 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-49/2/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

The Pauli Z gate exchanges the two Hadamard basis states:

Z ∣ + ⟩ = ∣ - ⟩

and

Z ∣ - ⟩ = ∣ + ⟩

. Multiplying the displayed gate by the Pauli X gate gives

U (θ) X = ∣ + ⟩ ⟨ 1∣ + e^{- i θ} ∣ - ⟩ ⟨ 0∣.

On the other hand,

e^{- i θ} Z U (- θ) = e^{- i θ} Z (∣ + ⟩ ⟨ 0∣ + e^{i θ} ∣ - ⟩ ⟨ 1∣) = e^{- i θ} ∣ - ⟩ ⟨ 0∣ + ∣ + ⟩ ⟨ 1∣.

The two operators agree, proving

U (θ) X = e^{- i θ} Z U (- θ) .

Also

U (θ) = H diag (1, e^{- i θ}) = J (- θ)

in the J gate convention. Keeping this angle sign explicit avoids confusing the two conventions for equatorial qubit measurements.

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