5
$\begingroup$

It is well known that the usage of the Gaussian basis set, in contrast to Slater basis set, leads relative simple semi-analytical expressions for the two electron repulsion integral $(ab|cd)$. Could you propose any other basis sets allowing simple computations of two-electron integrals? This is quite general questions meaning that these functions should not fit atomic orbitals well (the problem is formulated for wider class of potentials rather than for the Coulomb one only).

$\endgroup$
  • $\begingroup$ Plane wave, wavelet, although i am not sure if the integrals are simple $\endgroup$ – user26143 Nov 18 '13 at 14:36
  • $\begingroup$ One of the main objectives is the orthogonality of the basis functions. So you can choose any reasonable orthogonal basis. $\endgroup$ – ssavec Nov 19 '13 at 10:39
  • $\begingroup$ But, which of them gives analytical solution of the two-electron integral? For example, plane waves are ok. Independently of number of cites, they lead to Fourier transform of the rational function. Gaussian functions and plane wave, what else? $\endgroup$ – freude Nov 19 '13 at 12:08
  • $\begingroup$ Orthogonality of basis set (function) usually does not bring problem in computation. The issue arises when MO vs VB $\endgroup$ – user26143 Nov 19 '13 at 21:05
4
$\begingroup$

A few things come to mind: 1) my understanding is that (ab|cd) in a gaussian basis is exactly solved for arbitrary "angular momenta" using recurrence relations such as the Obara-Saika method. The O-S method also has ways of generating the requisite derivatives with respect to coefficients, without which it is difficult to minimize the energy or do anything practical.

2) On the question of alternative basis sets to the Gaussian, one simply needs a complete basis set (of which there are many), helps to be orthogonal (almost as many), has to describe the boundary conditions of vanishing at infinity or being periodic with unit cell (fewer), hopefully is cusp-like near the nuclei (very few), and integration has to be tractable without resorting to quadrature, since there are four indices (each with 3 dimensions) but also millions of these integrals to compute -- and that leaves you with the Gaussian.

$\endgroup$
  • 5
    $\begingroup$ Another good point for effective computations within the gaussian basis functions is that you can use fitting procedures $(ab|cd)\approx(ab|p)(q|cd)$ and transform four centres into three. $\endgroup$ – Martin - マーチン Jun 1 '14 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.