In order to relate results from quantum chemical calculations to experiment, it is essential to compute quantities that are directly available from measurements.

Properties describe the "response" of the molecular system to an external perturbation $\lambda$:

$ E(\lambda) = E(0) + \left(\frac{\partial E}{\partial \lambda}\right)_{0}\lambda + \frac{1}{2!}\left(\frac{\partial^2 E}{\partial \lambda^2}\right)_0\lambda^2 + \frac{1}{3!}\left(\frac{\partial^3 E}{\partial \lambda^3}\right)_0\lambda^3 + \ldots $

That is, to calculate any property of a molecule, it is necessary to take these derivatives by $\lambda$'s. There are two ways to do this:

  • Finite-differentiation techniques (so-called "finite-field" calculations).
  • The alternative to numerical differentiation is analytic differentiation.

So, for the finite-differentiation techniques we must know how the energy depends on the field $\lambda$ and calculate the finite differences:

$\frac{dE}{d\lambda} \approx \frac{E(\lambda + \delta\lambda) - E(\lambda)}{\Delta\lambda}$.

Let's consider Derivative studies in hartree-fock and møller-plesset theories by Pople at all

Formula from which to start for taking the first derivative is (22)

$ \frac{\partial E_{HF}}{\partial\lambda} = \sum_{\mu\nu} P_{\mu\nu} \left(\frac{\partial h_{\mu\nu}}{\partial\lambda}\right) + + \frac12 \sum_{\mu\nu\lambda\sigma} P_{\mu\nu}P_{\lambda\sigma} \frac{\partial}{\partial\lambda} (\mu\lambda|\nu\sigma) + \frac{\partial V_\text{nuclei}}{\partial\lambda} - \sum_{\mu\nu}W_{\mu\nu} \left(\frac{\partial S_{\mu\nu}}{\partial\lambda}\right) $

Then dances with a tambourine begin around how to take these derivatives caled coupled perturbed Hartree-Fock theory. To parametrize for this purpose the MO coefficient derivatives in the following way: $ c_{\mu i}(\lambda) = \sum_{p} c_{\mu p}(0) u_{pi}(\lambda) $

or for derivative

$ \frac{\partial c_{\mu i}(\lambda)}{\partial\lambda} = \sum_{p} c_{\mu p}(0) \frac{\partial u_{pi}(\lambda)}{{\partial\lambda}} $

coeffcients $u_{pi}(\lambda)$ as the parameters to be determined.

I still have some misunderstanding: where do these initial derivatives come from? Indeed, from the Hartree-Fock method, we obtain only numerical values of the orbital coefficients. But, as shown in the article, there you already need to know their derivatives with respect to the field. How to set these initial derivatives?

What is the meaning of the method of analytical derivatives using in all computational Chemistry soft?

The result of the QCh calculations is the value of ground state energy $E(0)$. Is it really necessary for the method to set some field value $\lambda$, or does the method do without them?

(There are a lot of formulas with derivatives in many textbooks, but the very zest of the method escapes me)

Algorithm (from article)

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  • $\begingroup$ May be able to provide an answer at some point, but just as a starting hint, you can get the initial MO coefficient derivatives for CP-SCF the same way you get the MO coefficients for SCF: you make some guess and then iterate to improve that guess. $\endgroup$
    – Tyberius
    May 22, 2022 at 20:55
  • 4
    $\begingroup$ For a variational wave function, like HF, you do not need the CPHF for the first derivative, only for the second derivative. The CPHF equations can be obtained by differentiating the SCF and orthonormality conditions. They become SCF like, and can be solved by the same iterative techniques as solving for the wave function (orbitals). $\endgroup$ May 23, 2022 at 12:26
  • $\begingroup$ @FrankJensen Thank you. But to calculate the second derivatives, we need perturbations in the first derivatives (dipole momentum, for example) and the use of excited orbitals, as I understand it. $\endgroup$
    – Sergio
    May 23, 2022 at 12:33
  • 1
    $\begingroup$ Second derivatives require the first order response of the wave function. This can be formulated as a sum over all excited state, but this is not how it is implemented. Rather, think of the response as the first order change in the wave function when you add the perturbation. $\endgroup$ May 24, 2022 at 10:53
  • $\begingroup$ Did the last comment by @FrankJensen help you solve this? Are you still working on it? Please update us! $\endgroup$ Feb 6, 2023 at 20:54


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