Van der Waals interaction. Consider two atoms a distance R apart. Because they are electrically neutral you might suppose there would be no force between them, but if they are polarizable there is in fact a weak attraction. To model this system, picture each atom as an electron (mass m, charge -e) attached by a spring (spring constant k) to the nucleus (charge + e), as in Figure 7.13. We’ll assume the nuclei are heavy, and essentially motionless. The Hamiltonian for the unperturbed system is

The Coulomb interaction between the atoms is

(a) Explain Equation 7.104. Assuming that and are both much less than R, show that

(b) Show that the total Hamiltonian plus Equation 7.105) separates into two harmonic oscillator Hamiltonians:

(c) The ground state energy for this Hamiltonian is evidently

Without the Coulomb interaction it would have been E₀ = ћω₀, where

show that

Conclusion: There is an attractive potential between the atoms, proportional to the inverse sixth power of their separation. This is the van der Waals interaction between two neutral atoms.
(d) Now do the same calculation using second-order perturbation theory. The unperturbed states are of the form Ψn₁(x₁),  Ψn₂(x₂), where Ψn(x) is a one-particle oscillator wave function with mass m and spring constant k; ΔV is the second-order correction to the ground state energy, for the perturbation in Equation 7.105 (notice that the first-order correction is zero).

Transcribed Image Text:

HO = 1 = P² + = kx² + ₁ 2m 2m P² + = kx₂². (7.103)