 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/10c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

∣0 x ⟩

∣0 x ⟩

and

∣1 \overset{x}{ˉ} ⟩

∣1 x ⟩

. Hence

CX ∣ β_{z x} ⟩ = \frac{∣0 ⟩ + ( - 1 ) ^{z} ∣1 ⟩}{2} \otimes ∣ x ⟩ = H ∣ z ⟩ \otimes ∣ x ⟩ .

Since the Hadamard gate satisfies

H^{2} = I

(H \otimes I) CX ∣ β_{z x} ⟩ = ∣ z x ⟩ .

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