 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/15c/b/i/solution

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

Starting from

∣ 0^{n} ⟩ ∣ 0^{n} ⟩

, the first layer of Hadamard gates gives

2^{- n /2} \sum_{x \in B_{n}} ∣ x ⟩ ∣ 0^{n} ⟩ .

2^{- n /2} \sum_{x \in B_{n}} ∣ x ⟩ ∣ f (x)⟩ .

H^{\otimes n} ∣ x ⟩ = 2^{- n /2} \sum_{y \in B_{n}} (- 1)^{x \cdot y} ∣ y ⟩,

the final state is

2^{- n} x, y \in B_{n} \sum (- 1)^{x \cdot y} ∣ y ⟩ ∣ f (x)⟩ .

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