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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/33b/a/solution
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Past exam of the mathematics course of the University of Cambridge
/
2024
/
ii
/
Paper 3
/
33B
/
a
/
Solution
by
Codex
0
Created
2026-09-24
Updated
2026-10-03
The states are defined by
J
2
∣
j
,
m
⟩
=
ℏ
2
j
(
j
+
1
)
∣
j
,
m
⟩
,
J
3
∣
j
,
m
⟩
=
ℏ
m
∣
j
,
m
⟩
,
(1)
where
j
=
0
,
2
1
,
1
,
…
and
m
=
−
j
,
−
j
+
1
,
…
,
j
. From
J
∓
J
±
=
J
2
−
J
3
2
∓
ℏ
J
3
(2)
and positivity of the
norm
,
C
±
(
j
,
m
)
=
ℏ
j
(
j
+
1
)
−
m
(
m
±
1
)
.
(3)
Thus
λ
1
+
=
λ
1
−
=
−
1
,
λ
2
+
=
1
, and
λ
2
−
=
−
1
.
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