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Double hybrid approximate functionals have "unoccupied" Kohn-Sham orbitals in their formulations due to their MP2 component; however, the "exact functional" depends only on the "occupied" K-S orbitals.

Thus the only way double hybrid approximations could approximate the "exact" functional is that the "exact" DFT density be theoretically invariant under MP2, or at least MPn where n tends to infinity.

This would mean that- should one apply MPn to a set of K-S orbitals, let n tend to infinity, and sum the squares of the resulting "MPn occupied orbitals"(like one theoretically does when trying to compute the density from an "unperturbed" K-S set), one should get the same density as the one computed without MPn being applied.

Is it mathematically true?

*I'm assuming exact functional here, not accounting for the fact that its exact form is unknown.

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    $\begingroup$ +1. You might find this paper relevant, but it only talks about MPn with n=155 at most (not n=infinity). Perhaps you'd like to ask the question "Does MPn converge to FCI as n -> infinity" in a separate question. But I think this question should probably just stick to MP2 (or MPn with finite 'n') since otherwise it seems that you're asking two different questions: whether it's true for finite n, and whether it's true for infinitely large n. $\endgroup$
    – Nike Dattani
    Oct 29, 2022 at 21:58
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    $\begingroup$ As for a more direct answer to the question: wouldn't the exact double-hybrid functional need to be different from the exact ordinary-DFT functional, because the former starts with MP2? You may also be interested in density-corrected DFT, in which a more accurate density is used: mattermodeling.stackexchange.com/a/1260/5 $\endgroup$
    – Nike Dattani
    Oct 29, 2022 at 22:04
  • $\begingroup$ I think that the more I think about this question, the more confused I get. What do you mean by "exact DFT density"? Do you mean the "exact physical density"? Why would the "exact double hybrid density" be different? $\endgroup$
    – Nike Dattani
    Oct 30, 2022 at 0:12
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    $\begingroup$ @NikeDattani the density one gets if one sums the squares of the occupied K-S orbitals $\endgroup$
    – Kanghun Kim
    Oct 30, 2022 at 5:36
  • $\begingroup$ I got confused by the word "exact". $\endgroup$
    – Nike Dattani
    Oct 30, 2022 at 12:34

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The question is false. There is no "exact" double hybrid density. The problem is that at the point you approximate the exact functional with a double hybrid, the functional is no longer exact. You can find the ground-state solution to the double hybrid functional, but it will not be the exact ground state density of the Schrödinger equation.

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  • $\begingroup$ Could you provide the piece of literature that (dis)proves it? $\endgroup$
    – Kanghun Kim
    Nov 17, 2022 at 9:32
  • $\begingroup$ This is indeed true if the XC part of the double hybrid is semilocal (since the resulting functional cannot describe dispersion exactly). However, if the XC term is allowed to take arbitrary non-local forms, then I don't see why a double hybrid cannot be exact, while a hybrid functional or pure functional can. In another comment I showed that the double hybrid energy and the density obtained from occupied KS orbitals cannot be both exact, but this can be circumvented by substituting the latter by the response density. $\endgroup$
    – wzkchem5
    Nov 17, 2022 at 12:11
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    $\begingroup$ @KanghunKim prove what? You made the claim that double hybrids are exact. This is false except in the trivial case where the weight of the DFT part is zero and the post-HF term is FCI@CBS. $\endgroup$
    – Susi Lehtola
    Nov 18, 2022 at 8:51
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    $\begingroup$ I mean 'There is no "exact" double hybrid' is true when the XC part is restricted to be semilocal. I now also realize that it's important to also stress that, what you said is also almost certainly true when the XC part is restricted to be closed form. One can as well come up with an infinite sequence of double hybrid functionals that approach exactness to any desired accuracy, so one can define a functional as the limit of this sequence. I have always assumed that this limit can also be called a double hybrid, but I guess you object this, and this is probably why we disagree. $\endgroup$
    – wzkchem5
    Nov 19, 2022 at 11:09
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    $\begingroup$ @SusiLehtola The limit of a sequence of approximations can be exact, even if all terms of the sequence are approximate. The problem is therefore: is the limit of a sequence of double hybrids also a double hybrid? I assumed that it is, but now I realized that this assumption was unfounded, since I haven't seen a definition of double hybrids that is rigorous enough to settle the question "whether the set of double hybrids is closed upon taking limits". So yes, I have used a terminology without being aware of its exact definition. Apologies. $\endgroup$
    – wzkchem5
    Nov 20, 2022 at 16:24

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