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The fidelity between a pure state $|\psi\rangle$ and an arbitrary mixed state $\rho$ is given by, $F(|\psi\rangle,\rho)=\sqrt{\langle\psi|\rho|\psi\rangle}$, which is stated to be equal to the square root of the overlap between $|\psi\rangle$ and $\rho$.

In what sense the term ${\langle\psi|\rho|\psi\rangle}$ is the overlap between $|\psi\rangle$ and $\rho$ ?


The fidelity between $\rho$ and $\sigma$ is given by $F(\rho,\sigma)=tr\sqrt{\rho^{1/2}\sigma\rho^{1/2}}$

Therefore, the fidelity between the pure states $|\psi\rangle$ and $|\phi\rangle$ is,

$F(|\psi\rangle,|\phi\rangle)=tr\sqrt{|\psi\rangle\langle\psi|\phi\rangle\langle\phi|\psi\rangle\langle\psi|}=tr\sqrt{\langle\psi|\phi\rangle\langle\phi|\psi\rangle|\psi\rangle\langle\psi|}=tr\sqrt{|\langle\psi|\phi\rangle|^2|\psi\rangle\langle\psi|}=\sqrt{|\langle\psi|\phi\rangle|^2}tr\sqrt{|\psi\rangle\langle\psi|}=|\langle\psi|\phi\rangle|$

The quantity $|\langle\psi|\phi\rangle|^2$ is the overlap between the pure states $|\psi\rangle$ and $|\phi\rangle$, in the sense that $|\langle\psi|\phi\rangle|$ is the component of $|\psi\rangle$ in the state $|\phi\rangle$.

The fidelity between the pure state $|\psi\rangle$ and the mixed state $\rho$ is,

$F(\psi,\rho)=tr\sqrt{|\psi\rangle\langle\psi|\rho|\psi\rangle\langle\psi|}=tr\sqrt{\langle\psi|\rho|\psi\rangle|\psi\rangle\langle\psi|}=\sqrt{\langle\psi|\rho|\psi\rangle}tr(|\psi\rangle\langle\psi|)=\sqrt{\langle\psi|\rho|\psi\rangle}$

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    $\begingroup$ I don't understand the question. Didn't you just write why the fidelity between a pure state and a general state has that form? $\endgroup$
    – glS
    Dec 9, 2022 at 20:58

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I'm also not quite sure what the question is but here's something. Let $\rho = \sum_x p(x) |\phi_x\rangle \langle \phi_x |$, i.e., $\rho$ can be seen as a probabilistic preparation of the states $|\phi_x\rangle$ with probabilities $p(x)$. Then $$ \begin{aligned} \langle \psi | \rho | \psi \rangle &= \langle \psi | (\sum_x p(x) |\phi_x\rangle \langle \phi_x |) | \psi \rangle \\ &= \sum_x p(x) \langle\psi |\phi_x\rangle \langle \phi_x | \psi \rangle \\ &= \sum_x p(x) |\langle \psi |\phi_x\rangle|^2 \end{aligned} $$ So you can see $\langle \psi | \rho | \psi \rangle $ as representing the average overlap with the preparation states that constitute $\rho$.

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