Polar decomposition of $\sqrt{\sqrt{\rho} \sigma \sqrt{\rho}}$

Question

By polar decomposition of a square invertible matrix $A$, I understand $A = |A| U$ for some unitary matrix $U$, where $|A| = \sqrt{A^\dagger A}$ with $\dagger$ denoting the conjugate-transpose operation. Now in Nielsen and Chuang's book, chapter 9, I find the following statement

$\dots$,apply the polar decomposition $\sqrt{\sqrt{\rho} \sigma \sqrt{\rho}} = \sqrt{\rho}\sqrt{\sigma} U$, $\dots$

But following the definition, if $A= \sqrt{\sqrt{\rho} \sigma \sqrt{\rho}}$, then $A^\dagger A = (\sqrt{\rho} \sigma \sqrt{\rho})^\dagger \sqrt{\rho} \sigma \sqrt{\rho} = \sqrt{\rho} \sigma \rho \sqrt{\rho}$, and the polar decomposition should read

$$\sqrt{\sqrt{\rho} \sigma \sqrt{\rho}} = \sqrt{\sqrt{\rho} \sigma \rho \sqrt{\rho}} U.$$

This seems different from what is mentioned. I'm surely missing something here!

Answer 1

Note that $\sqrt\rho\sigma\sqrt\rho$ is already positive semidefinite$^1$, so defining $A=\sqrt\rho\sigma\sqrt\rho$ leads to $|A|=A$. Not interesting.

Instead, let's define $A:=\sqrt\sigma\sqrt\rho$. Then $A^\dagger A=\sqrt\rho\sigma\sqrt\rho$ and $|A|=\sqrt{A^\dagger A}=\sqrt{\sqrt\rho\sigma\sqrt\rho}$. Thus, $|\sqrt\sigma\sqrt\rho|=\sqrt{\sqrt\rho\sigma\sqrt\rho}$ and by polar decomposition $\sqrt{\sqrt\rho\sigma\sqrt\rho}=\sqrt\sigma\sqrt\rho U$ for some unitary $U$.

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^{$^1$ To see this just consider a quadratic form with it.}