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  Baker Campbell Haudorff - Fermions

+ 3 like - 0 dislike

It is "well known" that the Baker Campbell Hausdorff formula involves the generating function for the Bernoulli numbers:  \(x/(1-e^{-x})\) -- see for example formula 1.3 of Shlomo Sternberg's book "Lie algebras"  http://www.math.harvard.edu/~shlomo/docs/lie_algebras.pdf -- What would the analogous formula be, for the odd part of a supersymmetric operator? Might it be \(x/(1+e^{-x})\)?  It seems vaguely plausible that it could be, by the following hand-waving argument: Fock space is the same thing as the "tensor algebra", and the Fock space for bosons would be the "symmetric algebra"; whereas for fermions, its the "exterior algebra" -- this is very well-known. Now, the symmetric algebra provides the set of basis vectors for the "universal enveloping algebra" of a Lie algebra -- it gives the same vector space, but its a different algebra (this is the PBW theorem); to get the correct algebra, one has to create something called the "algebra of symbols" or the "star algebra".  For the special case of the Lie algebra being the Heisenberg algebra, the star-product is more famously known as the "Moyal poduct". ... anyway... The BCH formula occurs during the formulation of how the star-product actually works -- there are some closed-form formulas giving this, the earliest one from Berezin.  It is here that the generating function for the Bernoulli numbers appears.  Now, all of this construction should go through, in exactly the same way, for the Lie superalgebras -- just keep track of the signs.  So ... what happens when one finally arrives at the equivalent of the Moyal product?  Is the sign flipped, as I write above?

If so, what does this imply for Fermi-Dirac statistics vs. Bose-Einstein statistics?  The superficial resemblance is just too painful to bear, and the pencil-scratching is just slightly too long and tedious to carry out... Is there anything to this?  Is it perhaps even well-known to those who know these things?  What is the physical interpretation of going from the variable x to E/kT?

asked Sep 25, 2016 in Mathematics by linas (85 points) [ revision history ]

http://scitation.aip.org/content/aip/journal/jmp/27/1/10.1063/1.527340 contains a BCH formula for a supergroup. This might be of interest in your context.

To check the validity of your conjecture, you could take one of the standard proofs of the BCH formula and rewrite it for the fermionic case....

Yes, I could do that, I was simply being lazy, posting the question, wondering perhaps if the answer was "well-known", before I plodded my way through. To amend/clarify the question, it should read "What is BCH for Jordan algebras?"

Its surely not widely known, so you don't waste time by trying to do it. 

OK, working on it.

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