 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/10e/b/i/solution

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

Since

∣ ψ_{2} ⟩ = - sin θ ∣0 ⟩ + cos θ ∣1 ⟩,

application of

X

gives

cos θ ∣0 ⟩ - sin θ ∣1 ⟩

, and subsequent application of

Z

gives

cos θ ∣0 ⟩ + sin θ ∣1 ⟩ = ∣ ψ_{1} ⟩ .

Thus $ZX ∣ ψ_{2} ⟩ = ∣ ψ_{1} ⟩$ .

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