/res-6-007-signals-and-systems-spring-2011/431b597316940ea786c72a16b8cd6371_MITRES_6_007S11_lec05.pdf |
6.581J / 20.482J
Foundations of Algorithms and Computational Techniques in Systems Biology
Professor Bruce Tidor
Professor Jacob K. White
6.581 / 20.482
6.581J / 20.482J
Foundations of Algorithms and Computational Techniques in Systems Biology
Professor Bruce Tidor
Professor Jacob K. White | https://ocw.mit.edu/courses/20-482j-foundations-of-algorithms-and-computational-techniques-in-systems-biology-spring-2006/43228505195bbb23403eddae2ba590ab_l07.pdf |
MATH 18.152 COURSE NOTES - CLASS MEETING # 6
18.152 Introduction to PDEs, Fall 2011
Professor: Jared Speck
Class Meeting # 6: Laplace’s and Poisson’s Equations
We will now study the Laplace and Poisson equations on a domain (i.e. open connected subset)
Ω ⊂ Rn. Recall that
(0.0.1)
The Laplace equation is
n
def∆ = ∑
1
i=... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
x, y, z , J3 t, x, y, z
(
)
(
))
)
)
)) is
is the electric field
is the magnetic induction
the current density
Maxwell’s equations are
(1.1.1)
(1.1.2)
∂tE − ∇ ×
∇ ×
+
tB
∂
B
−
= J,
=
,
E 0
∇ ⋅ E = ρ,
=
∇ ⋅
B 0.
−
Recall that ∇× is the curl operator,
=
Let’s look for steady-state solutions with ∂tE ∂tB 0. Then equation (... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
complex-valued function
analysis. Let z
= +
x iy (where x, y R) b
∈
(where u, v R
a complex number,
). We recall that f is said
e
∈
exists. If the limit exists, we denote it by f
A fundamen
x0
tal result of complex analysis is the following: f is differentiable at z0
if and only if the real and imaginary parts of f veri... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
Ω. We consider the PDE
(2.0.6)
∆u(x
) = ( )
f x ,
∈
x Ω,
supplemented by some boundary conditions. The following boundary conditions are known to lead
to well-posed problems:
MATH 18.152 COURSE NOTES - CLASS MEETING # 6
(1) Dirichlet data: specify a function g x defined on ∂Ω such that u ∂Ω x
(2) Neumann data: specify ... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
plemented by suitable boundary conditions. As in the case of the heat equation, we are able to
provide a simple proof based on the energy method.
Theorem 3.1. Let Ω ⊂ Rn be a smooth, bounded domain. Then under Dirichlet, Robin, or mixed
boundary conditions, there is at most one solution of regularity u C 2 Ω
C 1 Ω to t... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
10)
which implies that
(3.0.11)
and we can argue as
b
∫ w∇ ˆN w dσ = −α ∫ w2 dσ
≤ 0,
∂Ω
∂
Ω
efore conclude that w 0 in Ω.
= 0,
∫ ∣∇w∣2
Ω
≡
4
MATH 18.152 COURSE NOTES - CLASS MEETING # 6
∣
∇
ˆN w ∂Ω
Now in the Neumann case, we have that
ything
w is constant in Ω. But now we can’t say an
constant in Ω.
is that u v
= +
=... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
treat the case of general x by reducing it to
on R2. For a ball of radius r, we have that
the origin. We will work with polar coordinates (r, θ)
r dθ. Note also that along ∂Br 0 , we have that
the
≤
∂ru
corresponding
(
ˆN u, where N σ is the unit normal to ∂Br 0 . For any 0
)
(
<
r R
e define
, w
measure
ˆ
∇ ⋅
=
u N
( )... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
2π
∫
B1(0)
∆
u(y
) d2y.
But ∆u 0 since u is harmonic, so we have shown that
=
(4.0.17)
′
g
(r) =
0,
and we have shown (4.0.12b) for x 0.
=
T
o prove (4.0.12a), we use polar coordinate integration and (4.0.12b) (in the case x 0) to obtain
=
MATH 18.152 COURSE NOTES - CLASS MEETING # 6
5
(4.0.18)
u 0 R2
( )
We have now ... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
by harmonic functions. The property, known
as the strong maximum principle, says that most harmonic functions achieve their maximums and
minimums only on the interior of Ω. The only exceptions are the constant functions.
Theorem 5.1 (Strong Maximum Principle). Let Ω Rn be a domain, and assume that u C Ω
( )
verifies the... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
∣
( )∣ ( ) + ∫
z
z
u
B Br z
( )
/
2
u(y) d y} ≥
1
∣ ( )∣
B p
{∣Br
(z)∣u(z) + m(∣B(p)∣ − ∣Br(z)∣)} .
Rearranging inequality (5.0.22), we conclude that
(5.0.23)
u
(z) ≤ m.
Com
bining (5.0.21) and (5.0.23), we conclude that
(5.0.24)
u(x) = m
6
MATH 18.152 COURSE NOTES - CLASS MEETING # 6
holds for all points x ∈ B(p). Th... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
∣
uf x
( ) −
u
( )∣ ≤
g x
( )
∣
max f y
y ∂Ω
∈
−
)∣
(
g y .
Proof. We first prove the Comparison
we see that w solves
Principle.
Let
w
=
u
f
−
ug
. Then by subtracting the PDEs,
(5.0.26)
{
∆w
(
u x
=
) =
0,
f (x) − g
∈
x
Ω,
∈
0, x ∂Ω,
( )
x
≥
Since w is harmonic, since f x
not constant and that for every x
( ) − ( )
g x... | https://ocw.mit.edu/courses/18-152-introduction-to-partial-differential-equations-fall-2011/4336b9d0607ac33f430908fdf4cb78ba_MIT18_152F11_lec_06.pdf |
Chapter 2
Fine Structure
B. Zwiebach
c
(cid:13)
2.1 Review of hydrogen atom
The hydrogen atom Hamiltonian is by now familiar to you. You have found the bound state
spectrum in more than one way and learned about the large degeneracy that exists for all
states except the ground state. We will call the hydrogen atom Hami... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
0
=
me4
~2 =
mα2~2c2
~2
= α2 mc2 .
(2.1.4)
(2.1.5)
This states that the energy scale of hydrogen bound states is a factor of α2 smaller than
the rest energy of the electron, that is, about 19000 times smaller. We can thus rewrite the
possible energies as:
The typical momentum in the hydrogen atom is
En =
1
2 α2 mc2 1
n... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
� + 1) = n2
Xℓ=0
The states of hydrogen are shown in this energy diagram, which is not drawn to scale,
2.1. REVIEW OF HYDROGEN ATOM
27
...
n = 4
n = 3
n = 2
n = 1
S
ℓ = 0
...
P
ℓ = 1
...
D
ℓ = 2
...
F
ℓ = 3
...
N = 3 N = 2 N = 1 N = 0
N = 2 N = 1 N = 0
N = 1 N = 0
N = 0
The table features the commonly used notation wh... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
ℓ = 0 and m = 0. The normalized
wavefunction is
na0 Yℓ,m(θ, φ) ,
(2.1.11)
e−
−
(cid:19)
·
r
(2.1.12)
Comments:
ψ1,0,0(r) =
p
r
a0 .
e−
1
πa3
0
1. There are n2 degenerate states at any energy level with principal quantum number
n. This degeneracy explained by the existence of a conserved quantum Runge-Lenz
vector. For a... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
leading
to the Zeeman effect. These external fields are represented by extra terms in the
hydrogen atom Hamiltonian.
Let us now discuss two different choices of basis states for the hydrogen atom, both of
which include the electron spin properly.
Recall that, in general, for a multiplet of angular momentum j, we have stat... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
we are tensoring a a full ℓ multiplet to an s multiplet (here, of course,
When we form ℓ
s are eigenstates of ˆL2 and eigenstates of ˆS2, so ℓ and s are good
s = 1/2). All states in ℓ
(constant) quantum numbers for all j multiplets that arise in the tensor product. Each j
multiplet has states with quantum numbers (j, m... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
)
j=ℓ+
1
2 ⊕
L(ℓ)
j=ℓ
1
2
−
(2.1.18)
or more explicitly,
1
0
⊗
⊗
P 3
S 1
2
1
2 →
1
2 →
1
2 →
1
2 →
Thus, by the time we combine with electron spin, each ℓ = 0 state gives one j = 1
2 multiplet,
2 and j = 3
2 multiplets, each ℓ = 2 state gives j = 5
each ℓ = 1 state gives j = 3
2
multiplets, and so on. For hydrogen, the... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
. Since the electron is moving relative to the frame where we have
a static electric field, the electron also sees a magnetic field B. The spin-orbit coupling is
B of that magnetic field to the magnetic dipole moment µ of the electron.
the coupling
We have discussed before, in the context of the Stern-Gerlach experiment, ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
field is therefore represented by a Hamiltonian HB given by
HB =
µ
−
·
B =
e~
2mec
B .
σ
·
(2.2.5)
Our goal now is to show that this coupling, and its associated prediction of g = 2, arises
naturally from the non-relativistic Pauli equation for an electron.
Consider first the time-independent Schr¨odinger equation for a ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
the particle is charged and we couple it to external electromagnetic fields. The quantum
mechanical rule is that this inclusion can be taken care with the replacement
ˆp
ˆπ
ˆp
−
≡
→
q
c
A .
(2.2.12)
Here q is the charge of the particle and A is the external vector potential, a function
of position that becomes an operat... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
be thought of as derivatives acting on the spatially de-
pendent components of A. Moreover, the Ai’s being only functions of position, commute
among themselves and we have
[ πi , πj ] =
~
−
i
q
c
(∂iAj −
∂jAi) =
i~q
c
(∂iAj −
∂jAi) .
Therefore, back in (2.2.16)
(π
×
π)k = 1
2 ǫijk
i~q
c
(∂iAj −
∂jAi) =
i~q
c
ǫijk∂iAj =... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
-
magnetic fields, it is not a relativistic equation. As discovered by Dirac, to include relativity
one has to work with matrices and the Pauli spinor must be upgraded to a four-component
spinor. The analysis begins with the familiar relation between relativistic energies and
momenta
This suggests that a relativistic Ha... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
2)
(2.3.3)
mc, the
≪
34
CHAPTER 2. HYDROGEN ATOM FINE STRUCTURE
Expanding the right-hand side and equating coefficients one finds that the following must
hold
3 = β2 = 1 ,
αi, αj}
{
αi, β
}
{
The relations on the second and third lines imply that α’s and β’s can’t be numbers, because
they would have to be zero. It turns ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
V (r)
Ψ ,
(2.3.11)
i~ ∂Ψ
∂t
·
(cid:16)
h
e
c
(cid:17)
V (r) =
eΦ(r) =
−
e2
r
.
−
(2.3.12)
The great advantage of the Dirac equation (2.3.11) is that the corrections to the hydrogen
Hamiltonian H (0) can be derived systematically by finding the appropriate Hamiltonian H
that acts on the Pauli spinor χ. The analysis, can ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
indeed have
∼
≃
1
≃
p4
δHrel. =
−
8m3c2 ∼ −
α4mc2 .
For spin-orbit we first rewrite the term, using
1
r
dV
dr
=
1
r
d
dr
e2
r
(cid:19)
=
e2
r3 ,
−
(cid:18)
so that
For an estimate we set S
L
~2, r
δHspin-orbit ∼
·
e2
m2c2
∼
~2
a3
0
=
∼
α ~c
m2c2
~2
a3
0
δHspin-orbit =
S
L .
e2
2m2c2
1
r3
a0, and recall that a0 = ~
mc
~
... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
�p2
2m
+ V
H (0)
−
ˆp4
8m3c2
δHrel.
+
L
S
e2
·
r3
2m2c2
δHspin-orbit
+
π
2
e2~2
m2c2 δ(r)
.
δHDarwin
(2.4.21)
|
{z
}
| {z }
|
{z
}
|
{z
}
(2.3.15)
(2.3.16)
(2.3.17)
(2.3.18)
36
CHAPTER 2. HYDROGEN ATOM FINE STRUCTURE
We will study each of these terms separately and then combine our results to give the fine
structure of... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
)
n00,Darwin =
e2~2
2m2c2
1
0n3 = α4(mc2)
a3
1
2n3 .
This completes the evaluation of the Darwin correction
(2.4.23)
(2.4.24)
The Darwin term in the Hamiltonian arises from the elimination of one of the two two-
component spinors in the Dirac equation. As we will show now such a correction would arise
from a nonlocal c... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
�(u) denote the position dependent charge density, we have
V (r) =
d3u ρ(u)Φ(r + u) ,
(2.4.26)
Zelectron
where, as shown in the Figure, r + u is the position of the integration point, measured
relative to the proton at the origin. It is convenient to write the charge density in terms of
a normalized function ρ0:
e
ρ(u)... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
V (r) =
d3u ρ0(u)
(cid:16)
Z
V (r) +
∂iV
ui +
r
∂i∂jV
r
uiuj + . . .
.
(2.4.31)
e
(cid:12)
(cid:12)
(cid:12)
All r dependent functions can be taken out of the integrals. Recalling that the integral of
ρ0 over volume is one, we get
(cid:12)
(cid:12)
(cid:12)
(cid:17)
Xi
˜V (r) = V (r) +
∂iV
Xi
d3u ρ0 (u) ρi + 1
2
∂i∂jV
... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
35)
6 ∇
Z
To get an estimate, let us assume that the charge is distributed uniformly over a sphere of
radius u0. This means that ρ0(u) is a constant for u < u0
ρ0(u) =
3
4πu3
0
1,
0,
u < u0,
u > u0 .
(2.4.36)
The integral one must evaluate then gives
d3u ρ0(u)u2 =
Z
u0
4πu2du u2
4π
3 u3
0
=
u0
3
u3
0 Z
0
0
Z
u4du... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
.2. This
is clear because the perturbing operator p2p2 commutes with L2, with Lz, and with Sz.
The first operator guarantees that that the matrix for the perturbation is diagonal in ℓ, the
second guarantees that the perturbation is diagonal in mℓ, and the third guarantees, rather
trivially, that the perturbation is diag... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
ℓm +
n + V 2)
(cid:12)
(cid:12)
(cid:12)
inℓm
V 2
h
i
ψnℓm
E
.
(2.4.44)
40
CHAPTER 2. HYDROGEN ATOM FINE STRUCTURE
The problem has been reduced to the computation of the expectation value of V (r) and
V 2(r) in the ψnℓm state. The expectation value of V (r) is obtained from the virial theorem
that states that
n . For ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
extra equality to emphasize that the matrix elements depend only on n
and ℓ. We have already seen that in the full degenerate subspace with principal quantum
number n the matrix for δHrel is diagonal in the uncoupled basis. But now we see that in
each degenerate subspace of fixed n and ℓ, δHrel is in fact a multiple of ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
nℓmi
ℓmi
δHrel|
s|
nℓmk
ℓ mk
s i
2
ci|
|
nℓmi
h
ℓmi
δHrel|
s|
nℓmi
ℓmi
si
(2.4.50)
2f (n, ℓ) = f (n, ℓ) .
ci|
|
2.4.3 Spin orbit coupling
The spin-orbit contribution to the Hamiltonian is
δHspin-orbit =
e2
2m2c2
1
r3
L .
S
·
(2.4.51)
Note that δHspin-orbit commutes with L2 because L2 commutes with any ˆLi and any ˆSi.
... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
)
1
r3
nℓmℓ
(cid:12)
(cid:12)
(cid:12)
We need the expectation value of 1/r3 in these states. It is known that
nℓmℓ
.
(2.4.53)
1
ℓ + 1
2
n3a3
0ℓ
(ℓ + 1)
D
(cid:12)
(cid:12)
(cid:12)
Because of the mℓ independence of this expectation value (and its obvious ms independence)
1
the operator 1/r3 is a multiple of the identi... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
3)
ℓ
= 0 .
(2.4.56)
Since L vanishes identically acting on any ℓ = 0 state, it is physically reasonable, as we will
do, to assume that the spin-orbit correction vanishes for ℓ = 0 states. On the other hand
0, while somewhat ambiguous, is nonzero. We set
the limit of the above formula as ℓ
1
j = ℓ + 1
2 (the other possi... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
the coupled basis. The result,
therefore will give the shifts of the coupled states. Collecting our results (2.4.46) and (2.4.56)
we have
nℓjmj
D
δHrel + δHspin-orbit
(cid:12)
(cid:12)
(cid:12)
(E(0)
n )2
2mc2
=
−
3
(
nℓjmj
(cid:12)
4n
(cid:12)
(cid:12)
ℓ + 1
2
E
+
2n
j(j + 1)
−
ℓ + 1
2
ℓ
(cid:2)
ℓ(ℓ + 1)
(ℓ + 1)
3
4
−... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
1
−
1
f (j, ℓ)
≡
j(j + 1)
ℓ
−
ℓ + 1
2
3ℓ(ℓ + 1)
(ℓ + 1)
3
4
.
−
(2.4.60)
The evaluation of this expression in both cases gives the same result:
(cid:0)
(cid:1)
=
=
ℓ=j
1
2
−
ℓ=j+ 1
2
f (j, ℓ)
(cid:12)
(cid:12)
(cid:12)
f (j, ℓ)
(cid:12)
(cid:12)
(cid:12)
1
−
j
3(j
1
2
2 )(j + 1
2 )
j + 1
2
j(j + 1)
−
j
−
(cid:0)
(cid:1... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
2 −
h
3
=
i
α4(mc2)
−
1
2n4
h
n
j + 1
2 −
3
4
.
i
More briefly we can write
E(1)
nℓj,mj;fine =
α4mc2
−
Sn,j , with Sn,j ≡
·
1
2n4
h
n
j + 1
2 −
3
4
.
i
Let us consider a few remarks:
(2.4.62)
(2.4.63)
1. The dependence on j and absence of dependence on ℓ in the energy shifts could be
anticipated from the Dirac equation. ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
4 ≥
1
4 .
(2.4.64)
5. For a given fixed n, states with lower values of j get pushed further down. As n
increases splittings fall off like n−
3.
A table of values of Sn,j is given here below
1
n j Sn,j
1
2
1
2
1
8
5
128
2
3
3
2
1
2
3
2
5
2
1
128
1
72
1
216
1
648
The energy diagram for states up to n = 3 is given here (not... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
eter Zeeman (1865-1943)
discovered that atomic spectral lines are split in the presence of an external magnetic field.
For this work Zeeman was awarded the Nobel Prize in 1902. The proper understanding of
this phenomenon had to wait for Quantum Mechanics.
The splitting of atomic energy levels by a constant, uniform, ext... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
Bint associated with spin-
orbit coupling. This is the magnetic field seen by the electron as it goes around the proton.
We have therefore two extreme possibilities concerning the external magnetic field B of the
Zeeman effect:
(1) Weak-field Zeeman effect: B
Bint. In this case the Zeeman effect is small compared
with fine st... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
Fine Structure diagram. Degeneracies in this spectrum occur for different
values of ℓ and different values of mj.
To figure out the effect of the Zeeman interaction on this spectrum we consider the
matrix elements:
(2.5.7)
nℓ′jm′ji
Lz + 2Sz we see that δHZeeman commutes with L2 and with ˆJz. The
Since δHZeeman ∼
matrix ele... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
ˆJ if the following commutator holds for all values of
i, j = 1, 2, 3:
ˆJi , ˆVj
= i~ ǫijk ˆVk .
(2.5.10)
It follows from the familiar ˆJ commutators that ˆJ is a vector operator under ˆJ. Additionally,
if ˆV is a vector operator it has a standard commutation relation with ˆJ2 that can be quickly
confirmed:
(cid:3)
(cid... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
�J2 gives the same
eigenvalue when acting on the bra and on the ket. Therefore the matrix elements of the
right-hand side gives
k′; jm′j|
h
which implies that
( ˆV
ˆJ) ˆJ
·
k; jmji
|
= ~2 j(j + 1)
k′; jm′j|
h
ˆV
k; jmji
|
,
(2.5.14)
k′; jm′j|
h
ˆV
k; jmji
|
k′; jm′j|
= h
ˆJ) ˆJ
( ˆV
k; jmji
|
·
~2 j(j + 1)
.
(2.5.15)
T... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
calculated but it will introduce no mj dependence. In fact ˆS
ˆJ is a
scalar operator (it commutes with all ˆJi) and therefore it is diagonal in mj. But even more
is true; the expectation value of a scalar operator is in fact independent of mj! We will not
show this here, but will just confirm it by direct computation. ... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
corrections to the fine structure energy levels. Since all degeneracies
within j multiplets are broken and j multiplets with different ℓ split differently due to the
ℓ dependence of gJ (ℓ), the weak-field Zeeman effect removes all degeneracies!
2mc ≃
5.79
10−
×
Strong-field Zeeman effect. We mentioned earlier that when the Ze... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
eneracies among ℓ
2 states and degeneracies among such
multiplets with ℓ and ℓ′ different. This is illustrated in Figure 2.2.
⊗
Figure 2.2: Illustrating the degeneracies remaining for ℓ = 0 and ℓ = 1 after the inclusion of Zeeman
term in the Hamiltonian. Accounting for the spin of the electron there two degenerate state... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
the original H (0) degenerate spaces using the
uncoupled basis. But happily, it turns out that spin-orbit is diagonal in the more limited
degenerate subspaces obtained after the Zeeman effect is included. All these matters are
explored in the homework.
(2.5.29)
MIT OpenCourseWare
https://ocw.mit.edu
8.06 Quantum Phys... | https://ocw.mit.edu/courses/8-06-quantum-physics-iii-spring-2018/439fc15f7ed10dd76c69dc3f7fea600e_MIT8_06S18ch2.pdf |
8.04: Quantum Mechanics
Massachusetts Institute of Technology
Professor Allan Adams
2013 February 5
Lecture 1
Introduction to Superposition
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.
I want to start by describing to you, the readers, ... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
= B0ez splits electrons by color, such that
electrons of one color are sent in one direction toward a screen, and electrons of
the other color are sent in another direction toward the same large screen. The
electrons of each color always land in the same positions, and there are always
only 2 such landing locations... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
property is measured and all the electrons with a single
value of that property have the other property measured, the value of the other property is
found to be probabilistically evenly split. For example, if electrons that are known to all be
soft have their color measured, half will be black and half will be white... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
from the first two boxes, if hardness and color were interchanged, et
cetera. Apparently, the presence of the hardness box tampers with color, because without
the hardness box, repeatability would ensure that all the originally measured white electrons
would come out white again. This is suspicious!
You may be askin... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
might now be thinking that this just applies to the hardness and color of electrons.
Actually, every object has similar properties, including me, you, and a paper copy of these
notes! These properties hold true every time they are tried with new systems, though it is
simply easier to test these with electrons.
This... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
Every electron takes the hard path, so the color box input is always
hard. Then, feeding hard electrons into the color box yields half as black and half as white.
Indeed, this is empirically correct as well!
8.04: Lecture 1
7
Figure 9: Send in hard, split by hardness, recombine, measure color
Third, let us se... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
an effect
on electrons in the hard path, given that the hard path could ultimately be many millions
of kilometers away. Hence, we expect half as many electrons to exit, and all of them should
be white. Empirically, though, while the overall output is indeed down by half, once again
half of the electrons are white an... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
This is also true of molecules, bacteria, and other macroscopic
objects, though the effects are harder to detect. Physicists call such modes as being defined
by superposition, which for now means “we have no clue what is going on”.
In the context of our previous experiments, an initially white electron inside the appa... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
is really weird, but
true.
If all of this staggers your intuition, that is because your intuition was honed by throwing
spears, putting bread in a toaster, and playing with Rubik’s cubes, all of which involve things
things so big that quantum effects are not noticeable. Hence, you can safely and consistently
ignore... | https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/43a8712da99ace660cf042c1f1371b46_MIT8_04S13_Lec01.pdf |
Extrinsic Semiconductors
• Adding ‘correct’ impurities can lead to controlled domination of one carrier type
– n-type is dominated by electrons
– p-type if dominated by holes
• Adding other impurities can degrade electrical properties
Impurities with close electronic
structure to host
isoelectronic
hydrogenic
... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/43fea384070aa4c2f247637091f10d55_lecture_8.pdf |
nearly fully
ionized at room temperature
2 for Si: ~1020cm-3
– ni
– Add 1018cm-3 donors to Si: n~Nd
– n~1018cm-3, p~102 (ni
2/Nd)
• Can change conductivity drastically
– 1 part in 107 impurity in a crystal (~1022cm-3 atom density)
– 1022*1/107=1015 dopant atoms per cm-3
– n~1015, p~1020/1015~105
(cid:0) σ/σi~(p+n)/2n... | https://ocw.mit.edu/courses/3-225-electronic-and-mechanical-properties-of-materials-fall-2007/43fea384070aa4c2f247637091f10d55_lecture_8.pdf |
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