Abstract
The energy of interaction of two helium atoms near their van der Waals minimum has been calculated using a perturbation expansion. An SCF wave function is used to calculate the first-order energy and the second-order (dispersion) energy is taken from previous work. An energy minimum of -3·3 × 10-5 a.u. at a separation of 5·5 a.u. is obtained, in good agreement with experimental results.

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