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I often read that if some operator commutes with (say) the position operator $x$, it commutes with any function $f(x)$ of it. For analytical functions, this is obvious, but I was wondering if there was any way to prove that for a bigger set of functions.

Edit: Since that question has come up a lot, here is how I would define $f(x)$: I'd just say that for any function $f:\mathbb{R}\rightarrow \mathbb{C}$ you can define the operator $f:L_2 \rightarrow L_2, \psi \mapsto f \cdot \psi$ where $\cdot$ means point-wise multiplication. I'm aware that for operators other than the position operator this definition does not work, so maybe I'm asking only for the position operator. I'm not really familiar with functional analysis, so excuse my ignorance.

Tarik
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You can prove this theorem. (Last part of theorem 9.41 in my book Spectral Theory and Quantum Mechanics 2nd ed. 2017 Springer)

THEOREM Let $H$ be a complex Hilbert space, $A :D(A) \to H$ a generally unbounded selfadjoint operator, and $B:H\to H$ a bounded operator. The following facts are equivalent.

  • $BA x = ABx$ for every $x\in D(A)$;
  • $Bf(A)x = f(A)Bx$ for every $x\in D(f(A))$ and every Borel measurable map $f: \mathbb{R}\to \mathbb{C}$;
  • $B P(E) = P(E)B$ for every real Borel set $E$, where $P$ is the spectral measure of $A$.
  • $Be^{itA}= e^{itA}B$ for every $t\in \mathbb{R}$.

If $B$ is not bounded and everywhere defined, then the facts above are not equivalent in general. However if $B$ is selfadjoint and generally unbounded, there are similar equivalent properties (not so strong actually). See chapter 9 of my book quoted above.

Valter Moretti
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  • Thank you! But the position operator is not bounded, as far as I know. I looked into your book and it sounds really interesting and is definitely on my reading list. However, it seams really complicated, so until I find the time, could you maybe spell out how exactly that theorem works in case of unbounded hermitian operators such as the position operator? – Tarik Oct 04 '23 at 22:38
  • In my statement above $A$ is not bounded in general, so you can apply the result to $A= X$. – Valter Moretti Oct 05 '23 at 04:53
  • Right, sorry, I don't know how I could miss that. But could you maybe spell out how that works for unbounded, self-adjoint B? Because as far as I know almost every physical operator is unbounded, so your theorem would have little applications. – Tarik Oct 05 '23 at 20:40
  • Sorry,it would amount to a course on spectral theory.... – Valter Moretti Oct 05 '23 at 20:44
  • However, for unbounded operarors $U$ it is generally false that if it commutes (on suitably domains) with $X$ then it also commutes with $f(X)$. Nevertheless, this fact is quite irrelevant in quantum theories. What should commute for having some relevant physical consequences are the projector-valued measures associated to the operators, not the operators themselves. – Valter Moretti Oct 06 '23 at 05:59
  • Okay, thank you! Then I guess I'll just accept that. – Tarik Oct 09 '23 at 20:33