Why are single excitations ignored in the MP2 component of double hybrid functional calculations?

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.

Matter Modeling Asked on January 5, 2022

1 Answers

One Answer

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)

Answered by Tyberius on January 5, 2022

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