# In what sense is $\langle\psi|\rho|\psi\rangle$ the fidelity between the pure state $|\psi\rangle$ and the mixed state $\rho$?

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,

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

• 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?
– glS
Dec 9, 2022 at 20:58

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 $$\rho$$.