If the multiplication of two operators is hermitian, then will they commute?

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.

• 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? – march Jan 4 at 6:00
• In fact I saw the relationship in a book. I wanted to know if the opposite is true as well. – farhad mabrooki Jan 4 at 6:12
• This question is about math, not physics. The answer does not depend on any theory of physics. – 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.