 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/10c/a/solution

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

Immediately before the final Hadamard, the state is

\frac{1}{2} (∣0 ⟩ ∣ ψ ⟩ - i ∣1 ⟩ U ∣ ψ ⟩) .

The outcome-one component after that Hadamard is

(∣ ψ ⟩ + i U ∣ ψ ⟩) /2

. Thus, with

z = ⟨ ψ ∣ U ∣ ψ ⟩

p (1) = \frac{1}{4} ∥ ∣ ψ ⟩ + i U ∣ ψ ⟩ ∥^{2} = \frac{1}{2} (1 - Im z) .

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