1
Physics 316
Solution for homework 11
Spring 2005
Cornell University
I.
EXERCISE 1
A.
Consider the two dimensional wave equation for the vibration of a rectangular membrane,
∂
2
z
∂x
2
+
∂
2
z
∂y
2
=
1
v
2
∂
2
z
∂t
2
.
(1.1)
The membrane has a boundary at
x
= 0
,
y
= 0
and
x
=
L
x
,
y
=
L
y
at which it does not oscillate, i.e.
z
= 0
. Show
that
z
=
A
sin(
k
x
x
) sin(
k
y
y
) cos(
ωt
)
(1.2)
can be a solution for specific
k
x
,
k
y
, and
ω
.
To verify that the given form of
z
can be a solution, we substitute it into the given wave equation. After canceling
common terms, we find that
z
is a solution if
k
2
x
+
k
2
y
=
ω
2
v
2
.
(1.3)
The constraints from the fact that
z
= 0 on all the boundaries imply that
k
x
=
n
x
π
L
x
and
k
y
=
n
y
π
L
y
, with
n
x
, n
y
= 1
,
2
, ...
.
This leads to a constraint on
ω
:
ω
n
x
,n
y
=
πv
s
n
2
x
L
2
x
+
n
2
y
L
2
y
.
(1.4)
B.
Show for
L
x
=
L
y
that there can be different solutions
z
(
x, y, t
)
for the same frequency
ω
. This is called
degeneracy
.
For
L
x
=
L
y
=
L
, we can rewrite
ω
n
x
,n
y
as
ω
n
x
,n
y
=
πv
L
q
n
2
x
+
n
2
y
.
(1.5)
We can see that
ω
ab
=
ω
ba
, but the corresponding
z
(
x, y, t
) are different:
z
ab
=
A
sin
‡
aπ
L
x
·
sin
bπ
L
y
¶
cos(
ωt
)
z
ba
=
A
sin
bπ
L
x
¶
sin
‡
aπ
L
y
·
cos(
ωt
)
(1.6)
II.
EXERCISE 2
Show that the spherical finite well potential,
V
(
r
) =
V
0
H
(
r

R
)
with
H
(
x
) =
if
(
x>
0
,
1
,
0
)
, does not have a
spherically symmetric bound state if
V
0
<
¯
h
2
π
2
8
mR
2
.
We consider the spherical finite potential well
V
(
r
) =
V
0
H
(
r

R
).
In general, spherically symmetric states are
found with the radial part of the Schr¨
odinger equation,

¯
h
2
2
m
u
00
(
r
) +
V
(
r
)
u
(
r
) =
Eu
(
r
)
,
(2.1)
2
the actual wavefunction being given by Φ(
r
) =
u
(
r
)
r
.
We can solve (2.1) in the two domains
r < R
and
r > R
independantly and then make sure that
u
(
r
) and its first derivative are continuous at
r
=
R
:

¯
h
2
2
m
u
00
1
(
r
) =
Eu
1
(
r
)
(2.2)

¯
h
2
2
m
u
00
2
(
r
) =

(
V
0

E
)
u
2
(
r
)
.
(2.3)
Both solutions are already known:
u
1
(
r
) =
A
1
sin(
k
1
r
) +
B
1
cos(
k
1
r
)
(2.4)
u
2
(
r
) =
A
2
e
k
2
r
+
B
2
e

k
2
r
,
(2.5)
with
k
1
=
√
2
mE/
¯
h
and
k
2
=
p
2
m
(
V
0

E
)
/
¯
h
.
In order to have
u
1
(
r
)
r
well defined at the origin, we must have
u
1
(0) = 0, that is
B
1
= 0. Since we want Φ to be normalizable, we also need to put
A
2
= 0.
We now apply the continuity conditions:
u
1
(
R
) =
u
2
(
R
)
⇒
A
1
sin(
k
1
R
) =
B
2
e

k
2
R
(2.6)
u
0
1
(
R
) =
u
0
2
(
R
)
⇒
k
1
A
1
cos(
k
1
R
) =

k
2
B
2
e

k
2
R
.
(2.7)
Therefore, dividing (2.7) by (2.6), one gets
k
1
cot(
k
1
R
) =

k
2
.
(2.8)
0
π
2π
3π
1
A
B
C
k R
1
FIG. 1: Graphical solution of equation (2.10) for three different values of
k
V
.
A:
k
V
R <
π
2
(too small for any possible solution
for
k
1
R
).
B:
k
V
R
=
π
2
; this is the limiting value in order to get a solution.