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When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.