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PHYSICS 4455 — QUANTUM MECHANICS
Problem Set 5 — Solutions.
1.
Onedimensional problems.
Liboff Problem 3.23
Since the particle is free,
V
(
x
)=
0for
x
≤
0. So, the stationary Schrödinger equation for
this problem is
−
2
2
m
d
2
dx
2
ψ(
x
)=
E
ψ(
x
)
ψ
(
x
)+
2
m
2
E
ψ(
x
)=
0
The “rigid wall” at
x
=
0 translates into the boundary condition
ψ(
0
)=
0. Note that there is
no information here about a boundary condition at the other end. Following the discussion
of “wiggly” eigenfunctions in class, we should demand
E
>
0 in order to ensure that
ψ
and
ψ
have opposite signs. As a result, the solution will have the form
ψ(
x
)=
Ae
ikx
+
Be
−
ikx
with two (real) integration constants
A
and
B
. The boundary condition at
x
=
0 implies
A
+
B
=
0sothat
ψ(
x
)=
A
(
e
ikx
−
e
−
ikx
) =
Ā
sin
kx
Inserting
ψ(
x
)
into the Schrödinger equation results in
E
=
2
k
2
2
m
with no further conditions on
E
.Inparticular,
E
in this case is not quantized, i.e., the
allowed eigenvalues (the “spectrum”) are continuous. Also, the wave function is somewhat
pathological in that it cannot be normalized.
The timedependent solution
ψ(
x
,
t
)
is given by
ψ(
x
,
t
)=
Ae
i
ω
t
(
e
ikx
−
e
−
ikx
)
with
ω=
E
=
2
k
2
2
m
Liboff Problem 4.1
You can do this in two ways: First, you can of course write down the Schrödinger
equation for the onedimensional box on the interval
(−
a
/
2,
+
a
/
2
)
and solve it.
Alternatively, you can start from the eigenfunctions for the box on the interval
(
0,
a
)
,
namely,
ϕ
n
(
x
)=
2
a
sin
(
k
n
x
)
with
k
n
=
n
π/
a
,
n
=
1,2,.
.. andreplace
x
by
x
+
a
/
2. Note that the new function, called
ϕ
n
(
x
)
and defined by
ϕ
n
(
x
)≡ϕ
n
(
x
+
a
2
)=
2
a
sin
k
n
x
+
k
n
a
2
vanishes at
x
=−
a
/
2andat
x
=+
a
/
2, i.e., it satisfies the boundary conditions. Obviously, it
also satisfies the Schrödinger equation inside the box, namely,
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2
2
m
d
2
dx
2
ϕ
n
(
x
)=
2
k
n
2
2
m
ϕ
n
(
x
)=
E
n
ϕ
n
(
x
)
Using basic trigonometry, we can write
ϕ
n
(
x
)=
2
a
sin
(
k
n
x
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 Spring '09
 Mucciolo
 mechanics

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