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My textbook "Density Functional Theory of Atoms and Molecules" by Parr and Yang says that any N-representable density is derivable from a single determinantal wavefunction. A density $\rho$ is N-representable if it satisfies $$ \tag{1} \rho(\mathbf r) = \int d\sigma_1\int |\psi(\mathbf x_1, \mathbf x_2, ... \mathbf x_N)|^2 d^3\mathbf x_2 ... d^3\mathbf x_N $$ where $\psi(\mathbf x_1, \mathbf x_2, ... \mathbf x_N)$ is antisymmetric in exchanging any pair of its arguments, and $x \equiv (\mathbf r, \sigma)$. An antisymmetric wavefunction is in general a linear combination of Slater determinants. What I want to prove is that the above $\rho$ can also be written as $$ \tag{2} \rho(\mathbf r) = \sum_{\sigma=1}^2 \sum_{i=1}^{N_\sigma} |\phi_{i\sigma}(\mathbf r)|^2 $$ for some set of wavefunctions $\{\phi_i\}$. But I don't know where to go beyond this point.

Can someone help me to prove this statement?

EDIT: The Kohn-Sham DFT apparently unconditionally assumes that there exists a non-interacting system having ground state density that is identical to the exact ground state density of the original interacting system. Therefore, stated in a different way, my question can also be understood to ask about the existence of this reference system, does it really always exist for an arbitrary real electronic system? If it doesn't always exist, can one also specify which conditions does the original interacting system have to have in order for its non-interacting "twin" system to exist.

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  • $\begingroup$ +1. Thanks for bringing your question here :) Hopefully you'll get an answer quickly! Your Eq. 1 has just $x$ but the next line has $x_1$ instead. Which one is correct? $\endgroup$
    – Nike Dattani
    Jul 25, 2020 at 5:47
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    $\begingroup$ Well, $x_1$ or $x$ is just dummy variable, so I suppose it shouldn't matter mathematically. But I changed it anyway. $\endgroup$
    – nougako
    Jul 25, 2020 at 18:54
  • $\begingroup$ Even when they are dummy for calculations, are you using $\mathbf x$ instead $\mathbf r$? Because normally, $x$ is used to represent the x-axis, whereas $\mathbf r$ represents the vector position. $\endgroup$
    – Camps ♦
    Jul 25, 2020 at 18:59
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    $\begingroup$ @Camps $\mathbf{x}$ often means $\mathbf{r}$ (vector position) plus the spin. The user has already made that clear now. $\endgroup$
    – Nike Dattani
    Jul 25, 2020 at 19:27
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    $\begingroup$ @NikeDattani Oh well, I didn't see your last sentence before, which is an interesting reason to know how SE works. Ok then I will post the second one in a separate question but not in a few hours from now. $\endgroup$
    – nougako
    Jul 25, 2020 at 19:38

2 Answers 2

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The proof that (2) is the density arising from a Slater determinant wave function can be found in basically any quantum chemistry textbook.

(2) does NOT hold for multiconfigurational wave functions, since the one-particle density matrix becomes non-diagonal. You can make the density diagonal like (2) by switching to natural orbitals, but then you have fractional occupations for ALL orbitals, instead of just the occupied set of orbitals in one-determinant theories like Hartree-Fock and Kohn-Sham density fucntional theory.

edit: the edited question seems to be asking if exact DFT orbitals reproduce the full CI density, which has already been discussed in Orbitals in full CI and DFT with true functional

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    $\begingroup$ It would be interested if you could add some specific reference about the proof of equation (2). $\endgroup$
    – Camps ♦
    Jul 25, 2020 at 16:53
  • $\begingroup$ The proof for a single determinantal wavefunction is indeed ubiquitous, but what I am asking is in regard to the KS theory of DFT that assumes that a non-interacting reference system having identical ground state density always exist. I have modified my question to make this point clear. $\endgroup$
    – nougako
    Jul 25, 2020 at 19:26
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    $\begingroup$ @nougako I've edited the answer per your added question, which appears to be what was already discussed in mattermodeling.stackexchange.com/questions/1484/… $\endgroup$
    – Susi Lehtola
    Jul 26, 2020 at 9:12
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On the specific subject of the "EDIT", this was exactly the concern addressed by Mel Levy in the 1970s, e.g.

M. Levy, "Universal variational functionals of electron densities, first-order density matrices, and natural spin-orbitals and solution of the v-representability problem", PNAS 76 (12) 6062-6065 (1979); https://doi.org/10.1073/pnas.76.12.6062

The original Hohenberg-Kohn proof relied on several conditions, some of which were explicit (non-degenerate ground state) and some were implicit (N-representability, v-representability). Mel Levy's proof of the theorem is much more general and robust, but it is also rather longer and involved, which may be why people often still present the original, restricted proof.

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  • $\begingroup$ thank you for the link. I have actually seen another proof of HK theorem which relies on proof by construction but this proof involves some sort of iterative steps, if I remember it correctly, and I had hard time understanding it. I will have a closer look at your link, it looks like it involves standard quantum mechanical maths. $\endgroup$
    – nougako
    Nov 22, 2020 at 19:36

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