Canonical Bose-Einstein statistics

quetzalcoatl9
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I've been curious as to why Bose-Einstein statistics are always derived using the grand canonical partition function. Yes, I know it is easier, but there must also be an expression for the canonical ensemble. However, I was suprised that I have been unable to find it in the standard sources - so here is my own (troubled) derivation.

I start with the grand canonical partition function:

\sum^{0,1,..,M}_{\{n_k\}} \prod^{\infinty}_{k=1} e^{-\beta\left(\epsilon_k - \mu\right) n_k

where M is 1 for FD and M is infinity for BE stats.

I now impose the constraint of N=\sum_k n_k and wind up with:

\lambda^{N} \prod_{k=1}^{\infinty} \left(1 - e^{-\beta \epsilon_k} \right)^{\pm} = Z_{BE}^{FD}

why didn't the chemical potential go away? I was expecting to get the same expression, but without any lambda term out in front.

Any ideas? Anyone at least KNOW what the canonical expression IS (so that I can compare my answer)?
 
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everyone and his brother seem to be researching BE condensates these days, and yet no one is the least bit curious about this?
 
don't understand your notation
(1-e^x)^{\pm}?
what is to the power of +/-? as you know, you get either (1-e^x) or (1+e^x)
 
mjsd said:
don't understand your notation
(1-e^x)^{\pm}?
what is to the power of +/-? as you know, you get either (1-e^x) or (1+e^x)

no...it means you get "a" or "1/a"...

this is the standard way of writing Fermi-Dirac or Bose-Einstein statistics as combined in one expression, the notation isn't mine...
 
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