Since $A|x\rangle=(-1)^{f(x)}|x\rangle$, the <spectral decomposition> gives
$$
e^{iA}|x\rangle=e^{i(-1)^{f(x)}}|x\rangle.
$$
Replace the middle gate of the previous <compute-phase-uncompute construction> by
$$
e^{iZ}=\operatorname{diag}(e^i,e^{-i})=R_z(-2),
$$
a one-qubit <rotation about the z-axis>. Then
$$
\boxed{C^\dagger(I\otimes e^{iZ})C\,|x,0\rangle
=e^{i(-1)^{f(x)}}|x,0\rangle.}
$$
The circuit uses the gates of $C$, their inverses, and one fixed rotation; it returns the <quantum ancilla> to $|0\rangle$. The same construction implements $e^{itA}$ with middle gate $e^{itZ}$, an instance of <simulation of a computable diagonal Hamiltonian>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 324 / 3 / b / ii / Solution

{title2=$R_z(\theta)$}

The one-qubit <rotation gate> is $R_z(\theta)=\operatorname{diag}(e^{-i\theta/2},e^{i\theta/2})$. It implements the two relative phases needed in <simulation of a computable diagonal Hamiltonian>.

Rotation about the z-axis

Let $A|x\rangle=a(x)|x\rangle$, where a <reversible circuit> $C$ computes $a(x)$ into a clean register. Then applying $e^{-ita(x)}$ as a phase on that register and performing <uncomputation> implements $e^{-itA}$ on the data register. For $a(x)=(-1)^{f(x)}$, a Boolean output qubit suffices: $C^\dagger(I\otimes e^{-itZ})C$ acts as $e^{-itA}$ when that qubit starts in $|0\rangle$. This exact construction avoids a <Lie-Trotter product formula> because the eigenvalue is computed directly.

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 Incoming links: Simulation of a computable diagonal Hamiltonian