Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 3

In quantum binary hypothesis testing, hypothesis zero supplies ${\rho }_0$ with prior $p$ and hypothesis one supplies ${\rho }_1$ with prior $1-p$. A two-outcome POVM $\{Q,I-Q\}$ decides zero on outcome $Q$. The conditional errors areSymmetric testing minimizes the prior-weighted average error $p\alpha +(1-p)\beta$, equivalently maximizing the average success probability.

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Let $\Delta =p{\rho }_0-(1-p){\rho }_1$. The success probability of $Q$ isWrite the spectral decomposition $\Delta ={\Delta }_{+}-{\Delta }_{-}$. For every effect $0\le Q\le I$,with equality when $Q$ projects onto the positive spectral subspace, with arbitrary action on the kernel. Since $\mathrm{Tr}\Delta =2p-1$ and $\Vert \Delta {\Vert }_1=\mathrm{Tr}{\Delta }_{+}+\mathrm{Tr}{\Delta }_{-}$, the Holevo–Helstrom theorem follows:

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Put $c=\cos (\theta /2)$, $s=\sin (\theta /2)$, and $\zeta =e^{2\pi i/3}$. Averaging the three states cancels the off-diagonal phases:For $c,s>0$, the pretty good measurement isbecause ${\bar{\rho }}^{-1/2}|{\psi }_k⟩=|0⟩+{\zeta }^k|1⟩$. The matrices are positive and ${\sum }_k{M}_k=I$. At the endpoint values of $\theta$, the same formula is understood on the support of $\bar{\rho }$ and may be completed arbitrarily on its kernel.

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The Holevo optimality conditions say a POVM is optimal whenis Hermitian and $\Gamma -p_k{\rho }_k\ge 0$ for every $k$. Hereandwhose eigenvalues are $0$ and $2cs/3$. The pretty good measurement is therefore optimal. Its success probability is

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