Bidimensional quantum rotor

[p3rry]

Hello!

I need help with this typical quantum problem:

I have a quantum rotor in 2 dimensions. And a perturbation along the x direction:

Here's the unperturbed Sch equation:

-\frac{\hbar^{2}}{2M}\frac{\partial^{2}}{\partial \phi^{2}}\psi(\phi)=E\psi(\phi)

And here's the perturbation

H_{1}=-\epsilon \cos(\phi)

The text asks me about the eigenstates and their eigenvalues, I suppose it means at the first perturbative order.

I get involved into integrals that seems to be too complicated (I got it from a phd test in which a single exercise it's supposed not to take much time in calculations).

Thank you very much

P3rry

[Avodyne]

Start by writing down the unperturbed eigenvalues and eigenfunctions.

[p3rry]

The unperturbed eigenstates are:
\psi_{m}(\phi)=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{\m athrm{i}m\phi}

where m=0,1 \ldots
and the spectrum is
E_{m}=-\frac{\hbar^{2}m^{2}}{2M}

Now, as I said, I got problems in calculating the perturbed spectrum...

[Avodyne]

where m=0,1 \ldots
You're missing some of the states ...

[p3rry]

Sorry
where m=0,\pm1,\pm2 \ldots

[Avodyne]

OK, so states with positive m and negative m are degenerate. So you need to use degenerate perturbation theory, which means that you have to "diagonalize the perturbation in the degenerate subspace". Do you know how to do that?

[p3rry]

Ok, but I get 0 for every matrix element:

\left\langle m |H_{1}|m\right\rangle = \frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{d}\phi\mathrm {e}^{-im\phi}(-\epsilon \cos (\phi))\mathrm{e}^{im\phi}=0
and the off diagonal elements are equally 0
\left\langle m |H_{1}|-m\right\rangle = \frac{1}{2\pi}\int_{0}^{2\pi}\mathrm{d}\phi\mathrm {e}^{-im\phi}(-\epsilon \cos (\phi))\mathrm{e}^{-im\phi}=-\frac{\epsilon}{4\pi}\left\{\int_{0}^{2\pi}\mathrm {d}\phi\mathrm{e}^{-i(2m-1)\phi}+\int_{0}^{2\pi}\mathrm{d}\phi\mathrm{e}^{-i(2m+1)\phi}\right\{=0
Is that right?
I doubt I have to perform the calculation at the second order. What do you think?