# From spectrum/dispersion relation to the partition function

I know the spectrum/dispersion relation for a bosonic system.

$$E \left( \mathbf{k} \right) = \cdots$$

Is there a general method for writing down the partition function when the spectrum of the system is known?

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$Z = \sum_k \exp(-\beta E_k)$. Replace $\sum$ with $\int$ for continuous spectrum (which must have a defined measure too). – genneth Jan 7 '12 at 18:36
genneth: First of all, thanks for you reply. I came out with that very same idea but had some doubts about it. Don't I need to include something about the density of states or the Bose-Einstein distribution? I mean, the sum in $Z$ is a sum over all the microstates of the system, can I be confident that $E_{\mathbf{k}}$ for all allowed values of $\mathbf{k}$ spans over all the microstates, one and exactly one time? – zakk Jan 7 '12 at 18:49
you only need to make sure that each k labels only one state, and repetitions of energy for different k will take care of itself. – genneth Jan 8 '12 at 1:59

The definition of the partition function is $$Z = \sum_\mathbf{q} e^{-\beta E_\Sigma(\mathbf{q})} \qquad (1)$$ where
$\mathbf{q}$ is the set of quantum numbers describing the microscopical state of the system,
$E_\Sigma(\mathbf{q})$ is the energy of the system when it is in that microscopical state,
$\beta = 1/(k_B T)$

In your case $\mathbf{q}$ is the set of the values of the $\mathbf{k}$ vectors of the bosons: $$\mathbf{q} = (\mathbf{k}_1, \mathbf{k}_2, \ldots, \mathbf{k}_N).$$ Permutation of the particles does not produce new state since the bosons are indistinguishable. We will divide the sum by the number of permutations of the particles $N!$ to take this into account. This is like the states have fractional degeneracy $1/N!$.

The energy $E_\Sigma(\mathbf{q})$ is the sum of the energies of the particles: $$E_\Sigma(\mathbf{q}) = \sum_{i=1}^N E(\mathbf{k}_i)$$

So the sum (1) turns into a product of $N$ integrals over the $\mathbf{k}$ space: $$Z = \frac{1}{N!} \prod_{i=1}^N \int \frac{d^3\mathbf{k}_i}{(2\pi\hbar)^3} e^{-\beta E(\mathbf{k}_i)}$$

All the integrals are the same and we can omit the index $i$: $$Z = \frac{1}{N!(2\pi\hbar)^{3N}} \left(\int e^{-\beta E(\mathbf{k})} d^3\mathbf{k}\right)^N \qquad (2)$$

If there is spin degeneracy there will be additional factor $(2s+1)^N$, where $s$ is the spin of one particle.

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