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

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

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


    I have

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

    and also

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

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

    nrqed 3,106
    Science Advisor
    Homework Helper
    Gold Member

    Do you know the relation

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


    It's easy to prove. Just expand out.

    Now, use with [itex] C= B^{n-1} [/itex].
    , that is use [itex] [A,B^n] = B[A,B^{n-1}] + [A,B] B^{n-1} [/itex].
    Now, repeat this again on the first term using now [itex] C= B^{n-2} [/itex]. You will get a recursion formula that will give you the proof easily.
  4. 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!
  5. i also need to answer the same problem for my quantum physics course. thank you.
  6. Avodyne

    Avodyne 1,361
    Science Advisor

    Here's another way:

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

    Can you understand each step?

    Now repeat on the first term on the right. Keep going until you end up with [itex]B^n A[/itex] plus some other stuff. According to the statement of the problem, the other stuff should end up being [itex]n[A,B]B^{n-1}[/itex]. It's crucial that [itex][A,B][/itex] commutes with [itex]B[/itex] to get this final result.
  7. wohow! thanks Avodynefor the reply. got mixed up. my work all ended as a never ending subtraction of powers ofB. thanks for the idea.
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