 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-324/3/b/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Since

A ∣ x ⟩ = (- 1)^{f (x)} ∣ x ⟩

e^{i A} ∣ x ⟩ = e^{i (- 1)^{f (x)}} ∣ x ⟩ .

Replace the middle gate of the previous compute-phase-uncompute construction by

e^{i Z} = diag (e^{i}, e^{- i}) = R_{z} (- 2),

C^{†} (I \otimes e^{i Z}) C ∣ x, 0 ⟩ = e^{i (- 1)^{f (x)}} ∣ x, 0 ⟩ .

The circuit uses the gates of

C

, their inverses, and one fixed rotation; it returns the quantum ancilla to

∣0 ⟩

. The same construction implements

e^{i t A}

with middle gate

e^{i tZ}

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