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PHY3063 Spring 2006
R. D. Field
Final Exam Solutions
Page 1 of 9
April 25, 2006
PHY 3063 Final Exam Solutions
Problem 1 (35 points):
Consider an particle with mass
m
confined within an infinite square well defined by
V(x) = 0
for
0 < x < L
,
V(x) = +
∞
otherwise.
(a) (5 points):
Using Schrödinger’s equation calculate the
allowed stationary state eigenfunctions
ψ
n
(x)
, where the complete
wavefunctions are given by
h
/
)
(
)
,
(
t
iE
n
n
n
e
x
t
x
−
=
Ψ
ψ
.
Normalize
the eigenfunctions so that the probability of finding the particle somewhere in the box is one.
Answer:
)
/
sin(
2
)
(
L
x
n
L
x
n
π
=
Solution:
For the region outside of
0 < x < L
0
)
(
=
x
and inside the region
)
(
)
(
2
2
2
2
x
E
x
d
x
d
m
e
=
−
h
or
)
(
)
(
2
2
2
x
k
x
d
x
d
−
=
with
e
m
k
E
2
2
2
h
=
The most general solution is of the form
)
cos(
)
sin(
)
(
kx
B
kx
A
x
+
=
.
The boundary condition at x = 0 gives
0
)
0
(
=
=
B
and the boundary condition at x = L gives
0
)
sin(
)
(
=
=
kL
A
L
which implies that
n
kL
=
with n = 1, 2, 3,… Thus,
)
/
sin(
)
(
L
x
n
A
x
n
=
with
2
2
2
2
2
mL
n
E
n
h
=
.
The normalization is arrived at by requiring that
2
4
)
2
sin(
2
)
(
sin
)
/
(
sin
1
)
(
)
(
2
0
2
0
2
2
0
2
2
LA
y
y
n
LA
dy
y
n
LA
dx
L
x
n
A
dx
x
x
n
n
L
n
n
=
−
=
=
=
=
∫
∫
∫
+∞
∞
−
∗
Thus,
L
A
/
2
=
.
These states are called stationary because the probability density and all the
expectation values are independent of time.
(b) (5 points):
Show that the allowed energy levels of the system are,
E
n
= E
0
n
2
,
where
)
2
/(
2
2
2
0
mL
E
h
=
is the ground state energy and n = 1, 2, 3, ….
Why is
n = 0
excluded as
a possible energy level?
Solution:
We see from above that
0
2
2
2
2
2
2
E
n
mL
n
E
n
=
=
h
with
2
2
2
0
1
2
mL
E
E
h
=
=
.
The state with
n = 0
correstions to
ψ
0
(x) = 0
which is not normalizable.
(c) (10 points):
Consider the operator,
O = (x)
op
(p
x
)
op
(
i.e.
the product of the position operator
times the momentum operator).
Is the operator
O
hermitian?
Calculate the expectation value of
the operator
Q
for the n
th
stationary state (
i.e.
calculate <
ψ
n
O
ψ
n
>).
Answer:
O
is not hermitian and
2
/
h
i
xp
n
x
=
>
<
.
Solution:
We see that
x
x
x
x
xp
x
p
x
p
xp
O
≠
=
=
=
↑
↑
↑
↑
)
(
)
(
since (x)
op
and (p
x
)
op
do not commute.
Also,
V = +infinity
V = +infinity
Infinite Square Well
0
L
x
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View Full DocumentPHY3063 Spring 2006
R. D. Field
Final Exam Solutions
Page 2 of 9
April 25, 2006
()
2
4
2
cos
sin
4
1
)
sin(
2
2
2
)
/
2
sin(
2
2
)
/
cos(
)
/
sin(
2
)
(
)
(
)
(
)
)(
(
2
0
2
0
2
0
0
h
h
h
h
h
h
h
i
n
n
i
y
y
y
n
i
dy
y
y
n
L
L
n
L
i
dx
L
x
n
x
L
n
L
i
dx
L
x
n
L
x
n
x
L
n
L
i
dx
dx
x
d
x
x
i
dx
x
xp
x
xp
n
n
L
L
n
n
n
op
x
n
n
x
=
−
−
=
−
−
=
−
=
=
−
=
−
−
=
=
>
<
∫
∫
∫
∫
∫
+∞
∞
−
∗
+∞
∞
−
∗
π
ψ
where I used
θ
2
sin
cos
sin
2
1
=
and I let y = 2n
π
/L.
(d) (15 points):
Suppose the particle in this infinite square well has an initial wave function at
t = 0
given by
)
/
(
sin
)
0
,
(
2
L
x
A
x
=
Ψ
.
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 Spring '07
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