 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/15e/c/i/solution

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

N = 2^{m}

, the Hadamards first give

(H^{\otimes m} \otimes I) ∣0 ⟩^{\otimes m} ∣ ψ ⟩ = \frac{1}{N} \sum_{k = 0}^{N - 1} ∣ k ⟩ ∣ ψ ⟩ .

Since

U^{k} ∣ ψ ⟩ = e^{2 πik θ} ∣ ψ ⟩

and

θ = j / N

∣Φ ⟩ = \frac{1}{N} k = 0 \sum N - 1 e^{2 πijk / N} ∣ k ⟩ ∣ ψ ⟩ .

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