Particle in a ring. Correction to energy.

The problem for a single particle confined to a ring of fixed radius was solved using $H=-(\hbar^2)/(2mr^2)* (d^2/d\phi^2) = L^2/(2mr^2)$.

This Hamiltonian works just fine for the rotation of a molecule in a vacuum but has to be adjusted in order to model the rotation if, say the molecule were rotating in proximity to a solid surface with a surface charge. Assume that the effect of the surface is to add a potential that varies with molecular rotation angle as

$V(\phi) =b*\sin^2\phi$

where $b$ is an energy constant. Determine the correction to the energy resulting from this perturbation.

Given the Hamiltonian

$H=-\hbar^2/(2mr^2 )*(d^2/d\phi^2)$    and the Sch equation $H\Psi=E*\Psi$

The solutions are
$\Psi_n (\phi)=1/\sqrt{2\pi}*e^{in\phi}$   where $n$ is integer+or-including 0

And the unperturbed energies
$E_n=n^2*\hbar^2/(2mr^2 )$

If we have a perturbation

$V(\phi)=b*\sin^2\phi$

The corrections in the above energies are

$E_n’=(<\psi_n|V|\psi_n>)/(<\psi_n |\psi_n>)= 1/(2\pi)*\int_{0}^{2\pi}e^{-in\phi}*b\sin^2\phi*e^{in\phi}d\phi=$

$=b^2/(2\pi)*\int_0^{2\pi}\sin^2\phi*d\phi=(b^2/2\pi)*\pi=b^2$