Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 15E (Quantum Information and Computation)

The eigenphase Hadamard-test probability isThusor $\theta =\pm (2\pi {)}^{-1}\arccos (2p_0-1)(\mathrm{mod}1)$. This test alone cannot distinguish $\theta$ from $1-\theta$.

The state entering the controlled swap has lower-register states
$|{\varphi }_1⟩$ and $V|{\varphi }_2⟩$. The swap test therefore gives

Writing $N=2^m$, the Hadamards first giveSince $U^k|\psi ⟩=e^{2\pi ik\theta }|\psi ⟩$ and $\theta =j/N$,

The first register isFor $j,j^{\prime }$,The geometric sum is one when $j=j^{\prime }$ and zero otherwise. There are $N$ such orthonormal vectors in an $N$-dimensional space, so they form an orthonormal basis.

The operator $F$ maps the computational orthonormal basis $\{|j⟩\}$ to the orthonormal basis $\{|{\varphi }_j⟩\}$. It therefore preserves inner products and is surjective, so $F^{†}F=F{F}^{†}=I$ and $F$ is unitary.

First apply $F^{†}$, which sends $|{\varphi }_j⟩$ to $|j⟩$. Then measure in the computational basis. Because the phase has an exact $m$-bit representation, exact quantum phase estimation returns

The overlap isThe Helstrom-Holevo bound for two equiprobable pure states givesThe Breidbart basis is the eigenbasis of the difference of the two weighted state projectors, so its decision rule is precisely the Helstrom measurement and attains this bound.