Rotating Diatomic Molecules

Consider a dilute gas of diatomic molecules where the beginning and the end of the molecules are different, such as $H-D$ (Hydrogen-Deuterium) molecules. We focus here on the rotational degrees of freedom, and treat them quantum mechanically. From your QM text book (for example “Hydrogen atom”) the energy spectrum of the rotational degrees of freedom, i.e. the term

$H_R=\frac{1}{2I}\left [p_\theta^2+\frac{1}{2\sin^2(\theta)}p_{\phi}^2\right ]$

in the total Hamiltonian $H=p^2/2m +H_R$ for one molecule is of the form

$E_J=\frac{\hbar^2}{2I}j(j+1)=k_BT_R j(j+1)$

with each energy level (2j+1) fold degenerate, and with $T_R=\hbar^2/2Ik_B$ the characteristic crossover temperature between low temperature and (quantum) to high temperature (classical) behavior (as you will find out).

a) Why does this angular momentum occur in the hydrogen atom?

b) The canonical partition function factorizes in translational and rotational contributions for each molecule individually, as $Z_C=(z_T*z_n)^N$. Write the equation for the rotational contribution $z_R$ for each molecule. This is a sum over Boltzmann factors of the form $\exp(-j(j+1)T_R/T)$ each multiplied with the degeneracy of each quantum level, and with $T_R$ a characteristic temperature.

The general Hamiltonian is $\hat H=\hat p^2/2m +V$

For spherical coordinates $\hat p=\hat p_r+\hat p_{\theta}+\hat p_{\phi}$.  Since

$\hat p=-i\hbar\nabla$  and $\nabla =\frac{d}{d r}\hat r+\frac{1}{r}\frac{d}{d/\theta}\hat \theta+\frac{1}{r\sin\theta}*\frac{d}{d\phi}\hat \phi$.

It follows that $p^2=-\hbar^2\nabla^2=-\hbar^2\frac{d^2}{d r^2}-\frac{\hbar^2}{r^2}\frac{d^2}{d\theta^2}-\frac{1}{r^2\sin^2\theta}\frac{d^2}{d\phi^2}$

Hence the total Hamiltonian is written as

$\hat H=\frac{\hat p^2}{2m}+V=\left(-\frac{\hbar^2}{2m}\frac{d^2}{d r^2}+V \right)-\frac{\hbar^2}{2mr^2}\frac{d^2}{d\theta^2}-\frac{\hbar^2}{2mr^2}\frac{d^2}{d\phi^2}$

$\hat H =\left(\frac{\hat p_r^2}{2m}+V\right)+\frac{1}{2I}\left[ p_\theta^2+\frac{1}{\sin^2(\theta)}p_{\phi}^2\right ]$

if we define $\hat p_{\theta}^2=-i\hbar\frac{d}{d\theta}$  and $\hat p_{\phi}=-i\hbar\frac{d}{d\phi}$


The partition function for rotation is simply the sum over all possible states of the factors $\exp(-E_i/k T)$

Since any state $E_j$ is present $(2j+1)$ times in the sum (one says it is $g_j=2j+1$ degenerate) (same $j$ state is counted $2j+1$ times) the sum is

$Z_{rot}=\sum_j g_j\exp(-E_j/k T)=\sum_j (2j+1)\exp(-E_j/k T)$

The degeneracy of a level comes from the fact that the total wave function $\psi$ can be written in the form $\psi_{n l m}=R_{n l}(r)Y_l^m(\theta,\phi)$ and the Schrodinger equation can be uncoupled this way.

Since the degeneracy $(2j+1)$ of a level is given by how many ways it can rotate (by the angular part of the Hamiltonian) it follows that

$E_j=-\frac{\hbar^2}{2I} j(j+1)=k T_R*j(j+1)$  (for free motion when $V=0$).

Therefore we have

$Z_{rot}=\sum_j (2j+1) \exp\left(-\frac{j(j+1)T_R}{T}\right)$