Canonical Bose-Einstein statistics

[quetzalcoatl9]

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)?

 

[quetzalcoatl9]

everyone and his brother seem to be researching BE condensates these days, and yet no one is the least bit curious about this?

 

[mjsd]

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)

 

[quetzalcoatl9]

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...