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Commutator Identity

  1. Oct 5, 2010 #1


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    1. The problem statement, all variables and given/known data

    Prove the following identity:

    [tex]e^{x \hat A} \hat B e^{-x \hat A} = \hat B + [\hat A, \hat B]x + \frac{[\hat A, [\hat A, \hat B]]x^2}{2!}+\frac{[\hat A,[\hat A, [\hat A, \hat B]]]x^3}{3!}+...[/tex]

    where A and B are operators and x is some parameter.

    2. Relevant equations
    [tex] e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...[/tex]
    [tex] e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...[/tex]

    3. The attempt at a solution

    [tex] \hat B e^{-x \hat A} = \hat B - [\hat B, \hat A]x + \frac{[[\hat B, \hat A], \hat A] x^2}{2!}+...[/tex]

    It seems after I rearrange the commutation orders, the signs all become positive and this is the required result, so I know I must be doing something wrong. I think it has to do with how I'm multiplying out B into the series..

    i.e. [tex] \hat B (\hat A x)^2 = [\hat B, \hat A \hat A] x^2....??[/tex]

    or [tex] \hat B (\hat A x)^2 = [[\hat B, \hat A], \hat A] x^2...??[/tex]
  2. jcsd
  3. Oct 5, 2010 #2


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    Neither is correct.


    You won't have anything involving commutators until you multiply by both exponentials and collect terms in powers of the parameter [itex]x[/itex].
  4. Oct 5, 2010 #3


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    ahh I see it now I think..

    [tex]e^{x \hat A} \hat B e^{-x \hat A} = \left ( 1+\hat A x + \frac{1}{2} \hat A^2 x^2 + \frac{1}{6} \hat A^3 x^3+... \right ) \left ( \hat B - \hat B \hat A x + \frac{1}{2} \hat B \hat A^2 x^2 - \frac{1}{6}\hat B \hat A^3 x^3+... \right ) [/tex]

    So I multiply this out, collect terms in powers of x, and simplify to the commutator relations

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