 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/35b/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

[S_{i}, S_{j}] = i ℏ ε_{ijk} S_{k}, [S^{2}, S_{i}] = 0.

r_{i} = \frac{2 ⟨ S _{i} ⟩ _{ψ}}{ℏ} .

Purity gives

r_{x}^{2} + r_{y}^{2} + r_{z}^{2} = 1

, so the supplied sign determines

r_{y} = sgn ⟨ S_{y} ⟩_{ψ} 1 - r_{x}^{2} - r_{z}^{2} .

For

r_{z} \neq = - 1

, choose the overall phase so that the upper component is nonnegative. The normalized state is

∣ ψ ⟩ = \frac{1 + r _{z}}{2} ∣ + ⟩_{z} + \frac{r _{x} + i r _{y}}{2 ( 1 + r _{z} )} ∣ - ⟩_{z} .

r_{z} = - 1

, it is simply

∣ - ⟩_{z}

up to phase.

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