The Holevo optimality conditions say a POVM is optimal when
$$
\Gamma=\sum_kp_k\rho_kM_k
$$
is Hermitian and $\Gamma-p_k\rho_k\geq0$ for every $k$. Here
$$
\Gamma=\frac{c+s}{3}
\begin{pmatrix}c&0\\0&s\end{pmatrix}
$$
and
$$
\Gamma-\frac13\rho_k
=\frac{cs}{3}
\begin{pmatrix}
1&-\zeta^{-k}\\
-\zeta^k&1
\end{pmatrix}\geq0,
$$
whose eigenvalues are $0$ and $2cs/3$. The <pretty good measurement> is therefore optimal. Its success probability is
$$
\boxed{P_{\rm succ}=\frac13(c+s)^2
=\frac13(1+\sin\theta)}.
$$

Solution

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

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:
$$
\bar\rho=\frac13\sum_{k=0}^2|\psi_k\rangle\langle\psi_k|
=\begin{pmatrix}c^2&0\\0&s^2\end{pmatrix}.
$$
For $c,s>0$, the <pretty good measurement> is
$$
\boxed{M_k=\frac13
\begin{pmatrix}
1&\zeta^{-k}\\
\zeta^k&1
\end{pmatrix}}
$$
because $\bar\rho^{-1/2}|\psi_k\rangle=|0\rangle+\zeta^k|1\rangle$. The matrices are positive and $\sum_kM_k=I$. 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.

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

{title2=PGM}
{wiki}

For ensemble $\{p_j,\rho_j\}$ with average $\bar\rho=\sum_jp_j\rho_j$, the pretty good measurement has $M_j=\bar\rho^{-1/2}p_j\rho_j\bar\rho^{-1/2}$ on the support of $\bar\rho$.


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