Why does the classical continuous partition function blow up as $T \to 0$?

Question

At $T = 0$, we'd expect Entropy to be zero because there's only one microstate and the $\log(1) = 0$. However, when I take the limit as $T \to 0$ in the classical canonical ensemble, it goes to infinity.

For example, say we have the Hamiltonian of a simple harmonic oscillator: $$ \mathscr{H} = \frac{p^2}{2m} + \frac{m \omega^2x^2}{2} $$ with the partition function defined as: \begin{equation} Z = \frac{1}{N! \hbar^{3N}} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\beta \mathscr{H}(p, x)_j} dp dx \end{equation} we end up getting $$ Z = \frac{k_b T}{\hbar \omega} $$ and we can evaluate the entropy using: $$ S = k_b \ln{Z} + k_b T \frac{\partial \ln{Z}}{\partial T} $$ The term $\ln{Z}$ is bothering me because $$ \ln{\frac{k_b T}{\hbar \omega}} $$ goes to -infinity as $T\to 0$, whereas I'd expect $S \to 0$ as $T \to 0$. What am I missing here?

Answer 1

Your treatment of the system is purely classical and this prevents the possibility of obtaining a result consistent with the third principle of thermodynamics. A symptom of what is going wrong with classical statistical mechanics is that for a system of $N$ harmonic oscillators, equipartition theorem provides an average energy of $Nk_BT$. Therefore, the specific heat would be independent on temperature. However such a result is incompatible with the possibility of getting a finite value for the entropy at $T=0$: $$ S(T) - S(0) = \int_0^T dT^{\prime} \frac{Nk_B}{T^{\prime}} $$ diverges logarithmically at $T=0$, consistently with the direct calculation you reported.

A quantum treatment of the system solves the problem since the spacing between energy levels of the harmonic oscillator implies that below a temperature of the order of $\hbar \omega / k_B$ there is an progressive reduction of the specific heat, which goes to zero at $T=0$, thus eliminating the divergence of entropy and restoring the third principle.