QM perturbation theory: rigid diatomic molecule

In summary, the Hamiltonian for a rigid diatomic molecule is H_0 = {L^2 \over {2I}}where I is the moment of inertia of the molecule. The lowest four energy states of this system are 0, 1, 2, 3, corresponding to l=0, l=1, l=2, l=3. An external electric field is applied, leading to a perturbation H_1 = ED\cos\theta where E is the strength of a constant external electric field, D is the (fixed) dipole moment of the molecule, and \theta is the orientation of the dipole with respect to the direction of the external
  • #1
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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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  • #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?
 
  • #3
Pacopag said:

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].
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]!
(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?

H_1 is not a constant! It contains a cos theta! You simply have to do the angular integral and see what you get.
 
  • #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.
 
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  • #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.
 
  • #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.
 

1. What is QM perturbation theory?

QM perturbation theory is a mathematical method used in quantum mechanics to calculate the energy levels and properties of a system that is slightly perturbed from its original state. It is based on the assumption that the perturbation is small enough for the system to be treated as a sum of two parts: the unperturbed system and the perturbation.

2. How is QM perturbation theory applied to a rigid diatomic molecule?

In QM perturbation theory, the Hamiltonian of the system is expressed as a sum of the unperturbed Hamiltonian and a perturbation term. In the case of a rigid diatomic molecule, the unperturbed Hamiltonian represents the energy levels and wavefunctions of the molecule in the absence of any external forces, while the perturbation term accounts for the effects of external forces such as an electric or magnetic field.

3. What are the limitations of QM perturbation theory?

QM perturbation theory is only applicable to systems that can be treated as a sum of two parts, where one part is unperturbed and the other is a small perturbation. It also assumes that the perturbation is small enough that the system's energy levels and wavefunctions do not change significantly. Additionally, it is only accurate for low-order perturbations and may not be suitable for highly complex systems.

4. How does QM perturbation theory improve upon the basic quantum mechanical model?

The basic quantum mechanical model assumes that a system is in a single state at a time, which cannot account for the effects of external perturbations. QM perturbation theory allows for the inclusion of these perturbations, providing a more accurate description of the system's energy levels and properties.

5. Can QM perturbation theory be applied to other types of molecules?

Yes, QM perturbation theory can be applied to any type of molecule as long as it can be treated as a sum of two parts and the perturbation is small enough. It has been successfully used to study the energy levels and properties of various molecules, including polyatomic molecules and molecules with multiple electronic states.

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