# Separability of vibrational and rotational motion for diatomics

1. Dec 22, 2013

### tomothy

I'd like some help justifying the approximation that the vibrational and rotational motion of a diatomic molecule is separable.
For two atoms of masses m1 and m2 the full hamiltonian is
$H=-\hbar ^2 /2m_1 \nabla _1 ^2 - \hbar^2 /2m_2 \nabla _2 ^2 + V(r-r_0)$
Where V(r-r0) is the potential energy function, r0 is the equilibrium bond length and r is the atomic separation $r=|\textbf{x} _1- \textbf{x} _2|$. This is seperable into a centre of mass hamiltonian and one in terms of the reduced mass, in the molecular frame. The hamiltonian in the molecular frame is
$H_\mu = -\hbar ^2 /2\mu \nabla ^2 - V(|\textbf{x}|-r_0)$
From this point, I'm not sure how to show that the hamiltonian is approximately separable. I tried writing it in terms of the equilibrium displacement $x=r-r_0$ and then by saying in the approximation that $x>>r_0 , x/r_0 \approx 0$ so in the hamiltonian $r^2=r_0^2(1+x/r_0)^2\approx r_0^2$ but since $r=x+r_0$ is a variable and not a constant of motion, this seems like a dodgy approximation. Any help would be valued greatly!

2. Dec 22, 2013

### Jilang

I would try to split the last term into two terms.

3. Dec 22, 2013

### tomothy

This hamiltonian commutes with J^2 so it is separable into an angular part (the spherical harmonics) and an equation in r. The equation will only in general be solvable for j=0, if it is something like a harmonic potential or a morse potential. So the only way to get the rotational energy out is to throw use first order perturbation theory. So the rotational energy is a first order perturbation or vibrational energy?

4. Jan 3, 2014

### williamsn

Throw away the mathematics for 5 seconds and ask yourself what are the relative transition energies. This is the same reason that we can typically decouple vibrational and electronic transitions through the Born-Oppenheimer approx. Of course there can be ro-vibrational coupling in the same way that B-O fails at conical intersections, but for the most part the two transitions are at such different energy scales that they have low coupling strengths.