 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/15d/d/solution

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

\frac{1}{2} (∣0 ⟩ ∣ ψ ⟩ + e^{i θ} ∣1 ⟩ ∣ ψ ⟩) .

The final Hadamard gate gives

\frac{1}{2} [(1 + e^{i θ}) ∣0 ⟩ + (1 - e^{i θ}) ∣1 ⟩] ∣ ψ ⟩ .

Hence the Hadamard test returns one with probability

Pr (1) = \frac{∣1 - e ^{i θ} ∣ ^{2}}{4} = \frac{1 - cos θ}{2} = sin^{2} \frac{θ}{2} .

U ∣ ψ ⟩ = ∣ ψ ⟩

, then

θ = 0

and the outcome is zero with certainty. If

U ∣ ψ ⟩ = - ∣ ψ ⟩

, then

θ = π

2 π

and the outcome is one with certainty. The single measurement therefore distinguishes the two promised cases exactly.

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