 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-3/33d/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

z

∣ ↑ ⟩ = (1, 0)^{T}

and

∣ ↓ ⟩ = (0, 1)^{T}

s_{i} = (ℏ/2) σ_{i}

[σ_{i}, σ_{j}] = 2 i ϵ_{ijk} σ_{k}

, so

[s_{i}, s_{j}] = i ℏ ϵ_{ijk} s_{k}, s^{2} = \frac{ℏ ^{2}}{4} i \sum σ_{i}^{2} = \frac{3 ℏ ^{2}}{4} I .

The second identity is the correct

s (s + 1) ℏ^{2}

s = 1/2

s_{x} ∣ ↑ ⟩ = \frac{ℏ}{2} ∣ ↓ ⟩, s_{x} ∣ ↓ ⟩ = \frac{ℏ}{2} ∣ ↑ ⟩, s_{y} ∣ ↑ ⟩ = \frac{i ℏ}{2} ∣ ↓ ⟩, s_{y} ∣ ↓ ⟩ = - \frac{i ℏ}{2} ∣ ↑ ⟩ .

Thus

s_{x}

and

s_{y}

flip the spin-

z

basis states, with different complex phases.

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