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I am trying to learn by myself quantum computing and information and I have a very simple question concerning the demonstration of the following equality: $F( \rho , \sigma) = |\langle \psi_{\rho} | \psi_{\sigma}\rangle|^2$ when $\rho$ and $\sigma$ are density matrix of a pure state.

Indeed Fidelity of quantum states' definition is:
$$ F(\rho,\sigma) = \left(\text{tr}\sqrt{\sqrt{\rho}\sigma\sqrt{\rho}}\right)^2 $$

But in the following proof of the formula in the pure case this is what I can read on Wikipedia: Fidelity formula for pure states from Wikipedia
Where does the square root on $|\psi_{\rho}\rangle$, colored in yellow upside, have gone as we know that for a projector $\rho$ is not necessarily equal to $\rho^{1/2}$?

Moreover why does the expression underlined in red is equal to one?
Part of the proof of the formula for pure states
Again as far as I know in the cas of a pure state we have: $Tr(\rho^2)=Tr(\rho)=1$ and not $Tr(\rho^{1/2}) = 1$?

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2 Answers 2

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When a state is pure, its density matrix is a projector. The eigenvalues of a pure density matrix are either$^1$ 1 or 0. Hence, taking the square root will give you the same eigenvalues, $\sqrt{1} = 1,\sqrt{0} = 0$; and hence, the square root of a pure density matrix is the same density matrix.

$$\therefore \rho^2 = \rho = \sqrt{\rho}\,,$$

$^1$: It only has a single (+1) eigenvalue, and all others are 0 as the trace is always 1 for $\rho$.


Spectral theorem says; for any matrix $A$ with eigenvalues $\{\lambda_i\}n$ and eigenvectors $\{|\lambda_i\rangle \}$,

$$f(A) = \sum_i f(\lambda_i) | \lambda_i \rangle \langle\lambda_i |\,,$$

where $$A = \sum_i \lambda_i | \lambda_i \rangle \langle\lambda_i|\,,$$ is known as the spectral decomposition.


So, for pure $\rho$, $$\rho = |\psi \rangle \langle \psi | \,.$$ As you can observe, this $\rho$ is already in a spectral form, where the summation is only over the single term and $\lambda =1$, i.e., $\rho = (1) \cdot |\psi \rangle \langle \psi |$

$$ \begin{align} \therefore \sqrt{\rho} &= \sqrt{1}|\psi \rangle \langle \psi |\\ &= (1) \cdot|\psi \rangle \langle \psi |\\ &= |\psi \rangle \langle \psi |\\ &=\rho \end{align} $$

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    $\begingroup$ First thank for your help. So if I understand you well $\rho^2=\rho=\rho^{1/2}$ is true because $\rho$ is a special projector operator (indeed the last equality doesn't all for all projector). More precisely it is true because $\rho$ in plus of being a projector matrix it is too a Hermitian matrix. Is it correct? Thk you. ----PS: btw if it is true where can I find a prove that: $\rho^2=\rho=\rho^{1/2}$ for $\rho$ a Hermitian projector. $\endgroup$
    – X0-user-0X
    Sep 12, 2023 at 12:28
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    $\begingroup$ For any projector eigenvalues are 0 and 1, so any positive power of projector is projector itself $\endgroup$
    – EvgeniyZh
    Sep 12, 2023 at 12:42
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    $\begingroup$ @X0-user-0X I wrote the answer in a hurry yesterday. I’ll try to edit it and make it more comprehensive sometime later today and include the proof aswell. You can prove that simply by using the spectral theorem. $\endgroup$
    – FDGod
    Sep 12, 2023 at 13:34
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    $\begingroup$ @X0-user-0X A matrix can have multiple square roots. $I$ itself is a valid square root. But the $Q$ given in that question is another root. However, that $Q$ has negative eigenvalues, but we care only about positive eigenvalues. Mostly, when we say $A^{\frac{1}{2}} = B$, we talked about when B is positive-semidefinite. This is the same reasoning we use that for a number, $\sqrt{4} = 2$, but if $4 = a^2$, then $a=\pm2$. ____ TL;DR: Its a notational choice. When we see the square root of a matrix, we usually only refer to the positive-semidefinite root, just as we do for scalars. $\endgroup$
    – FDGod
    Sep 12, 2023 at 14:35
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    $\begingroup$ @FDGod Thank a lot for your answer and the time you ve taken to write it $\endgroup$
    – X0-user-0X
    Sep 12, 2023 at 15:24
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  • For any pure state $\rho=|\psi\rangle\!\langle\psi|$ we have $\rho=\rho^2=\sqrt\rho$.
  • For any pair of pure states $\rho=|\psi\rangle\!\langle\psi|$, $ \sigma=|\phi\rangle\!\langle\phi|$ we have $\rho\sigma=|\psi\rangle\!\langle\phi|\,\, \langle\psi|\phi\rangle$, and thus $$\rho\sigma\rho = |\psi\rangle\!\langle \psi| \, \langle\psi|\phi\rangle\, \langle\phi|\psi\rangle = |\psi\rangle\!\langle\psi| \,\, |\langle\psi|\phi\rangle|^2.$$
  • If $A$ is any matrix that is a scalar multiple of a rank-1 projection, meaning $A=\alpha |u\rangle\!\langle u|$ for some $\alpha\in\mathbb{C}$ and ket $|u\rangle$, then $\operatorname{tr}\sqrt A=\sqrt\alpha$. It follows that $$\operatorname{tr}\sqrt{\rho\sigma\rho} = |\langle\psi|\phi\rangle|.$$
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    $\begingroup$ Thank a lot for your answer $\endgroup$
    – X0-user-0X
    Sep 12, 2023 at 15:22

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