While reading some papers on Quantum Information theory, I observed that several theorems required working with square roots of non-negative operators. So I thought of proving some operator results so as to get comfortable. So far I've been successful, but I am having some trouble proving the following assertion.
If $A$ and $B$ are non-negative bounded operators on a Hilbert space over $\mathbb{C}$, then $A$ and $B$ commute iff $\sqrt{A}$ and $\sqrt{B}$ commute.
I was able to show $\sqrt{A}$,$\sqrt{B}$ commute $\Rightarrow$ $A$,$B$ commute via $$AB = \sqrt{A}\sqrt{A}\sqrt{B}\sqrt{B}=\sqrt{A}\sqrt{B}\sqrt{A}\sqrt{B}=\cdots=\sqrt{B}\sqrt{B}\sqrt{A}\sqrt{A}$$
As for the reverse implication, I did the following $$AB=BA \iff \sqrt{A}\sqrt{A}\sqrt{B}\sqrt{B}=\sqrt{B}\sqrt{B}\sqrt{A}\sqrt{A}$$
This expansion didn't get me anywhere. So I thought of proving $\sqrt{AB}=\sqrt{A}\sqrt{B}$. By uniqueness of square root, $\sqrt{AB}=\sqrt{BA}=\sqrt{B}\sqrt{A}$ and I would be done. I'm not sure how to prove this though. Squaring rhs gives $(\sqrt{A}\sqrt{B})^2=\sqrt{A}\sqrt{B}\sqrt{A}\sqrt{B}$. But that looks just as convoluted.
I'd appreciate any hints on how to proceed.
If $A$ and $B$ commute, then $B$ commutes with $p(A)$ for all polynomials $p$. Since $A^\frac{1}{2}$ is a limit of polynomials of $A$, it follows that $B$ commutes with $A^\frac{1}{2}$. By the same reasoning, it follows that $A^\frac{1}{2}$ commutes with $B^\frac{1}{2}$.