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I am looking for a proof that the quantum fidelity $$F(\rho, \sigma) = \left(\text{tr} \sqrt{\sqrt{\rho}\sigma\sqrt\rho}\right)^2$$ is commutative, i.e. $F(\rho, \sigma) = F(\sigma, \rho)$.

I have been able to find some proofs (such as here); however, they are quite involved and force the reader to understand different topics (such as measure theory).

Is there more straightforward proof of the fact?

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    $\begingroup$ Please give the exact reference (e.g. chapter, page, equation)... $\endgroup$ Mar 15, 2023 at 8:03

2 Answers 2

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If the operators $\rho$ and $\sigma$ act in a finite-dimensional space, then you can easily see that all nonzero eigenvalues of the operators $\sqrt{\rho}\sigma\!\sqrt{\rho}$ and $\sqrt{\sigma}\rho\!\sqrt{\sigma}$ coincide. Indeed, let's consider the eigenvector $\Psi$ of the operator $\sqrt{\rho}\sigma\!\sqrt{\rho}$ corresponding to a non-zero eigenvalue $\lambda$. Then the following equality holds $$ \sqrt{\rho}\sigma\!\sqrt{\rho}\ \Psi = \lambda \Psi \qquad (1) $$ From this equality we proceed to $$ \lambda \sqrt{\sigma}\!\sqrt{\rho}\ \Psi = \sqrt{\sigma}\!\sqrt{\rho}\ \sqrt{\rho}\sigma\!\sqrt{\rho}\ \Psi = \sqrt{\sigma}\rho \sqrt{\sigma}\ \sqrt{\sigma}\!\sqrt{\rho}\ \Psi.\quad (2) $$ Vector $\Phi = \sqrt{\sigma}\!\sqrt{\rho}\ \Psi$ is nonzero, otherwise $\lambda$ would be zero according to (1). Therefore equality (2) means that $\Phi$ is the eigenvector of the operator $\sqrt{\sigma}\rho \sqrt{\sigma}$ corresponding to the eigenvalue $\lambda$. I think it is now obvious that the operators $\sqrt{\sqrt{\sigma}\rho \sqrt{\sigma}}$ and $\sqrt{\sqrt{\rho}\sigma\!\sqrt{\rho}}$ have the same trace. Two operators have the same eigenvalues, then their square roots have the same eigenvalues, then the traces of the square roots of these operators are equal.

I believe that in the case of an infinite-dimensional Hilbert space, a similar consideration is possible when $\rho$ and $\sigma$ are density matrices.

Addition. In the case where degenerate eigenvalues are present, additional analogous consideration is required to ensure that the operators $\sqrt{\rho}\sigma\!\sqrt{\rho}$ and $\sqrt{\sigma}\rho\!\sqrt{\sigma}$ have equal numbers of each of the eigenvalues.

Addition 2. Let's consider the case of degenerate eigenvalues. Suppose that the operator $\sqrt{\rho}\sigma\sqrt{\rho}$ has exactly $k$ linearly independent eigenvectors $\Psi_i$, $i = 1,\ldots,k$ corresponding to the eigenvalue $\lambda$. In this case, for any numbers $a_i$, $i =1,\ldots,k$, $\sum_{i=1}^k|a_i| \neq 0$ we have that the vector $\Psi(a) = \sum_{i=1}^k a_i \Psi_i$ is also an eigenvector of the operator $\sqrt{\rho}\sigma\sqrt{\rho}$, corresponding to the eigenvalue $\lambda$. According to the previous consideration, all vectors $\Phi(a) = \sqrt{\sigma}\sqrt{\rho} \Psi(a)$ are (nonzero) eigenvectors of the operator $\sqrt{\sigma}\rho\sqrt{\sigma} $ corresponding to the eigenvalue $\lambda$. Taking into account the arbitrariness in the choice of the coefficients $a$, the latter fact means that the vectors $\Phi_i = \sqrt{\sigma}\sqrt{\rho} \Psi_i$ are $k$ linearly independent eigenvectors of the operator $\sqrt{\sigma}\rho\sqrt{\sigma}$. This reasoning can also be carried out the other way, whence it follows that the operators $\sqrt{\rho}\sigma\sqrt{\rho}$ and $\sqrt{\sigma}\rho\sqrt{\sigma}$ have exactly the same number of eigenvectors corresponding to the eigenvalue $\lambda$.

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  • $\begingroup$ You state that since $\sqrt\sigma\rho\sqrt\sigma$ and $\sqrt\rho\sigma\sqrt\rho$ have the same trace, their square roots do as well. Why? Also, in case of degenerate eigenvalues, I cannot see how we can save the proof :) $\endgroup$ Mar 15, 2023 at 23:56
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    $\begingroup$ @SlowerPhoton I have edited my answer. $\endgroup$
    – Gec
    Mar 16, 2023 at 16:20
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  1. Define square roots $R:=\sqrt{\rho}$ and $S:=\sqrt{\sigma}$. Both $R,S$ are semi-positive definite bounded operators. Define the bounded operator $T:= SR$. Notice that $T^{\dagger}=RS$.

  2. The definition of quantum fidelity then becomes the trace $$ \sqrt{F(\rho, \sigma)} ~=~ {\rm tr}|T|, \tag{1}$$ because the absolute value operator $|T|$ satisfies $$ |T|~=~\sqrt{T^{\dagger}T}.\tag{2} $$

  3. The polar decomposition yields $$ T~=~U|T|\qquad \Leftrightarrow \qquad|T|~=~U^{\dagger}T,\tag{3}$$ where $U$ is a partial isometry with $$ {\rm ker} U~=~{\rm ker} T .\tag{4}$$

  4. This leads to $$ |T^{\dagger}|~=~U|T| U^{\dagger}, \tag{5}$$ and in turn OP's sought-for relation $$ \sqrt{F(\sigma,\rho)} ~=~ {\rm tr}|T^{\dagger}| ~=~ {\rm tr}|T| ~=~\sqrt{F(\rho, \sigma)} \tag{6}$$ by cyclicity of the trace.

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