Solving the Schrödinger equation for a rotating triatomic linear molecule This source is showing that solving the Schrödinger equation for a triatmoic linear molecule yields the same formula for the rotationaI quantum states $BJ(J+1)$ as for dipoles.
For dipoles, the total rotational energy can be expressed as the total length of the molecule $R$ and a particle with reduced mass $\mu$ which makes it possible to express the coordinates in the Schrodinger equation specifically for only one particle with that reduced mass $\mu$. The sources I used for solving the equation and deriving the energylevel formula is found here and here
I am clueless how the Schrödinger equation is solved for triatomic linear molecules because there are more variables that can not be reduced the same way as mentioned in these two sources. I would therefore expect to be seperate coordinates for each atom in the Schrodinger equation, as well as more than one bond length.
How is it derived and is it actually possible to do so?
 A: 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
$$\mathbf{R} = \frac{\sum_i m_i\mathbf{r}_i}{\sum_i m_i},$$
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:
$$I = \sum_i m_i\left[\mathbf{n}\times(\mathbf{r}_i-R)\right]^2.$$
We then obtain its energy as
$$E = \frac{\mathbf{P}^2}{2\mu} + \frac{\mathbf{L}^2}{2I}.$$
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.
