 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-3/10d/a/solution

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

Apply a Pauli X gate and then a Hadamard gate to the third qubit, and apply Hadamard gates to the first two. This prepares

(\frac{1}{2} \sum_{x \in B_{2}} ∣ x ⟩) ∣ - ⟩, ∣ - ⟩ = \frac{∣0 ⟩ - ∣1 ⟩}{2} .

U_{f} (\frac{1}{2} \sum_{x} ∣ x ⟩ ∣ - ⟩) = \frac{1}{2} \sum_{x} (- 1)^{f (x)} ∣ x ⟩ ∣ - ⟩ = ∣ f ⟩ ∣ - ⟩ .

The answer qubit is unentangled and may be ignored. If all three qubits must be restored to the displayed register form, applying

H

and then

X

to the answer qubit maps

∣ - ⟩

back to

∣0 ⟩

. Thus the first register is exactly $∣ f ⟩$ after one use of $U_{f}$ .

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