15. Time Dependent Approximation Methods
The Goal
• All approximation methods have assumed
the Hamiltonian is constant in time
H(t)
• Now we will consider H = H(t)
• To make things simple, let the time
dependence take place only over a time T
HI
– Typically t = 0 to t = T
• Sometimes want to deal with H that always
T
varies with time
– Have it vary from t = – T/2 to t = T/2
– Take limit T  
H I  I  EI  I
• Assume we have already found eigenstates of HI and HF:
H F  F  EF  F
• Assume it starts in state |I
  0   I
• Evolve it forward to time T using Schrödinger
P  I  F    F  T 
• What is probability it ends in |F?
HF
t
2
15A. The Sudden Approximation
Short Time = No Change
• Suppose H changes quickly from HI to HF, T small
• We must solve Schrödinger:
d
i
 t   H t   t 
dt
H
HF
H(t)
I
• Since T is small, use Taylor expansion:
2


2
 T     0   T
 t 
 12 T
 t 
2
t

0
t
t
1
  I  T H  0   I  O T 2 
 I
i
P  I  F    F  T 
2
t
t 0
T

PI  F   F I
2
Comments On the Sudden Approximation
PI  F   F I
2
• In general, though F and I are both eigenstates, they are eigenstates of
different Hamiltonians
1
When is this approximation valid?
 T    I  T H  0   I 
i
• The leading term we dropped was O(TH/)
• H is of order the energies, so this is small if TE
• Can show, only differences in energies matter, so really need T E
• Probabilities are always pure numbers
– If there are any dimensionful combinations in your answer, your answer
is wrong
– Answers should have a numerical answer  1
Sample Problem
A particle is initially in the ground state of a Harmonic
oscillator potential with frequency(t), where(t)
changes from 0 to 20 very quickly (T 0 << 1).
What is the probability it remains in the ground state?
20
0
(t)
2
• The probability is: 2
PI  F   F I
T
t
P  0  0   0 0
14
 m0 x 2 
 m0 
• The initial state has wave function
 I  x  

 exp  
2 
  

• The final state has same wave function,
14
2
with 0 changed to 20:


2
m

m

x

0 
0
 F  x  

 exp  
• The overlap is
  




2
m

2
0

0 0    F*  x  I  x  dx 
exp

3
m

x
 2  dx
0





2m0

2
2 2

3m0
3

P  0  0  
2 2
3
 94.28%
15B. The Adiabatic Approximation
Instantaneous Eigenstates
•
•
•
•
Suppose H changes slowly from HI to HF, T large
At each moment, find eigenstates of Hamiltonian H  t   i  t   Ei  t   i  t 
Eigenstates are always orthonormal, therefore

t   i t   1

i
Take time derivative:
d
0
 i  t   i  t    i  t   i  t    i  t   i  t   2 Re  i  t   i  t 
dt
• Imaginary part is undetermined:
– The eigenvalue equation for |i(t) is ambiguous up to a phase
– By multiplying |i(t) by a time-dependent phase, we can make it
whatever we want
• We can choose imaginary part to vanish too, so
 i t   i t   0


State Vector in Terms of Eigenstates
  t    ci  t   i  t 
• Let’s write |(t) in terms of our eigenstates
i
d
• Substitute into
i
 t   H t   t 
Schrödinger’s Equation:
dt
i  ci  t   i  t   ci  t   i  t     ci  t  H  t   i  t    ci  t  Ei  t   i  t 
i
i
i
• Take inner product with one state n(t)|:
i
 c  t    t    t 
i
i
n
i
i cn  t   i
 ci  t   n  t   i  t     ci  t  Ei  t  
 c t   t   t 
i
n
i
i
 cn  t  En  t 
i
cn  t   i
cn  t  En  t    ci  t   n  t   i  t 
1
i
n
t   i t 
Pulling Out the Large Phase
cn  t   i
cn  t  En  t    ci  t   n  t   i  t 
1
i
• Because Hamiltonian is changing slowly, eigenstates are also changing slowly
– Expect second term to be small
• Solution to just first term is pretty easy
t
1




c
t

b
t
exp

i
E
t
dt






n
n
n
• Let us define bn by
0


• Substitute into differential equation:
bn  t  exp  i

t
1
1







E
t
dt

i
b
t
E
t
exp

i
E
t
dt








n
n
0 n
0 n



t
1
1

 i bn  t  En  t  exp i  En  t   dt    bi  t   n  t   i  t  exp  i

 i

0
1
t

bn  t    bi  t   n  t   i  t  exp i
i
1
1



E
t
dt


i
0

t
  E t  E t  dt
t
0
n
i
Integrating the Equation

bn  t    bi  t   n  t   i  t  exp i
i
• The term with i = n is absent because
• Integrate to get the total change in bn(t):

bn T   bn  0     exp i
T
i n
0
1
1
  E t  E t  dt
t
n
0
i
 n t   n t   0
  E t   E t  dt b t   t   t 
t
0
n
i
i
n
i
dt
Derivatives are small because H is slowly changing
• But integration is over long time!
• But it’s multiplied by a phase that is constantly changing
• So it should be small, just need to prove it
t
• For each value of i and n,
1
    En  t   Ei  t  dt 
0
define the phase angle  by
• Note that
 i  t  dt  d i  t  dt dt  d i   d d
• We therefore have:
 T 
bn T   bn  0     ei bi    n   d i   d d
in
0
The Approximation
bn T   bn  0    
in
 T 
0
ei bi    n   d i   d d
iei b      d   d  T 

i
n
i
0


bn T   bn  0      T 

d
i

i  n 
bi    n   d i   d  d 
ie

d
 0

• If we assume that (T) is large, then derivatives are small
• Every term inside {} should be small, so
bn T   bn  0
• Recall that c  t   b  t  exp  i 1 t E  t   dt 
n
n
0 n


• Integrate
by parts:
cn T   bn  0  exp  i

1

T
0
En  t  dt   cn  0  exp  i


cn T   cn  0 
2
2
1

T
0
En  t  dt 

Now We Tie It All Together
cn T   cn  0 
2
• Recall what cn is:
2
  t    ci  t   i  t 
i
• Initial state assumed:
• We want probability
P  I  F    F  T 
 t  0   I
2
cn  0   nI
 cF T   cF  0    FI2
2
2
P  I  F    FI
• It stays in the same state, of a different Hamiltonian
• Normally, ground state  ground state, first excited  first excited, etc.
• We explicitly assumed (T) is large

1
  E t  E t dt
t
0
n
i
• If we let E be typical energy splitting, then we are assuming
• This is exact reverse criterion of sudden approximation
1
T E
T E
1
Sample Problem
A particle is initially in the ground state of a Harmonic
oscillator potential with frequency(t), where(t)
changes from 0 to 20 very slowly (T 0 << 1). What
is the probability it remains in the ground state?
20
0
P  I  F    FI
• The probability is:
P  0  0   0,0
P  0  0  1  100%
(t)
T
t
Level Crossings
• Sometimes, as the Hamiltonian evolves, eigenstates will have energies that
approach each other
T E
• Adiabatic approximation applies only if we always have
• Normally, the difference E will not vanish, because there will be mixing that
causes the states to mix heavily as Hamiltonian changes
• If there is mixing between the states, then we follow the general rule
– Ground state  ground state, etc.
• Suppose we have a symmetry operator T that T , H  t    0
commutes with the Hamiltonian at all times
T , n   , n
• Then eigenstates will be eigenstates of both T and H:
H  , n  E ,n  , n
• As time evolves, eigenvalue under T will stay the same
• The lowest energy state with the initial eigenvalue will become the lowest
energy state with the same eigenvalue, etc. ,0  ,0 , ,1  ,1 , etc.
• You can have level crossings between
states with different values of .
Sample Problem
An electron in spin state |+ at t = –  is in a region with time-dependent magnetic field
B  t   btzˆ  Bx xˆ . What state will it be in at t = +  if: (a) eBx2 >> mb (b) Bx = 0
• The relevant Hamiltonian is:
Bx = 0
Bx  0
ge
ge
ge  bt Bx 
H
B S 
 bt z  Bx x  


4m
2m
4m  Bx bt 
• Eigenvalue graph as a function of time:
• Two spin states are mixed around t = 0 if Bx  0
– For characteristic time T ~ Bx/b
• For part (a), the low energy state will remain the low
energy states in adiabatic approximation:   
• Valid if
T E ~  e Bx m Bx b 
bm eBx2
• For part (b), the Hamiltonian always commutes with Sz
• Therefore, Sz eigenvalue will never change
  
15C. Time-Dependent Perturbation Theory
Setting Up the Problem
H t   H0  W t 
• Consider a Hamiltonian with a small time-dependent part
• Assume we know the eigenstates and eigenvalues of H0
• W(t) is assumed to be small
H 0 n  En n
  t    cn  t  n
• These states are a complete, orthonormal basis
• We can write the state vector in terms of these states
n
• Assume we start in state |I and want probability we end in state |F
  t  0   I
cn  0   nI
P  I  F   F   T 
P  I  F   cF  T 
2
2
Schrödinger’s Equation
  t    cn  t  n
H 0 n  En n
n
• We now write out Schrödinger’s equation
d
i
  t    H 0  W  t     t 
dt
d
i  cm  t  m   cm  t   H 0 m  W  t  m    cm  t   Em m  W  t  m 
m
m dt
m
• Take inner product with n|
d
i  cm  t   mn   cm  t   Em mn  n W  t  m 
m dt
m
d
i
cn  t   En cn  t    n W  t  m cm  t 
dt
m
Pulling Out the Large Phase
•
•
•
•
d
i
cn  t   En cn  t    n W  t  m cm  t 
dt
m
If W is small, we expect second term to be small
In absence of second term, solution to first is cn ~ exp(–iEnt/)
iE t
We therefore define: cn  t   bn  t  e n
Substitute in: d
bn  t  eiEnt   Enbn  t  eiEnt   n W  t  m bm  t  e iEmt
i
dt
m
d
iEnt
i e
bn  t   Enbn  t  eiEnt  Enbn  t  e iEnt   n W  t  m bm  t  e iEmt
dt
m
d
1
bn  t    i   n W  t  m bm  t  eiEnt iEmt
dt
m
Wnm  t   n W  t  m
• Define the frequency difference   En  Em
nm
and matrix element:
• We therefore have:
d
1
bn  t    i  Wnm  t  bm  t  einmt
dt
m
Integrating and Working on Answer
d
1
bn  t    i  Wnm  t  bm  t  einmt
dt
m
T
1
• Integrate to time T:
bn T   bn  0     Wnm  t  bm  t  einmt dt
i m 0
iEnt
 bn  0   nI
c
0


and
c
t

b
t
e
n 
nI
n 
n 
• Recall:
1
bn T    nI 
i
• Therefore
m
0
Wnm  t  bm  t  einmt dt
P  I  F   cF T   bF T 
• Recall:
• Define:

T
2
SFI  bF T 
• Then we have:
2
P  I  F   S FI
1
S FI  bF T    FI 
i

n
T
0
2
WFn  t  bn  t  eiFnt dt
Making a Series Expansion
T
1
S FI  bF T    FI    dtWFn  t  eiFnt bn  t 
i n 0
• Substitute this expression into itself repeatedly
1
S FI   FI 
i
1
  FI 
i

T
0

0
n
dtWFI  t  e
1
S FI   FI 
i

1
i
dtWFn  t  e
T

3

T
0
iFI t
1
i 
dtWFI  t  e
 
n

iFnt
m
T
0
2
iFI t
1

 nI  i


n

dtWFn  t  e
0
i
iFn t

inmt 



dt
W
t
e
b
t

nm  
m  

0
m

iFn t



t
0
2

n
T
0
dtWFn  t  e
dt Wnm  t   e
inmt 
• We have a perturbative expansion for the transition
dt Wnm  t   einmt   mI 
t
0
m
1
2
t
dtWFn  t  e
T
P  I  F   S FI

iFnt
t
0

t
0
dt WnI  t   einI t 
dt WmI  t   eimI t  

Comments on Perturbative Expansion
1
S FI   FI 
i

1
i 
3

T
0
dtWFI  t  e
 
n
0
m
P  I  F   S FI
T
2
iFI t

dtWFn  t  e
1
i 
iFn t

t
0
2

n
T
0
dtWFn  t  e
dt Wnm  t   e
inmt 

iFnt
t
0

t
0
dt WnI  t   einI t 
dt WmI  t   eimI t  
Wnm  t   n W  t  m
• We almost never interested in cases where F = I
– Ignore first term
2
1 T
P  I  F   2  WFI  t  eiFI t dt
0
• If we include only the
second term, then we get
• Not hard to show that SFI is a unitary matrix
– Closely related to U(T,0) being a unitary operator
nm 
En  Em
*
S
S
 FI F I   FF 
I
*
S
S
 FI FI    II 
F
Sample Problem
A hydrogen atom is in state |1,0,0 when it is subjected to a temporary electric field
given by E  E0 zˆ exp   t 2 2T 2  . What is the probability it transitions to state |n,l,m ?
2
1 T
• Electric field comes from scalar potential and/or
iFI t
P  I  F   2  WFI  t  e dt
0
E  U  A t
vector potential
• We can use:
U   E0 Z exp   t 2 2T 2 
• This adds perturbation W t  eU  E eZ exp  t 2 2T 2



0
• Matrix
2
2
W
t

eE
n
,
l
,
m
Z
1,
0,
0
exp

t
2
T




element is
FI
0
• Normally, run time integral from 0 to T, but in this case, from – to + 
• We have:
2 2
2
2 2
2
12 T 2
e
E

2
2

e E0
0
FI
P  2 n, l , m Z 1, 0, 0  e t 2T iFI t dt  2 n, l , m Z 1, 0, 0 T 2 e 2

• Non-zero only for l = 1, m = 0
2 e2 E02 2 FI2 T 2
T e
n, l , m Z 1, 0, 0
• Vanishes if T = 0 (sudden approximation) P 
2
• Vanishes if T =  (adiabatic approx.)
2
15D. Harmonic Perturbations
Harmonic Perturbations
• It is very common to have the perturbation look
W  t   W cos t 
like a sine or a cosine
W  t   12 W  eit  e it 
W  t   W sin t 
• These can be rewritten:
W  t   21i W  eit  e it 
• Define a harmonic perturbation as one of the form:
W  t   Weit  W †eit
– Assume   0
1 T
• We will use our perturbation expansion,
S FI   WFI  t  eiFI t dt
i 0
but assume F  I, and only keep first order
• We actually want to think of the perturbation as always present
– Make t run from – T/2 to +T/2; later take limit T  
• Define WFI as
WFI  F W I
WFI  t   WFI eit  WFI† eit
• Then we have
1T
1
2
iFI t
 it
† it
• Substitute in: S 
W
e

W
e
e
dt


FI
FI
FI
1T


2
i
Large Effects Near Resonance
• Do the integrals:
1
S FI 
i
2

i
 W
1T
2
 12 T
FI
e
 it
† it
FI
W e
e
iFI t
1
dt 
i
i FI  t
i FI  t
1T
2
WFI e

WFI† e



 i FI    i FI      12 T
1


sin  12 FI    T 
sin



2 FI    T 


†
 WFI
WFI







FI
FI




• In general, if W is small, expect these terms to be small
– Unless we have a small denominator!
• We will approximate two cases:
1
sin
FI    T 


2
W
2

FI
– Where FI   (energy increasing)
S FI 
i
FI  
– Where FI  – (energy decreasing)
• In each of these cases, we will only keep
2WFI† sin  12 FI    T 
S FI 
track of the large term, and ignore the other
i
FI  
Energy Increasing Case – Large T limit
2WFI sin  12 FI    T 
S FI 
i
FI  
2
2 1

sin
 2 FI    T  
4
W
2


• We want transition probability: P I  F  S
FI

 FI  2 

2
• Consider limit T  
FI   




• Look at function in {}’s:
• As T  , it looks like a delta function
FI T  400
2 1
sin  2 FI    T 
lim
 A FI   
2
T 
FI   
2 1
• To find the coefficient A,
 sin  2 FI    T 

 d   A    d
 FI 
integrate both sides over all  
2


FI   
• Let Maple do the work,
1
or use contour integration
2 T  A
2
sin 2  12 FI    T  1
2
P
I

F

2

T
W
 FI   


FI
lim


T






FI
2
2
T 
FI   
Some Reminders on Delta-Functions
P  I  F   2
T WFI  FI   
2
2

f  xi 
• General integral
f  x    g  x    


of a delta function:
xi  g 1  0  g   xi 
• This allows you to
  x  xi 
rewrite delta functions   g  x   
xi  g 1  0 g   xi 
• In particular,
  k  x  x0    k 1  x  x0 
• We therefore have P  I  F   2
• Recall that: FI 
• So
T WFI   FI   
1
2
E f  Ei
P  I  F   2
T WFI   EF  EI   
1
2
Transition Rates
P  I  F   2
T WFI   EF  EI   
1
2
• Note that probability is proportional to T
– The longer you wait, the more likely you are to have a transition
• Define the rate  as the probability per unit time
• This formula assumed EF > EI, we still need to do the other case
• Then put the formulas together:
 2

I  F   

2
1
1
WFI   EF  EI    if EF  EI
2
† 2
FI
W
  EF  EI    if EF  EI
• Note energy is not conserved
– Not surprising for a non-constant Hamiltonian
• Note the energy change is 
– This energy is being added/subtracted from a classical background source
– Later we’ll realize this is photon energy
Reverse Transitions
 2 1 WFI 2   EF  EI    if EF  EI

I  F   
1
† 2
WFI   EF  EI    if EF  EI

2
• Consider a case where EF > EI, and we consider F  I
  F  I   2
1
† 2
IF
W
  EI  E F   
WIF†  I W † F  F W I
  F  I   2
1
*
 WFI*
WFI   EF  EI   
2
• Rate for forwards and backwards transitions
F  I   I  F 
are the same!
• Implies that in a thermal background, excited/ground states will eventually
become equally populated
• We’ll correct this false impression when we quantize EM fields
What to Do With the Delta Functions
•
•
•
•
•
 2 1 WFI 2   EF  EI    if EF  EI

I  F   
1
† 2
WFI   EF  EI    if EF  EI

2
How should we handle the delta functions?
– Naively, we would always get zero or infinity
dP
The frequency might be changing with time
P   dt

dt
There might be many frequencies present
– We’ll use this in a later chapter
There may be multiple final states that need to be added up, which can be
converted to an integral
– For example, final states are free particles
– Deal with this in next section
If experiment is finite time, you can simply not take the limit T  
– It ends up not quite a delta function
Sample Problem
A harmonic oscillator with mass m and angular frequency 0 is in the ground state |0 at
t = 0, and is subjected to a perturbation W  t   AX sin t  . The angular frequency of
the driving perturbation is gradually adjusted to increase, so  = t. To first order in
perturbation theory, what state can it transition to, and what is the corresponding
probability?
  I  F   2 1 WFI   EF  EI   
• Since in the ground state,
it can only go up in energy
it
† it
W
t

We

W
e


• Need to write perturbation in the form
2
sin t  
1
2i
e
it
 e it 
W  t   A  12 iXe it  12 iXeit 
• We need the matrix element:
WFI  F W I  F 12 iAX 0  12 iA
  0  1  2
1
2m
W  12 iAX
F  a  a†  0  12 iA
A
A2
3
1
  0   
  2 0  2 0    
4m0
8m0
2
2m
 F1
Sample Problem (2)
A harmonic oscillator with mass m and angular frequency 0 is in the ground state |0 at
t = 0, and is subjected to a perturbation W  t   AX sin t  . The angular frequency of
the driving perturbation is gradually adjusted to increase, so  = t. To first order in
perturbation theory, what state can it transition to, and what is the corresponding
probability?
 A2
  0  1 
  0   
4m0
• This is the rate at any given time
• We integrate this rate to get the total probability
 A2
  0  t  dt
P     0  1 dt 

4m0
 A2
P  0  1 
4m0 
Sample Problem
A hydrogen atom in its ground state is in a region of rapidly oscillating electric field in
the z-direction, Ez = E0cos(t), where  is much larger than the binding energy of
hydrogen. What is the rate at which it will dissociate the electron?
• As usual, we have
  I  F   2
• Initial state is ground state of hydrogen:
1
WFI   EF  EI   
2
I  1, 0, 0 , I  r  
1
 a03
e  r a0
3/2
• EF will be large, so large that the final
F  k , k  r    2  eik r
electron will effectively be a free electron
• The electric potential is easily found
U   E0 z cos t 
• This lets us get the perturbation:
W  t   eU  eE0 z cos t   12 eE0 z  e it  eit 
W  12 eE0 Z
• We now need the matrix element
WkI  k W I
 12 eE0 k Z 1,0,0
Sample Problem (2)
A hydrogen atom in its ground state is in a region of rapidly oscillating electric field in
the z-direction, Ez = E0cos(t), where  is much larger than the binding energy of
hydrogen. What is the rate at which it will dissociate the electron?
WkI  eE0 k Z 1,0,0 
1
2

4
ieE0

2a03 k z
ieE0
2


2 2a03 k z


0

2
0
eE0
2  2 

a
3/2
 d re
3
3
0
 ik r
ze
 r a0


3
 ik r  r a0
d
r
e
e

2a03 k z
ieE0
4 2
d  r dr  e  ikr cos  r a0 d cos  
0
2
1
1

ik
ik
ir 2 dr ikr  r a0 ikr  r a0  ieE0  


e

e
2
2
3 k
kr
 8a0 z   ik  1 a0   ik  1 a0  
7
4 a0

i

1

ieE
k
32
a
3
0 z
0


eE
2
a

0
0
2
2
3
2 2

k z  a 2k 2  1
 8a03 k z  k 2  1 a02 
  a0 k  1
0
ieE0
Sample Problem (3)
A hydrogen atom in its ground state is in a region of rapidly oscillating electric field in
the z-direction, Ez = E0cos(t), where  is much larger than the binding energy of
hydrogen. What is the rate at which it will dissociate the electron?
• After a lot
of work:
•
•
•
•
WkI 
ieE0 k z 32a07
  a k  1
2
0
2
3
  100  k   2
Ek 
2
k 2 2m
1
WkI   Ek  E1   
2
The energy of the final states is
Decay rate to a particular k is:

64e 2 E02 k z2 a07  2 k 2
 k  

 E1   
6
The states |k are continuum states
2 2

  a0 k  1  2m
We can go into any |k to dissociate

2 2
2 2
1


64e 2 E02 a07 2
k
k
dk
k
3
2
   d k  k  
d  cos  d cos  

 E1   
6

0

1
2
2

 2m

0  a0 k  1
2 2 7 4
2 2
256e E0 a0 k m
256
me
E0


6
2
2 2
k
3 3 a05 k 9
3  a0 k  1
15E. EM Waves and the Dipole Approximation
The Perturbation
• Consider the effects of an external electromagnetic field on an atom
• We start with the Hamiltonian:
2
e
 1 


H  
P

e
A
R
,
t

eU
R
,
t

B
R
,
t

S



j
j 
j 
j 
j   Va  R1 , R 2 ,

m

j 1  2m
N
•
•
•
•
, RN 
The potential Va contains all the internal interactions of the atom with itself
The unperturbed Hamiltonian is just:
1 N 2
H0 
Pj  Va  R1 , R 2 , , R N 

Assume we know the eigenstates |a
2m j 1
Assume the external fields are small
– Only keep first order in A
N
e 


W
t

A
R
,
t

P

B
R
,
t

S

eU
R
,
t
     j  j  j  j 
• The perturbation will be:

j 
m

• We implicitly assumed PA = AP j 1 
– Works if A = 0
– We’ll justify this assumption shortly
The Waves
e 

W  t      A  R j , t   Pj  B  R j , t   S j   eU  R j , t 

j 1  m
• We now need to make an electromagnetic wave
ik r it
* *  ik r  it
A
r
,
t

A
ε
e

A
  0
• Multiple ways to do this, but one
0ε e
way (Coulomb gauge) is to use:
U  r, t   0
*
ε

ε
1
• The polarization vector  is a normal vector perpendicular to k:
  ck
• Note that this satisfies A = 0
ε k  0
• Moves at speed c:

ik r it
* *  ik r  it
E
r
,
t


A
r
,
t





i

A
ε
e

i

A
εe
• The electric and
0
0
t
magnetic fields are:
ik r it
*
*
 ik r  it
B
r
,
t


A
r
,
t

iA
k

ε
e

iA
k

ε
e








0
0
• So we have:
e N
ik R j it
*  ik R j it
*
*



W  t    A0e
ε

P

i
k

ε

S

A
e
ε

P

i
k

ε
 S j 




j
j
0
j


m j 1
N
• Compare with:
W  t   Weit  W †eit
e
ik R j
ε  Pj  i  k  ε   S j 
W  A0  e
• So we have:
m j 1
N


The Dipole Approximation
N
e
ik R
W  A0  e j ε  Pj  i  k  ε   S j 
m j 1
• We will need matrix elements F W I
• Atomic states have size ~ a0
• We will typically be working with waves with energy of order the binding
energy of the atom
 a0 1a0 ke e 2 a0
ke e 2
1
   137
k  R j ~ ka0
• We therefore will have
c
a0 c
c
c
• Therefore, kRj << 1
k  ε   S j
k
1
• The relative size of the second
~
~ ka0 ~   137
εP
a0
term compared to the first is:
• We make the electric dipole approximation:
N
e
W  A0  ε  Pj
m j 1
e
WE1  A0ε  P
m
Working With the Dipole Approximation
e
WE1  A0ε  P
m
  I  F   2
2
F W I
2
 FI   
• There is a better way to work out this matrix element. Consider:
1
2i
 1

i
2
2


 ij P j   P
 H0 , R    Pj  Va ,  Ri     Pj , R i  

m
2m j i
i
 2m j
 2m j i
• It follows that:
e
F WE1 I  A0ε  F P I  e mi A0ε  F  H 0 , R  I
m
m
ei
ei
 A0ε  F  H 0 R  RH 0  I  A0  EF  EI  ε  F R I  ieA0FI F ε  R I
• Our rate is therefore:
 E1  2
e A0 
2 2
2
2
FI
F ε  R I
2
 FI   
Rewrite in Terms of Intensity
• You can’t measure the vector potential A, just the fields E and B
E  i A0εeik r it  i A0*ε*e ik r it
B  iA0  k  ε  eik r it  iA0*  k  ε*  e ik r it
• Most commonly, you measure the intensity , the power per unit area
• Computed in E&M as the time-averaged magnitude of the Poynting Vector:
*
*
* *


A
A
ε

k

ε

A
A


1
0 0ε  k  ε 
 0 0

S
EB


0
0  A02ε   k  ε  e2ik r  2it  A0*2ε*   k  ε*  e2ik r  2it 


• Time average and expand out the triple cross products:

2
S 
A0  ε  ε*  k  ε*  k  ε    ε  ε*  k  ε  k  ε*  
0
2
2
• Use that  is normalized and orthogonal to k
S 
A0 k
0
2
• The intensity is therefore
2 k A0
I 
0 I
2

0
A0 
• And we therefore have
2 k
Putting it All Together
A0
2
0 I

2 k
 E1  2
e A0 
2 2
2
2
FI
• Define the dipole matrix element rFI  F R I
• Speed of light,
1
4 ke
1
c

0  2  2
from E&M:
 0 0
c
c 0
• Recall ck  
• Substitute in:
• Fine structure constant:
F ε  R I
A0
2
2
 FI   
2 ke I
 2
c k
A0
4 2 ke e2FI2
2
 E1 
I
ε

r
FI  FI   
2 2
c 
ke e 2

c
• Delta function assures that
 = FI, so this simplifies to:
 E1  4 2
1
I ε  rFI  FI   
2
2
2 ke

I
2
c
Dealing With the Dirac Delta Function
 E1  4 
2
•
•
•
•
1
rFI  F R I
I ε  rFI  FI   
2
How do we deal with the delta function?
Often, incident light has a range of frequencies: I   I   d
Logically, should be called d/d, but it isn’t
Other measures of
2
I  f   2 I   , I    
I  
intensity are related:
2 c
• Replace  by
an integral:
 E1  4 
1
• So we have:
 E1  4 
1
2
2
ε  rFI
2
 I    
I FI  ε  rFI
FI
2
   d
Averaging Over Polarization, And Direction
 E1  4 2
1
I FI  ε  rFI
2
• Formula assumes we know the polarization 
rFI
• If incident light is unpolarized, need to average
the two polarizations perpendicular to k
ε
• For example, let’s assume rFI is real

• Let  be angle between k and rFI
k
ε
• One polarization can be chosen in the plane of k and rFI
• The other is perpendicular to both k and rFI
2
2
1
• The rate, averaged over
1
 unpol  4  I FI  2 ε  rFI  ε   rFI
polarizations will be:
• If direction of light (or orientation
2
2
1
 unpol  2  I FI  rFI sin 2 
of atom) is random, we can also
average over angles
2
2
1
4
 random  3   I FI  rFI
1
 random  4   unpol d 

2

Comments on Electric Dipole Rate
 E1  4 
2
1
I FI  ε  rFI
2
rFI  F R I
• If you know the polarization, you don’t need all components of rFI
– Just calculate the component you need: F ε  R I
Only certain states have non-zero matrix elements
• Non-zero only if lF  lI  1
• Because operator R is parity -1, initial and final states must have
opposite parity
– This is automatically assured by l = 1 for Hydrogen, but not for
other atoms
• If l  1, then electric dipole can’t cause transition
– But higher order interactions can
Sample Problem
An electron is trapped in the ground state of a cubical box of size a3, with a
corner at the origin. It is then bathed in light moving in the z-direction,
polarized in the x-direction, and having intensity function (). In the dipole
approximation, (a) which states can the electron transition into, (b) what are the
corresponding frequencies , and (c) what are the transition rates
2
2
2
2
2
• We first need the eigenstates
nmp , Enmp 
n

m

p


2
2ma
and eigenergies of the electron
8
in the box:
  nx    my    pz 
nmp  r   3 sin 
 sin 
 sin 

• Polarization is ε  xˆ
a
a
a
a

 
 

• We need matrix element ε  r  nmp ε  R 111  nmp X 111
FI
a
a
8 a
  nx    x 
  my    y 
  pz    z 
 3  x sin 
 sin 
 dx 0 sin 
 sin 
 dy 0 sin 
 sin 
 dz
0
a
 a   a 
 a   a 
 a   a 
2
2
2

• Only states possible are
n even
8 an  n  1
  m1 p1  
|n11 with n even
0
n odd


Sample Problem (2)
(b) what are the corresponding frequencies , and (c) what are the transition rates
111  n11
Enmp 
2
2
2ma
2
2
2
n

m

p


2
 E1  4 
2
1
ε  rFI 
I FI  ε  rFI
2
• The frequencies are (n is even):
FI 
1
 EF  EI 
• The rate is:
1 2 2 2
2
2

n

2

3

n
 1


2
2 
2ma
2ma
  111  n11  
256 n 2 a 2
 2  n 2  1
4
I FI 
8an

2
n
2
 1
2
15F. Beyond the Dipole Approximation
Higher Order Terms
N
e
ik R
W  A0  e j ε  Pj  i  k  ε   S j 
m j 1
• Recall:
• We dropped second term and kRj because they were smaller by factors of ka0 ~ 
N
• So, keep them to
e
W  A0  1  ik  R j  ε  Pj  i  k  ε   S j 
one more order
m j 1
• First term leads to
N
e
electric dipole term
 A0  ε  Pj  i  k  R j  ε  Pj   i  k  ε   S j 
m j 1
• For the second term,
 k  ε    R j  Pj    k  R j  ε  Pj    k  Pj  ε  R j 
use this identity:
• Rewrite as k  R ε  P  1 k  R ε  P  1 k  P ε  R  1  k  ε   R  P

j

j

2

j

j
 
2
j

j

2

j
j

N
ie
• New
W  A0   12  k  R j  ε  Pj   12  k  Pj  ε  R j    k  ε    12 R j  Pj  S j 
terms:
m j 1
Electric Quadrupole and Magnetic Dipole
N
ie
W  A0   12  k  R j  ε  Pj   12  k  Pj  ε  R j    k  ε    12 R j  Pj  S j 
m j 1
• Last term:

N
j 1
R j  Pj  S j     12 L j  S j   12 L  S
N
1
2
j 1
• Magnetic dipole term:
F WM 1 I 
ie
A0  k  ε   F
m
 12 L  S  I
• We’ll set this aside and let you work on it with homework
• The remaining term is the electric quadrupole term:
F WE 2
N
ie
I 
A0  F  k  R j  ε  Pj    k  Pj  ε  R j  I
2m j 1
Simplifying the Electric Quadrupole Term
F WE 2
N
ie
I 
A0  F  k  R j  ε  Pj    k  Pj  ε  R j  I
2m j 1
• As before, we can use a clever commutator to simplify:
N
 1 N 2



H 0 ,  k  R j  ε  R j   
Pi  Va ,   k  R j  ε  R j 



j 1
j 1
 2m i 1

2
N N   Pi ,  k  R j    ε  R j  
N N
1
1



2



Pi ,  k  R j  ε  R j  

 


2m i 1 j 1   k  R   P 2 ,  ε  R  
2m i 1 j 1
j  i
j 

 k  Pi   ε  R j   i N
2i N N


k  Pj  ε  R j    k  R j  ε  Pj 






ij 
2m i 1 j 1   k  R j   ε  Pi  
m j 1


N
ie im
• So we have:
F WE 2 I 
A0  F  H 0 ,  k  R j  ε  R j  I
2m
j 1
N


Simplifying the Electric Quadrupole Term (2)
N
ie im
F WE 2 I 
A0  F  H 0 ,  k  R j  ε  R j  I
2m
j 1
N

e
• Let H0 act to the
F WE 2 I 
A0  EF  EI   F  k  R j  ε  R j  I
2
left or to the right,
j 1
N
as before
  12 eA0FI  F  k  R j  ε  R j  I
j 1
• Now get the rate:
 E 2  2
2
2
F WE 2 I
 12  e2 A0
2
 FI   
  k  R ε  R  
N
FI2
2
j 1
F
j
j
2
 FI   
I
• As usual, convert to intensity
A0
2
2 ke

I
2
c
E 2
 ke e 

I
2 2
c 
2
2
2
FI
  k  R ε  R  
N
j 1
F
j
j
I
2
 FI   
Putting It Together
E 2
 ke e 

I
2 2
c 
2
2
2
FI
  k  R ε  R  
N
j 1
F
j
j
I
2
 FI   
• Use fine structure constant
ke e 2

c
• As before,  = FI because of the delta function
• Integrate over  I  I   d 

• Final answer:
E 2 

2
I FI   F  k  R j  ε  R j  I
N
j 1
2
Comments on E2 and M1 transition
WE1 ~ F  ε  R  I
WE 2 ~ F  k  R  ε  R  I
WM 1 ~ F
 12 L  S  I
• Electric quadrupole term2 tends to be down by factor of (ka0)2 ~ 2
compared to electric dipole
• Same can be proven about the magnetic dipole contribution
The electric quadrupole term can be shown to be a rank-2 spherical tensor
• Means that |l| = 0, 1 or 2
Both electric quadrupole and magnetic dipole commute with parity
• Connect only same parity states
• Unlike electric dipole, which reverses parity
• Means no interference with electric dipole
Magnetic dipole operator commutes with H0
• Only connects states that are split by smaller effects
– Spin orbit
– Hyperfine
– External B-field
Sample Problem
An electron is trapped in the ground state of a cubical box of size a3, with a
corner at the origin. It is then bathed in light moving in the z-direction,
polarized in the x-direction, and having intensity function (). In the electric
quadrupole approximation, (a) which states can the electron transition into, (b)
what are the corresponding frequencies , and (c) what are the transition rates
2
2
2
2
2
• We first need the eigenstates
nmp , Enmp 
n

m

p


2
2ma
and eigenergies of the electron
8
in the box:
  nx    my    pz 
nmp  r   3 sin 
 sin 
 sin 

• Polarization is ε  xˆ
a
a
a
a

 
 

• We need matrix element nmp  ε  R  k  R  111  k nmp XZ 111
a
a
8k a
  nx    x 
  my    y 
  pz    z 
 3  x sin 
 sin 
 dx 0 sin 
 sin 
 dy 0 z sin 
 sin 
 dz
0
a
 a   a 
 a   a 
 a   a 
2
2
4 2
2
2

• Only states possible are
n, p even
64 a knp  n  1  p  1
  m1  
|n1p with n,p even
0
otherwise


Sample Problem (2)
(b) what are the corresponding frequencies , and (c) what are the transition rates
111  n1 p
E 2   
2
1
I  ε  R  k  R  F 
I FI  F  ε  R  k  R  I
2
Enmp 

2
4
n
64a 2 knp
2
 1
2
2ma
• The rate is:
  111  n1 p  
1024 n p  n  p  2 
2
2
2
 m c  n  1  p  1
2
2 2
2
4
2
p
2
 1
2
2
2
n

m

p

2 
• The frequencies are (n, p are even):
2

1 2 2 2
1
2
2
2

n

p
 2

n

p

1

3
FI   EF  EI 

2 
2 
2ma
2ma
• k = /c
2
2
4
2
I FI 
2
15G. Time-Dependent Perturbation Theory
With a Constant Perturbation
Why?
• Imagine having a perturbation of the form H  H 0  W
W  t   W Wmn  t   Wmn  m W n
• We can use same formulas as before with
Why would you want to use this formalism?
• Consider a particle scattering off a potential. In this formalism:
– It starts in a plane wave
– Then it scatters
– Then it is a plane wave again
• Consider a particle decaying, say +  +
– First it’s a pion
– Then it decays
– Then it’s a muon plus neutrino
Time Symmetric Notation
• Previous S FI   FI  1
i
formula:

•
•
•
•
1
i

3

dtWFI  t  e
T
0
 
n
m
T
0
iFI t

dtWFn  t  e
1
i
iFn t


t
0
2

n
T
0
dtWFn  t  e
dt Wnm  t   e
inmt 

iFnt
t
0

t
0
dt WnI  t   einI t 
dt WmI  t   eimI t  
1   0 ,
All matrix elements have no time-dependence
    
Make integrals time symmetric, from –T/2 to +T/2
0   0 .
Enforce limits on integration using the Heaviside function:
We have:
T /2
WFnWnI T /2
WFI T /2
iFn t inI t 
iFI t


S FI   FI 
dt
e

dt
dt

t

t
e
e



2



i T /2
n
 i  T /2 T /2

n
m
WFnWnmWmI
i 
3

T /2
T /2
dt 
T /2
T /2
dt  
T /2
T /2
dt   t  t    t   t   eiFnt einmt eimI t  
• Not enough integrals for you? Don’t worry, we’ll add more!
Rewriting the Heaviside Function

i
1
e
d
1   0 ,
• We wish to prove the
    lim
    
 0 2 i    i
following theorem:

0   0 .
We will use the technique of contour integration
• The path of integration in the complex plane of an integral is independent of the
path it takes from its starting point to its ending point, provided the function is
analytic (can take multiple derivatives of it)
• The function being integrated is analytic everywhere except  = i
– We say there is a pole at  = i
– Don’t cross the pole!
Strategy for doing the integrals:
• Figure out where in the complex plane the ei is small
– If  > 0, then when Im() > 0, i has large negative real part, and ei is small
– If  < 0, then when Im() < 0, i has large negative real part, and ei is small
• Add a half-circle at infinity to “close the loop”
• Distort the loop into a tiny circle, easy to integrate
• Do the integral
Case 1:  > 0

1
ei d
lim
 0 2 i    i

The current path of integration is along the
real axis of the complex  plane
If you add more integration in the upper
half plane, it doesn’t add anything
We can add this to the path of integration with impunity
Now, we can distort the path as long as we
don’t cross the pole at  = i
And while we’re at it, we can move the pole to the origin
Now do integral assuming it is a tiny circle of radius 
Can write  =  ei
Take limit   0
•
•
•
•
•
•
•
•
1
2 i

e d
i

2
1

2 i 0
exp  i ei  d  ei 
e
i

i
2 i 
2
0
Im()
1

exp  i e  d
2
i

Re()

2
0
d  1
Case 2:  < 0
Im()

1
ei d
lim
 0 2 i    i

• The current path of integration is along the
real axis of the complex  plane
• If you add more integration in the lower
half plane, it doesn’t add anything
• We can add this to the path of integration
with impunity
• Now, we can distort the path as long as we
don’t cross the pole at  = i
• The integral is now away from the pole
1
• It is a finite number over a tiny contour 2 i
• For the two cases, we therefore have
1
e d 1   0
lim


 0 2 i
  i 0   0


i

Re()

ei d
0
  i

1
ei d
    lim
 0 2 i    i

More Integration!
WFI
S FI   FI 
i

n

T /2
T /2
dt e
i 
m

WFnWnI

i 
2
n
WFnWnmWmI
3
iFI t

T /2
T /2
dt 
T /2
T /2
T /2
T /2
dt  
T /2
T /2
dt 
T /2
T /2
dt   t  t   eiFnt einI t 
dt   t  t    t   t   eiFnt einmt eimI t  

• We rewrite all Heaviside functions using our integral:
WFI
S FI   FI 
i
 
n
m
i  t t 

T
/2
T
/2

W
W
e
d iFnt inI t 

iFI t
Fn nI
dt  dt  
e e


2
T /2 dt e  lim



T
/2

T
/2

0
  i
 n  i  2 i
T /2
WFnWnmWmI

 i   2 i 
3
1
ei d
    lim
 0 2 i    i

2
T /2
T /2
dt 
T /2
T /2
dt  
T /2
T /2
dt  


e
i  t t 
d  e   d  iFnt inmt  imI t 
e e e


  i     i
i  t  t 



Do Most of the Time Integrals
WFI
S FI   FI 
i
 
n
m
i  t t 

T /2
T /2
 e
W
W
d iFnt inI t 

iFI t
Fn nI

dt  dt 
e e

2
T /2 dt e  lim

T /2
T /2
   i
 0
 n  i  2 i
T /2
WFnWnmWmI

i
2

i
  
3
2
T /2
T /2
dt 
T /2
T /2
dt  
T /2
T /2
dt  


e
i  t  t 
d  e   d  iFnt inmt  imI t 
e e e



  i
   i
• We want to simplify in the approximation T  
• Do integrals for all times except t using the identity:
WFI
S FI   FI 
i
 
n
m

T /2
T /2
WFnWnmWmI
i 
3 2
i
dt e
iFI t
 W W
 lim  Fn 2 nI
 0
 n  i  i



eit  dt   2  
T /2 i   t
d
 Fn 




e
dt


   i nI T /2

T /2 i   t
d  d 
 Fn 












e
dt 
  mI
T /2
nm
   i    i 




i  t t 



Do All the  Integrals
WFI
S FI   FI 
i
 
n
m

T /2
T /2
WFnWnmWmI
i 
3 2
i
dt e
iFI t
 W W
 lim  Fn 2 nI
 0
 n  i  i
T /2 i   t
d
 Fn 




e
dt


   i nI T /2

T /2 i   t
d  d 
 Fn 












e
dt 




nm
mI
   i    i
T /2

• Use the -functions to do all the  integrals:
W
S FI   FI  FI
i
 
n
m

T /2
T /2
dt e
WFnWnmWmI
i
 
2
iFI t
 WFnWnI
1
 lim 
 0
 n i    nI  i
1
1
nm  mI  i mI  i

T /2
T /2
e

T /2
T /2
e
i Fn nI t
i Fn nm mI t
dt
dt 






Simplify Some More
WFI
S FI   FI 
i
 
n
m

T /2
T /2
dt e
WFnWnmWmI
i
 
• Use definition of nm:
2
iFI t
 WFnWnI
1
 lim  
 0
 n i    nI  i
1
1
nm  mI  i mI  i
nm 
En  Em

T /2
T /2
e

T /2
T /2
e
i Fn nI t
i Fn nm mI t
dt
dt 



• The t integrals all end up the same: Fn  nI  FI , Fn  nm  nI  FI
• Multiply through the – by the  denominators
1
S FI   FI 
i


WFnWnI
WFnWnmWmI

 
WFI  lim
 
0
n m  EI  Em  i  EI  En  i

 n EI  En  i



T /2
T /2






dt eiFI t
The Transition Matrix



WFnWnI
WFnWnmWmI


 
 
WFI  lim

0
n m  EI  Em  i  EI  En  i 

 n EI  En  i



T /2
• Define the transition matrix FI as the expression in {}’s
 dt eiFI t
• Note that since we are taking   0, we might as well replace  by  T /2
1
S FI   FI 
i

WFnWnI
WFnWnmWmI
TFI  WFI  lim  
 

 0
 n  EI  En  i  n m  EI  Em  i  EI  En  i 
•
•
•
•



sin  12 FI T 
2
We have
dt e
  FI  TFI

T /2
i
FI
Do the last time
integral, but must be careful before taking limit T  
2
The probability of a transition is given by
P  I  F   S FI
Assuming the final state is distinct from
2 1
the initial state, we therefore have
4
2 sin  2  FI T 
P  I  F   2 TFI
FI2
1
S FI   FI  TFI
i
T /2
iFI t
Fermi’s Golden Rule
PI  F  
4
2
TFI
2
sin 2  12 FI T 
FI2
• We need to take limit T  . We found before: lim
T 
• So we have:
2
P  I  F   2 2 TFI T  FI 
sin 2  12 FI T 
FI2
 12  T  FI 
• Define the rate  as the probability per unit time
• You can put one factor of -1 into the delta function, where it becomes an .
• We now have
2
1
  I  F   2
TFI   EF  EI 
Fermi’s Golden Rule
•
•
•
•
The remaining delta-function is usually handled by integrating over final states
2
This yields a formula like:
1
  I  F   2
TFI   EI 
Where  is the density of states
This formula is also called Fermi’s golden rule, but I don’t find it useful
Comments on Fermi’s Golden Rule
  I  F   2
1
TFI   EF  EI 
2

WFnWnI
WFnWnmWmI
TFI  WFI  lim  
 

 0
 n  EI  En  i  n m  EI  Em  i  EI  En  i 



• Most often, just keep first term
  I  F   2 1 WFI   EF  EI 
What to do with the ’s?
• Only relevant if you go beyond first order
• If none of the intermediate states match the initial state energy, set  = 0
• If the intermediate states do match energy, much care is required
What to do with the final delta function?
• Often have continuum states for the final state
• Sum over these final states
– For continuum states, that means an integral 
f  xi 
• When you have an integral over a delta
 f  x    g  x   x 
1
g   xi 
i g  0
function, don’t forget the derivative
2
Sample Problem
A spin up electron is trapped in the ground state of a spherical box of radius a
centered at the origin; however, this box only traps spin up electrons. There is also
a small perturbation W = x that connects spin up and spin down electrons. Find
the rate at which this electron escapes due to the perturbation.
• H0 is the infinite spherical square well for spin up; free particles for spin down
• The
2 2
1
r 
100  , E100 
, 100  r   
sin   for r  a
2
initial state:
2ma
r 2 a
 a 
• The final state:
 2 2k 2
3/2
k  , Ek 
, k  r     2  eik r
2m
• We need matrix elements

1
r 
3
 ik r 1
  x   d re
sin  
WFI  F W I   k   x 100  
3
r
2 a
 a 
8
2
2
a r dr
a


  r  1 ikr cos
  r  sin  kr 
 2
d 
sin    e
d  cos   
rdr sin  


0
0

1
0
r
4 a
 a
 a 
 a  kr
Sample Problem (2)
…Find the rate at which this electron escapes due to the perturbation.
a


  r  sin  kr 

WFI 
rdr sin  

0
k a
 a
 a  kr
r 
0 dr sin  a  sin  kr 
• Use first order approximation, FI = WFI, and then Fermi’s Golden Rule
2
2
2
2
a

2


2


r

1


2
  I  F   2
TFI   EF  EI  
dr sin   sin  kr   
 k  2 
2

0
k a
a 
 a 
 2m 
• The delta-function will assure that
a
1
dr
sin

r
a
sin
kr





k = /a, which makes the integral simple: 0
2a
• Final states were not normalized
 2 a  2  2  2 
I  k 
   k  2 
• But we need to add up (integrate)
2
2 k  2m 
a 
over all possible k values anyway:
1

2
2
2
2
2
2


2

ma



a

k


k
3
2
2

  I  kd k
d   
 k  2   2 dk  2 a 


k
2
2m 
a  k
m 
2
2

0

2 ma


a
Sample Problem
Find the “rate” at which a particle of mass  with wave number kI
scatters from a weak potential V(r) into some new wave number kF.
P2
• The Hamiltonian is a free
H
V R
2
Hamiltonian plus a potential
P2
• Since the potential is weak, we’ll treat it as a perturbation H 0  2 , W  V  R 
• We first need to
2 2
1
k
ik r
find eigenstates of
k , k  r  
e , Ek 
3/2
2
 2 
the free Hamiltonian
• These states are continuous states, not properly normalized
– We’ll deal with this problem later
• We’ll calculate the transition TFI  WFI   F V I   2 3 eik I k F rV  r  d 3r

matrix to leading order only
2
1
• Then substitute in
  I  F   2
TFI   EF  EI 
Fermi’s Golden Rule
2
 2 2
5 1
i  k I k F r
3
2 
  2 
V  r  d r    k F  k I 
e
 2

Sample Problem (2)
Find the “rate” at which a particle of mass  with wave number kI
scatters from a weak potential V(r) into some new wave number kF.
2

1
3
2
2 
  k I  k F    2 
V  r  d r    k F  k I 
e
 2

• Because we used continuum states for the final states, they weren’t normalized
• We don’t want to go into a particular state, we want to go to arbitrary kF
• To handle this, integrate over all possible final kF values
2
2

5 1
i  k I k F r
3
3
2
2 
   2 
V  r  d r    k F  k I 
 d kF  e
 2

5

i  k I k F r
2
2

i  k I k F r
1
2
3
2
2 
  2 
V  r  d r    kF  kI 
 d F 0 kF dkF  e
 2

2
2

5 1 2
 kI
i  k I k F r
3
i  k I k F r
3
   2 
kF 2  d   e
V r  d r 
d  e
V r  d r
5 3 
kF
 2 
5
2
Sample Problem (3)
Find the “rate” at which a particle of mass  with wave number kI
scatters from a weak potential V(r) into some new wave number kF.
2
k
i  k I k F r
3

d  e
V r  d r
5 3 
 2 
3/2 ik r
I  r    2  e I
• The initial states weren’t normalized
• Going back to chapter 14, we dealt with this before by finding
k

I
the flux (particle density times velocity) for the initial particles

• We then defined the cross-section
3
2
  2    k I
i  k I k F r
3
 
d  e
V r  d r
5 3 

kI  2 
• Differential cross-section is the same expression not integrated over angles
• Identical to first Born
2
d
2
i  k I k F r
3
 2 4 e
V r  d r
approximation
d  4
2