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I want to calculate the derivative of the overlap matrix $S$ with regard to atomic positions. For this derivative I expect a dimension of (3$\times$natm$\times$nao$\times$nao).

Using the pyscf grad.rhf.get_ovlp() method I receive a derivative of size (3$\times$nao$\times$nao). Furthermore I receive non-zero terms on elements belonging to the same atom. Last but not least the matrix I receive is not even symmetric. How can I use pyscf to get the derivative I need?

EDIT: This is a minimal example

from pyscf import scf, grad, gto mol = gto.Mole() mol.build( atom = '''O -0.93940 0.01570 -0.66740 H -0.00994 -0.84002 -0.83652 H 0.10473 0.75287 -0.86054''', basis = 'minao') grad_mol = grad.rhf.Gradients(scf.RHF(mol)) dS_dr = grad_mol.get_ovlp(mol)

In particular I expect dS_dr[0:5,0:5] to be zero, as the basis functions are centered on the same atom. But there are 4 non-zero terms in this block, [0,2], [1,2], [2,0] and [2,1]. Also the elements I expect to be non-zero are not symmetric e.g. [0,6] = -[6,0].

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    $\begingroup$ +1 and welcome. Maybe you can try PySCFad (PySCF with automatic derivatives). Your question will likely get closed if you don't provide the input/output file. By providing these, people can get the derivatives matrix to work (with the problems that you mentioned fixed) and put the new input/output file in an answer. $\endgroup$ Commented May 31 at 0:53
  • $\begingroup$ @NikeDattani-NoFreeTime Thank you very much, I have added a code example and discussed my expectations. $\endgroup$ Commented May 31 at 8:21
  • $\begingroup$ I can't really make sense of it. You can try to manually get the numerical derivative of $\mathbf{S}$ for a simpler system and compare elements between the matrices. I have successfully used numerical derivatives of integrals to implement a SCF gradient, albeit in FORTRAN (and the code is really ugly and supports only STO-6G s functions). $\endgroup$ Commented Jun 5 at 5:20

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The derivative of the overlap matrix

$$ S_{\mu \nu} = \langle \mu | \nu \rangle $$

with respect to a nuclear coordinate $R_i$ is given by

$$ {\rm d}S_{\mu \nu} / {\rm d} R_{i} = \Bigg\langle \frac {{\rm d}\mu} {{\rm d}R_i} \Bigg| \nu \Bigg \rangle + \Bigg\langle \mu \Bigg| \frac {{\rm d}\nu} {{\rm d}R_i} \Bigg \rangle. $$

Note that ${\rm d}\mu / dR_i$ vanishes unless $\mu$ is centered on atom $i$. Now, if you write

$$ D_{\mu \nu; i} = \Bigg\langle \mu \Bigg| \frac {{\rm d}\nu} {{\rm d}R_i} \Bigg \rangle $$

you get

$$ {\rm d}S_{\mu \nu} / {\rm d} R_{i} = D_{\mu \nu;i} + D_{\nu \mu;i} $$

Since the derivative integral $D$ trivially vanishes unless the derivative operator acts on the coordinate of the basis function in question, you can pack D as an $ 3 \times N_{\rm atom} \times N_{\rm bf} \times N_{\rm bf} $ array. This is what I assume PySCF is doing.

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  • $\begingroup$ You are totally right! Pyscf does in fact pack the derivative according to your $D$. In my calculations I missed the product rule of the derivative... Thank you! $\endgroup$ Commented Jun 18 at 9:01

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