Canonical ensemble near absolute zero

Question

In the canonical ensemble for an ideal gas of $N$ bosons, the partition function for $T\to 0$ scales like $$Z\sim e^{-\beta\epsilon_0N},$$ when $\epsilon_0$ is the lowest (non-degenerate) single particle energy niveau, $\beta = 1/k_B T$ and the Boltzmann constant $k_B$. The energy in this limit is $E=\epsilon_0 N$ and the entropy can be obtained by

$$S=k_B(\log Z+\beta E),$$

which is independent of $T$ and identical zero in this case.

What is the next order approximation of the partition function with respect to $T$, such that there remains a temperature dependency of the entropy?

EDIT:

The general partition function of the $N$-particle system for arbitrary $T>0$ is the restricted sum

$$Z(\beta,N)=\sum_{\{N=n_0+n_1+...\}}\, x_0^{n_0}\,x_1^{n_1}\,...$$

with $x_i=e^{-\beta \epsilon_i}$, for $i=0,1,2,...$ Note, that $i$ is the quantum number and $\epsilon_i<\epsilon_{i+1}$, for all $i$.

Answer 1

First, let's consider the case of distinguishable particles. Consider a two-level system where each of the $N$ particles is either in a state with energy $\epsilon_0$ or $\epsilon_1$. The partition function is $$ Z = \left(\text{e}^{-\beta \epsilon_0} + \text{e}^{-\beta \epsilon_1}\right)^N $$ It follows that the free energy is $$ F = - \beta^{-1} \ln Z = - \beta^{-1} N \ln \left(\text{e}^{-\beta \epsilon_0} + \text{e}^{-\beta \epsilon_1}\right) $$ and the average internal energy $$ \bar{E} = - \frac{\partial \ln Z}{\partial \beta} = N \frac{\epsilon_0\text{e}^{\beta \epsilon_1} + \epsilon_1 \text{e}^{\beta \epsilon_0}}{\text{e}^{\beta \epsilon_0} + \text{e}^{\beta \epsilon_1}} $$

The entropy is $$ T S = \bar{E} - F $$

Defining $\Delta = \epsilon_1 - \epsilon_0 > 0$ to be the spacing between energy levels we find after some rearrangements that $$ \frac{S}{k_{\text{B}}} = N \left[ \frac{\beta \Delta}{1+\text{e}^{\beta \Delta} } + \ln \left( 1 + \text{e}^{-\beta \Delta} \right) \right] $$

As expected the entropy depends only on the difference $\Delta$ between the energy levels and is extensive (i.e., proportional to $N$). In the limit $\beta \Delta \rightarrow \infty$ (or $T \rightarrow 0$) the entropy goes to zero as before, while when $\beta \Delta \rightarrow 0$ (or $T \rightarrow \infty$) the entropy approaches $N \ln{2}$ as expected for a completely disordered two-state system.

Second, let's consider the case of indistinguishable bosons. The partition function is now $$ Z = \sum_{n=0}^N x_0^n \, x_1^{N-n} = \frac{x_0^{N+1} - x_1^{N+1}}{x_0-x_1} $$ where $x_i = \text{e}^{- \beta \epsilon_i}$ is the Boltzmann factor of the $i$-th energy level. The calculation of $F$, $\bar{E}$ and $S$ works the same as before. The expressions get longer now but they are not too complicated, which makes them perfect for something like Mathematica. For the entropy I get $$ \frac{S}{k_{\text{B}}} =\frac{\beta \Delta \left(-e^{\beta \Delta }+N \left(-e^{\beta \Delta (N+2)}\right)+(N+1) e^{\beta \Delta (N+1)}\right)+\left(e^{\beta \Delta }-1\right) \left(e^{\beta \Delta (N+1)}-1\right) \log \left(e^{\beta \Delta (N+1)}-1\right)}{\left(e^{\beta \Delta }-1\right) \left(e^{\beta \Delta (N+1)}-1\right)}-\log \left(e^{\beta \Delta }-1\right) $$ For $N=1$ this expression is the same as the obtained before, as expected. In the low-temperature limit $\beta\Delta \rightarrow \infty$ the entropy goes to zero; in the high-temperature limit $\beta\Delta \rightarrow 0$ the entropy is $\ln (N+1)$.