Quantum fidelity commutativity proof

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?

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

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 $$\lambda$$.

• 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 :) Mar 15, 2023 at 23:56
• @SlowerPhoton I have edited my answer.
– Gec
Mar 16, 2023 at 16:20
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.