 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/33d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

[S_{z}, S_{\pm}] = \pm ℏ S_{\pm}, [S^{2}, S_{\pm}] = 0.

Thus

S_{\pm} ∣ s, σ ⟩

σ \pm 1

. Its squared norm follows from

S_{\mp} S_{\pm} = S^{2} - S_{z}^{2} \mp ℏ S_{z} :

∥ S_{\pm} ∣ s, σ ⟩ ∥^{2} = ℏ^{2} [s (s + 1) - σ (σ \pm 1)] .

Choosing the conventional positive phases gives the spin ladder operator formula

S_{\pm} ∣ s, σ ⟩ = ℏ s (s + 1) - σ (σ \pm 1) ∣ s, σ \pm 1 ⟩ .

For

s = 1

, in the ordered basis

∣1, 1 ⟩, ∣1, 0 ⟩, ∣1, - 1 ⟩

S_{+} = 2 ℏ 000100010, S_{-} = S_{+}^{†} .

Since

S_{x} = (S_{+} + S_{-}) /2

S_{x} = \frac{ℏ}{2} 010101010 .

The incident state is the spin coherent state obtained by rotating

∣1, 1 ⟩

through

θ

about the

y

axis. Its amplitudes in the

S_{z}

⟨ 1, 1∣ n, + ⟩ = cos^{2} \frac{θ}{2}, ⟨ 1, 0∣ n, + ⟩ = \frac{s i n θ}{2}, ⟨ 1, - 1∣ n, + ⟩ = sin^{2} \frac{θ}{2},

up to an irrelevant phase convention. The Born rule therefore gives

\frac{N _{+}}{N _{-}} = \frac{cos ^{4} ( θ /2 )}{sin ^{4} ( θ /2 )} = cot^{4} \frac{θ}{2} .

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