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

  1. Jun 22, 2006 #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. Jun 22, 2006 #2


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    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. Sep 28, 2009 #3
    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. Jan 27, 2011 #4
    i also need to answer the same problem for my quantum physics course. thank you.
  6. Jan 27, 2011 #5


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    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. Jan 30, 2011 #6
    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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