Show that [A,B^{n}]=nB^{n-1}[A,B]

[dimensionless]

I'm having trouble figuring out the following commutator relation problem:

Suppose A and B commute with their commutator, i.e., [B,[A,B]]=[A,[A,B]]=0. Show that

[A,B^{n}]=nB^{n-1}[A,B]

I have

[A,B^{n}] = AB^{n} - B^{n}A

and also

[A,B^{n}] = AB^{n} - B^{n}A = ABB^{n-1} - BB^{n-1}A

I don't know where to go from here. I'm not positive the above relation is correct either.

[nrqed]

I'm having trouble figuring out the following commutator relation problem:

Suppose A and B commute with their commutator, i.e., [B,[A,B]]=[A,[A,B]]=0. Show that

[A,B^{n}]=nB^{n-1}[A,B]

I have

[A,B^{n}] = AB^{n} - B^{n}A

and also

[A,B^{n}] = AB^{n} - B^{n}A = ABB^{n-1} - BB^{n-1}A

I don't know where to go from here. I'm not positive the above relation is correct either.

Do you know the relation

[A,BC] = B[A,C] + [A,B] C

?

It's easy to prove. Just expand out.

Now, use with C= B^{n-1} .
, that is use [A,B^n] = B[A,B^{n-1}] + [A,B] B^{n-1} .
Now, repeat this again on the first term using now C= B^{n-2} . You will get a recursion formula that will give you the proof easily.

[Frogs4U2]

Help! I need to do this same exact problem. But I just don't understand what to do so I was wondering if you wouldn't mind showing the steps that you explained to do in order to get the final solution. Thank you so much!

[angie_liamzon]

i also need to answer the same problem for my quantum physics course. thank you.

[Avodyne]

Here's another way:

AB^n = (AB)B^{n-1}
=(BA+[A,B])B^{n-1}
=BAB^{n-1} + [A,B]B^{n-1}

Can you understand each step?

Now repeat on the first term on the right. Keep going until you end up with B^n A plus some other stuff. According to the statement of the problem, the other stuff should end up being n[A,B]B^{n-1}. It's crucial that [A,B] commutes with B to get this final result.

[angie_liamzon]

wohow! thanks Avodynefor the reply. got mixed up. my work all ended as a never ending subtraction of powers ofB. thanks for the idea.