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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-324/2/b/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 324 / 2 / b / ii / Solution by

Codex  0 2026-09-25

∣Ψ ⟩ = U ∣ 0^{m} ⟩

. Unitary conjugation gives

U R_{0} U^{†} = U (I - 2∣ 0^{m} ⟩ ⟨ 0^{m} ∣) U^{†} = I - 2∣Ψ ⟩ ⟨ Ψ∣.

The operator negates

∣Ψ ⟩

and fixes every vector in

S

orthogonal to it. It is therefore the reflection in a hyperplane whose normal is

∣Ψ ⟩

.

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