Please, does anyone knows whether $\operatorname{tr} |AB| = \operatorname{tr} |BA|$ for general trace operators $A$ and $B$ in Hilbert spaces ? Do you have a counter-example or a proof ?
Thanks
ps : I know $\operatorname{tr} AB = \operatorname{tr} BA$
Consider the (randomly chosen) $2 \times 2$ matrices $$ A = \pmatrix{0 & 2\cr 2 & 3\cr},\ B = \pmatrix{1 & 0\cr -3 & -2}$$ I get $$|AB| = \sqrt{(AB)^* AB} = \frac{1}{\sqrt{17}} \pmatrix{31 & 22\cr 22 & 20}$$ $$ |BA| = \sqrt{(BA)^* BA} = \frac{1}{\sqrt{5}} \pmatrix{4 & 8\cr 8 & 26}$$ with traces $3\sqrt{17}$ and $6 \sqrt{5}$ which are different.
For a very easy example, take $$A=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\ \ B=\begin{bmatrix} 0&1\\0&0\end{bmatrix}.$$ Then $$ AB=B,\ \ BA=0. $$ Thus, as $|AB|=|B|=I-A$, we have $$0=\operatorname{Tr}(|BA|)\ne1=\operatorname{Tr}(|AB|).$$
