1. ## Prove commutator [A,B^n]=nB^(n-1)[A,B]

Let A and B be two observables that both commute with their commutator [A,B].

a) Show, e.g., by induction, that [A,Bn]=nBn-1 [A,B].

Prove for n=1
[A,B1]=1B1-1 [A,B].
[A,B]=B0[A,B]=[A,B]

Show that it is true for n+1
[A,Bn+1]=[A,BnB]=Bn[A,B]+[A,Bn]B

I am not sure how to continue. Could someone give me a hint what I should do next?

Thank you.

2. ## Re: Prove commutator [A,B^n]=nB^(n-1)[A,B]

From $[A,B^{n+1}]=[A,B^nB]=[A,B^n]B+B^n[A,B]$, apply the induction hypothesis, the fact that $B$ commutes with $[A,B]$ and standard ring operations.

3. ## Re: Prove commutator [A,B^n]=nB^(n-1)[A,B]

What you need is an earlier result:

$[A,BC] = [A,B]C + B[A,C]$ (this shows that $\delta_A: R \to R$ given by $\delta_A(B) = [A,B]$ is a derivation on your ring $R$).

Explicitly:

$A(BC) - BC(A) = ABC - BCA = ABC - BAC + BAC - BCA = (AB - BA)C + B(AC - CA)$

Then, taking $C = B^n$, we have:

$[A,B^{n+1}] = [A,B]B^n + B[A,B^n] = B^n[A,B] + B[A,B^n]$ (from the fact that $B$ and thus any power of $B$, commutes with $[A,B]$)

$= B^n[A,B] + B(nB^{n-1}[A,B])$ (from our induction hypothesis)

$= B^n[A,B] + nB^n[A,B] = (n+1)B^n[A,B]$ (distributive law)

This is just the usual "power rule" from first-year calculus generalized to the derivation $\delta_A$: it says that:

$\delta_A(B^n) = nB^{n-1}\delta_A(B)$, or as might look more familiar: $d(x^n) = nx^{n-1}dx$ (commutativity is a non-issue in the real numbers).

4. ## Re: Prove commutator [A,B^n]=nB^(n-1)[A,B]

Thank you both so much. I did not know that B commutes with [A,B]. I could not find that in the notes. It all makes perfect sense now.
Just saw that B commutes with [A,B] in the problem statement. Well, this is what kills me, I overlook something. Thanks again.

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