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I'm reading Ostlund's Modern Quantum Chemistry. In Appendix A, the kinetic energy integral is evaluated using the Gaussian Basis functions to be

$$ \left(A\left| -\frac{1}{2}\nabla^2 \right| B\right) = \alpha\beta/(\alpha + \beta)[3 - 2\alpha\beta/(\alpha + \beta) |\mathbf{R}_A - \mathbf{R}_B|^2][\pi/(\alpha + \beta)]^{3/2} \\ \times \exp [-\alpha\beta/(\alpha + \beta)|\mathbf{R}_A - \mathbf{R}_B|^2] \tag{A.11}\label{kin-en.int} $$

So, in the integral evaluation the Gaussian functions themselves are not used, but in the computer program he is evaluating the integral using

T11=T11+T(A1(I),A1(J),0.0D0)*D1(I)*D1(J)

The function T() is calculating the equation above \eqref{kin-en.int}.

I can't understand why he is multiplying by D1 and D2 which are the Gaussian functions themselves. $$g_\mathrm{1s}(\alpha) = (2\alpha/\pi)^{3/4}\mathrm{e}^{-\alpha\mathbf{r}^2}$$

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    $\begingroup$ Just a comment, if you are looking to write your own integral code, I highly recommend taking a look at the following link joshuagoings.com/2017/04/28/integrals $\endgroup$ – Erik Kjellgren Feb 4 at 16:12
  • $\begingroup$ Thank you Erik, it is very helpful $\endgroup$ – SssunnN Feb 8 at 7:55
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The equation for the Kinetic Energy matrix element that you quote is for two unnormalised 1s Gaussians. The d factors contain the normalisation factors and the contraction coefficients - look more carefully at the code, you have got it slightly wrong what the d's represent.

Talking of the code please, Please don't use that as a model for your own Fortran - that style is getting on for 50 years out of date!

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  • $\begingroup$ Thank you Ian for your answer, I'm using another java version, I just wanted to quote the exact line from the book. $\endgroup$ – SssunnN Feb 4 at 9:21
  • $\begingroup$ He is using T11,T22 and T12. I understand that T11 is the kinetic energy of the first atom, T22 is for kinetic energy of the second atom, what is T12 used for?. Thank you $\endgroup$ – SssunnN Feb 10 at 9:00
  • $\begingroup$ Best to ask a separate question but (without looking at the book) think matrices is my guess $\endgroup$ – Ian Bush Feb 10 at 9:13

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