Let $\Delta=\rho-\sigma=\Delta_+-\Delta_-$ be the <positive-negative decomposition of a Hermitian operator>. Since $\operatorname{Tr}\Delta=0$,
$$
\operatorname{Tr}\Delta_+
=\operatorname{Tr}\Delta_-
=\frac12\|\Delta\|_1.
$$
For $0\leq P\leq I$,
$$
\operatorname{Tr}(P\Delta)
\leq\operatorname{Tr}(P\Delta_+)
\leq\operatorname{Tr}\Delta_+.
$$
Equality is attained by the projector onto the positive eigenspace of $\Delta$. This proves the <variational characterization of trace distance>
$$
\boxed{D(\rho,\sigma)
=\max_{0\leq P\leq I}\operatorname{Tr}[P(\rho-\sigma)]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 4 / a / ii / Solution

For the binary <POVM> $\{E,I-E\}$ and equal priors,
$$
p_{\rm succ}
=\frac12\operatorname{Tr}(E\rho_1)
+\frac12\operatorname{Tr}[(I-E)\rho_2]
=\frac12\left[1+\operatorname{Tr}E(\rho_1-\rho_2)\right].
$$
Maximizing with the <variational characterization of trace distance> gives the equal-prior <Holevo–Helstrom theorem>
$$
\boxed{p_{\rm succ}^{\max}
=\frac12[1+D(\rho_1,\rho_2)]}.
$$

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 323 / 4 / b / i / Solution

For density operators $\rho,\sigma$,
$$
D(\rho,\sigma)=\max_{0\leq P\leq I}\operatorname{Tr}[P(\rho-\sigma)].
$$
The maximum is attained by the projector onto the positive spectral subspace of $\rho-\sigma$.


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 Incoming links: Variational characterization of trace distance