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For probability distributions $p,q$ on the same finite alphabet, the Bhattacharyya coefficient measures their square-root overlap:The vectors $(\sqrt{p_i}{)}_i$ and $(\sqrt{q_i}{)}_i$ have unit Euclidean norm, so the Cauchy-Schwarz inequality gives $0\le B\le 1$. Equality $B=1$ holds exactly when the distributions agree; $B=0$ holds exactly when they assign positive probability to disjoint outcomes. For $p=(0.9,0.1)$ and $q=(0.6,0.4)$, $B=\sqrt{0.54}+0.2\approx 0.934847$. This overlap is not the probability that two independent draws agree, which here is $0.58$.

The fidelity of commuting mixed states reduces to this coefficient. More generally, the measurement characterization of quantum fidelity minimizes it over outcome distributions of a common measurement in quantum mechanics, connecting classical overlap to the unsquared quantum fidelity.

For all finite-dimensional density operators, including pure states and mixed states, the unsquared quantum fidelity and trace distance obeyIn the squared quantum fidelity convention this is $1-\sqrt{F_{\mathrm{sq}}}\le D\le \sqrt{1-F_{\mathrm{sq}}}$. The bounds translate a high quantum fidelity into a small operationally meaningful trace distance. If $F\ge 1-\epsilon$ with $0\le \epsilon \le 1$, then $D\le \sqrt{2\epsilon -{\epsilon }^2}$; if $D\le \delta$, then $F\ge 1-\delta$. The upper bound is attained by the pure-state trace distance and fidelity identity. For the fidelity of commuting mixed states example, the bounds read $0.0652\lesssim 0.3\lesssim 0.3551$. ^[ref]

A POVM $\{E_i\}$ gives outcome probability distributions $p_i=\mathrm{Tr}(\rho E_i)$ and $q_i=\mathrm{Tr}(\sigma E_i)$ by the Born rule. Their classical overlap, the Bhattacharyya coefficient, is $B(p,q)={\sum }_i\sqrt{p_i{q}_i}$. Fuchs and Caves showed that minimizing this overlap over all POVMs gives the unsquared quantum fidelity:The same POVM is applied to either possible input quantum state. Each measurement in quantum mechanics has $B(p,q)\ge F$; some measurement attains equality, including when the density operators have zero eigenvalues. Thus quantum fidelity quantifies the smallest classical overlap that observation can reveal. For fidelity of commuting mixed states, measurement in a common eigenbasis attains the minimum. This operational interpretation compares statistics without assigning a probability that two mixed states are equal. ^{[ref] [ref]}