Solving the Schrödinger equation for a rotating triatomic linear molecule

What is important here is that the molecule is a) rigid (no vibrational degrees of freedom), and b) linear. Thus, the only two things that can happen to the molecule is 1) being displaces as a whole, and 2) rotating about its center-of-mass.

The position of the center-of-mass can be determined as $$mathbf{R} = frac{sum_i m_imathbf{r}_i}{sum_i m_i},$$ where the sum is over all the particles in the molecule. Its dipole moment (in respect to the center-of-mass) is $$mathbf{D} = sum_i q_i(mathbf{r}_i-R),$$ Finally one can calculate in the same way its momen of inertia in respect to the axis passing through the center of mass and perpendicular to the molecule: $$I = sum_i m_ileft[mathbf{n}times(mathbf{r}_i-R)right]^2.$$ We then obtain its energy as $$E = frac{mathbf{P}^2}{2mu} + frac{mathbf{L}^2}{2I}.$$

Note that this derivation is A) completely classical, and B) applicable to a rigid and linear molecule with any number of atoms (for a non-rigid molecule one has to account for vibrations, whereas for non-linear molecules the rotational part of the Hamiltonian is more complex). Transition to the quantum case means replacing $mathbf{P}$ and $mathbf{L}$ by their operators and possibly adding the spin term.