PHY4604 Fall 2007
Problem Set 7 Solutions
Department of Physics
Page 1 of 14
PHY 4604 Problem Set #7 Solutions
(Total Points = 115)
Problem 1 (30 points):
The state of a two particle system is described by the wave function
)
,
,
(
2
1
t
r
r
r
r
Ψ
.
The time evolution is given by Schrödinger’s equation
t
i
H
∂
Ψ
∂
=
Ψ
h
with
)
,
,
(
2
2
2
1
2
2
2
2
2
1
1
2
t
r
r
V
m
m
H
r
r
h
h
+
∇
−
∇
−
=
where
2
2
2
2
2
2
2
k
k
k
k
z
y
x
∂
∂
+
∂
∂
+
∂
∂
=
∇
for k = 1,2.
For timeindependent potentials, we obtain a complete
set of solutions of the form
h
r
r
r
r
/
2
1
2
1
)
,
(
)
,
,
(
iEt
e
r
r
t
r
r
−
=
Ψ
ψ
, where
)
,
(
2
1
r
r
r
r
satisfies the time
independent Schrödinger equation
E
r
r
V
m
m
=
+
∇
−
∇
−
)
,
(
2
2
2
1
2
2
2
2
2
1
1
2
r
r
h
h
.
Typically the interaction potential depends only on the vector
2
1
r
r
r
r
r
r
−
=
(
i.e.
the separation
between the two particles).
Suppose
)
(
)
,
(
2
1
r
V
r
r
V
r
r
r
=
and we change variables from
1
r
r
and
2
r
r
to
2
1
r
r
r
r
r
r
−
=
and
)
/(
)
(
2
1
2
2
1
1
m
m
r
m
r
m
R
+
+
=
r
r
r
(
i.e.
the centerofmass vector).
(a) (5 points)
Show that
r
m
R
r
r
r
r
)
/
(
1
1
μ
+
=
and
r
m
R
r
r
r
r
)
/
(
2
2
−
=
, where
)
/(
2
1
2
1
m
m
m
m
+
=
is
the “reduced mass”.
Solution:
We see that
r
m
r
m
m
r
r
m
r
m
r
m
r
m
R
m
m
r
r
r
r
r
r
r
r
2
1
2
1
1
2
1
1
2
2
1
1
2
1
)
(
)
(
)
(
−
+
=
−
+
=
+
=
+
and hence
r
m
R
r
m
m
m
R
r
r
r
r
r
r
1
2
1
2
1
)
(
+
=
+
+
=
.
Also,
r
m
r
m
m
r
m
r
r
m
r
m
r
m
R
m
m
r
r
r
r
r
r
r
r
1
2
2
1
2
2
2
1
2
2
1
1
2
1
)
(
)
(
)
(
+
+
=
+
+
=
+
=
+
and hence
r
m
R
r
m
m
m
R
r
r
r
r
r
r
2
2
1
1
2
)
(
−
=
+
−
=
.
(b) (5 points)
Show that
r
R
m
∇
+
∇
=
∇
r
r
r
)
/
(
2
1
and
r
R
m
∇
−
∇
=
∇
r
r
r
)
/
(
1
2
, where
)
/(
2
1
2
1
m
m
m
m
+
=
is the “reduced mass”.
Solution:
Let
)
,
,
(
Z
Y
X
R
=
r
and
)
,
,
(
z
y
x
r
=
r
then
)
/(
)
(
2
1
2
2
1
1
m
m
x
m
x
m
X
+
+
=
and
2
1
x
x
x
−
=
and we see that
x
r
x
R
x
m
m
m
x
X
m
m
m
x
x
x
X
x
X
x
)
(
)
(
)
(
2
1
1
2
1
1
1
1
1
1
∇
+
∇
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∂
∂
+
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
=
∇
.
Also,
)
/(
)
(
2
1
2
2
1
1
m
m
y
m
y
m
Y
+
+
=
and
2
1
y
y
y
−
=
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Problem Set 7 Solutions
Department of Physics
Page 2 of 14
and we see that
y
r
y
R
y
m
m
m
y
Y
m
m
m
y
y
y
Y
y
Y
y
)
(
)
(
)
(
2
1
1
2
1
1
1
1
1
1
∇
+
∇
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∂
∂
+
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
=
∇
Also,
)
/(
)
(
2
1
2
2
1
1
m
m
z
m
z
m
Z
+
+
=
and
2
1
z
z
z
−
=
and we see that
z
r
z
R
z
m
m
m
z
Z
m
m
m
z
z
z
Z
z
Z
z
)
(
)
(
)
(
2
1
1
2
1
1
1
1
1
1
∇
+
∇
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∂
∂
+
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
=
∇
.
Hence,
r
R
r
R
m
m
m
m
∇
+
∇
=
∇
+
∇
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∇
r
r
r
r
r
2
2
1
1
1
μ
.
Similarly,
x
r
x
R
x
m
m
m
x
X
m
m
m
x
x
x
X
x
X
x
)
(
)
(
)
(
2
1
2
2
1
2
2
2
2
2
∇
−
∇
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∂
∂
−
∂
∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
=
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
=
∇
.
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 Spring '07
 FIELDS
 mechanics, Atom, Photon, X1, Excited state, ground state

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