Past exam of the atomic and Optical Physics course of the University of Cambridge / 2012 / Paper 1 / 2 / b / Solution

The first Hadamard gate maps the prepared superposition $|+⟩=(|1⟩+|2⟩)/\sqrt{2}$ to $|1⟩$. The phase gate leaves that state unchanged, and the last Hadamard gate returns it to $|+⟩$. ThereforeThis is a fully polarized pure state, not an unpolarized ensemble. Its Bloch vector lies along the rotation axis of the combined operation. There is no Ramsey interferometry fringe as $\alpha$ varies, because the state during phase accumulation occupies only one basis level. Quantum coherence in the final state alone does not guarantee phase sensitivity to the specified sequence.