Analytical derivative of nuclear-electron attraction integrals over gaussian functions

Question

I'm currently writing a program which evaluates the first derivatives of molecular integrals (over gaussian basis sets) with respect to the cartesian coordinates of the nucleus. I'm somewhat stuck with the derivative of the nuclear-electron attraction integral, i.e.: $$ \frac{\partial V_{\mu\nu}}{\partial X_A}=\frac{\partial }{\partial X_A} \left\langle \mu| \hat V |\nu\right\rangle $$ where $\mu$ and $\nu$ are Gaussian-Type Orbitals (GTOs) and $\hat V $ is the operator: $$ \hat V = -\sum_A \frac{Z_A}{|r-R_A|} $$ As was previously discussed on this forum, the upper derivative can be expressed as a sum of derivatives applying the chain rule: $$ \frac{\partial }{\partial X_A} \left\langle \mu| \hat V |\nu\right\rangle=\left\langle\frac{\partial \mu}{\partial X_A}|\hat V|\nu\right\rangle+\left\langle \mu|\frac{\partial \hat V}{\partial X_A}|\nu\right\rangle+\left\langle \mu|\hat V|\frac{\partial \nu}{\partial X_A}\right\rangle$$ The first and latter terms are relatively easy to evaluate, since the derivative of a Gaussian-Type Orbital with respect to one of the coordinates of the point where it's centered is a sum of GTOs with higher and lower angular momentum. The evaluation of the middle term doesn't seem so obvious. According to Szabo & Ostlund (pg. 442), the derivative of the $\hat V$ operator is: $$ \frac{\partial \hat V}{\partial X_A}=-Z_A\sum_i \frac{X_i-X_A}{r_{iA}^3} $$ where $\mu$ and $\nu$ are Gaussian-Type Orbitals (GTOs) So we are dealing with a new type of molecular integral: $$ \left\langle \mu\left|-Z_A\sum_i \frac{X_i-X_A}{r_{iA}^3}\right|\nu\right\rangle $$ After searching for some time, I couldn't find any reference which discuss this topic. Could you provide some insight on the evaluation of this particular integral? Thanks in advance.

Answer 1

Like two-electron integrals, the nuclear attraction integrals are evaluated using the Laplace transform

$$ \frac 1 {r_{12}} = \frac {2} {\sqrt{\pi}} \int_{0}^{\infty} {\rm d}t e^{-t^2r_{12}^2} $$

It would seem that using this transform also makes the derivative integrals easy to evaluate.

Answer 2

The original Obara-Saika paper contains formulae for this in the "Electric field and electric field gradient Integrals" section of the appendix towards the end of the paper

Answer 3

The direct computation of the derivative of the nuclear-electron attraction operator can be simply avoided. Remember that in the McMurchie-Davidson algorithm, the Coulomb integral is reduced to a form, which requires contraction of the E and R tensors. To get the derivative to an atom center, it just needs to change the index of the R tensor (see Molecular electronic-structure theory (T. Helgaker, P.Jorgensen, J.Olsen), chapter 9.9.3, p.376).