Do discrete symmetries in molecules leave statistical "artifacts" like the Equipartition Theorem?

Question

In statistical mechanics we have the equipartition theorem which can derive heat capacity simply from degrees of freedom. For example, diatomic gases have an increase heat capacity of $ \frac{7}{2} N k_B T $.

I am studying Molecular Orbital Theory and there is a big emphasis on discrete symmetries and their effects on electron configuration. I was wondering if you can find statistical effects of the discrete symmetries of molecules in such a way that you could experiment on something like the heat capacity and deduce that the molecule in question has a given symmetry.

For example, in $ AH_2 $ molecules, if they bend, then we have an additional degree of freedom that could contribute to the equipartition heat capacity. This is still a continuous symmetry that was broken, but we could imagine discrete ones like $ ACH_3 $. If A is a hydrogen atom, then we have methane and a $ E 8C_3 3C_2 6S_4 6\sigma_d $, otherwise, we just get $ E 2C_3 3\sigma_v $ symmetry (from here).

Does this increase in symmetry leave any statistical artifacts that experiment could show?

Answer 1

Please correct me if I am wrong!!

After thinking for a while, I believe that you need a thermodynamic control parameter (think pressure, volume, magnetic field, etc.) that breaks the symmetry.

Consider a system with a phase space $ \Lambda $ that is symmetric under group operations $ g \in G $. If you calculate the partition function, you notice it can be broken up,

$$\begin{align} Z &= \int_\Lambda dz~e^{-\beta \mathcal{H}(q, p; A)} \\ &= \sum_{g \in G} \int_{\Lambda ~/~ G} e^{-\beta ~\mathcal{H}\big(T_g(q), ~T_g(p) ~;~ T_g(A)\big)} \\ &= \sum_{g \in G} \int_{\Lambda ~/~ G} e^{-\beta ~\mathcal{H}\big(q, p ~;~ T_g(A)\big)} \\ &= \sum_{g \in G} Z_g \end{align}$$

where $ A $ is a statistical control variable like pressure or volume.

Computing some thermodynamic quantity would be simply done by,

$$\begin{align} A &= \frac{\partial \log Z}{\partial A} \\ &= Z^{-1} \sum_{g \in G} \frac{\partial Z_g}{\partial A} \end{align}$$

If $ A $ doesn't transform under the group action, then thermodynamics will not have an artifact.

Working on an example