Abstract

A momentum space treatment shows that perturbation equations for the H(1s)–H(1s) van der Waals interaction can be exactly solved in their Schrödinger forms without invoking any variational methods. Using the Fock transformation, which projects the momentum vector of an electron from the three-dimensional hyperplane onto the four-dimensional hypersphere, we solve the third order integral-type perturbation equation with respect to the reciprocal of the internuclear distance R. An exact third order wave function is found as a linear combination of infinite number of four-dimensional spherical harmonics. The result allows us to evaluate the exact dispersion energy E6R−6, which is completely determined by the first three coefficients of the above linear combination.

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