PHY4604 Fall 2007
Problem Set 5 Solutions
Department of Physics
Page 1 of 9
PHY 4604 Problem Set #5 Solutions
Problem 1 (20 points):
Use separation of variables in
Cartesian
coordinates to solve the infinite
cubical well (or “particle in a box”):
⎩
⎨
⎧
∞
+
=
0
)
,
,
(
z
y
x
V
otherwise
L
z
L
y
L
x
<
<
<
<
<
<
0
,
0
,
0
(a) (10 points)
Find the stationary states and the corresponding energies.
Answer:
)
/
sin(
)
/
sin(
)
/
sin(
8
)
,
,
(
3
L
z
n
L
y
n
L
x
n
L
z
y
x
z
y
x
n
n
n
z
y
x
π
ψ
=
and
)
(
2
2
2
2
2
2
2
z
y
x
n
n
n
mL
E
+
+
=
h
Solution:
In this case the V(x,y,z) = V(x)+V(y)+V(z) and we look for stationary state solutions
of the form
)
(
)
(
)
(
)
,
,
(
z
Z
y
Y
x
X
z
y
x
=
and Schrödinger’s equation becomes
E
Z
z
V
dz
Z
d
m
Z
Y
y
V
dy
Y
d
m
Y
X
x
V
dx
X
d
m
X
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
)
(
2
1
)
(
2
1
)
(
2
1
2
2
2
2
2
2
2
2
2
h
h
h
and hence
x
E
X
x
V
dx
X
d
m
X
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
)
(
2
1
2
2
2
h
y
E
Y
y
V
dy
Y
d
m
Y
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
)
(
2
1
2
2
2
h
z
E
Z
z
V
dz
Z
d
m
Z
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
)
(
2
1
2
2
2
h
where the E
i
is constants and E
x
+E
y
+E
z
= E.
The problem breaks into three one dimensional
infinite wells and we know that the solution to the one dimensional problem is
)
/
sin(
2
)
(
L
x
n
L
x
X
x
n
x
=
with
2
2
2
2
2
x
n
n
mL
E
x
h
=
where n
x
= 1, 2, 3, …
)
/
sin(
2
)
(
L
y
n
L
y
Y
y
n
y
=
with
2
2
2
2
2
y
n
n
mL
E
y
h
=
where n
y
= 1, 2, 3, …
)
/
sin(
2
)
(
L
z
n
L
z
Z
z
n
z
=
with
2
2
2
2
2
z
n
n
mL
E
z
h
=
where n
z
= 1, 2, 3, …
Thus,
)
/
sin(
)
/
sin(
)
/
sin(
8
)
(
)
(
)
(
)
,
,
(
3
L
z
n
L
y
n
L
x
n
L
z
Z
y
Y
x
X
z
y
x
z
y
x
n
n
n
n
n
n
z
y
x
z
y
x
=
=
,
and
)
(
2
2
2
2
2
2
2
z
y
x
n
n
n
n
n
n
mL
E
E
E
E
z
y
x
+
+
=
+
+
=
h
.
(b) (5 points)
Call the energies E
1
, E
2
, E
3
, …, in order of increasing energy.
Find E
1
, E
2
, E
3
, E
4
,
E
5
, and E
6
.
Determine the degeneraries (
i.e.
the number of different states with the same
energy).