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

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

    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 3,048
    Science Advisor
    Homework Helper

    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 1,276
    Science Advisor

    Here's another way:

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