Intermediate QMI
Midterm 3 Solutions
Problem 1
Given the raising and lowering operators
S
`
±
=
S
`
x
±
Â
S
`
y
and that
S
`
±
†
=
S
`
°
.
ü
Part a
We have
B
S
`
z
,
S
`
±
F
=
B
S
`
z
,
S
`
x
±
Â
S
`
y
F
=
B
S
`
z
,
S
`
x
F
= Â
—
S
`
y
±
Â
B
S
`
z
,
S
`
y
F
= Â
—
S
`
x
=
±
—
S
`
x
+ Â
—
S
`
y
=
±
—
J
S
`
x
±
Â
S
`
y
N
therefore
B
S
`
z
,
S
`
±
F
=
±
—
S
`
±
ü
Part b
We have
S
`
z
B
S
`
±
j
,
m
^F
=
:B
S
`
z
,
S
`
±
F
+
S
`
±
S
`
z
>
j
,
m
^
¬
B
S
`
z
,
S
`
±
F
=
±
—
S
`
±
S
`
z
j
,
m
^
=
—
m
j
,
m
^
=
B
±
—
S
`
±
+
—
m S
`
±
F
j
,
m
^
S
`
z
B
S
`
±
j
,
m
^F
=
—
H
m
±
1
L
S
`
±
j
,
m
^
Therefore,
m
£
=
m
+
1 for
S
`
+
j
,
m
^
and
m
£
=
m

1 for
S
`

j
,
m
^
.
ü
Part c
We have
Z
j
,
m
S
`
°
S
`
±
j
,
m
^
=
Z
j
,
m
S
`
±
†
S
`
±
j
,
m
^
=
S
`
±
j
,
m
^
2
Computing another way, we take
S
`
°
S
`
±
=
J
S
`
x
°
Â
S
`
y
N J
S
`
x
±
Â
S
`
y
N
=
S
`
x
2
±
Â
S
`
x
S
`
y
°
Â
S
`
y
S
`
x
+
S
`
y
2
=
J
S
`
x
2
+
S
`
y
2
N
=
S
`
2

S
`
z
2
±
Â
B
S
`
x
,
S
`
y
F
Â
—
S
`
z
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