If something commutes with $x$, it commutes with any $f(x)$

Question

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.

Answer 1

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.