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Looking for a proof regarding Baker-Campell-Hausdorff

  1. Sep 16, 2013 #1
    Hello!
    Is there any nice proof that if [A,B] belongs to the same vectorspace as A and B, then C is in the same vectorspace to all orders, given that
    [tex]e^Ae^B = e^C[/tex]
    ?
    It is obvious to the second order but at higher orders it seems as if terms will cancel but I cant prove it.

    Thanks!
     
  2. jcsd
  3. Sep 16, 2013 #2

    fzero

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    Yes, it's actually not enough for ##C## to be in the same vector space. The quantities ##A,B,\[A,B\]## and higher commutators actually close as a Lie algebra. This is still a vector space, but might be much smaller than the original vector space. ##C## will be in this Lie algebra. I don't have a reference for a detailed proof at hand, but the wiki discussion has an outline based on what is called Friedrichs' theorem.
     
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