Physics 742 – Graduate Quantum Mechanics 2
Solutions to Final Exam, Spring 2013
Due Monday, May 6 by 5 PM
You may use (1) class notes, (2) former homeworks and solutions (available online), or
(3) any math references, such as integral tables, Maple, etc, including any routines I
provided for you. If you cannot do an integral or sum, or if you cannot solve an equation,
try to go on, as if you knew the answer. Some useful integral appear on the last page.
Feel free to contact me with questions.
Work: 758-4994
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Cell:
407-6528
1. [30] An electron of mass m lies in a region with no scalar potential U  0 and a
vector potential A  a  yxˆ  xyˆ  , where a is constant.
(a) [5] Demonstrate that this vector potential actually results in vanishing
electric and magnetic fields. Hence we are solving a trivial problem (a free
particle) in a difficult way.
The electric and magnetic fields are just given by

A 0,
t
 A A 
 A A   A A 
B    A  xˆ  z  y   yˆ  x  z   zˆ  y  x 
z 
x   x
y 
 z
 y


 


  
 xˆ   ax   yˆ  ay   zˆ  ax  ay   0 .
y 
 z 
 z   x
E  U 
(b) [4] Write the Hamiltonian explicitly. Convince yourself that the three
momentum operators P, though they commute with each other, do not all
commute with the Hamiltonian.
The Hamiltonian is
H
2
1 2
ge
1 2
1 
2
π  eU  R, t  
B  R, t   S 
π 
P
eaY
P
eaX



 Pz2  .




x
y

2m
2m
2m
2m 
The presence of the operators X and Y guarantees that this expression will not commute
with Px and Py , though it will of course commute with Pz . Since spin plays no role, we
will ignore it henceforth.
(c) [7] In contrast, convince yourself that all three canonical momenta π
commute with each other and commute with the Hamiltonian. Hence they
can all be simultaneously diagonalized. Let the three eigenvalues of π be k ,
so π k  k k . What are the eigenvalues of the Hamiltonian in terms of k ?
The three operators are
 x  Px  eaY ,  y  Py  eaX ,  z  Pz .
It is clear that  z commutes with  x and  y , but do  x and  y commute with each other?
We need to check:
 x ,  y    Px  eaY , Py  eaX   ea  Px , X   ea Y , Py   iea  iea  0 .
It is then obvious that since all three π ’s commute with each other, and H  π 2 2m , H
will also commute with all three π ’s. Hence we can diagonalize them simultaneously. If
we pick our states so that
π k  k k
then
 k  k , so E   2 k 2 .
π2
k 
H k 
2m
2m
2m
2
(d) [8] Show explicitly that the wavefunction   x, y, z   exp  ik  r  i xy 
actually satisfies all three equations π  k if the constant  is chosen
appropriately.
We simply substitute it in and see if we can get it to work out. We want:
 

 eay  exp  ik  r  i xy    k x   y  eay  ,
k x   x   Px  eay   
 i x

 

k y   y   Py  eax   
 eax  exp  ik  r  i xy    k y   x  eax  ,
 i y

k z   z  Pz 
 
exp  ik  r  i xy   k z .
i z
The last equation is automatically true. The others will be true if we pick   ea , or
  ea  , and our eigenstates take the form
  x, y, z   exp  ik  r  ieaxy   .
(e) [6] Show that by a suitable choice of gauge transformation function  , we
can simultaneously turn our wave functions into the usual form   eik r and
eliminate the spurious vector potential A .
Under a gauge transformation, the scalar potential, vector potential, and wave
function should change to


 
,
t
t
A  A  A    a  xyˆ  yxˆ    ,
U U U 
      exp   ie    exp  ik  r  ieaxy   ie   .
We are trying to get the last two terms in the exponential for the wave function to vanish,
which suggests trying
0   ieaxy   ie  , or
   axy .
We then find

0
t
A  a  xyˆ  yxˆ     axy   a  xyˆ  yxˆ   a  xyˆ  yxˆ   0 ,
U  
   exp  ik  r  ieaxy   ieaxy    exp  ik  r  .
So it worked perfectly.
2. [20] Two particles in a one-dimensional infinite square well with allowed region
 a  x  a . In this region, the wave function is   x1 , x2   Nx1 x2  x1  x2  where
N is a normalization constant. The  simply means you are doing two problems
at once, so it might be + or it might be -. In any subsequence computation, you
must consider both possibilities, so you may have two distinct answers.
(a) [6] What is the correct normalization constant N?
The wave function must be normalized, so that if we square it and integrate over
both coordinates, we get 1. So we have
1  N 2  dx1  dx2 x12 x22  x1  x2   N 2  dx1  dx2  x14 x22  x12 x24  2 x13 x23 
a
a
a
a
2
 N 2  dx1  13 x14 x23  15 x12 x25  12 x13 x24 
a
a
 N 2  152 x15 a 3  152 x13 a 5 
N  a 4
15
8
a
a
a
a
a
a
x2  a
x2  a
 N 2  dx1  32 x14 a 3  52 x12 a 5 
a
a
 N 2  154 a8  154 a8   158 N 2 a8 ,
.
The normalization is the same in both cases.
(b) [5] What is the probability that particle one has positive position, x1  0 ?
We integrate x2 over all possible values, while we integrate x1 over positive values.
So we have
a
a
a
P  x1  0   N 2  dx1  dx2 x12 x22  x1  x2   N 2  dx1  23 x14 a 3  52 x12 a 5 
2
a
0
N
2

2
15
0
x a  152 x a
5 3
1
3 5
1

a
0
 158 a
8

2
15
a8  152 a8   158  154  12 .
The probability is the same in both cases.
(c)[5] What is the probability that both particles have positive position x1 , x2  0 ?
This time we restrict both integrals appropriately.
P  x1 , x2  0   N 2  dx1  dx2 x12 x22  x1  x2   N 2  dx1  13 x14 x23  15 x12 x25  12 x13 x24 
a
a
0
0
2
a
x2  a
0
x2  0
 N 2  dx1  13 x14 a 3  15 x12 a 5  12 x13 a 4   N 2  151 x15 a 3  151 x13 a 5  81 x14 a 4 
a
0
8


15
8
a
P  x1 , x2  0  
31
64
, P  x1 , x2  0  
1
15
a 
8
1
15
a  a
8
1
8
8
1
64

1
8
 
1
8
15
64

16 15
64
a
0
,
.
In this case, there is a big difference between the symmetric + case and anti-symmetric –
case.
(d) [4] Suppose the particles are identical particles. What would be the
appropriate sign for the  if they are both bosons? If they are both fermions?
Assume that any spin state would be symmetric, so their spin state looks like
,  .
If they are bosons, the total state must be symmetric, and since the spin part is
symmetric, the space part must be symmetric as well; that is,   x1 , x2     x2 , x1  . This
only works with the plus (+) sign.
If they are fermions, the total state must be anti-symmetric, and since the spin part
is symmetric, the space part must be ant-symmetric; that is,   x1 , x2     x2 , x1  . This
only works with the minus (-) sign.
3. [30] A single particle of mass m is in a Harmonic oscillator with frequency  ,
but we aren’t exactly sure which state it is in. At t = 0, it is put into one of the
two states
a 
1
3
0 
2
3
1
b 
or
0 
2
3
1
3
1 .
However, it is not known which one, and there is equal probability for both.
(a) [6] Working in the restricted basis of just the state 0 and 1 , write the
2  2 state operator   0  at t = 0. Check that   0  is Hermitian and
Tr    0    1 .
1
2
Since each state has equal probability, their probabilities must be
the state operator is
  0   12  a  a  12  b  b 
1
 
2 
1
3
2
3
2
3
2
3
 1
 
 2


2
3
2
3
2
3
1
3
1
2

 

 
 
1
3
2
3
1
2
2
3
2
3
1
2





 1

2 
2
3 
1
3

2
3
1
3

each. Hence






1 
3 
2
3

.


This is obviously Hermitian (real on the diagonal, and off-diagonal are equal real
numbers) and has Tr    0    12  12  1 .
(b) [9] What is the Hamiltonian in this basis, as a 2  2 matrix? Let the 2  2
state matrix   t  at arbitrary time t take the form
a b
.
c d 
 
Using the equivalent of Schrödinger’s equation for the state vector, find
differential equations for the time derivatives a , b, c, d .
Since we are working in the basis of the eigenstates of the Hamiltonian, the Hamiltonian
will be diagonal and will have diagonal values just equal to the eigenvalues, so
 12 0 
H   
.
3
0 2 
The equivalent of Schrödinger’s equation in terms of the state operator is
 1 0   a b   a b   12 0  
1
d 1
  H ,       2 3  


3 
dt i
i
 0 2   c d   c d   0 2  
  1 a 12 b    12 a 32 b 
  23
,
 
i  2 c 32 d  i  12 c 32 d 
 a b 
 0 b
 i 


.

 c 0 
 c d 
This then reduces to the four uncoupled differential equations
a  d  0 , b  ib , c  ic .
(c) [6] Solve the differential equations you found in part (b) subject to the
boundary condition you found in part (a).
Clearly, a and d will remain unchanged, but it is easy to see that the solutions to
the c and b equation will take the form
b  t   b  0  eit 
eit , c  t   c  0  e  it 
2
3
2
3
e it ,
so that the density matrix will be

 t   


2
3
1
2
2
3
e it
eit 
.
1

2

(d) [9] The momentum operator in this restricted basis takes the form
P
m  0 i 


2 i 0 
Find the expectation value P at all times for this state operator.
The expectation value of any operator is just given by
P  tr   P  
m  0 i  
tr 

2
 i 0  
1
2
2
3
e  it
2
3
eit  
m  i
tr 
 
1


2
2


2
3
1
2
e  it
i
 12 i 

i 32 eit 
 13 m ieit  ie  it   13 m i cos t   sin t   i cos t   sin t  
  23 m sin t 
4. [20] A particle of mass m in three dimensions has potential
V r    r   r .
where  is a positive constant. Attempt to estimate the energy of the ground
state for this potential using the variational principle, using the trial wave
function   r   exp   12 Ar 2  .
We first need to calculate   ,  P 2  , and  V  , which are given by
    d r exp   Ar
3
2
 P   P
2

2
  4  r e
2  Ar 2
dr  4
0
  d r i exp   Ar
3
1
2
2

2
  32 
2 A3/ 2
 2
1
2
A
3 2
 
 
 A

   d r rˆ exp   12 Ar 2 
r
2
3/2
,
2
3

2
  52 
2
2 2
 Ar 2 2 

 4   r dr Are
 4  A  r 4 e  Ar dr  4  2 A2


2 A5/2
0
0

2
2
 2 34  2  A  32   2  A ,
 V     d r r exp   Ar
3

2
  4  r e
3  Ar 2
0
dr  4
  2
2A
2
 2 A2 .
We can then find the expectation value of the energy as a function of A, using
3/2
 H
1  1
  A   1 3 2  2 
2

 P    V    

 2 
E  A 

   2m
A
A 
     2m 2
3 2 A 2
.


4m
A
To use the variational principle, we must find the minimum of this function,
which we find by setting the derivative equal to zero, which gives us
d
3 2

E  A 
,

dA
4m
 A3
4 m
 A3  2 ,
3
0
 4 m 
A

 3  
2/3
.
Substitethis back into the equation for E  A  , to estimate the energy
3 2  4 m 
E
4m  3 2  
2/3
2  4 m 



 A  3 2  
1/3
31/3 21/3  2/3 2/3 1

 2  1 
 1/3m1/3
1/3
3  6 2  2 

 .
2 m 
5. [35] A particle of mass m is confined in a two-dimensional infinite square well
given by
 0 if 0  x  a and 0  y  a ,
V  x, y   
otherwise

(a) [5] What are the exact eigenstate wave functions and energies for this
potential? Which is/are the ground state(s)? Which is/are the first excited
state(s)?
The eigenstates are simply products of eigenstates for the infinite square well in
either direction. If we label the states as i, j , then the eigenstates are
2   ix    jx 
 ij  x, y   sin 
 sin 
.
a
 a   a 
The energy of these eigenstates is just the sum of the energies in each direction, so
Eij 
 22
2ma
2
i
2
 j2  .
The ground state is unique, 1,1 , while the first excited state is degenerate, since 1, 2
and 2,1 have exactly the same energy.
(b) [15] To this potential is added a small perturbation
x 
 y 
W  x, y   w cos 
 cos 

 a 
 a 
Find the ground state wave function to first order in w, and the ground state
energy to second order in w.
It is helpful to find the general matrix elements n, m W i, j , which is given by
4
  nx    my 
x 
  y    ix    jy 
dx  dy sin 
 sin 
 cos 
 cos 
 sin 
 sin 

2 
a 0 0
 a   a 
 a 
 a   a   a 
a
n, m W i, j  w
a
4
  nx    ix 
x 
  my    jy 
 y 
sin 
 sin 
 cos 
 dx  sin 
 sin 
 cos 
 dy
2 
a 0  a   a 
 a  0
 a   a 
 a 
4 a
a
 w 2   n ,i 1   i ,n 1  1,i  n   m, j 1   j ,m 1  1, j  m   14 w n i ,1 m  j ,1
a 4
4
a
a
w
Hence W connects only states to states that differ on both labels by exactly one. This
means that the ground state is connected only to the state 2, 2 , so the energy is, to
second order
E11  11  1,1 W 1,1 

 22
1,1   i , j
2
2
w4
 11  0 
1,1   2,2
 2  2 w2 ma 2
w2 16


.
 2  2 ma 2  4 2  2 ma 2 ma 2 48 2  2

ma 2

i , j 1,1
i, j W 1,1
The state vector is given by
 1,1  1,1 

i, j i, j W 1,1
1,1   i , j
i , j 1,1
w4
ma 2 w
 1,1 
2, 2  1,1 
2, 2 .
1,1   2,2
12 2  2
Written as a wave function, this means
2   x    y  maw
 2 x   2 y 
 1,1  x, y   sin   sin 
  2 2 sin 
 sin 
.
a
 a   a  6 
 a   a 
(c) [15] Find the first excited state wave functions to leading order, and the first
excited state energies to first order in w.
This part is harder, in that we have degenerate states, and hence need degenerate
perturbation theory, but also easier, because we are asked only for the state wave
functions to leading order, and the energies to first order. This means we need only work
out i, j  W i, j for the two states we are working with (four matrix elements in all).
Fortunately, we can quickly calculate all the matrix elements to give us the W matrix:
 1, 2 W 1, 2
W  
 2,1 W 1, 2
1, 2 W 2,1  w   0,1 0,1 1,11,1  w  0 1 
.
 

2,1 W 2,1  4  1,11,1  0,1 0,1  4  1 0 
We’ve encountered this matrix so often we instantly know the eigenstates and
eigenvalues:
 
1 1
1
  , with eigenvalues  4 w .

1
2 
The eigenvalues of W give us the first correction to the energies, so we have states
 
1
2
 1, 2
 2,1  , E 
5 2  2 w
 .
2ma 2 4
Written as wave functions, these states are, to leading order,
   x, y  
2    x   2 y 
 2 x    y  
sin 
 sin 
  sin 
 sin 
 .

a   a   a 
 a   a 
6. [15] A particle of mass m with wave number k is incoming in the +z direction,
and scatters from a weak potential of the form V  r   V0 e   r , where V0 and 
are positive constants. Find the Fourier transform of  d 3r V  r e iK r for arbitrary
K , and the differential and total cross-section for scattering from this potential
in the first Born approximation.
Since the potential is spherically symmetric, assume for the moment that K is in
the z-direction, so we have
 d r V  r e
3
 iK r
2
1
0
1
 V0  d 3r e   r iK r  V0  d  d cos 
   iK cos  
1
 4 V0 

2 iV0
K
1
3
d cos  


0
r 2 e   r iKr cos dr
2 V0
2 cos 1
   iK cos   cos 1
iK

 2 iV0  4  iK 
8 V0 
1
1



.

2
2
2
2
2 2
    iK     iK   K   2  K 2 


K


We now proceed directly to the differential cross-section, using the fact that
K 2  2k 2 1  cos   , where now  is the angle between the initial and final directions of
the particle:
2
d
m
 2 4
d  4 
2

 d r V  r e
3
 iK r
2


16m 2V02  2
m  8 V0  
 2 4

4
4     2  K 2 2 
4   2  K 2 


2
16m 2V02  2
 4   2  2k 2  2k 2 cos  
4
We then simply integrate this over all angles to get the total cross-section:
 
16m 2V02  2
d
d 
d
4

2
0
d  d cos    2  2k 2  2k 2 cos  
1
4
1


3 cos 1
32 m 2V02  2 1
16 m 2V02  2  1
1
2
2
2
2
2
cos
k
k







 cos 1 34k 2   6 2 2 3 
6k 2
4
   4k  

64 m 2V02  4k 2 16k 4 
1 2 

.
3 
3 4 

 4   2  4k 2  

7. [15] An electron is in a magnetic field B whose magnitude is constant but its
direction changes. In this problem we are concerned only with the spin of the
electron, so that at all times the Hamiltonian (approximating g = 2) is just
H  eB  S m . At early times, the magnetic field is given by B  B0 zˆ , but then it
rotates by 120, until it is given be B  B0

3
2

xˆ  12 zˆ at late times.
(a) [7] What are the eigenvalues and eigenstates of the spin Hamiltonian at early
times? At late times? You may find it helpful to look at the solutions to
problem 8.3.
At early times, the Hamiltonian is H  eB0 S z m , which has eigenstates  and
eigenvalues
E  
eB0 
.
2m
At late times, the Hamiltonian is
eB0
m
H

3
2
S x  12 S z

This is the operator S  cos  S z  sin  S x for   120 , as defined in problem 8.3, and it
has eigenstates
  cos  12     sin  12     cos  60    sin  60   
1
2
 
   sin  12     cos  12      sin  60    cos  60    
3
2
 ,
  12  .
3
2
Its eigenvalues are exactly the same as before, E   eB0  2m .
(b) [8] Suppose the electron starts in the more positive energy state at t = 0.
What is the probability of it ending in each of the late eigenstates at late
times, if the magnetic field is rotated slowly? Now redo the problem if the
magnetic field is rotated quickly.
If you rotate the magnetic field slowly, we can use the adiabatic approximation,
which means that more positive eigenstate will remain the more positive eigenstate, so
P    
 1,
P    
  0.
If you rotate the magnetic field quickly, then the sudden approximation is appropriate,
and we have
P    

 
2
P    

 
2
         ,
          

2
3
2
1
2
3
2
1 2
2
2
1
2
1
4
3
2
2
3
4
.
8. [25] A system consists of a superposition of photon states with different numbers
of photons, all having the same wave number q  qxˆ and the same polarization
ε q  yˆ ,
i n 1
  N  2 n, q ,  ,
n 1 n

where N is a real normalization constant. Some useful sums appear at the end of
this problem; however, some of the sums involved cannot be done analytically.
Do as much as you can on each part, but don’t let it bother you if you can’t quite
complete all the sums
(a) [8] Find the normalization factor N.
We need to have
   i m 1
m, q, 
1     N 
2
 m 1 m
2
N

2


 i 
90
2
i
m2n2
m 1 n 1
N
m 1 n 1
   i n 1

   2 n, q,  
  n 1 n

m, q,  n, q,   N

2


m 1 n 1
 i 
m 1 n 1
i
m2n2
1 4 2

N ,
4
90
n 1 n

 mn  N 2 
.
(b) [17] Find the expectation value of the electric field  E  r   at an
arbitrary place r.
We simply insert the electric or magnetic field operators. The only terms that will
contribute from E  r  will be those where k matches q, since adding or subtracting some
other photon cannot give non-zero results. We have
E r  

90
4
90
4


m 1
 i 
m 1
cq
2 0V

 i 
m 1
cq
 4 yˆ
2 0V

cq
 4 yˆ
2 0V

90
cq
 4 yˆ
2 0V

90
k,

90

m, q, 
m2
 n 1
k
i
 ik r †
ik r
*
i  εk e ak  εk e ak  2 n, q, 
2 0V
n 1 n


m 1 n 1


m 1 n 1



m 1 n 1
m, q,   ε q e aq  ε e
iqr
m2


m 1
 i 
m 1 n
i
2
mn
 i 
m 1 n
i
2
mn
 i 
2
2
m 1 n
2
mn
i
2
*
q
 iqr
  in

a    2 n, q ,  
 n 1 n

†
q
m, q,   eiqx aq  e  iqx aq†   n, q,  

m, q,  eiqx n n  1, q,   e  iqx n  1 n  1, q, 
e
iqx

n m, n 1  e  iqx n  1 m ,n 1 .

It is easiest to continue by using the Kronecker deltas to do the n sum on the first term,
but to do the m sum on the second term, so that
E r 
cq
 4 yˆ

2 0V
90


90
4

1
cq iqx  iqx 
e  e 

3/ 2 2
2 0V
n
n 1  n  1
yˆ
180
4
 iqx   i m 1 i m 1 m  1  iqx   i n i n n  1 
e 
e 

2
2
2
n 1
 n  1 n 2 
 m 1 m  m  1


n 1
1
 n  1
3/2
n
2
yˆ
cq
cos  qx  .
2 0V
I know of no way to do the remaining sum analytically, but it is perfectly reasonable to
do it numerically.
> evalf(180*sum(1/n^2/(n+1)^(3/2),n=1..infinity)/Pi^4);
E  r   0.791859yˆ

1
 ,

n 1 n

1
   3  1.202 ,

3
n 1 n


 1
n 1
n
n 1


n 1
 1
n3
n
cq
cos  qx  .
2 0V
1 2
 ln 2 ,  2 
,
6
n 1 n



n 1
 1
n 1
n2
1 4
   3  0.802 ,  4 
,
90
n 1 n

3
4



n 1
2
12
,
 1
n4
n 1

7 4
.
720
Note: Problem 9 will be treated more like a homework problem than a final exam
problem. Although you are not allowed to ask friends how to solve the problem, you are
welcome and encouraged to seek my help in doing it.
9. [10] An electron is trapped in a 3D harmonic oscillator potential,
H  P 2 2m  12 m02 R 2 . It is in the quantum state nx , n y , nz  2,1, 0 .
(a) [6] Calculate every non-vanishing matrix element of the form
nx , ny , nz R 2,1, 0 where the final state is lower in energy than the initial
state.
The three coordinate operators can be written in the form Ri   2m0  ai  ai†  ,
so
R 210   2m0  xˆ  ax  ax†   yˆ  a y  a †y   zˆ  az  az†   210
  2m0  xˆ 2 110  xˆ 3 310  yˆ 200  yˆ 2 220  zˆ 211  .
We are interested in decay, which implies we want a lower energy state, so there are only
two possible final states that work
110 R 210  xˆ

m0
200 R 210  yˆ
and

2m0
(b) [4] Calculate the decay rate   210  nx , ny , nz  for this decay in the dipole
approximation for every possible final state.
The decay rate is given by   I  F   43 IF3 rFI
2
c 2 . In each case, the
frequency difference is simply 0 , so we have
  2,1, 0  1,1, 0  
403 
4 02
402 
2 02
,
2,1,
0
2,
0,
0
.







3c 2 m0
3mc 2
3c 2 2m0
3mc 2
Definite Integrals:
For each of the following formulas, m, n and p are assumed to be positive integers.

 Ax n
 e x dx 
0
  n  1
if Re  A   0,
An 1

 Ax
n
 e x dx 
2
0
  n  1  n !,   12    ,   32  
a
1
2
  n21 
n 1
2A 2
 ,   52  
if Re  A   0,
3
4
.
 2a n if n is odd,
  nx 
 dx  0
if n is even,
a 

 sin 
0
a
  nx 
 dx  0
a 
 cos 
0
  nx    mx 
  nx 
  mx 
0 cos  a  cos  a  dx  0 sin  a  sin  a  dx  12 a nm
a
a
2na   n 2  m 2  if n  m is odd
  nx 
  mx 
sin
cos

dx
0  a   a  
0
if n  m is even

a
a
a
  nx    mx 
  px 
 sin 
 cos 
 dx   n ,m  p   m, n  p   p , n  m 
4
a   a 
 a 
 sin 
0
a
a
  mx 
  px 
  nx 
 cos 
 dx   n ,m  p   m ,n  p   p , n  m 
 cos 
4
a 
 a 
 a 
 cos 
0