Fidelity of quantum states

Remarks

Example: let $\rho = \lvert \psi \rangle\langle \psi \rvert$ be pure. Then $\sqrt\rho = \rho$ and $\sqrt\rho\, \sigma \sqrt\rho = \langle \psi \rvert\sigma\lvert \psi \rangle\, \rho$, so $F(\rho, \sigma) = \langle \psi \rvert\sigma\lvert \psi \rangle$. If $\sigma = \lvert \phi \rangle\langle \phi \rvert$ is pure too, then $F = \abs{\ip{\psi}{\phi}}^2$.

Non-example: the overlap $\tr(\rho\sigma)$ is not the fidelity for mixed states. For $\rho = \sigma = I/2$ on $\C^2$, $\tr(\rho\sigma) = 1/2$, while $F(\rho, \sigma) = (\tr (I/2))^2 = 1$.

Standard properties: $0 \le F(\rho, \sigma) \le 1$; $F(\rho, \sigma) = 1$ if and only if $\rho = \sigma$; $F(\rho, \sigma) = F(\sigma, \rho)$. Equivalently $F(\rho, \sigma) = \norm{\sqrt\rho \sqrt\sigma}_1^2$, where $\norm{A}_1 = \tr\sqrt{A^* A}$ is the trace norm. Some authors call the square root $\sqrt{F}$ the fidelity, so the convention must be checked.