= Pure-state trace distance and fidelity identity
{title2=$D=\sqrt{1-F^2}$}
For normalized <pure states> $\rho=|\psi\rangle\langle\psi|$ and $\sigma=|\varphi\rangle\langle\varphi|$, the <trace distance> and <quantum fidelity> satisfy
$$
D(\rho,\sigma)=\frac12\|\rho-\sigma\|_1=\sqrt{1-|\langle\psi|\varphi\rangle|^2}=\sqrt{1-F^2}=\sqrt{1-F_{\mathrm{sq}}}.
$$
The last two expressions distinguish the <unsquared quantum fidelity> from the <squared quantum fidelity>. Identical <pure states> give $D=0$, and <orthogonal> <pure states> give $D=1$. The equality need not hold for <mixed states>; the <Fuchs-van de Graaf inequalities> give bounds in that case.
Brief proof: for two distinct <pure states>, their two-dimensional span contains the nonzero part of $\rho-\sigma$. This restriction has <trace> zero and <determinant> $-(1-F^2)$, hence <eigenvalues> $\pm\sqrt{1-F^2}$. Its <trace norm> is twice the positive <eigenvalue>, giving <trace distance> $D=\sqrt{1-F^2}$. For identical <pure states>, $\rho-\sigma=0$ directly.
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