 Solution

ID: past-exam-of-the-introduction-to-quantum-information-course-of-the-university-of-oxford/2015/3/a/solution

Use the conventional phase gate

P_{φ} = diag (1, e^{i φ})

0, 1

. The single-qubit Hadamard gate and the successive quantum states are

H = \frac{1}{2} (11 1 - 1), ∣0 ⟩ H \frac{∣0 ⟩ + ∣1 ⟩}{2} P_{φ} \frac{∣0 ⟩ + e ^{i φ} ∣1 ⟩}{2} .

The final Hadamard gate recombines the two components, giving

∣ ψ_{out} ⟩ = \frac{1 + e ^{i φ}}{2} ∣0 ⟩ + \frac{1 - e ^{i φ}}{2} ∣1 ⟩ = e^{i φ /2} (cos \frac{φ}{2} ∣0 ⟩ - i sin \frac{φ}{2} ∣1 ⟩) .

The leading factor in the last expression is a global phase. The Born rule therefore yields

P_{0} (φ) = \frac{1 + e ^{i φ}}{2}^{2} = \frac{1 + cos φ}{2} = cos^{2} \frac{φ}{2} .

φ = 0

the output is certainly zero, whereas at

φ = π

it is certainly one. The phase gate changes no intermediate outcome probabilities; the final Hadamard gate converts a relative quantum phase into an observable quantum interference fringe.

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