Parition function for Boson gas with two quantum numbers

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SUMMARY

The discussion focuses on calculating the partition function for a system of non-interacting Bosons with single-particle energies defined by the equation εp,m = p²/2m + αm. The partition function is expressed as ZN = ∑{n(𝑝,m)} exp(-β ∑𝑝,m εp,mn(𝑝,m)). The grand canonical formalism is applied to derive Q = ∑N=0 eβμN ∑{n(𝑝,m)} exp(-β ∑𝑝,m εp,mn(𝑝,m)). The final expression for the average occupation number is derived as ⟨n(𝑝,m)⟩ = 1/(z-1eβεp,m - 1), confirming the Bose-Einstein distribution with dependencies on both quantum numbers.

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Parition function for Boson "gas" with two quantum numbers

Let's say that we have a system of non-interacting Bosons with single-particle energies given by,

\epsilon_{p,m} = \frac{p^2}{2m} + \alpha m

where m = -j, ... ,j

and we want to calculate the partition function of this system. To do this, you would write,

Z_N = \sum_{\{n(\vec{p},m)\}} \exp(-\beta \sum_{\vec{p},m} \epsilon_{p,m}n(\vec{p},m))

Where n(\vec{p},m) are the occupation numbers. From there, you would use the grand cannoncial formalism, and have that,

Q = \sum_{N=0} e^{\beta \mu N} \sum_{\{n(\vec{p},m)\}} \exp(-\beta \sum_{\vec{p},m} \epsilon_{p,m}n(\vec{p},m))

Assuming I haven't made any mistakes yet (and if I have PLEASE point them out!) I'm not sure how to evaluate this when there is double sums involved since, n(\vec{p},m) can certainly be degenerate...

I'm thinking I can just write this as,

Q = \sum_ {\{n(\vec{p},m)\}} \prod_{\vec{p},m} \exp(-\beta(\epsilon_{p,m} - \mu )n(\vec{p},m))

and then proceed as normal, but I'm really not sure... any stat-wizards out there want to help me out?
 
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So I think what I wrote above makes sense, so proceeding:

Peform the sum over \{n(\vec{p} ,m)\},
Q = \prod_{\vec{p},m} [ 1 - \exp(-\beta(\epsilon_{p,m} - \mu )) ]^{-1}
and,
\ln (Q) = -\sum_ {\vec{p},m} \ln( 1 - \exp(-\beta(\epsilon_{p,m} - \mu )) )

and so,

\langle n(\vec{p},m) \rangle = \frac{\partial \ln (Q) }{\partial ( \beta \epsilon_{p,m} )} = \frac{1}{z^{-1}e^{\beta\epsilon_{p,m}} - 1}

So we get essentially the same Bose-Einstein distribution, but now the average occupation number depends on both quantum numbers. Feel free to comment if you think this makes sense.
 
Last edited:
Yes it makes sense, but to use the canonical partition function first and then sum over the particle numbers for the grand-canonical one is pretty complicated. It's much simpler you start from the grand-canonical statistical operator right away,
\hat{R}=\frac{1}{Z} \exp[-\beta(\hat{H}-\mu \hat{N})], \quad Z=\mathrm{Tr} \exp[-\beta(\hat{H}-\mu \hat{N})].
Here \hat{N} can be any conserved number (or charge) operator. Then you simply sum over all possible Fock states (occupation-number states) with n(\epsilon_{p,m}) \in \mathbb{N}_0, leading precisely to the result you've given in #2.
 
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