Calculating Eigenstates and Eigenvalues of a 2D Quantum Rotor with Perturbation

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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
 
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Start by writing down the unperturbed eigenvalues and eigenfunctions.
 
The unperturbed eigenstates are:
\psi_{m}(\phi)=\frac{1}{\sqrt{2\pi}}\mathrm{e}^{\mathrm{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...
 
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p3rry said:
where m=0,1 \ldots
You're missing some of the states ...
 
Sorry
where m=0,\pm1,\pm2 \ldots
 
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?
 
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?
 
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