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.

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    $\begingroup$ 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? $\endgroup$ Jul 13, 2023 at 16:32
  • $\begingroup$ And please use MathJax! $\endgroup$ Jul 13, 2023 at 16:34
  • $\begingroup$ @TobiasFünke yeah, understood. I do not know about MathJax. Would try learning it! $\endgroup$ Jul 14, 2023 at 17:43
  • $\begingroup$ Here is a tutorial. $\endgroup$ Jul 14, 2023 at 22:02

1 Answer 1


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 $$(\operatorname{tr}[\sqrt{\rho\sigma\rho}])^2 = (\langle \psi|\sigma|\psi\rangle^{1/2})^2 = \langle \psi|\sigma|\psi\rangle.$$

  • $\begingroup$ Thank you so much! $\endgroup$ Jul 14, 2023 at 17:42
  • $\begingroup$ If this answered your question, consider to "accept" it. See e.g. here. And since none of your other question has been accepted either, also consider it for those... $\endgroup$ Jul 14, 2023 at 22:02
  • $\begingroup$ @TobiasFünke okay! $\endgroup$ Jul 16, 2023 at 11:02

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