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?
Thanks in advance!
 A: 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.
