Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 58 / 4 / c / Solution

At this stage $A$ is an arbitrary unitary operator; do not yet assume it prepares the displayed uniform state from zero. Put $|a⟩=A|0⟩$, $p=e^{i\varphi }$ and $t=e^{i\theta }$. The phase on the zero basis state can be written using a rank-one orthogonal projection as $S_0^{\varphi }=I+(p-1)|0⟩⟨0|$, henceThe marked-state phase sends $|{\psi }_1⟩$ to $t|{\psi }_1⟩$ and leaves $|{\psi }_0⟩$ unchanged. ThereforeandThis is the complete general action. Unless $|a⟩$ lies in their span, that two-dimensional space need not be invariant. The PDF uses $S_f^{\theta }$ in $Q$; the TeX conversion's replacement by $S_0^{\theta }$ is a transcription error and would describe a different operation.