During an OO-RI-ωB97X-2-D2 calculation on ORCA, the non-dispersion-corrected OO-DHDF part of the calculation is performed, the D2 dispersion correction energy added to the resulting OO-DHDF energy, and then the properties (including the relaxed MP2 density, and therefore the fully relaxed orbitals) calculated. This means that the fully relaxed orbitals (and their energies) are dependent of the D2 correction iff the relaxed MP2 density depends on the total, OO-DHDF-D2, energy.

This was done on a small scale on a model system, namely gas-phase water at the OO-RI-ωB97X-2-D2/aug-pcSseg-2/aug-cc-pwCVQZ(C)/AutoAux(J; ORCA does not support RIJK in OO-RI-MP2) level of theory, the D2 keyword being added separately(ORCA does not support ωB97M(2), which would definitely have "dispersion corrections" in the resulting orbitals in its OO-RI-ωB97M(2) form); the very large molecule I'm actually going to analyse would be likely to have a lot of NCIs and "artificially broken symmetries" in it and hence the OO-RI-ωB97X-2-D4/aug-pcSseg-2/aug-cc-pwCVQZ(C)/AutoAux(J) level of theory would be used(checked the literature; pcSseg-2 already gives at worst chemical accuracy for core-LUMO excitation energies of small molecules at the TD-B97-1 level of theory; no prizes for guessing how small the core-valence correlation errors, mandated as part of OO-RI-MP2 calculations on ORCA, would be, given that B97-1 is actually not good enough for most purposes), the necessary parameters for D4 being taken from those for ωB97X.

My main question now follows- would adding D4 to the mix (per my recipe) have meaningful effects on the orbital energies, shapes, the electrostatic potential, NBO populations, and/or other "fully relaxed" properties?


1 Answer 1


The DFT-D family of corrections, including DFT-D2, DFT-D3 and DFT-D4, are energy corrections that depend on the atomic coordinates but not the wavefunction. (The DFT-D4 energy does depend on atomic charges, and the latter can be calculated from a wavefunction theory; however, to facilitate the calculation of nuclear gradients, one typically chooses another wavefunction theory that is considerably cheaper than the one used in calculating the "dispersion-free" electronic energy, for example a semiempirical method. Most of the time, one uses even coarser methods such as EEQ, which does not even involve wavefunctions.) DFT-D will only contribute to a property if the property is an energy derivative of (and only of) atomic coordinates, such as nuclear gradients or Hessians. Once you take a derivative with respect to a non-geometric parameter (e.g. suppose you want to calculate the geometric derivative of dipole moments, which is the mixed second-order derivative of the energy with respect to atomic coordinates and the electric field), the DFT-D contribution vanishes completely.

For the properties that you explicitly mentioned, it is even easier to see that DFT-D does not contribute, since they involve only the wavefunction, and DFT-D does not contribute to the wavefunction. Whether DFT-D may have an influence on "other fully relaxed properties" is not so clear-cut, since nuclear gradients and Hessians are, in some sense, fully relaxed properties too. In any case, it should be easy to determine whether DFT-D contributes to a property using the fact that DFT-D only contributes to the energy and only depends on atomic coordinates.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .