 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/15c/d/i/solution

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

After the first Hadamard gate, the joint state is

\frac{1}{2} (∣0 ⟩ + ∣1 ⟩) ∣ ψ ⟩ ∣ ϕ ⟩ .

The controlled SWAP gate then gives

\frac{1}{2} (∣0 ⟩ ∣ ψ ⟩ ∣ ϕ ⟩ + ∣1 ⟩ ∣ ϕ ⟩ ∣ ψ ⟩) .

\frac{1}{2} [∣0 ⟩ (∣ ψ ⟩ ∣ ϕ ⟩ + ∣ ϕ ⟩ ∣ ψ ⟩) + ∣1 ⟩ (∣ ψ ⟩ ∣ ϕ ⟩ - ∣ ϕ ⟩ ∣ ψ ⟩)] .

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