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QM perturbation theory: rigid diatomic molecule

  1. May 31, 2008 #1
    1. The problem statement, all variables and given/known data
    The Hamiltonian for a rigid diatomic molecule is
    [tex]H_0 = {L^2 \over {2I}}[/tex]
    where [tex] I [/tex] is the moment of inertia of the molecule.
    (a) What are the lowest four energy states of this system?
    (b) An external electric field is applied, leading to a perturbation
    [tex]H_1 = ED\cos\theta[/tex]
    where [tex]E[/tex] is the strength of a constant external electric field, [tex]D[/tex] is the (fixed) dipole moment of the molecule, and [tex]\theta[/tex] is the orientation of the dipole with respect to the direction of the external electric field.
    Find the change in the ground state's energy to second order in perturbation theory.

    2. Relevant equations
    First order correction is [tex]\Delta E^{(1)} = <\psi |H_1|\psi >[/tex].
    Second order correction is [tex]\Delta E^{(2)} = \sum_{\psi \neq \psi '} {{|<\psi '|H_1|\psi >|^2}\over{E_\psi^{(0)}-E^{(0)}_{\psi '}}}[/tex]

    3. The attempt at a solution
    (a) If we can regard the moment of inertia [tex] I [/tex] to be constant, the we can just use the spectrum of [tex]L^2[/tex]. So the lowest four unperturbed energies are
    [tex]0, {\hbar^2 \over I}, {\hbar^2 \over{3I}}, {\hbar^2 \over{6I}}[/tex], corresponding to [tex]l=0,l=1,l=2,l=3[/tex].

    (b) This is where things get weird for me. The ground state eigenfunction is just that for the [tex]L^2[/tex] operator. That is, the lowest spherical harmonic [tex]Y_0^0[/tex]. Also, the perturbation [tex]H_1[/tex] seems to be just a constant. So shouldn't the first order correction just shift the whole spectrum up by an amount [tex]H_1[/tex], while the second-order correction vanishes? Here I have used the orthogonality of the eigenfunctions (i.e. spherical harmonics).

    Can someone please check if I'm doing this correctly?
    Last edited: May 31, 2008
  2. jcsd
  3. Jun 5, 2008 #2
    I am way out of practise on the formalism, but I think I have some idea of how the picture is supposed to go. They tell you to find the first four energy states: I believe these are:

    i) the spin zero state

    ii) THREE spin-one states: up, down, and zero.

    The up and down states are basically clockwise and counterclockwise about the z axis. The "zero" state combines with the other two states to give you spin about the x or y axis; all together, the three spin-one states form a set of basis states for arbitrary orientation of the spinning molecule.

    The perturbation they are looking for has to have mostly the (i) state mixed with the "middle" state from (ii) to give you some preference to align with the external (presumably z-axis) electric field.

    I wonder if this is helpful?
  4. Jun 5, 2008 #3


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    How did you get 1/3 and 1/6:confused:
    The eigenvalues of [tex]L^2[/tex] are [tex]l(l+1) \hbar^2[/tex]!
    H_1 is not a constant! It contains a cos theta! You simply have to do the angular integral and see what you get.
  5. Jun 6, 2008 #4
    Right. It should be 3 and 6 instead of 1/3 and 1/6.
    Ok. I'll give it a try and let you know. Thank you.
    Last edited: Jun 6, 2008
  6. Jun 6, 2008 #5
    My answer is:
    First order correction vanishes.
    Second order correction is [tex]-{I \over 3}\left( {ED\over {\hbar}} \right)^2[/tex].
    Does this sound reasonable? Particularly the correction being negative?
    I don't really feel like texing my whole solution, but I will if you'd like.
  7. Jun 6, 2008 #6
    Actually, mathematically it makes sense to me if I look at the formula for the second order correction. Is this typical that a perturbation lowers the energy? Sorry if this is a silly question. I've never done any perturbation theory before. This is my first try.
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