 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/10d/a/solution

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

Let

P = ∣ ψ ⟩ ⟨ ψ ∣

. Since

∣ ψ ⟩

is normalized,

P^{2} = P = P^{†}

A = I - 2 P

is Hermitian and

A^{†} A = A^{2} = I - 4 P + 4 P^{2} = I .

Therefore $A$ is unitary.

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