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

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!

• – glS
Commented Aug 1 at 14:34
• also you're missing a square root in the post I think. Your $A$ isn't compatible with your expression for $A^\dagger A$. The point here is to observe $|\sqrt\rho\sqrt\sigma|=\sqrt{\sqrt\sigma \rho \sqrt\sigma}$
– glS
Commented Aug 1 at 14:35

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 $$U$$.
$$^1$$ To see this just consider a quadratic form with it.
• Thanks @Adam, I think with your definition of $A$, the polar decomposition should read $\sqrt{\sigma}\sqrt{\rho} = \sqrt{ \sqrt{\rho} \sigma \sqrt{\rho}} U$, not the other way around. Commented Aug 1 at 14:20