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## Homework Statement

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.

## Homework 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]

## 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?

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