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Bidimensional quantum rotor

  1. Oct 23, 2009 #1

    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:

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

    And here's the perturbation

    [tex]H_{1}=-\epsilon \cos(\phi)[/tex]

    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

  2. jcsd
  3. Oct 23, 2009 #2


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    Start by writing down the unperturbed eigenvalues and eigenfunctions.
  4. Oct 24, 2009 #3
    The unperturbed eigenstates are:

    where [tex]m=0,1 \ldots [/tex]
    and the spectrum is

    Now, as I said, I got problems in calculating the perturbed spectrum...
    Last edited: Oct 24, 2009
  5. Oct 24, 2009 #4


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    You're missing some of the states ...
  6. Oct 24, 2009 #5
    where [tex]m=0,\pm1,\pm2 \ldots[/tex]
  7. Oct 24, 2009 #6


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    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?
  8. Oct 25, 2009 #7
    Ok, but I get 0 for every matrix element:

    [tex]\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[/tex]
    and the off diagonal elements are equally 0
    [tex]\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[/tex]
    Is that right?
    I doubt I have to perform the calculation at the second order. What do you think?
    Last edited: Oct 25, 2009
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