- #1

- 537

- 1

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:

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

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

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

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

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)?

I start with the grand canonical partition function:

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

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

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

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

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)?

Last edited: