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Let $P_{+}$ be the maximizing projection in the variational characterization of trace distance. Use the binary POVM $\{P_{+},I-P_{+}\}$ and its measurement channel. The difference between its two output probability vectors iswhere $D=D(\rho ,\sigma )$ and the second equality uses $\mathrm{Tr}(\rho -\sigma )=0$. Since trace distance between diagonal density operators is half the L1 norm of their probability-vector difference,Thus a binary quantum measurement can preserve the distinguishability of this particular pair exactly. This is the trace-distance-preserving binary measurement, which may depend on the two states.

Choose an orthonormal basis $\{|j⟩{\}}_{j=1}^d$ of the input space and define operators from the input to the $k$-dimensional outcome register byThese give a Kraus representation:The POVM completeness relation givesTherefore $\Phi$ is a completely positive map and is trace preserving: it is a quantum channel representing a deterministic measurement channel.

For two density operators $\rho ,\sigma$, let $P_{+}$ project onto the positive spectral subspace of $\rho -\sigma$. The binary POVM $\{P_{+},I-P_{+}\}$ has output probability difference $(D,-D)$, where $D=D(\rho ,\sigma )$. Its measurement channel therefore preserves this pair's trace distance exactly. The optimizing quantum measurement depends on the pair; a single fixed measurement need not preserve every pair's distance.