How to Prove [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.