 Solution

ID: past-exam-of-the-atomic-and-optical-physics-course-of-the-university-of-cambridge/2012/paper-1/2/a/solution

Following the pure state through the first two gates gives

(∣1 ⟩ + e^{i α} ∣2 ⟩) / 2

. After the last Hadamard gate it is

∣ Ψ_{f} ⟩ = e^{i α /2} [cos (α /2) ∣1 ⟩ - i sin (α /2) ∣2 ⟩] .

Using the qubit polarization convention defined above,

P_{f} = (0, - sin α, cos α), p_{2} = sin^{2} (α /2) .

The output remains pure and has unit Bloch vector length. The population oscillation is the Ramsey interferometry fringe: the first gate supplies quantum coherence, and the last turns the acquired relative quantum phase into a measurable population difference.

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