Two-Particle Rigid-Rotor

The rigid-rotor is a system of two particles for which the distance between them $\vert\bar{r}\vert= d$ is constant. The Hamiltonian of the system is described by Eq. (41), where the first two terms are equal to zero, and $E_{\mu}= \frac{\hbar^2}{2 \mu d^2} l(l+1) +V(d)$, with $\psi_{\mu}=Y_l^m(\theta, \phi)$.

The moment of inertia of a system of particles is $I_{\zeta} \equiv \sum_{i=1}^{2} m_i r_i^2$, where $m_i$ is the mass of particle $i$ and $r_i$ is the particle distance to the $\zeta$ axis.

Exercise 30: Prove that $I=\mu d^2$ for the two-particle rigid rotor, where $\mu=\frac{m_1m_2}{m_1+m_2}$, $d= r_2-r_1$, and $\zeta$ is an axis with the center of mass of the system and is perpendicular to the axis that has the center of mass of both particles. Assume that the center of mass lies at the origin of coordinates, and that the $x$ axis has the center of mass of both particles in the system.

The rotational energy levels of the rigid rotor are:

$$E_{\mu}= \frac{\hbar^2}{2 I}l (l+1), \hspace{.1cm} l=0, 1, 2,...$$ (42)

These energy levels usually give a good approximation of the rotational energy levels of diatomic molecules (e.g., the HCl molecule).