It is proven that if two operators $\hat{X}$ and $\hat{Y}$ commute, then the multiplication of them will be hermitian, i.e. if $\hat{X}\hat{Y}=\hat{Y}\hat{X}$, then $\left(\hat{X}\hat{Y}\right)^\dagger=\hat{X}\hat{Y}$.

My question is that is the opposite true as well? In other words, if the multiplication of two operators is hermitian, then will they commute? If yes, I need a proof.

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    $\begingroup$ The identity operator commutes with every other operator, including non-Hermitian ones. Therefore, the first statement is false. I suspect the second is false as well. Perhaps you meant to say that if two Hermitian operators commute, then their product is Hermitian? $\endgroup$ – march Jan 4 at 6:00
  • $\begingroup$ In fact I saw the relationship in a book. I wanted to know if the opposite is true as well. $\endgroup$ – farhad mabrooki Jan 4 at 6:12
  • $\begingroup$ This question is about math, not physics. The answer does not depend on any theory of physics. $\endgroup$ – G. Smith Jan 4 at 6:46

The statement of your assertion is a little off, but essentially yes. The correct statement is that two Hermitian operators must commute if their product is also Hermitian. The proof is entirely straightforward as a Hermitian product implies $XY=(XY)^\dagger$ but $$ (XY)^\dagger=Y^\dagger X^\dagger=YX $$ using that $X$ and $Y$ are both Hermitian themselves. Hence, $XY=YX$.

So, in fact the full statement of the theorem would be given two Hermitian operators $X$ and $Y$, the operators commute if and only if their product is also Hermitian.


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