# Prove $[A,B^n] = nB^{n-1}[A,B]$

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.

• I'm confused. Isn't $[A,B] = AB - BA$? – Prahar Sep 22 '13 at 19:27

It seems that the question (v1) is caused by the fact that there are two different notions of the commutator:

1. 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.

2. 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.

• I got it, you have to use the identity [A,BC] = [A,B]C + [B,A]C coupled with what you said above. I am so foolish. Thanks! – user28823 Sep 22 '13 at 19:46
• if $[A,B]$ doesn't commute with $B$, then does it go before or after $B$? – psitae Jun 11 '18 at 16:38
• Eqs. ($\ast$) & ($\ast\ast$) do generically not hold in either ordering (or even linear combinations thereof) if $[A,B]$ doesn't commute with $B$. – Qmechanic Jun 11 '18 at 17:21

Maybe you could show this relation in an other way:

$[A,B^n]=[A,B^{n-1}B]=B^{n-1}[A,B]+[A,B^{n-1}]B=...$

Then you take $[A,B^{n-1}]B$ and repeat the process:

$[A,B^{n-1}]B=[A,B^{n-2}B]B=B^{n-2}[A,B]B+[A,B^{n-2}]BB$

So $B$ commute with $[A,B]$

$[A,B^{n-1}]B=B^{n-1}[A,B]+[A,B^{n-2}]B^2$

Repeat for $n$ steps

$[A,B^n]=B^{n-1}[A,B]+B^{n-1}[A,B]+...+B^{n-1}[A,B]=nB^{n-1}[A,B]$

Here's another way to prove this relationship, by induction:

1. Verify that the statement holds for $n = 1$

$$[A,B^1]=1 \cdot B^{1-1} [A,B] = [A,B]$$

1. Show that, if the formula holds for $n=k ~ \big(I\big)$, then it also holds for $n=k+1$, using the identity $[X, YZ] = [X,Y]Z + Y[X,Z] ~\big(II\big)$ and the fact that $B$ commutes with $[A,B] ~\big(III\big)$

\begin{align} [A,B^{(k+1)}] &= [A,B^k B] \\ & = [A,B^k]B + B^k[A, B] ~\big(II\big)\\ & = k B^{k-1}[A,B]B + B^k[A, B] ~\big(I\big) \\ & = k B^{k-1}B[A,B] + B^k[A, B] ~\big(III\big) \\ & = k B^k[A,B] + B^k[A, B] \\ & = (k + 1) B^k[A,B] \\ & = (k + 1) B^{(k+1)-1}[A,B] \\ \end{align}

1. Since both the basis and the inductive step have been performed, by mathematical induction, the statement holds for all natural numbers $n$. $\Box$