= Solution
At this stage $A$ is an arbitrary <unitary operator>; do not yet assume it prepares the displayed uniform state from zero. Put $|a\rangle=A|0\rangle$, $p=e^{i\phi}$ and $t=e^{i\theta}$. The phase on the zero basis state can be written using a rank-one <orthogonal projection> as $S_0^\phi=I+(p-1)|0\rangle\langle0|$, hence
$$
AS_0^\phi A^{-1}=I+(p-1)|a\rangle\langle a|.
$$
The marked-state phase sends $|\psi_1\rangle$ to $t|\psi_1\rangle$ and leaves $|\psi_0\rangle$ unchanged. Therefore
$$
\boxed{Q|\psi_1\rangle=-t\left[|\psi_1\rangle+(p-1)|a\rangle\langle a|\psi_1\rangle\right],}
$$
and
$$
\boxed{Q|\psi_0\rangle=-\left[|\psi_0\rangle+(p-1)|a\rangle\langle a|\psi_0\rangle\right].}
$$
This is the complete general action. Unless $|a\rangle$ 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.
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