Partition function for continuous spectrum

[HomogenousCow]

Let's say that we have a one-particle Hamiltonian that admits only a continuous spectrum of eigenvalues $E(k)=\alpha k^2$ parameterized by asymptotic momentum $\mathbf{k}$ (assuming the eigenfunctions become planewaves far from the origin), would the partition function then be $$Z=\int d^{3}k e^{-\beta \alpha k^2}$$? This feels odd to me because it would imply that all continuum spectras have the same partition function. Is this true or have I forgotten about something?

 

[Demystifier]

You have forgotten that $Z$ is a function of the variable $\beta$, while $\alpha$ is a parameter. Different kinds of particles have different $\alpha$.

 

[vanhees71]

This is the one-particle partition sum for a free non-relativistic particle. Of course, $\alpha=1/(2m)$. For a classical ideal gas of $N$ partices you simply have $Z_N=Z^N/N!$, where I borrowed the $1/N!$ from the quantum theoretical indistinguishability of particles; this is the minimal input from quantum mechanics you need to avoid inconsistencies like Gibbs's paradoxon.

For the full quantum-mechanical treatment of the ideal gas you need to take into account the bosonic or fermionic nature of the particles. That's best done in the occupation-number representation and by first using a finite volume. You can take any boundary conditions you like, because those get unimportant in the infinite-volume/thermodynamical limit). Choosing periodic boundary conditions with the volume being a cube of length, $L$ your single-particle momentum spectrum is $\vec{k} \in 2 \pi \hbar/L \mathbb{Z}$, and in the grand-canonical ensemble you get:

Fermions:

There for each $\vec{k}$ and each spin you have $N(\vec{k},\sigma) \in \{0,1 \}$. So your partition sum is
$$Z(\beta,\mu)=\prod_{\vec{k},\sigma} \sum_{N=0}^1 \exp[N(-\beta \vec{k}^2/(2m) + \beta \mu)] = \prod_{\vec{k}} [1+\exp(-\beta \vec{k}^2/(2m)+\beta \mu)]^{2s+1}.$$
Where $s$ is the spin of the particle (from relativistic QFT we know it must be half-integer for fermions).

Bosons:

For bosons the calculation is the same but for each $(\vec{k},\sigma)$ the occupation numbers $N(\vec{k},\sigma) \in \mathbb{N}_0$. This gives
$$Z(\beta,\mu)=\prod_{\vec{k},\sigma} \sum_{N=0}^{\infty} \exp[N(-\beta \vec{k}^2/(2m) + \beta \mu)] = \prod_{\vec{k}} [1-\exp(-\beta \vec{k}^2/(2m)+\beta \mu)]^{-(2s+1)}.$$
Note that here you necessarily must have $\mu<0$, because otherwise the geometric series wouldn't converge (at least for $\vec{k}=0$).

The thermodynamic/infinite-volume limit is simple for fermions. You consider the logarithm of $Z$, and the resulting sum over the momenta can be approximated by an integral:
$$\sum_{\vec{k}} \simeq L^3 \int \mathrm{d}^3 k/(2 \pi \hbar)^3,$$
because in a momentum-volume element $\Delta^3 k$ there are $\Delta^3 k L^3/(2 \pi \hbar)^3$ single-particle momentum eigenstates.

This leads to
$$\ln Z=\frac{(2s+1) V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 k \ln[1+\exp(-\beta \vec{k}^2/(2m)+\beta \mu)].$$
For bosons the thermodynamic limit is subtle, because the ground state must be considered separately. Before you go the infinite-volume limit, keeping the discrete momenta there is no problem. You just get a sum, and if you try to fix the particle number at a given temperature by choosing an appropriate $\mu<0$, you can get any particle number you want, because for $\mu \rightarrow -0^{+}$ the particle number diverges, i.e., you can accommodate as many particle you like at any given $T$. In the limit $T \rightarrow 0$ you get that all particles are in the ground state (Bose-Einstein condensate).

If you go over to the infinite-volume limit, the total particle number does not diverge anymore for $\mu \rightarrow -0^+$. So if you naively approximate the sum with the integral you miss the zero-mode contribution. So the correct approximation is to take the zero-mode out and only for the rest you approximate with the integral. Then you get
$$\ln Z=-(2s+1) \ln[1-\exp(\beta \mu)] -\frac{(2s+1)V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \ln [1-\exp(-\vec{k}^2/(2m)+\mu \beta].$$
To get the (average) particle number you have to set $\mu \beta=\alpha$, differentiate wrt. $\alpha$ and then set $\alpha=\mu \beta$ again. This gives
$$N=\frac{2s+1}{\exp(-\mu \beta)-1} + \frac{(2s+1) V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 k \frac{1}{\exp(\beta \vec{k}^2/(2m)-\mu \beta)-1}.$$
The correct infinite-volume limit now is to divide this by $V$ to get the particle-number density $n=N/V$. Then for a given $T$ you keep $n$ fixed and take $V \rightarrow \infty$. Then there are two cases

(a) "Normal state"

$$n < \frac{(2s+1) V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 k \frac{1}{\exp(\beta \vec{k}^2/(2m))-1} = n_{>}. \qquad (*)$$
Then you choose $\mu<0$ such that
$$n=\frac{(2s+1) V}{(2 \pi \hbar)^3} \int_{\mathbb{R}^3} \mathrm{d}^3 k \frac{1}{\exp(\beta \vec{k}^2/(2m)-\mu \beta)-1}.$$
Making then $V \rightarrow \infty$ at fixed $\beta$ and $\mu$, the contribution from the zero mode obviously vanishes.

(b) "Condensate state"

If (*) is not fulfilled, then you set $\mu=0$ in the integral. In the zero-mode contribution you make $V \rightarrow \infty$ and $\mu \rightarrow -0^+$ such that this expression stays constant at
$$n_0=n-n_{>}.$$

This is a state, where a macroscopically relevant portion of the particles occupies the ground state at $\vec{k}=0$, i.e., you have a Bose-Einstein condensate in addition to a "normal" (degenerate) Bose gas. For $T \rightarrow 0$ all particles occupy the ground state.

This problem with the particle-number density never occurs of course for fermions, because there you can choose $\mu \in \mathbb{R}$ arbitrarily, and thus you can choose $\mu$ always such that the integral at a given $T$ gives any particle-number density you like by adjustint $\mu$.