Looking for a proof regarding Baker-Campell-Hausdorff

  • Thread starter Kontilera
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  • #1
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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!
 

Answers and Replies

  • #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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