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Do van der Waals corrected DFT functionals include Keesom and Deby corrections too or just London dispersion? Because when I look at these methods in detail, I see that there is just an attempt to correct London dispersion correction and no mention to the remaining two (Deby and Keesom) van der Waals correction. What is the reason? If there is just London dispersion correction, then why are they called van der Waals corrections?

Another question that I could not find an answer to is: Are Keesom and Deby forces already included in semilocal and local density functionals compared to long-range London dispersion interactions?

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    $\begingroup$ +1 Welcome to Matter Modelling SE! I am not an expert on DFT, but my guess is that Keesom and Debye forces are accounted for in local DFT because they are kind of static polarization/electrostatic effects. Dispersion is not included because it is modelled by charge transfer, which cannot be done by local/semilocal functionals. $\endgroup$
    – S R Maiti
    Jun 21 at 13:42
  • $\begingroup$ @ShoubhikRMaiti Dispersion is a long-range correlation effect, arising from the double excitation \Psi_{ij}^{ab}, where the orbitals i and a are localized on one molecule and j and b are localized on the other molecule. There is a dipole-dipole Coulomb interaction between the occupied-virtual pairs (ia) and (jb), which explains the long-rangedness of the dispersion interaction. But the excitations i->a and j->b are both local, so there is no charge transfer, thus dispersion is not modeled by charge transfer. $\endgroup$
    – wzkchem5
    Jun 22 at 10:01
  • $\begingroup$ @ShoubhikRMaiti Charge transfer is described by the single excitations \Psi_i^b, where i and b are far apart. They cannot be reliably described by TDDFT when local and semilocal functionals are used (although they can be described by SCF), which is due to the DFT exchange being short-ranged. This is also a famous pitfall of many functionals, but a different one than the pitfall of describing dispersion, the latter of which being due to the DFT correlation being short-ranged. $\endgroup$
    – wzkchem5
    Jun 22 at 10:05
  • $\begingroup$ @wzkchem5 I see, thanks. I didn't know about the single vs double excitation difference. $\endgroup$
    – S R Maiti
    Jun 22 at 11:49
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In computational chemistry, the term "van der Waals interaction" tend to refer to the London term only; the Keesom term (electrostatic interactions between two freely-rotating permanent dipoles) is generally included in the electrostatic interaction, while the Debye term (interaction between a permanent dipole and an induced dipole) is usually considered as a type of induction interaction and is treated on the same footing as the ion-induced dipole force. Both the Keesom and the Debye terms are described excellently by most density functionals, with caveats that are detailed below.

To compute the Keesom term, the most rigorous way is to perform a rotational average over the interacting molecules, which may require either a molecular dynamics simulation or a Monte-Carlo sampling; the simple "geometry optimization + frequency + single point (+ counterpoise correction)" workflow cannot be guaranteed to capture the Keesom term correctly when the molecules can rotate quite freely, but if the molecules have a relatively fixed orientation with respect to each other (as is often the case for molecules that are not very small), then the rotational average is not necessary, and the usual quasi-RRHO scheme (as implemented in, e.g., ORCA) usually suffices to give accurate results. In more layman terms, if you don't do rotational averaging, the electronic energy does not accurately capture the Keesom term (and usually overestimates its magnitude), but the thermal correction U(T)-U(0), which you obtain from your frequency calculation, largely corrects the error of the Keesom term, so that if you add the U(T)-U(0) and ZPE terms to the electronic energy to give the energy at finite temperature U(T), you actually have the Keesom term at hand already. Note however that some programs (e.g., Gaussian) uses the RRHO scheme, instead of the quasi-RRHO scheme, in calculating the partition functions; in this cases the Keesom contribution may be poorly reproduced, and the way out is to use third-party tools like GoodVibes (https://github.com/bobbypaton/GoodVibes) or Shermo (http://sobereva.com/soft/shermo/) to recalculate the partition functions using quasi-RRHO. In any case, even if you do fail to compute the Keesom term accurately, it is not a failure of the functional, but a failure of (quasi-)RRHO.

The Debye term depends on the permanent dipole of the polarizer molecule and the polarizability of the polarized molecule. Both are accurately reproduced by most functionals if and only if the basis set is sufficiently diffuse. So again, even if the Debye term is calculated wrong, it's not the functional to blame.

For the London dispersion term, things are wholly different. The London interaction is a long-range correlation interaction that decays like $1/R^6$. By contrast, all local and semilocal functionals, as their name suggests, have a correlation contribution that decays at least exponentially. It is because of this reason that they cannot describe the London dispersion term adequately. Within the semilocal framework, one can parameterize the functional such that it describes the London interaction when the molecules are not very far apart (e.g. the M06 and M06-2X functionals), but as the molecules become progressively far from each other, any parameterization must eventually fail and underestimate the London interaction, because the exponential decay is bound to kick in. Thus, the simplest physically correct way to restore the London interaction is to introduce diatomic attraction potentials with built-in $1/R^6$ behavior, which is exactly the approach used by DFT-D. More robust but costly alternatives (such as the VV10 correction) also exist.

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