 Solution

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

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

∣ x_{0} ⟩

. Since that amplitude in

∣ ψ ⟩

1/2

I_{x_{0}} ∣ ψ ⟩ = ∣ ψ ⟩ - ∣ x_{0} ⟩ .

Also,

∣ ψ ⟩ ⟨ ψ ∣ x_{0} ⟩ = \frac{1}{2} ∣ ψ ⟩ .

A ∣ ψ ⟩ = - ∣ ψ ⟩

and

A ∣ x_{0} ⟩ = ∣ x_{0} ⟩ - ∣ ψ ⟩

, so

A I_{x_{0}} ∣ ψ ⟩ = A (∣ ψ ⟩ - ∣ x_{0} ⟩) = - ∣ x_{0} ⟩ .

The global minus sign has no observable effect. Prepare

∣ ψ ⟩ = H^{\otimes 2} ∣00 ⟩

, query

I_{x_{0}}

once, apply the fixed operator

A

x_{0}

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