Commutator Identity

  1. 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. gabbagabbahey

    gabbagabbahey 5,009
    Homework Helper
    Gold Member

    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. 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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