In the original paper by Grimme introducing double hybrid functionals (also summarized in just 3 paragraphs here), it says:

"[As] opposed to nonempirical versions of KS-PT2 [19], the single excitations contribution is neglected, i.e., we implicitly assume for all Fock matrix elements $f_{ia}=0$."

What is the reason for neglecting these single-excitations?
Please accept my apologies if this question is too trivial, since my knowledge of how double-hybrid (or single hybrid!) functionals work, began and ends with my browsing of the original paper to write the above-mentioned 3 paragraph summary, fewer than 2 months ago.

  • $\begingroup$ My guess is: the first implementation was either in Gaussian (via iops) or using Turbomole. In both, you basically just run a MP2 calculation with a slightly different input and scale down the result (I've done that long time ago). Later, someone somewhere probably tried including the singles, but did not get significantly better results. $\endgroup$
    – TAR86
    Commented Jul 25, 2020 at 6:38
  • 1
    $\begingroup$ My gut reaction is that it is because in the formulation of MP2 where one uses a shifted Fock operator, the first-order correction is equal to zero. If one does not use the shifted Fock operator, the first-order correction reclaims the HF energy. So, either way, one sort of knows what the first-order correction is without any calculation. This is appealing for a functional in the sense that one can choose complete exact exchange and complete MP2 correlation and recover the MP2 energy exactly. I'm not a DFT expert though, so this is just speculation. $\endgroup$
    – jheindel
    Commented Aug 3, 2020 at 17:56
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    $\begingroup$ @jheindel but I would assume that only holds if you have a HF reference. DFT orbitals / the DFT density is not a HF solution so it has non-vanishing singles..? $\endgroup$ Commented Oct 8, 2020 at 22:28
  • $\begingroup$ @ShoubhikRMaiti Are you saying that the singles contribution to the energy will be 0? The first half of the quoted sentence implies that it's not true for "non-empirical versions of KS-PT2". $\endgroup$ Commented Apr 10, 2021 at 20:44

1 Answer 1


For calculating the MP2 energy from the Hartree-Fock, as mentioned in the comments, single excitations don't contribute due to Brillouin's theorem, which states that the Hamiltonian matrix elements between the optimized HF ground state and any singly excited determinant are explicitly zero.

There is no such restriction if we were to compute a perturbed energy for a DFT wavefunction, so in principle single excitations should contribute. However, when Perturbation Theory-Kohn Sham (PT-KS) was initially attempted¹, it produced spuriously low energies, with the error being traced to the single excitation energy. They attribute this to the single excitation energy being negative definite and having occupied-virtual orbital energy differences in the denominator, which tend to be underestimated by DFT.

Since double hybrid DFT is already semiempirical due to the parameter fitting for the HF exchange, DFT xc, and MP2 correlation, it made sense to neglect these single excitations that had already proven problematic rather than proceed with the nonempirical form of the PT contribution.

  1. P. Mori-Sanchez, Q. Wu, and W. Yang, J. Chem. Phys. 123, 062204 (2005)

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