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Commutator [A^n,B] = ?

  1. Oct 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that [AnB] =nAn-1[A,B] for integrer n , assume [A,[A,B]]=0=[B,[A,B]]

    2. Relevant equations
    [A,B]=AB-BA


    3. The attempt at a solution
    Does anyone know how i should go to manipulate the exponent n ? I have tried to search but found nothing about a commutator like this properties.
     
  2. jcsd
  3. Oct 14, 2012 #2
    Use the Principle of Mathematic induction and the commutator rule:
    [tex]
    \left[ A \, C, B \right] = A \, \left[ C, B \right] + \left[A, B \right] \, C
    [/tex]
     
  4. Oct 14, 2012 #3
    you can play as long as you ant with this formula ,but you will never be able to get what is asked except if you differentiate somehow ,becsuse you need have one n ,that is not exponent of A or B
     
  5. Oct 14, 2012 #4
    You will also need to prove as an intermediate step that:
    [tex]
    \left(\forall n \in \mathbb{N} \right) \left[ A^{n}, \left[ A, B \right] \right] = 0
    [/tex]
    For this, use the principle of mathematical induction again and what you are given in the problem.
     
  6. Oct 14, 2012 #5

    gabbagabbahey

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    Homework Helper
    Gold Member

    By this logic, the only way to get [itex]3x^2[/itex] from [itex]x^3[/itex] is to differentiate. Does that really seem like a reasonable thing to claim? (What happens if you multiply [itex]x^3[/itex] by [itex]\frac{3}{x}[/itex], for example?

    Instead of differentiating your commutator somehow (Have you even defined how to differentiate an operator? Are your operators differentiable?), I suggest you follow DickFore's advice and use a proof by induction. I'd start by looking at a fairly simple case like [itex]n=2[/itex].
     
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