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Let's say I have 2 density operators $A$ and $B$. Now, here is what I am trying to calculate: $$\newcommand{\tr}{\operatorname{trace}} \tr(A \sqrt{B} A \sqrt{B}). $$ I saw that this trace can be rewritten as: $$ \tr(A \sqrt{B} A \sqrt{B}) = \tr\Bigg(\Big(\sqrt{\sqrt{B}} A \sqrt{\sqrt{B}}\Big)^2\Bigg). $$

I was wondering, which property of trace is being used here. I do not think the cyclic property would help, would it?

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This follows from the cyclicity of trace.

\begin{align} \text{tr}\left(\left(\sqrt{\sqrt{B}}A\sqrt{\sqrt{B}}\right)^2\right) &= \text{tr}\left(\sqrt{\sqrt{B}}A\sqrt{\sqrt{B}}\sqrt{\sqrt{B}}A\sqrt{\sqrt{B}}\right)\\ &= \text{tr}\left(\sqrt{\sqrt{B}}A\sqrt{B}A\sqrt{\sqrt{B}}\right)\\ &= \text{tr}\left(A\sqrt{B}A\sqrt{\sqrt{B}}\sqrt{\sqrt{B}}\right)\\ &= \text{tr}\left(A\sqrt{B}A\sqrt{B}\right) \end{align}

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  • $\begingroup$ nice! I missed the idea... $\endgroup$ – Hasan Iqbal Oct 16 at 17:35

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