Pure-target lower bound on trace distance
= Pure-target lower bound on trace distance
{title2=$D(\rho,P)\geq1-F(\rho,P)^2$}
For a <density operator> $\rho$ and <pure state> $P=|\psi\rangle\langle\psi|$, the <diagonal absolute-sum bound for the trace norm> in a basis containing $|\psi\rangle$ gives $D(\rho,P)\geq1-\langle\psi|\rho|\psi\rangle=1-F(\rho,P)^2$. The <quantum fidelity> is unsquared. Block-diagonal states with respect to the target and its complement attain equality.