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In my previous question, I was basically asking whether the results of the double hybrid using the exact XC functional are the same as those of just the exact XC functional.

Even that sentence is hard to grasp, so I'm going to ask something more simple- is there an analogue of the K-S theorem for double hybrids, where the "double" part "climbing the ladder" from MP2 to MPn to CISD(T) to full CI, AND the "hybrid" part simply being the exact XC, that yields the exact density of the ground state of the system of interest?

UPDATE: I was talking strictly about mathematical rigour, ignoring facts like how "the exact XC is unknown".

UPDATE 2: To make everything clear- given a set of exact K-S orbitals, does applying MP2, CI(etc) on these orbitals (as if they were H-F orbitals) and summing their squares give the exact ground-state density as well, mathematically?

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There is no Kohn-Sham theorem. There are the three Hohenberg-Kohn theorems that demonstrate that the density and the potential are two ways to approach the ground state problem: the potential defines the density, and the density defines the potential (up to an additive constant).

The Kohn-Sham approach is to remove the need for a kinetic energy functional by approximating the kinetic energy with the Hartree-Fock kinetic energy of the Kohn-Sham orbitals. The difference between the many-body kinetic energy and the single-determinant kinetic energy is added to the exchange-correlation energy.

The question is the same misunderstanding as in Is an "exact" double hybrid density the same as the "exact" DFT density?. The exact functional is not a double hybrid. You can build various density functional approximations (DFAs), but they will not give you the exact solution. The functional error is commonly decomposed into "functional error", which the difference of the approximate energy evaluated at the exact density to the exact energy functional at the exact density, and the "density driven error", which is the contribution of the density relaxation driven by the functional error.

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    $\begingroup$ @wzkchem5 your claim is tantamount to claiming that there is an exact double hybrid functional. By the Hohenberg-Kohn theorems, this would be the exact functional. no? $\endgroup$
    – Susi Lehtola
    Commented Nov 18, 2022 at 8:53
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    $\begingroup$ Yes. But the same functional can be written in more than one equivalent way, so while the exact functional can be written as a hybrid or pure functional, this does not prove that it cannot also be written as a double hybrid functional, which has very different Kohn-Sham orbitals and orbital energies, but still exact energies and exact (response) densities. If, however, you argue that a functional is only exact when the sum of norm squares of Kohn-Sham orbitals equals the exact density, then I agree with you that a double hybrid cannot be the exact functional. $\endgroup$
    – wzkchem5
    Commented Nov 18, 2022 at 11:22
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    $\begingroup$ @wzkchem5 please prove that the exact functional can be written as a hybrid or pure functional. $\endgroup$
    – Susi Lehtola
    Commented Nov 19, 2022 at 10:46
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    $\begingroup$ Let's first agree on the definition of a pure functional - do you agree that any functional where the XC energy depends solely on the density and not directly on the KS orbitals counts as a pure functional? If yes, do you further agree that a pure functional does not need to have a closed form expression? If also yes, then the proof is trivial. If however you think that only closed form functionals count as hybrid or pure functionals, then I agree that the exact functional almost certainly cannot be a hybrid or a pure functional. $\endgroup$
    – wzkchem5
    Commented Nov 19, 2022 at 10:59
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    $\begingroup$ "pure" functional is a misnomer since that would make the exact functional non-pure. The better term for what is usually meant by that term is (semi-)local functional. The exact functional is a non-local functional of the electron density. $\endgroup$
    – Susi Lehtola
    Commented Nov 23, 2022 at 11:02

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