Homework 1: # 4.1, 4.2, 4.3, 4.4, 4.8, 4.10, 4.12
Michael GoodJan 20, 2005
Problem 4.1Calculate the three lowest energy levels, together with their degeneracies, forthe following systems (assume equal mass distinguishable particles):a. Three noninteracting spin
12
particles in a box of length
L
.b. Four noninteracting spin
12
particles in a box of length
L
.Solution:In a box we have three dimensions. So the energy for one particle is
E
1
n
x
,n
y
,n
z
= (
n
2
x
+
n
2
y
+
n
2
z
)
π
2
¯
h
2
2
mL
2
If the particles are distinguishable, the composite wave functions are products, as explained in an example by Griﬃths, section 5.1. Therefore the energyof three particles are
E
T
=
E
1
+
E
2
+
E
3
This is
E
T
=
3
i,j
n
2
ij
π
2
¯
h
2
2
mL
2
where the index
i
represents the particle, and
j
represents the dimension. Notethat there are 9 distinct
n
’s. The
n
’s are nonzero integers, so the lowest energyis where all the
n
values are equal to 1. That is just 1
2
∗
9 = 9
E
1
=9
π
2
¯
h
2
2
mL
2
The second lowest energy is where only one of the
n
values is 2 and the restare still 1. That is 2
2
+ 8 = 12. So
E
2
=12
π
2
¯
h
2
2
mL
2
1
The third lowest energy is where two of the
n
’s have value 2 and rest arestill 1. That is 2
2
+ 2
2
+ 7 = 15. So
E
3
=15
π
2
¯
h
2
2
mL
2
The degeneracy is found by multiplying the number of spatial functions bythe number of spin functions. There are 8 diﬀerent spin functions.
 − −−
,
 − −
+
,
 −
+
−
,

+
−−
,

+ +
−
,
 −
++
,

+
−
+
,

+ ++
or 2
3
= 8 combinations. The spatial functions are found by looking at thepossibilities for the
n
’s. For
E
1
there is only one possibility, that is, with all the
n
values equal to 1. For
E
2
there are 9 possibilities, where each of the 9 valuesof
n
could be 2 where the rest are 1. For
E
3
its a little more diﬃcult, becausetwo of the
n
’s may have the value 2 so we use the combination formula ’9 choose2’.
N
!(
N
−
n
)!
n
!=9!(9
−
2)!2!= 36possibilities for
E
3
. Multiplying the spatial possibilities times the spin possibilities gives the degeneracy.
E
1
→
1
∗
8 = 8
E
2
→
9
∗
8 = 72
E
3
→
36
∗
8 = 288For four particles we have 4
∗
3 = 12 distinct
n
’s. So for the lowest energy,all of them are 1:
E
1
=12
π
2
¯
h
2
2
mL
2
For the second lowest energy we have 2
2
+ 11 = 15:
E
2
=15
π
2
¯
h
2
2
mL
2
For the third lowest energy we have 2
2
+ 2
2
+ 10 = 18:
E
3
=18
π
2
¯
h
2
2
mL
2
There are 2
4
= 16 spin possibilities. There is only 1 spatial possibility for
E
1
. There are 12 spatial possibilities for
E
2
. And ’12 choose 2’ possibilities for
E
3
12!(12
−
2)!2!= 662
The degeneracies are therefore
E
1
→
1
∗
16 = 16
E
2
→
12
∗
16 = 192
E
3
→
66
∗
16 = 1056Problem 4.2Let
T
d
denote the translation operator (displacement vector
d
);
D
(ˆ
n,φ
), therotation operator (ˆ
n
and
φ
are the axis and angle of rotation, respectively);and
π
the parity operator. Which, if any, of the following pairs commute? Why?
ã T
d
and
T
d
(
d
and
d
in diﬀerent directions).
ã D
(ˆ
n,φ
) and
D
(ˆ
n
,φ
) (ˆ
n
and ˆ
n
in diﬀerent directions).
ã T
d
and
π
.
ã D
(ˆ
n,φ
) and
π
.Solution:a.
T
d
and
T
d
commute as can be seen graphically by translating a vector intwo directions in diﬀerent orders or algebraically(where
r
and
d
are vectors) by
T
d
φ
(
r
) =
φ
(
r
+
d
)
→ T
d
T
d
φ
(
r
) =
φ
(
r
+
d
+
d
)
T
d
φ
(
r
) =
φ
(
r
+
d
)
→ T
d
T
d
φ
(
r
) =
φ
(
r
+
d
+
d
)Therefore[
T
d
,
T
d
]
φ
(
r
) = 0and[
T
d
,
T
d
] = 0because
φ
(
r
) can be any function of vector
r
.b.
D
(ˆ
n,φ
) and
D
(ˆ
n
,φ
) do not commute. As we know that rotations alongdiﬀerent axis do not commute and
D
acts on
φ
to cause a rotation. This can bedone by rotating a book around two diﬀerent axis.c.
T
d
and
π
do not commute. This can be done graphically by translatinga vector, then mirroring it, and see if that point matches up if you mirror thevector then translate it. Algebraically,3
πφ
(
r
) =
φ
(
−
r
)
→ T
d
πφ
(
r
) =
φ
(
−
r
+
d
)
T
d
φ
(
r
) =
φ
(
r
+
d
)
→
π
T
d
φ
(
r
) =
φ
(
−
r
−
d
)Therefore[
T
d
,π
]
φ
(
r
)
= 0[
T
d
,π
]
= 0d.
D
(ˆ
n,φ
) and
π
commute. Graphically by rotating then mirroring, andchecking to see if that matches up with mirroring then rotating. Algebraically,
D
φ
(
r
) =
φ
(
x
)
→
π
D
φ
(
r
) =
φ
(
−
x
)
πφ
(
r
) =
φ
(
−
r
)
→ D
πφ
(
r
) =
D
φ
(
−
r
) =
φ
(
−
x
)Therefore[
D
,π
]
φ
(
r
) = 0[
D
,π
] = 0Problem 4.3A quantummechanical state Ψ is known to be simultaneous eigenstate of twoHermitian operators
A
and
B
which anticommute,
AB
+
BA
= 0What can you say about the eigenvalues of
A
and
B
for stat Ψ? Illustrate yourpoint using the parity operator (which can be chosen to satisfy
π
=
π
−
1
=
π
†
)and the momentum operator.Solution:Taking a look at the eigenvalues we see
Aψ
=
aψ Bψ
=
bψB
(
Aψ
) =
B
(
aψ
) =
abψA
(
Bψ
) =
A
(
bψ
) =
baψ
And we have
ABψ
=
−
BAψ
4