# Prove that the state fidelity for pure states has the form $\langle\psi|\sigma|\psi\rangle$

For two density operators $$\rho$$ and $$\sigma$$, the fidelity is given by $$F(\rho,\sigma)=\left(\mathrm{Tr} \sqrt{\rho^{\frac{1}{2}}\sigma\rho^{\frac{1}{2}}}\right)^2$$For a pure state $$\rho=|\psi\rangle\langle \psi|$$, fidelity is given by $$F = \langle \psi| \sigma |\psi\rangle$$ However I have not been able to show this. I do not know how to find out $$\rho^{\frac{1}{2}}$$ for $$\rho = |\psi\rangle\langle \psi|$$. Any clue regarding how to proceed would be helpful.

• Hint: $\rho^{1/2}$ is defined through the spectral representation of $\rho$. Or, here equivalently, as the unique positive operator for which $\rho=\rho^{1/2}\rho^{1/2}$. Now since for a pure state $\rho^2=\rho$ -what do you conclude? Jul 13, 2023 at 16:32
• And please use MathJax! Jul 13, 2023 at 16:34
• @TobiasFünke yeah, understood. I do not know about MathJax. Would try learning it! Jul 14, 2023 at 17:43
• Here is a tutorial. Jul 14, 2023 at 22:02

If $$\rho$$ is a pure state, then $$\rho^2=\rho$$, and thus also $$\sqrt\rho=\rho$$. Furthermore, if $$\rho=|\psi\rangle\!\langle\psi|$$, then $$(\rho\sigma\rho)^{1/2}=\langle \psi|\sigma|\psi\rangle^{1/2} \rho,$$ hence $$(\operatorname{tr}[\sqrt{\rho\sigma\rho}])^2 = (\langle \psi|\sigma|\psi\rangle^{1/2})^2 = \langle \psi|\sigma|\psi\rangle.$$