 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/35c/a/solution

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

The spin components satisfy

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

s = 0, \frac{1}{2}, 1, \frac{3}{2}, \dots

∣ s, σ ⟩

, where

σ = - s, - s + 1, \dots, s .

On this basis,

S^{2} ∣ s, σ ⟩ = ℏ^{2} s (s + 1) ∣ s, σ ⟩, S_{z} ∣ s, σ ⟩ = ℏ σ ∣ s, σ ⟩ .

Each Cartesian component is Hermitian,

S_{i}^{†} = S_{i},

and

S_{+}^{†} = S_{-}

. Also

tr S_{z} = ℏ \sum_{σ = - s}^{s} σ = 0.

Rotational equivalence, or the off-diagonal form of

S_{x}, S_{y}

in this basis, gives

tr S_{x} = o p er a t or nam e t r S_{y} = o p er a t or nam e t r S_{z} = 0.

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