= Fuchs-van de Graaf inequalities
{c}
{title2=$1-F\leq D\leq\sqrt{1-F^2}$}
For all finite-dimensional <density operators>, including <pure states> and <mixed states>, the <unsquared quantum fidelity> and <trace distance> obey
$$
1-F(\rho,\sigma)\leq D(\rho,\sigma)\leq\sqrt{1-F(\rho,\sigma)^2}.
$$
In the <squared quantum fidelity> convention this is $1-\sqrt{F_{\mathrm{sq}}}\leq D\leq\sqrt{1-F_{\mathrm{sq}}}$. The bounds translate a high <quantum fidelity> into a small operationally meaningful <trace distance>. If $F\geq1-\varepsilon$ with $0\leq\varepsilon\leq1$, then $D\leq\sqrt{2\varepsilon-\varepsilon^2}$; if $D\leq\delta$, then $F\geq1-\delta$. 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 $0.0652\lesssim0.3\lesssim0.3551$. https://arxiv.org/pdf/quant-ph/9712042[Fuchs and van de Graaf, Cryptographic Distinguishability Measures for Quantum Mechanical States, Theorem 1, equation 46; original preprint 1997, journal publication 1999]{ref}
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