Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 324 / 2 / b

In the ordered orthonormal basisof $\mathcal{S}$, the Pauli Z gate acts asIt fixes the line $|g{⟩}^{\perp }\cap \mathcal{S}$ and negates $|g⟩$, so it is the reflection in a hyperplane orthogonal to the good state.

Set $|\Psi ⟩=U|0^m⟩$. Unitary conjugation givesThe operator negates $|\Psi ⟩$ and fixes every vector in $\mathcal{S}$ orthogonal to it. It is therefore the reflection in a hyperplane whose normal is $|\Psi ⟩$.

Put $\sqrt{p}=\sin \theta$ and $\sqrt{1-p}=\cos \theta$, where $0<\theta <\pi /2$. In the ordered basis $(|b⟩,|g⟩)$, the product of the two reflections isThe leading minus sign is a physically irrelevant global phase. The remaining rotation matrix rotates the good-bad plane through $2\theta$, whereEquivalently, one iterate changes the initial angle $\theta$ to $3\theta$.

Attach a quantum ancilla prepared with amplitudeon $|1⟩$; this is possible because $p>1/4$. Mark a state as good only when both the original success qubit and this ancilla equal one. The enlarged state's good probability is $p{c}^2=1/4$, so its good amplitude is $1/2=\sin (\pi /6)$. One amplitude amplification iteration rotates the angle from $\pi /6$ to $3\pi /6=\pi /2$. It therefore prepares $|{\psi }_1⟩$ exactly, after which the flag and ancillary qubits may be discarded. This is an instance of exact amplitude amplification.