The <variational characterization of trace distance> turns a <matrix> distance into the best possible binary experiment. For <density operators> $\rho,\sigma$ on the same <finite-dimensional vector space>,
$$
D(\rho,\sigma)=\max_{0\leq P\leq I}\operatorname{Tr}[P(\rho-\sigma)]
=\max_{0\leq P\leq I}|\operatorname{Tr}[P(\rho-\sigma)]|.
$$
An operator $P$ with $0\leq P\leq I$ is an effect of a <POVM>: both $P$ and $I-P$ are <positive operators>. The <Born rule> assigns the first outcome <probabilities> $\operatorname{Tr}(P\rho)$ and $\operatorname{Tr}(P\sigma)$. Their signed gap is $\operatorname{Tr}(P\Delta)$, where $\Delta=\rho-\sigma$. Since $\operatorname{Tr}\Delta=0$, replacing $P$ by $I-P$ reverses this gap, explaining the equality with the absolute-value maximum.

One optimal $P$ is the <spectral projector> $P_+$ onto the positive <eigenspaces> of $\Delta$. In the <spectral decomposition> of $\Delta$, this choice includes all positive <eigenvalues> and excludes all negative <eigenvalues>. Their positive sum equals the <trace distance> by the <spectral-parts formula for trace distance>. Any zero <eigenspace> can be assigned to either outcome without changing the gap. This is the effect selected by the equal-prior <Holevo–Helstrom theorem>. https://cs.uwaterloo.ca/~watrous/TQI/TQI.3.pdf[Watrous, The Theory of Quantum Information, Theorem 3.4, equations (3.16)–(3.19)]{ref}

The same optimum has a classical description. For any finite-outcome <POVM> $\{E_i\}$, let $p_i=\operatorname{Tr}(E_i\rho)$ and $q_i=\operatorname{Tr}(E_i\sigma)$. Then
$$
D(\rho,\sigma)=\max_{\{E_i\}}\frac12\sum_i|p_i-q_i|
=\max_{\{E_i\}}\max_A\left|\sum_{i\in A}(p_i-q_i)\right|.
$$
The expression $\frac12\sum_i|p_i-q_i|$ is the <total variation distance> of the entire outcome <probability distributions>; the expression for $A$ is the gap for one event. Grouping outcomes in $A$ gives the single effect $P_A=\sum_{i\in A}E_i$. Conversely, an effect $P$ gives the binary <POVM> $\{P,I-P\}$, whose two <probability> gaps are opposites, so their <total variation distance> is $|\operatorname{Tr}(P\Delta)|$. Thus the optimum over all <POVMs> is already attained by one binary <projective measurement>. https://cs.uwaterloo.ca/~watrous/TQI/TQI.3.pdf[Watrous, The Theory of Quantum Information, §3.1.1, Proposition 3.5 and the discussion following Theorem 3.4]{ref}

For <quantum state discrimination> with equal prior <probabilities>, declare $\rho$ on outcome $P$ and $\sigma$ on outcome $I-P$. Directly averaging the two conditional success <probabilities> gives
$$
P_{\mathrm{success}}(P)=\frac12\operatorname{Tr}(P\rho)
+\frac12\operatorname{Tr}((I-P)\sigma)
=\frac12\left(1+\operatorname{Tr}(P\Delta)\right).
$$
Optimizing this expression gives $(1+D)/2$. For the diagonal example in <trace distance>, $P=|0\rangle\langle0|$ gives a gap $0.8-0.3=0.5$. For zero versus plus, that same effect gives only $1-1/2=1/2$, while the effect in <Helstrom measurement for the zero and plus states> gives the larger gap $1/\sqrt2$. The <trace-distance-preserving binary measurement> depends on the pair of preparations; a fixed measurement <basis> does not generally attain the <trace distance> for every pair.


 Variational characterization of trace distance (source code)