Solution to Problem 4.4
4.4 [15] A particle of mass m lies in a one-dimensional infinite square well in the region
[0,a]. At t = 0, the wave function in the allowed region is
x   t  0   30 a 5  ax  x 2  .
(a) [8] Write the wave function in the form   t  0    n cn n
where n are the energy eigenstates. Check that the wave function is properly
normalized, both in the original coordinate basis, and in the new basis, either
analytically or numerically.
The integral we need is
2 30
  nx 
cn   n   t  0     x    x, t  0  dx 
ax  x 2  sin 
 dx
5 
a a 0
 a 
0
a
a
*
n
a
2
3
2 15  a
  nx  a
  nx  2a
  nx  
2
ax  x  cos 
 3 

  2 2  a  2 x  sin 
  3 3 cos 

a  n
 a   n
 a   n
 a 0

3 3

4 15
4 15 
n
  8 15  n
n






cos
1
1
1






  3 n3 
 
 3 n3 
0

if n odd,
if n even.
So our initial state is
  0    cn n 
n
8 15
n .
3 3
n
n odd

In the original basis, normalization is given by
a
 0   0   x  0
2
a
dx 
0
a
2
30
30
ax  x 2  dx  5   a 2 x 2  2ax3  x 4  dx
5 
a 0
a 0
a
30
 5  13 a 2 x3  12 ax 4  15 x5   1.
0
a
In the new basis, the normalization condition is
  0    0    cn* n
n
c
m
m
m   cn*cm n m   cn 
2
n
m
n
960

6
1
n
n odd
6

960
6
1.0014 
 1.0000.
I did the sum numerically, keeping only three terms, or you can do it analytically using Maple
> sum(1/(2*n+1)^6,n=0..infinity);
(b) [5] Find the expectation value of the Hamiltonain H for this wave function at t = 0,
both in the original coordinate basis and the eigenstate basis. Check that they are
the same, either analytically or numerically.
In the original basis, we simply use the Hamiltonian, but in the allowed region the
potential vanishes, so
1
 2 30
d2
30 2
2
2
P2 
ax

x
ax

x
dx


 dx 2 
 ma5
2m
2m a 5 0
a
H 

a
2
  ax  x  dx 
0
30 2 1 2 1 3 a
 ax  3 x  0
ma 5 2
5
.
ma 2
2
In the final basis, we have
H    0  H   0    cn* n H  cm m   Em cn*  cm n m   cn En
2
n

n odd
m
960   n
480
 4 2
6 6
2
n 2ma
 ma

2
2
2
2
1
n
n odd
4
n

m
n
480
4.9999
1.0147  
.
4
2 
 ma
ma 2
2
2
I did the final sum numerically, including ten terms. Obviously, it worked pretty well.
(c) [2] Write the state vector   t  at all times.
At arbitrary time, the wave function is
  t    cn n exp   iEn t   
n
 i 2 n 2 
8 15

exp
 3 3 n   2ma 2 t  .
n odd  n

