= Solution
The first <Hadamard gate> maps the prepared <superposition> $|+\rangle=(|1\rangle+|2\rangle)/\sqrt2$ to $|1\rangle$. The <phase gate> leaves that state unchanged, and the last <Hadamard gate> returns it to $|+\rangle$. Therefore
$$
\boxed{|\Psi_f\rangle=|+\rangle,\qquad \vec P_f=(1,0,0),\qquad p_2=\tfrac12.}
$$
This is a fully polarized <pure state>, not an unpolarized ensemble. Its <Bloch vector> lies along the <rotation> axis of the combined operation. There is \b[no <Ramsey interferometry> fringe as $\alpha$ varies], because the state during phase accumulation occupies only one basis level. <Quantum coherence> in the \i[final] state alone does not guarantee phase sensitivity to the specified sequence.
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