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

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?

• Are you sure about your formula for the entropy? $S =\frac{\partial(T\ln Z)}{\partial T}$, but it doesn't make any difference to your observation – innisfree Feb 20 '19 at 4:57
• Good catch! Have a typo, will fix that now. – Drew Lilley Feb 20 '19 at 5:05
• I suspect that at $T\to 0$, the system is always found in the ground state $p = x = 0$, such that the entropy $-\int dp dx P(x, p)\ln P(x, p)$ is something like $-\int dp dx \delta(x)\delta(p) \ln(\delta(x)\delta(p))$ which diverges to negative infinity. So to me the result makes sense. I think your hunch that it should equal zero is because you are thinking of discrete variables, not continuous ones. – innisfree Feb 20 '19 at 5:06
• I apologize, I don't understand the argument. I'm new to Statistical mechanics -- what are the integral equations you're using there? What temperature then should correspond to zero entropy? – Drew Lilley Feb 20 '19 at 5:57
• The integral equations are the definition of the entropy in terms of probability density functions on the phase space – innisfree Feb 20 '19 at 6:19

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.