How Does a Negative First Order Energy Indicate Van der Waals Attraction?

[Niles]

Homework Statement

Hi all

Please take a look at this link: http://books.google.dk/books?id=bA9...e is an attractive potential between"&f=falseI have a question for the above-mentioned case with van der Waals interaction: In my QM-book (Griffiths, problem 6.31), we look at a system of two neutral atoms a distance R apart. Treating the Coulomb-interaction as a perturbation, we have the following values:

E_unperturbed = [itex]\hbar \omega_0[/itex]
E_{perturbation, total} = E_unperturbed + E_{first order}

Thus we have

E_{perturbed, total}-E_unperturbed = E_{first order} < 0.

Now the author concludes that this implies an attractive potential between the atoms, and this is called the van der Waals interaction. I don't understand why we are able to conclude that?

 

[NateD]

Homework Equations Eunperturbed = \hbar \omega_0Eperturbation, total = Eunperturbed + Efirst orderThe Attempt at a Solution The difference between the total perturbed energy and the unperturbed energy represents the total potential energy due to the Coulomb interaction. If this potential energy is negative, then it means that the Coulomb interaction is attractive, which is also known as the van der Waals interaction. This is why we can conclude that there is an attractive potential between the atoms.