PHY4604 Fall 2007
R. D. Field
Department of Physics
Page 1 of 7
Exam 1 Solutions
PHY 4604 Exam 1 Solutions
(Total Points = 100)
Problem 1 (20 points):
Circle true or false for following (2 point each).
(a) (True or False)
One of the “breakthroughs” that lead to quantum mechanics was the idea of
associating differential operators with the dynamical variables.
(b) (True or False)
Solutions of Schrödinger’s equation of the form
)
(
)
(
)
,
(
t
x
t
x
φ
ψ
=
Ψ
correspond to states with definite energy
E
.
(c) (True or False)
Solutions of Schrödinger’s equation of the form
)
(
)
(
)
,
(
t
x
t
x
=
Ψ
correspond to states in which the probability density
2

)
,
(

)
,
(
t
x
t
x
Ψ
=
ρ
is independent of time.
(d) (True or False)
The wave function
Ψ
(x,t)
must vanish in a region of infinite potential.
(e) (True or False)
If
P
op
is the parity operator,
P
op
ψ
(x) =
ψ
(x)
, then
P
op
2
= P
op
.
(f) (True or False)
In quantum mechanics particles can enter the “classically forbidden” region
where V
0
> E (
i.e.
KE < 0).
(g) (True or False)
The operator
A
op
A
↑
op
is hermitian.
(h) (True or False)
If
A
op
and
B
op
are hermitian then
A
op
B
op
is also hermitian.
(i) (True or False)
The commutator operator
[(p
x
)
op
,(x
2
)
op
]
is equal to
x
i
h
2
−
.
(j) (True or False)
In positionspace the commutator operator
[(p
x
)
op
,sin(kx)]
is equal to
)
cos(
kx
i
h
−
.
Problem 2 (40 points):
Consider an electron with mass
m
e
confined
within an infinite square well defined by
V(x) = 0
for
0 < x < L
,
V(x) = +
∞
otherwise.
(a) (2 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
, and normalize the eigenfunctions
V = +infinity
V = +infinity
Infinite Square Well
0
L
x
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentPHY4604 Fall 2007
R. D. Field
Department of Physics
Page 2 of 7
Exam 1 Solutions
so that the probability of finding the electron somewhere in the box is one.
Answer:
)
/
sin(
2
)
(
L
x
n
L
x
n
π
ψ
=
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.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '07
 FIELDS
 mechanics, Sin, dx, Department of Physics, R. D. Field

Click to edit the document details