Van Der-Waals Forces and Perturbation Theory for Atom Interactions

[MathematicalPhysicist]

Homework Statement

The van der waals forces describe the forces between neutral atoms, separated by a distance much larger than the typical size of a single atom, as we saw in class.
Therefore we describe the interaction between the atoms by: $$W=W_{dd}+W_{dq}+W_{qd}+W_{qq}+...$$ i.e W is the energy sum which has contributions from dipole-dipole, quadrapule-dipole, quadrapule-quadrapule and so.
In class we were shown that the first order correction due to dipole dipole interaction $$W_{dd}$$ vanishes, and we solved this with second order perturbation theory.
But this is actually justified if first order correction of all other contriubtions to W vanish as well. Show that this is the case here.

Homework Equations

The Attempt at a Solution

I know that I need to find the quadrapule tensor Q for two atoms, but not sure how does this look like? i.e I know the general equation for the potential of a quadrapule and the euqation for the quadrapule itself, but don't know how to write Q explictly.
P.s
I am using Cohen-Tanoudji notation and his coverage of Vand Der Waals forces, in Vol 2, page 1130 complement Cx1.

Thanks in advance.

 

[Jhero]

Thank you for your post. I am glad to see that you are interested in further exploring the van der Waals forces and their interactions between neutral atoms.

To answer your question, the quadrapole tensor Q for two atoms can be written explicitly as:

Q = α1 * α2 * (3 * r1 * r2 - r^2 * δij)

Where α1 and α2 are the polarizabilities of the two atoms, r1 and r2 are the position vectors of the two atoms, and δij is the Kronecker delta. This expression takes into account both the quadrapole moment of each atom and their relative positions.

Using this expression, we can then calculate the first order correction for the quadrapole-dipole interaction, W_{qd}, and show that it also vanishes. This is because the first order correction term involves the dot product of the quadrapole and dipole moments, which is equal to zero for two neutral atoms. Therefore, the first order correction for all other contributions to W also vanish, justifying the use of second order perturbation theory.

I hope this helps clarify your doubts. If you have any further questions, please do not hesitate to ask.