Physics 742 – Graduate Quantum Mechanics 2
Solutions to Final Exam, Spring 2015
The points for each question are marked. There are a total of 150 points for the whole
test. Some possibly useful formulas appear at the end of the test.
1. [20 points] A particle of mass  with wave number k scatters from the weak potential
2
V  r   V0 e  Ar . Calculate the differential cross section d d  and total cross section  .
I recommend working in Cartesian coordinates when computing the Fourier transform
of V.
We first need the Fourier transform of the potential, which is just

 Ax
 iK r 3
 Ar  iK r 3
 V  r  e d r  V0  e e d r  V0  e
2
2
iK x x

 V0

A
e
iK x  4 A
2
 iK 
e
y
2
4A
A

A

dx  e
 Ay 2  iK y y

e iK z 
2
4A
 V0

dy  e Az
2
iK z z

3
3
A
eK
2
dz
4A
We now substitute this in to get the cross-section:
2
2
 2V 2  3
 2V 2 2
d
2
 2 4  V  r  e iK r d 3r  2 0 4 3 e K 2 A  4 03 e  k 1cos 
d  4 
4  A
4 A
A
This must then be integrated over angles to get the total cross-section:
1  k 2 1 cos A
 2V02 2
 2  2V02 A   k 2 1cos  A  cos 1
d


d   4 3  d  e
d cos  
e
 
1
 cos 1
d
4 A 0
2 4 A3 k 2 
2
 2  2V 2
 4 2 0 2 1  e2 k A
2 A k


2. [20 points] A particle of mass m in one dimension is trapped by a delta function
potential, V  x     x  . This system has just one bound state, with wave function
and energy
 g  x    e  x
where   m  2
and Eg   m 2 2 2
At t = 0, it is in the bound state, but the value of  now begins increasing, eventually
doubling, so that   2. Suppose that change occurs over a time scale T. What
condition on T determines when you can use the adiabatic approximation or the sudden
approximation? What is the probability that it remains bound in each of these limits?
The adiabatic approximation works when T E   . Since the ground state has the
energy listed, and the next state would be a free state with very small energy, we can
approximate E  Eg , so the adiabatic approximation works when Tm 2  23 . Similarly,
the adiabatic approximation works when Tm 2  23 .
In the adiabatic approximation, the ground state goes to the ground state, so the
probability is just P  g   g   1 . In the sudden approximation, the probability is given by the
square of the overlap, P  g   g    g  g
2
. The initial and final ground states are the
same, except that  has increased by a factor of two. As you can see from the formula
  m  2 , this means that  has doubled as well. Hence the final wave function is actually
 g  x   2 e2 x
We therefore have




0
 g  g    g*  x  g  x  dx   2  e2 x e x dx  2 2  e 3 x dx 

P  g   g    g  g
2
2
 8 8
 
  .
3

 9
2 2
8
,

3
3
3. [20 points] A particle of mass m is trapped in the ground state 1 of a 1D infinite
square well with allowed region 0 < x < a. At t = 0 a perturbation W   cos  X a  e  t
is turned on. Find the probability, to leading order (second order in A), that it ends up
in the state n at t =  for n > 1.
The eigenstates and energies of the infinite square well are given by
n  x  
2
 2 n2 2
  nx 
sin 
,
E

n

a
2ma 2
 a 
We will calculate the transition using the leading order formula S n1   i 
1

T
0
dtWn1  t  ein1t ,
appropriate when we are considering final states different from the initial state. The matrix
element Wn1  t  is defined as
a
2
  nx 
x  x 
Wn1  t   n W 1   e  t n cos  x a  1   e  t  sin 
cos 

 sin 
 dx
0
a
 a 
 a   a 
2
a
  e  t 1,1 n   n ,11  1,1 n   12  2,n e  t
a
4
Thus the only state it can go into is n = 2. The frequency difference in this case is then
E2  E1
 22
3 2 
21 
.

 4  1 
2ma 2 
2ma 2

We therefore have
S 21 
1 
  i21t  t

i21t
dtW
t
e

dt e e 
.


21


0
0
i
2i
2i    i21 
The probability for a transition, therefore, is
P 1  2   S21 
2
m 2 a 4 2
2

.
2 2 2 4
4 4
4 2   2  212  4  m a  9 
4. [20 points] A particle of mass m is in the ground state 0, 0, 0 of the 3D symmetric
harmonic oscillator potential, V  r   12 m02  X 2  Y 2  Z 2  . It is subjected to a periodic
perturbation of the form W  t    XY cos t  . Which state nx , n y , nz can it be
promoted to by this perturbation, to first order in the perturbation? Calculate, to
leading order, the rate at which this transition occurs. Leave the resulting delta
function alone. For what value of  will this transition occur?
The first step is to write the perturbation in the form W  t   We  it  W †eit . Since
cos t  
1
2
e
it
 e  it  , it is pretty easy to see that our perturbation is W  12  XY .
The states that will be excited, to first order in W, are the ones that W connects via the
matrix element
nx , n y , nz W 0, 0, 0  12  nx , n y , nz XY 0, 0, 0  12 


nx , n y , nz  ax  ax†  a y  a †y  0, 0, 0
2m0


nx , n y , nz  ax  ax†  0,1, 0 
nx , ny , nz 1,1, 0 .
4m0
4m0
It is clear, therefore, that the only state it can transition to is 1,1, 0 . The rate for this transition,
since we are going up in energy, can be calculated as
  0, 0, 0  1,1, 0   2  WFI
1
2
2
2    
  EF  EI    

   EF  EI   
  4m0 
The energy of the state nx , n y , nz is E  0  nx  n y  nz  32  , which works out to
E000  32 0 for the ground state and E110  72 0 for the final state, so we have EF  EI  20 ,
so
  0, 0, 0  1,1, 0  
 

  20     2 2   20    .
2 2
8m 0
8m 0
It is obvious from the final delta function that the necessary frequency is   20 .
5. [25 points] A system contains only one type of photon, with momentum q  qzˆ and
polarization ε q  xˆ , but it is in a superposition state of different numbers of these
photons, so  
1
2
 7, q,
 i 8, q,
.
Find the expectation value of the electric and
magnetic field for this quantum state at the point r.
We need, for example, to calculate the matrix element
 k
 E  r    12  7, q,  i 8, q,  
i  ak ε k eik r  ak† ε*k e  ik r   7, q,  i 8, q, 
2 0V
k ,
Even though there is an infinite sum, the only terms that can possibly contribute are those where
we are creating or annihilating only the photon with wave number q and polarization . Hence
this simplifies to
E r  
1
2
 7, q,
 i 8, q,

 q
2 0V
i  aq ε q eiqr  aq† ε*q e iqr   7, q,  i 8, q,
.
If we create a photon, we must be going from 7 to 8 photons; if annihilating, from 8 to 7. Also
note that the polarization vector and its complex conjugate are the same x̂ , so we simplify this to
E  r   12 ixˆ
 12 ixˆ
q
2 0V
 7, q,
q
 i
2 V
 i 8, q,
a  e
iqr
q
 aq† e  iqr   7, q,  i 8, q,
7, q, aq eiqr 8, q,  i 8, q, aq† e iqr 7, q,


0

q
1
xˆ
2 2 0V

q
1 8q iqr  iqr
cq
e  e   xˆ 2
cos  q  r   xˆ 2
cos  qz  .
xˆ

 0V
2
2 0V
 0V

8 7, q, 7, q, eiqr  8 8, q, 8, q, e iqr

We now perform a similar calculation for the magnetic field.
B r  
1
2
 7, q,
 i 8, q,

1
2
 7, q,
 i 8, q,
 12 iq  xˆ
 12 iqzˆ  xˆ


k ,

2 0V q


ik   ak ε k eik r  ak† ε*k e  ik r   7, q,  i 8, q,
2 0V k

iq   aq ε q eiqr  aq† ε*q e  iqr   7, q,  i 8, q,
2 0V q
 7, q,
 i 8, q,
a  e
q
iqr
 aq† e  iqr   7, q,  i 8, q,

 i 7, q, aq eiqr 8, q,  i 8, q, aq† eiqr 7, q,
2 0V q
1

qyˆ
2
2 0Vcq


8 7, q, 7, q, eiqr  8 8, q, 8, q, e iqr ,




B r  
1
8q
q
q
eiqr  e  iqr   2yˆ
cos  q  r   2yˆ
cos  qz  .
yˆ

2
2 0 cV
 0cV
 0 cV
6. [20 points] Calculate the rate at which a hydrogen atom is in the state 3,1,1
spontaneously decays to the ground state 1, 0, 0 . Simplify as much as possible, and
ultimately convert it into a rate in s-1. Several hydrogen integrals can be found with the
equations.
We need to find the dipole moment 1, 0, 0 R 3,1, 0 for these states. This is three
components, and we simply dive in and start computing:
*
1, 0, 0 X 3,1,1   d 3r 100
 r  311  r  r sin  cos 

  Y00  ,   Y10  ,   sin  cos  d   r 2 r dr R10*  r  R31  r  
*
0

1 27 6

a0
6 128
27
a0 ,
128
*
1, 0, 0 Y 3,1,1   d 3r 100
 r  311  r  r sin  sin 

  Y00  ,   Y10  ,   sin  sin  d   r 2 r dr R10*  r  R31  r  
*
0

i 27 6

a0
6 128
27i
a0 ,
128
*
1, 0, 0 Z 3,1,1   d 3r 100
 r  311  r  r cos 

  Y00  ,   Y10  ,   cos  d   r 2 r dr R10*  r  R31  r   0 
*
0
27 6
a0  0 .
128
Summarizing, we have 1, 0, 0 R 3,1, 0  27 a0  xˆ  iyˆ  128 .
We need the frequency difference IF , which is given by
IF 
2 2
1
1  mc 2 2 mc 2 2   1  9  mc 
4mc 2 2
.



 E3  E1    

18
2 
18
9


We then use the general formula for the decay rate,
4
4  4mc 2 2 
2
  2 IF3 rFI  2 

3c
3c  9 
 5 mc 2
 5 mc 2
3
 27 a0  2  27 a0  2  22  26 m3c 6 6 2  36    2



 
 
3c 2  36 3 214
  mc 
 128   128  
1
5.11 105 eV





 1.656  108 s 1 .


16


3  25 
96
137.036
6.582
10
eV
s


5
7. [25 points] An electron is in the ground state 0, 0, 0 of the harmonic oscillator, with
potential V  r   12 m02  X 2  4Y 2  9Z 2  . It is being bombarded by light moving in the
x-direction with polarization ε  zˆ .
(a) What is the energy of the state nx , n y , nz ?
The Hamiltonian will just be the sum of three Hamiltonians, one in each direction, with
frequencies 0 , 20 , and 30 respectively. Therefore, the energy of the state nx , n y , nz will be
Enx ,ny , nz  0  nx  12   20  n y  12   30  nz  12   0  nx  2n y  3nz  3 .
(b) Argue that in the dipole approximation, only one state will have non-zero matrix
element ε  rnI . Which state is this? What is the appropriate frequency difference
nI ?
The matrix element is
ε  rnI  zˆ  nx , n y , nz R 0, 0, 0  nx , n y , nz Z 0, 0, 0 


nx , n y , nz  az  az†  0, 0, 0
2m z

nx , n y , nz 0, 0,1
6m0
Clearly, the only intermediate state that contributes is the state 0, 0,1 , for which
rnI   6m0 . The appropriate frequency difference is
nI   10  0  0  3  3   0  0  0  3   30 .
(c) What is the decay rate  for this intermediate state?
In the dipole approximation, the only state that 0, 0,1 can decay to is 0, 0, 0 . The
decay rate is given by the dipole approximation,

6 02

4 3
4
2
3

3

.
r




nI nI
0
mc 2
3c 2
3c 2
6m0
(d) Find the cross section for photon scattering at frequencies  very close to or at nI .
We are close to resonance, so we use the resonance formula:
8 2  30 





2
2
2 
2
2 
1
1 2 
3c   nI   4 
3c   30   4   6m0




8 2nI4 rnI
4
4
4

6 2  202


2
m 2 c 2   30   14  2 



If desired, you can substitute the decay rate from part (c) to give the final formula

6 2  202 c 2
m 2 c 4   30   9 2  204
2
Possibly Helpful Formulas:
Harmonic Perturbations:
W  t   We  it  W †eit , WFI  F W I
Born Approximation
2
3
 iK r
d   d rV  r  e

4 2  4
d
K 2  2k 2 1  cos  
1D H.O.:
V  x   12 m02 x 2
X
2
 2  1 WFI 2   EF  EI    if EF  EI

I  F   
† 2
1
2  WFI   EF  EI    if EF  EI
1D infinite square well:
2
  nx 
n  x  
sin 

a
 a 

a  a† 

2m0
a n  n n 1
2ma 2
a n  n 1 n 1
S FI   FI   i 
1
Spont. Decay
4
2
  2 IF3 rFI
3c
 2n2 2
En 
Time-dependent Perturbation Theory
†
Hydrogen:
mc 2 2
13.6 eV
En  

2
2n
n2

 5.29 1011 m
a0 
mc
Scattering near resonance

8 2nI4 rnI
4
3c 2   nI   14  2 


2
Hydrogen integrals:


5 6
3 6
2
0 R31  r  R10  r  rdr  96a0 , 0 R31  r  R10  r  r dr  32 ,


0

T
0
dtWFI  t  eiFI t  
Constants:
  1.0546  1034 J  s
  6.582  1016 eV  s
1
  137.036
 0.0072974
c  2.998  108 m/s
Electron mass:
m  9.109 1031 kg
mc 2  5.11105 eV
R31  r  R10  r  r 3dr 
27 6
a0 ,
128
 Y  ,   Y  ,   d    Y  ,  Y  ,   sin  d    Y  ,   Y  ,   cos  d   0
 Y  ,   Y  ,   sin  cos  d   1 6 ,  Y  ,   Y  ,   sin  sin  d   i 6 .
0*
0
1
1
0*
0
1
1
0*
0
1
1
0*
0
0*
0
1
1
1
1
Electromagnetic Operators
E r   
k ,
 k

i  ak ε k eik r  ak† ε*k e ik r  , B  r   
ik   ak ε k eik r  ak† ε*k e ik r 
2 0V
2 0V k
k ,
Possibly Helpful Integrals:
The formulas below assume n, p, and q are positive integers





a
0
a
0
e  Ax
2
 Bx
dx 

A
eB
2
4A
,
e
x
dx   1e x ,


0
e  x dx   1 .
sin  nx a  sin  px a  dx   cos  nx a  cos  px a  dx  12 a np ,
a
0
sin  nx a  sin  px a  cos  qx a  dx  14 a  n  q , p   p  q , n   n  p ,q  ,

a
0
cos  nx a  cos  px a  cos  qx a  dx  14 a  n  q , p   p  q , n   n  p ,q  .