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

1. Mar 12, 2015

### rogeralms

1. The problem statement, all variables and given/known data

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].

3. The attempt at a solution
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. Mar 12, 2015

### TSny

When using induction, you want to show that the relation holds for n+1 assuming it holds for n.

Last edited: Mar 13, 2015
3. Mar 13, 2015

### rogeralms

Thank you but I know how induction works. That is why I set up the problem with n+1.

I need a hint on how to manipulate the commutators to show this is true, that [A,Bn+1]=(n+1)Bn [A,B].

4. Mar 13, 2015

### TSny

You're showing the relation is true for n+1 assuming it is true for n. You have not yet used the assumption that it is true for n. Note on the right hand side you have a term [A,Bn]B. Simplify this term.

5. Mar 15, 2015

### rogeralms

I overlooked the part in the problem statement that said that A and B both commute.

Then we have
Assume [A,Bn]=nBn-1[A,B] is true
Show that it is true for n+1 [A,Bn+1]=(n+1)Bn[A,B] is true.
[A,Bn+1]=[A,BnB]=Bn[A,B]+[A,Bn]B=Bn[A,B]+B[A,Bn]
=Bn[A,B]+BnBn-1[A,B]=Bn[A,B]+nBn[A,B]=(n+1)Bn[A,B]

6. Mar 15, 2015

### TSny

A and B are not assumed to commute. You are given that A commutes with [A, B]. Also, B commutes with [A, B]. But A does not necessarily commute with B.

OK. One thing that you might consider is that in getting to the end of your first line you used [A,Bn]B= B[A,Bn]. It might not be clear to the reader why this is justified. You can avoid this by rewriting [A,Bn]B in a different way using the assumption that [A,Bn]=nBn-1[A,B] is true.