# Prove the fidelity equals $F( \rho , \sigma) = |\langle \psi_{\rho} | \psi_{\sigma}\rangle|^2$ for pure states

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:
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?

Again as far as I know in the cas of a pure state we have: $$Tr(\rho^2)=Tr(\rho)=1$$ and not $$Tr(\rho^{1/2}) = 1$$?

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}

• 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. Sep 12, 2023 at 12:28
• For any projector eigenvalues are 0 and 1, so any positive power of projector is projector itself Sep 12, 2023 at 12:42
• @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. Sep 12, 2023 at 13:34
• @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. Sep 12, 2023 at 14:35
• @FDGod Thank a lot for your answer and the time you ve taken to write it Sep 12, 2023 at 15:24
• 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 $$\operatorname{tr}\sqrt{\rho\sigma\rho} = |\langle\psi|\phi\rangle|.$$
• Thank a lot for your answer Sep 12, 2023 at 15:22