As far as I understood in quantum mechanics two operators can commute even though they are not functionally independent, which means that they can depend on the same independent variable.

On the other hand I am not 100% sure about this, I've never seen two operators that depend on the same variable commuting between them. If my statement is correct, do you have any example of such operators?

  • 1
    $\begingroup$ What do you mean exactly here? Can you give examples? $\endgroup$ Sep 12, 2023 at 11:45
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    $\begingroup$ Presumably OP means e.g. $[AA,AAA]=0$. $\endgroup$
    – Qmechanic
    Sep 12, 2023 at 11:57

1 Answer 1


Actually, if you restrict attention to the finite-dimensional version of the problem, it is the other way around: two self-adjoint operators commute if and only if the are polynomials of the same operator.

To prove the "hard part" of this, a simple argument would be the following: two commuting self-adjoint operators share a common eigenbasis $(v_i)_{i \in I}$ for some set $I$, choose any family $(\lambda_i)_{i \in I}$ of pairwise distinct real numbers. Now take the two Lagrange polynomials that take the $\lambda_i$'s to the eigenvalues of the two operators and you're good.

The same should be true for self-adjoint operators on infinite-dimensional operators, but there are functional analytic problems to solve...


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