Commutator Problem: Show [A,Bn] = cnBn-1

[Calu]

Homework Statement

Let the commutator [A,B] = cI, I the identity matrix and c some arbitrary constant.

Show [A,Bⁿ] = cnB^n-1

Homework Equations

[A,B] = AB - BA

The Attempt at a Solution

So I have started off like this:

[A,Bⁿ] = ABⁿ - BⁿA = cI

I'm not sure where to go from here.

 

[HallsofIvy]

I think you are misunderstanding the question. [A,B^n] = AB^n - B^nA = cI is NOT in general true. You seem to be thinking that "AB- BA= cI" is to be true for all A, B. It is not. In this problem AB- BA= cI is true for this specific A and B.

You are told that AB- BA= cI. So [A, B^2]= AB^2- B^2A= AB^2- BAB+ BAB- B^2A= (AB- BA)B+ B(AB- BA)= cB+ Bc= 2cB. etc. Use proof by induction.

 

[Dick]

Homework Statement

Let the commutator [A,B] = cI, I the identity matrix and c some arbitrary constant.

Show [A,Bⁿ] = cnB^n-1

Homework Equations

[A,B] = AB - BA

The Attempt at a Solution

So I have started off like this:

[A,Bⁿ] = ABⁿ - BⁿA = cI

I'm not sure where to go from here.

For any operators you have the identity $[D,EF]=[D,E]F+E[D,F]$ that's handy and it's easy to prove. The case $n=1$ is obvious, so now try $n=2$. Write $[A,B^2]$ as $[A,BB]$. For the general case think about induction.