 Variational characterization of trace distance

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,An operator $P$ with $0\le P\le I$ is an effect of a POVM: both $P$ and $I-P$ are positive operators. The Born rule assigns the first outcome probabilities $\mathrm{Tr}(P\rho )$ and $\mathrm{Tr}(P\sigma )$. Their signed gap is $\mathrm{Tr}(P\Delta )$, where $\Delta =\rho -\sigma$. Since $\mathrm{Tr}\Delta =0$, 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. ^[ref]

The same optimum has a classical description. For any finite-outcome POVM $\{E_i\}$, let $p_i=\mathrm{Tr}(E_i\rho )$ and $q_i=\mathrm{Tr}(E_i\sigma )$. ThenThe expression $\frac{1}{2}{\sum }_i|p_i-q_i|$ 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 $|\mathrm{Tr}(P\Delta )|$. Thus the optimum over all POVMs is already attained by one binary projective measurement. ^[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 givesOptimizing this expression gives $(1+D)/2$. For the diagonal example in trace distance, $P=|0⟩⟨0|$ 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 $1/\sqrt{2}$. 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.