Is there any paper that proves/disproves that the occupied orbitals obtained using full Configuration Interaction (CI) are the same as the orbitals obtained using DFT (assuming we use the exact functional, which is not found yet)?


FCI doesn't really have "occupied orbitals" in the mean-field sense, since all orbitals become fractionally occupied. What one usually does is look at natural orbitals, which one gets by diagonalizing the one-particle density matrix; then each spin-orbital has some occupation number $f_n \in [0,1]$.

However, DFT isn't really aiming to reproduce the orbitals but rather the electron density, i.e., the diagonal part of the one-particle density matrix. There has also been a lot of unclarity in this question: again, Kohn-Sham DFT uses a set of $N$ fully occupied spin-orbitals, whereas the exact solution i.e. FCI has $K\gg N$ fractionally occupied orbitals, that should yield the same total density at every point in space.

This question has been discussed recently by Mayer et al in a short letter, J. Chem. Theory Comput. 13, 3961 (2017). The authors point out that FCI and exact DFT can never be equivalent in a finite basis set; the equivalence only holds in an infinite and complete basis.

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  • $\begingroup$ +1. I think the asker couldn't have hoped for anything better than this answer! $\endgroup$ – Nike Dattani Jul 9 at 16:10

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