I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.
If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.
$$\therefore AB=BA$$
$$n=1; [A,B^1] = (1)B^0[A,B] = e$$ This statement is certainly true. however moving on to $n = 2$ I find...
$$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$
Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:
$$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$
Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance.
Answer
It seems that the question (v1) is caused by the fact that there are two different notions of the commutator:
One for group theory: $$\tag{1} [A,B] ~:=~ ABA^{-1}B^{-1}$$ (or sometimes $[A,B] := A^{-1}B^{-1}AB$, depending on convention), which is relatively seldom used in physics.
One for rings/associative algebras: $$\tag{2} [A,B]:=AB-BA,$$ which is the definition usually used in physics. (This latter definition (2) generalizes to a supercommutator in superalgebras.)
The identity
$$\tag{*} [A,B^n] ~=~ nB^{n-1}[A,B]$$
holds in the latter sense (2), if $[[A,B],B]=0$. (It is not necessary to demand $[A,[A,B]]=0$.) More generally, for a sufficiently well-behaved function $f$, we have
$$\tag{**} [A,f(B)] ~=~ f^{\prime}(B)[A,B], $$
if $[[A,B],B]=0$.
The group commutator (1) is dimensionless, which (among other things) makes the identity (*) unnatural to demand for group commutators.
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