= Fidelity of commuting mixed states
{title2=$F=\sum_i\sqrt{p_iq_i},\quad D=\tfrac12\sum_i|p_i-q_i|$}
When two <density operators> are <commuting elements>, choose a common <orthonormal eigenbasis> and write $\rho=\operatorname{diag}(p_i)$, $\sigma=\operatorname{diag}(q_i)$. Their <quantum fidelity> becomes the classical overlap of the two <probability distributions>, while their <trace distance> becomes <total variation distance>:
$$
F(\rho,\sigma)=\sum_i\sqrt{p_iq_i},\qquad F_{\mathrm{sq}}(\rho,\sigma)=\left(\sum_i\sqrt{p_iq_i}\right)^2,\qquad D(\rho,\sigma)=\frac12\sum_i|p_i-q_i|.
$$
This follows by computing the <positive square root of an operator> entry by entry in the common <eigenbasis>. For the <qubit> <mixed states> $\rho=\operatorname{diag}(0.9,0.1)$ and $\sigma=\operatorname{diag}(0.6,0.4)$,
$$
F=\sqrt{0.54}+\sqrt{0.04}=\sqrt{0.54}+0.2\approx0.934847,\qquad
F_{\mathrm{sq}}=0.58+0.4\sqrt{0.54}\approx0.873939,\qquad D=0.3.
$$
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.
\Image[/commuting-state-fidelity.png]
{title=Commuting mixed-state probabilities, square-root overlap contributions to unsquared fidelity F, squared fidelity Fsq, and trace distance D}
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 $\sqrt{1-F^2}\approx0.3551>D$: the <pure-state trace distance and fidelity identity> is an equality for <pure states>, rather than a formula for arbitrary <mixed states>.
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