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A tractable Baker-Campbell-Hausdorff formula

  1. Aug 24, 2011 #1
    1. Let A and B be two matrices, and [itex]\lambda[/itex] be a continuous parameter.



    2. Now, define a function [itex]f(\lambda) \equiv e^{\lambda A}e^{\lambda B}[/itex]. We need to show that [itex]\frac{df}{d\lambda} = \left\{A + B + \frac{\lambda}{1!}[A, B] + \frac{\lambda^2}{2!}[A, [A, B]] + ... \right \}f[/itex]

    Once this is shown, setting [itex]\lambda = 1[/itex], and [itex][A, [A, B]] = [B, [A, B]] = 0[/itex] gives us a Baker-Campbell-Hausdorff formula.


    3. I had shown this result quite a while ago, but now I have forgotten completely what I had done. This time, I tried differentiating [itex]f(\lambda)[/itex] w.r.t the argument, and then using the commutation was able to get the first two terms on the R.H.S., but thereafter I got stuck. The very minimal hint would be all that I need.

    Thank you!
     
    Last edited: Aug 24, 2011
  2. jcsd
  3. Aug 28, 2011 #2

    diazona

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    Without seeing your exact steps I can't say much, but you may need to expand out an exponential or two and work out some commutators term-by-term.
     
  4. Aug 28, 2011 #3
    I solved it. Thanks anyway!
     
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