Let $t=e^{iz}$ and $|u\rangle=U|b\rangle=p|\xi\rangle+q|\phi\rangle$. The first factor in $G$ is
$$
(1-t)|u\rangle\langle u|-I.
$$
Since $R(\xi,z)|u\rangle=pt|\xi\rangle+q|\phi\rangle$, the coefficient of $|\phi\rangle$ in $G|u\rangle$ is
$$
q\left[(1-t)(p^2t+q^2)-1\right].
$$
It vanishes when
$$
(1-t)(p^2t+q^2)=1,
$$
or, using $p^2+q^2=1$,
$$
t+t^{-1}=\frac{p^2-q^2}{p^2}.
$$
A unit-modulus solution $t=e^{iz}$ exists exactly when the right-hand side lies in $[-2,2]$. The upper bound is automatic, while the lower bound is
$$
q^2\leq3p^2.
$$
Thus exact preparation by one application of $G$ is possible precisely when
$$
\boxed{p\geq\frac12}
\qquad\text{or equivalently}\qquad
\boxed{q\leq\sqrt3\,p}.
$$
One may choose $z$ so that $\cos z=(p^2-q^2)/(2p^2)$. Then $G|u\rangle$ has no $|\phi\rangle$ component and, by unitarity, equals $|\xi\rangle$ up to phase. This is a <phase-matched amplitude amplification> step.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 324 / 2 / iv / Solution

Phase-matched amplitude amplification replaces the two sign-flip reflections by selective phase rotations. Choosing their phases from the known initial overlap can rotate the state exactly onto the target even when an integer number of ordinary Grover rotations would overshoot it.


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