Free particle with Coulomb Perturbation

[PhysChem176]

Homework Statement

This is a question I have about something stated in a textbook without much explanation. From Richard D. Mattuck's "A guide to Feynman Diagrams in the Many-Body Problem" Appendix A.1 pg 337

"for example consider the Coulomb interaction between two electrons in a metal. This has the form:

V(r_i,r_j) = e^2 / abs(r_i-r_j)

If the ground state energy of the system is calculated using this as the perturbation, we obtain the result:

E_o = E_o^(0) + E_o^(1) + inf + inf + ..."

The reason for this is of course because the Coulomb interaction is not a small perturbation and perturbation theory won't work in this case. This is what I've always been told in class anyways. I figure though I should figure it out mathematically

Homework Equations

First order energy perturbation:
[tex]
E_{o}^{(1)} = \left\langle n^{(0)} | H^1 | n^{(0)} \right\rangle
[/tex]

where H^1 in this case is the Coulomb potential

Second order energy perturbation:
[tex]
E_{o}^{(2)} = \sum_{m \neq n} \frac {\left| \left\langle m^{(0)} | H^1 | n^{(0)} \right\rangle \right|^2}{E_n^{(0)}-E_m^{(0)}}
[/tex]

(sorry about the switch from no-tex to tex I figured out how the notation worked)

The Attempt at a Solution

I'm assuming since the non-perturbed Hamiltonian to be that for the KE of the free particle
[tex]
\frac{p^2}{2m}
[/tex]
and to be completely separable for the two electrons. So no problem and I get [tex] E_o^{(0)}[/tex]

Now for the first order perturbation I'm a bit stuck since the wavefunction should be the plain wave wavefunction. How does the perturbation act on that wavefunction? If it was just a constant and got pulled outside (which it shouldn't be) then I'll get the integral over all space of a constant which diverges, but the answer which the book quotes states that the first order perturbation should converge and only the second and above order terms diverge.

Now assuming that the book is right and the first order term does converge then what part of the second order term causes it to diverge? Is it the energy denominator? It shouldn't be since the sum across the denominator does not contain the term where n = m (we're assuming there's no degeneracy here, is that a correct assumption?). Is it the fact that the plane wave wavefunction has a finite value at r = 0? This would cause it to diverge, but then the first order term would also suffer a divergence would it not?

Any help would be greatly appreciated. Thanks.

 

[Jhero]

Thank you for your question regarding the Coulomb interaction between two electrons in a metal. You are correct in your understanding that the Coulomb interaction is not a small perturbation and perturbation theory will not work in this case. This is because the Coulomb potential is not a small term in comparison to the non-perturbed Hamiltonian, which is the kinetic energy of the free particles.

In terms of your attempt at a solution, you are correct that the wavefunction for the first order perturbation should be the plain wave wavefunction. The way the perturbation acts on this wavefunction is through the potential term, which in this case is the Coulomb potential. This potential term will change the shape and amplitude of the wavefunction, resulting in a different energy value.

In terms of the first order perturbation converging while the second and above order terms diverge, this is due to the fact that the first order term only involves the ground state wavefunction, while the second order term involves a sum over all possible excited states. This means that the second order term will have contributions from states that have a higher energy than the ground state, resulting in a divergence.

In conclusion, you are correct in your understanding that the Coulomb interaction is not a small perturbation and perturbation theory will not work in this case. I hope this helps to clarify your doubts. If you have any further questions, please do not hesitate to ask.