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PHY4604 Fall 2007
R. D. Field
Department of Physics
Page 1 of 16
Exam 2 Solutions
PHY 4604 Final Exam
Take Home Exam (Total Points = 100)
Problem 1 (20 points):
(4 points each).
(1a)
An electron is in a onedimensional infinite square well with zero potential energy in the
interior and infinite potential energy at the walls and size
L
has a ground state energy
E
0
.
If the
size of the square well is decreased to
L/2
what is the new ground state energy? (
circle the
correct answer
)
(a) 0
(b) E
0
/2
(c) E
0
/4
(d) 2E
0
(e) 4E
0
Solution:
The groud state energy is
0
2
2
2
2
/
2
2
2
0
4
)
2
/
(
2
2
E
L
m
mL
E
L
L
=
⎯
⎯→
⎯
=
→
π
h
h
.
(1b)
Which of the following angular momentum states has the smallest lower bound on the
product (
Δ
L
x
)×(
Δ
L
y
)? (
circle the correct answer
)
(a)
l
= 2 m = 2
(b)
l
= 2 m = 1
(c)
l
= 2 m = 0
(d)
l
= 2 m = 1
(e)
l
= 2 m = 2
Solution:
We know that from the uncertainty principle that
>
<
≥
Δ
Δ
]
,
[
B
A
i
B
A
and hence
>
<
=
>
<
≥
Δ
Δ
z
y
x
y
x
L
L
L
i
L
L
h
]
,
[.
The <L
z
> is zero in the state
l
= 2 m = 0 which is the smallest.
(1c)
A “glue ball” is a particle constructed entirely from gluons?
Gluons carry intrinsic spin 1
(
i.e.
they are vector particles).
Which of the following correspond to possible “glue ball” spins?
(
circle all the correct answers
)
(a) 0
(b) 1
(c) 2
(d) 3/2
(e) 1/2
Solution:
3 × 3 = 5 + 3 + 1 but cannot form 3/2 or ½ from any number of spin 1 particles even if
you add in orbital angular momentum.
(1d)
Consider the angle,
θ
, between the angular momentum vector and the zaxis for the
following states.
For which state is the angle
θ
the smallest? (
circle all the correct answers
)
(a)
l
= 2 m = 2
(b)
l
= 2 m = 1
(c)
s
= 3/2 m = 3/2
(d)
s
= 1/2 m = 1/2
(e)
l
= 3 m = 2
Solution:
We know that
)
1
(
cos
+
=
l
l
m
θ
or
)
1
(
cos
+
=
s
s
m
and hence
State
cos
θ
l
=2 m = 2
0.816
l
= 2 m = 1
0.408
s =3/2 m = 3/2
0.775
s = ½ m = 1/2
0.577
l
= 3 m = 2
0.577
(1e)
What is
4 × 3 × 2 × 1
in SU(2)?
Solution:
4 × 3 × 2 × 1 = 4 × (4 + 2) = 4 × 4 + 4 × 2 = 7 + 5 + 3 + 1 +5 + 3 = 7 + 5 + 5 +3 + 3 + 1
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View Full Document PHY4604 Fall 2007
R. D. Field
Department of Physics
Page 2 of 16
Exam 2 Solutions
Problem 2 (40 points):
Consider the case of two
noninteracting
particles
both with mass
m
in a onedimensional infinite square well given by
V(x) = 0
for
0 < x < L,
and
V(x) =
∞
.
For one particle we know that the
stationary states of Schrödinger’s equation are given by
h
/
)
(
)
,
(
t
iE
n
n
n
e
x
t
x
−
=
Ψ
ψ
and
0
2
E
n
E
n
=
, and n is a positive integer and
2
2
2
0
2
mL
E
h
π
=
and where
)
/
sin(
2
)
(
L
x
n
L
x
n
=
.
For two (
noninteracting
)
particles we look for a solution of the form
h
h
/
2
1
/
2
1
2
1
)
(
)
(
)
,
(
)
,
,
(
iEt
iEt
e
x
x
e
x
x
t
x
x
−
−
=
=
Ψ
with
E
m
p
m
p
x
x
=
+
2
)
(
2
)
(
2
2
2
1
.
(a) (4 points):
Show that
)
(
)
(
)
,
(
2
1
2
1
x
x
x
x
β
α
αβ
=
, is a solution to the two particle non
interaction Schrödinger equations, where
α
and
β
are positive integers and
ψ
α
(x)
and
ψ
β
(x) are
the one particle stationary state solutions.
Show that the allowed energy levels are given by
0
2
2
)
(
E
E
+
=
.
This solution corresponds to the case where the two particles are
distinguishable.
Define the following three probabilities.
Let
LL
P
be the probability of finding
both particles in the left 1/2 of the box as follows:
∫∫
=
2
/
0
2
/
0
2
1
2
2
1

)
,
(

LL
LL
dx
dx
x
x
P
.
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This note was uploaded on 05/29/2011 for the course PHY 4064 taught by Professor Fields during the Spring '07 term at University of Florida.
 Spring '07
 FIELDS
 mechanics, Energy, Potential Energy

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