- #1
carvas
- 6
- 0
1. Prove that [tex] [A,B^n] = nB^{n-1}[A,B] [/tex]
Given that: [tex] [[A,B],B] = 0 [/tex]
My Atempt to resolution
We can write that:
[tex] [[A,B],B] = [A,B]B-B[A,B] = 0 [/tex]
So we get that: [tex] [A,B]B = B[A,B] [/tex]
After some working several expansions, and considering that [tex] [X,YZ] = Y[X,Z] + [X,Y]Z [/tex]
I arrived at this expression:
[tex] [A,B^n] = B^{n-1}[A,B]+B^{n-2}[A,B]+[A,B^{n-2}]B^2 [/tex]
But from here I'm a bit lost on how to get the desired result.
So, could anyone help me?
Thanks a lot!
Given that: [tex] [[A,B],B] = 0 [/tex]
My Atempt to resolution
We can write that:
[tex] [[A,B],B] = [A,B]B-B[A,B] = 0 [/tex]
So we get that: [tex] [A,B]B = B[A,B] [/tex]
After some working several expansions, and considering that [tex] [X,YZ] = Y[X,Z] + [X,Y]Z [/tex]
I arrived at this expression:
[tex] [A,B^n] = B^{n-1}[A,B]+B^{n-2}[A,B]+[A,B^{n-2}]B^2 [/tex]
But from here I'm a bit lost on how to get the desired result.
So, could anyone help me?
Thanks a lot!