Physics 742 – Graduate Quantum Mechanics 2
Solutions to Chapter 15
3. [15] A particle of mass m is initially in the ground state of a harmonic oscillator with
frequency . At time t = 0, a perturbation is suddenly turned on of the
form W  t   AXe  t . At late times ( t   ), the quantum state is measured again.
(a) [10] Calculate, to second order in A, the amplitude S n 0 that it ends up in the state
n , for all n (most of them will be zero).
We will need matrix elements of the form n W  t  m  A n X m e  t , which are given
by
Ae  t n X m  Ae  t


n  a  a †  m  Ae  t
2m
2m


m n ,m 1  n m ,n 1 .
To second order, then, our amplitudes are
S00   00 

1
1
0 W  t  0 dt 
2

i 0
 i 
t

n
0 W  t  n ei0 nt dt  n W  t   0 ein 0t dt 
0
0
 1
A2 

2 
  2m
 1
A2    i 
1 
 1
1
,



4m 2 2    i   2   i 
4m    2   2 
S10  10 
A



2


t
0
0
  t  i t
  t  it 
 e e dt  e e dt   1 

A2
e  t e  it e  i t  1 dt

2m     i  0
2

1
1
1 W  t  0 ei10t dt 
2

i 0
 i 


n
0
t
1 W  t  n ei1nt dt  n W  t   0 ein 0t  dt 
0


A
A

eit t dt 
,

i 2m 0
i 2m    i 
S20   20 


1
1
2 W  t  0 ei20t dt 
2

i 0
 i 
A2 

2 
  2m



2


0
t


n
0
2eit t dt  eit t  dt   
0
t
2 W  t  n ei2 nt dt  n W  t   0 ein 0t  dt 
0

A2
eit t eit t  1 dt

2m  i    0

1
1 
A2


,


2
2m    i   2    i    i 
2 2m    i 
S n 0  0, n  2 .
A
2
(b) [5] Calculate, to at least second order, the probability that it ends up in the state n .
Check that the sum of the probabilities is 1, to second order in A.
We now simply square the amplitudes we found previously, and we find

A2    i   
A2    i  
A2
2



P  0  0   S00  1 
1
1
,



2
2
2
2
2m    2   2 
 4m         4m       
A2
2
P  0  1  S10 
,
2m   2   2 
P  0  2   S20 
A4
2
8 m     
2
2
2
2

2 2
.
To second order in A, the first two expressions add to 1 and the third is zero.
4. [10] A particle of mass m is in the ground state 1 of an infinite square well with
allowed region 0  X  a . To this potential is added a harmonic perturbation
W  t   AX cos t  , where A is small.
(a) [6] Calculate the transition rate  1  n  for a transition to another level. Don’t let
the presence of a delta function bother you. What angular frequency  is necessary
to make the transition occur to level n = 2?
We first rewrite this perturbation in the form W  t   12 AX  eit  e  it  . Our interaction
is thus W  12 AX , and the matrix elements we require are W1n  12 A n X 1 .
The eigenstates and energies of the unperturbed square well are
 n  x 
2
 2n2 2
  nx 
E
sin 
,
.

n

2ma 2
a
 a 
Maple is happy to do the integrals for us. We find the matrix elements we want are
A
A 2
2 Aan 
n
  nx    x 
n X 1    x sin 
1   1  .
 sin 
 dx   2 2
2 

2
2 a0
 a   a 
  n  1
a
Wn1 
> assume(n::integer);
> integrate(A/a*sin(Pi*n*x/a)*sin(Pi*x/a)*x,x=0..a);
The rate then is given by
2
n
8a 2 A2 n 2 1   1    2  2

2
2

 
Wn1   En  E1    
n 2  1    .
 1  n  


4
2

 2ma

 3   n 2  1
For the transition to n = 2, we need the delta function to vanish, which requires a frequency of
  3 2  2ma 2 .
(b) [4] Now, instead of keeping a constant frequency, the frequency is tuned
continuously, so that at t = 0 the frequency is 0, and it rises linearly so that at t = T it
has increased to the value  T   2 2  ma 2 . The tuning is so slow that at any given
time, we may treat it as a harmonic source. Argue that only the n = 2 state can
become populated (to leading order in A). Calculate the probability of a transition
using P 1  2     1  2  dt .
T
0
The maximum frequency achieved is 4/3 times larger than required to make this
transition, enough for the 1 to 2 transition, but not enough for anything higher. We therefore
need only consider the transition to n = 2. The frequency as a function of time is
 t  
2 2 t
ma 2T
To find the total transition probability, we then simply integrate the transition rate
P 1  2   
T
0
T
 3 2  2 2 2  2t  128a 2 A2 ma 2T 64ma 4 A2T
128a 2 A2

 1  2  dt 




81 3  0  2ma 2 ma 2T 
81 3  2 2  2
81 5 3