Given $[A,B]=0$ then $[A,f(B)]=0$

Question

Given $A$ and $B$ two operators that commute ($[A,B]=0$) then $A$ commutes with an arbitrary function of $B$

I recently saw this property of the commutators on a quantum mechanics course. We didn't get any proof on the property and we didn't get told wether this propery is true for any function $f(B)$ or if it needs to fulfil some conditions.

I've tried to prove it on my own but I have no idea on how to approach the problem as I haven't taken any course on operator theory.

I'd appreciate if anyone could prove this here and if anyone could tell me if this is true for any function.

Thanks in advance.

Answer 1

I fear that by asking this here instead of PSE, you are inviting people to overthink it. Almost certainly your context assumes $f(x)$ is well behaved, analytic, etc, and is assumed to amount to its Taylor expansion around the origin. So, by linearity of the commutator, you are supposed to prove that A commutes with all monomials $B^n$.

You can prove this by induction, given the identity $$ [A,BC]=[A,B]C + B[A,C]. $$ Now take $C=B^n$ and apply induction.

It's not meant to be rocket science; this is a routine identity you'll be using more than once a week.