CHAPTER
8
1.
The solutions are of the form
ψ
n
1
n
2
n
3
(
x
,
y
,
z
)
=
u
n
1
(
x
)
u
n
2
(
y
)
u
n
3
(
z
)
where
u
n
(
x
)
=
2
a
sin
n
π
x
a
,and so on. The eigenvalues are
E
=
E
n
1
+
E
n
2
+
E
n
3
=
h
2
π
2
2
m
a
2
(
n
1
2
+
n
2
2
+
n
3
2
)
2.
(a) The lowest energy state corresponds to the lowest values of the integers
{
n
1
,
n
2
,
n
3
}, that is, {1,1,1)Thus
E
ground
=
h
2
π
2
2
ma
2
×
3
In units of
h
2
π
2
2
ma
2
the energies are
{1,1,1}
3
nondegenerate)
{1,1,2},(1,2,1},(2,1,1}
6
(triple degeneracy)
{1,2,2},{2,1,2}.{2,2,1}
9
(triple degeneracy)
{3,1,1},{1,3,1},{1,1,3}
11
(triple degeneracy)
{2,2,2}
12
(nondegenrate)
{1,2,3},{1,3,2},{2,1,3},{2,3,1},{3,1,2},{3,2,1}
14 (6-fold degenerate)
{2,2,3},{2,3,2},{3,2,2}
17
(triple degenerate)
{1,1,4},{1,4,1},{4,1,1}
18
(triple degenerate)
{1,3,3},{3,1,3},{3,3,1}
19
(triple degenerate)
{1,2,4},{1,4,2},{2,1,4},{2,4,1},{4,1,2},{4,2,1}
21 (6-fold degenerate)
3.
The problem breaks up into three separate, here identical systems. We know that the
energy for a one-dimensional oscillator takes the values
h
ϖ
(
n
+
1
/2
)
, so that here the
energy eigenvalues are
E
=
h
ϖ
(
n
1
+
n
2
+
n
3
+
3
/2
)
The ground state energy correspons to the
n
values all zero. It is
3
2
h
ϖ
.
4.
The energy eigenvalues in terms
of
h
ϖ
with the corresponding integers are
(0,0,0)
3/2
degeneracy 1
(0,0,1) etc
5/2
3
(0,1,1) (0,0,2) etc
7/2
6
(1,1,1),(0,0,3),(0,1,2) etc
9/2
10
(1,1,2),(0,0,4),(0,2,2),(0,1,3)
11/2
15
(0,0,5),(0,1,4),(0,2,3)(1,2,2)