# Condition of the product of 2 operators being an observable [duplicate]

I'm trying to understand a bit the conditions of operators commuting, or themselves being an observable.

Here I have the operator $$\hat{A}$$ which has Eigenvalues $$-1,+1$$ and Eigenstates $$|u_1\rangle, |u_2\rangle$$ and operator $$\hat{B}$$ that also has Eigenvalues $$-1,+1$$ and Eigenstates which are linear combinations of $$|u_1\rangle, |u_2\rangle$$.

The question is whether $$\hat{A}\hat{B}$$ corresponds to an observable?

Here is my thought process: If it is the case then the product must be Hermitian. Now I now that operators commute only if their product is Hermitian but I don't know if that works the other way around. My idea is that since commuting operators share eigenstates and that $$\hat{B}$$ somewhat uses the Eigenstates of $$\hat{A}$$, then they commute and the product is an observable.

What do you think?

Many thanks!

• Does this answer your question? Conjugate complex of linear operators in quantum mechanics Jan 14 at 17:54
• It answers it partially. I still don't know how to determine if the operators commute. My opinion is since their eigenstates are linked then they are. @TobiasFünke Jan 14 at 18:01
• I don't understand; I thought you asked whether or not it holds that two hermitian operators commute if and only if their product is hermitian, no, which is what I read from "I don't know if that works the other way around"? Jan 14 at 18:03
• If you look above : "The question is whether $\hat{A}\hat{B}$ corresponds to an observable?" Jan 14 at 18:07
• "Here is my thought process: If it is the case then the product must be Hermitian. Now I now that operators commute only if their product is Hermitian but I don't know if that works the other way around" Jan 14 at 18:09