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?


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 $i$.


1 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 $\ln (N+1)$.

  • $\begingroup$ As far as I understand, in addition you take the next energy level into account. The idea seems promising. However, some attention has to be paid because your partition function is true only for distinguishable particles. The case of indistinguishable particles (Bosons) is more difficult. $\endgroup$
    – user56224
    Jul 5, 2018 at 4:37
  • $\begingroup$ @user8153, "As expected the entropy ... is extensive (i.e., proportional to N)" For two-level N Boson particles system the entropy limit ($T\rightarrow\infty$) will be $\ln(N+1)$, and of course is not "extensive" i.e. not proportional to $N$. Physical extensivity assumes proportionality to the number of particles $N$ in the thermodynamic limit ($N\rightarrow\infty$, $G\rightarrow\infty$, $N/G=const$, where $G$ -the number of system levels $\endgroup$ Jul 5, 2018 at 18:08
  • $\begingroup$ You are correct that my answer is valid only for distinguishable particles; I had missed the "bosons" in the original questions. I have updated the answer to warm the reader. $\endgroup$
    – user8153
    Jul 5, 2018 at 18:32
  • $\begingroup$ @AlekseyDruggist I added the case of undistinguishable bosons. $\endgroup$
    – user8153
    Jul 5, 2018 at 23:56
  • $\begingroup$ @user8153, indeed, the case of two-level boson system is not so complicated, The partition function is actually the sum of the geometric progression $\endgroup$ Jul 6, 2018 at 7:37

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