 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/35a/c/solution

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

Using the states from part (a),

∣ χ ⟩ = \frac{1}{2} [- s ∣ ↑↑ ⟩ + c ∣ ↑↓ ⟩ - c ∣ ↓↑ ⟩ - s ∣ ↓↓ ⟩] .

The Born rule therefore gives

outcome (+ ℏ/2, + ℏ/2) (+ ℏ/2, - ℏ/2) (- ℏ/2, + ℏ/2) (- ℏ/2, - ℏ/2) probability \frac{1}{2} sin^{2} (θ /2) \frac{1}{2} cos^{2} (θ /2) \frac{1}{2} cos^{2} (θ /2) \frac{1}{2} sin^{2} (θ /2) .

After the equal-sign outcomes the state is respectively

∣ ↑↑ ⟩

∣ ↓↓ ⟩

S^{2}

2 ℏ^{2} .

The opposite-sign product states are each a nontrivial linear combination of the singlet and the

m = 0

triplet, so they are not eigenstates of

S^{2}

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