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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.

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  • $\begingroup$ +1 bur have you considered using automatic derivatives, or the analytic derivatives code from open source packages? $\endgroup$ Aug 22, 2023 at 1:51

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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.

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    $\begingroup$ Nit pick about "are evaluated" - there are other methods, you can use a Fourier transform for instance which makes particular sense for the Nuclear Attraction integrals as it is a convolution between the charge distribution due to the basis function overlap and the potential due to the atomic point charge. $\endgroup$
    – Ian Bush
    Aug 23, 2023 at 17:45
  • $\begingroup$ @IanBush sure but that's not how it's done in McMurchie-Davidson, Obara-Saika, or the Rys quadrature algorithms that all rely on the above expansion. Of course it doesn't matter at the end how you come to the solution, the point being that one doesn't even try to compute the integral for the original definition but instead makes a transformation... $\endgroup$ Aug 24, 2023 at 18:08
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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

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  • $\begingroup$ Why the downvote? When I have time and energy I hope to enter the complicated MathsJax here, but what else can I do to improve this? $\endgroup$
    – Ian Bush
    Aug 23, 2023 at 20:53
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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).

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Feb 8 at 13:23

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