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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!

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    $\begingroup$ Does this answer your question? Conjugate complex of linear operators in quantum mechanics $\endgroup$
    – Tobias Fünke
    Jan 14 at 17:54
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    $\begingroup$ 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 $\endgroup$
    – bsaoptima
    Jan 14 at 18:01
  • $\begingroup$ 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"? $\endgroup$
    – Tobias Fünke
    Jan 14 at 18:03
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    $\begingroup$ If you look above : "The question is whether $\hat{A}\hat{B}$ corresponds to an observable?" $\endgroup$
    – bsaoptima
    Jan 14 at 18:07
  • $\begingroup$ "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" $\endgroup$
    – Tobias Fünke
    Jan 14 at 18:09

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I am also new to mathematical quantum mechanics. During the study, I found the distinction between Observable and the mathematical concept corresponding to an Observable, self-adjoint Operator, to be very important.

So, since you didn't clarify what do you mean by Observable, I really need further clarification in order to answer that.

But your thought process is as far correct. A and B are operators representing observables for themselves, since they commute they share the same set of eigenvectors which can span the entire space. Since the product is hermitian (self adjoint once more) the eigenvalue is real. So, if I understand correctly your actual question would be, the self adjoint Operator (AB), does it necessarily always represent a physical observable right?

BR

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  • $\begingroup$ a side remark on your question of how to determine whether two operators commute, you observe the commutator of them say AB-BA, and you let it act on any test function and see whether the result gives you 0. If so, then they commute. $\endgroup$
    – Tianjiao Li
    Jan 14 at 19:02

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