Question

Prove that commutator [a,b^n] =b^j [a,b] b^n-j-1

Answer 1

Answer:

am trying to show that [A,Bn]=nBn−1[A,B][A,Bn]=nBn−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−1B−1=e[A,B]=ABA−1B−1=e where e is the identity element of the group.

∴AB=BA∴AB=BA

n=1;[A,B1]=(1)B0[A,B]=en=1;[A,B1]=(1)B0[A,B]=e

This statement is certainly true. however moving on to n=2n=2 I find...

[A,B2]=AB2A−1B−2=ABBA−1B−1B−1=BBAA−1B−1B−1[A,B2]=AB2A−1B−2=ABBA−1B−1B−1=BBAA−1B−1B−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: