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Commutator Algebra proof

  1. Dec 29, 2011 #1
    1. Prove that [tex] [A,B^n] = nB^{n-1}[A,B] [/tex]

    Given that: [tex] [[A,B],B] = 0 [/tex]

    My Atempt to resolution

    We can write that:
    [tex] [[A,B],B] = [A,B]B-B[A,B] = 0 [/tex]

    So we get that: [tex] [A,B]B = B[A,B] [/tex]

    After some working several expansions, and considering that [tex] [X,YZ] = Y[X,Z] + [X,Y]Z [/tex]

    I arrived at this expression:

    [tex] [A,B^n] = B^{n-1}[A,B]+B^{n-2}[A,B]+[A,B^{n-2}]B^2 [/tex]

    But from here I'm a bit lost on how to get the desired result.
    So, could anyone help me?

    Thanks a lot!
     
  2. jcsd
  3. Dec 30, 2011 #2

    vela

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    Prove it using induction.
     
  4. Jan 1, 2012 #3
    yes, i've tried that, but i cant get to the desired result...

    could you help me?

    thx again
     
  5. Jan 1, 2012 #4

    vela

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    Show us what you have so far.
     
  6. Jan 1, 2012 #5
    what i have is the last expression in my first post.
    so, by induction, and starting from this expression, i cannot get what i want to prove.
     
  7. Jan 1, 2012 #6

    vela

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    Do you know how to do a proof by induction?
     
  8. Jan 1, 2012 #7

    Fredrik

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    This result contradicts the formula you say that you're using. (Think X=A, Y=Bn-2, Z=B2).
     
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