Van der Waals Forces in Helium

A variational method of calculating perturbation energy has been applied to the first two terms of a rapidly converging series expansion of the mutual potential energy of two helium atoms. The result is compared with the much simpler second-order perturbation scheme of Margenau, which it tends to support. The same variational method was then used with the perturbing potential in closed form. This involves the calculation of integrals not previously evaluated. The result validates the previous expansion of the potential. We obtain, for the van der Waals interaction of two unexcited helium atoms: $\epsilon =-\left(\frac{A}{R^6}\right)\{1+\frac{6.26}{R^2}+\frac{51.2}{R^4}+\cdots \} \mathrm{ergs},$ where $R$ is the atomic separation with the first Bohr radius (0.5282 angstrom) as unit distance. The complexity of the integrations requires the use of simple exponential wave functions of the Hartree type; the resulting value of the coefficient $A$ is therefore poor, but this coefficient is well known from other work. It yields a polarization molecule with a dissociation energy of 0.30×${10}^{-16\mathrm{}}$ erg. The bracketed expression giving the relative contributions of the higher order terms is the chief contribution of the present method of attack.