- #1
Pacopag
- 197
- 4
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?
Last edited: