 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/10c/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

∣ - ⟩ = (∣0 ⟩ - ∣1 ⟩) / 2

. Applying the oracle with this answer qubit produces phase kickback:

U_{f} ∣ x ⟩ ∣ - ⟩ = (- 1)^{f (x)} ∣ x ⟩ ∣ - ⟩ .

Inside

A

, the

X

gates on both search qubits conjugate the phase oracle for

11

. Hence

A ∣ x_{1} x_{2} ⟩ ∣ - ⟩ = (- 1)^{[x_{1} x_{2} = 00]} ∣ x_{1} x_{2} ⟩ ∣ - ⟩ = I_{0} ∣ x_{1} x_{2} ⟩ ∣ - ⟩,

where

I_{0} = I - 2∣00 ⟩ ⟨ 00∣

The first oracle similarly acts on the search register as

I_{x_{0}} = I - 2∣ x_{0} ⟩ ⟨ x_{0} ∣

. Reading the remaining gates from right to left, the search-register operation after its initial Hadamards is therefore

- Q = H^{\otimes 2} I_{0} H^{\otimes 2} I_{x_{0}},

while the answer qubit remains in

∣ - ⟩

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