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When I consider the grid positions $r$ in a molecular system, I have a matrix, where each row can be regarded as an N-dimensional vector and each column can be regarded as a G-dimensional vector. In the following figure, N is the number of orbitals and G is the number of grid positions.

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In this case, the elements of $i$th N-dimensional vector is the values of orbital functions on the position $r_i$; the elements of $j$th G-dimensional vector is the values of $j$th orbital function.

In this matrix, I believe that there are two kinds of interactions. (1) The interactions between different distances (red arrow). (2) The interactions between different orbitals (blue arrow) .

The DFT concept is based on the non-interacting electrons moving in the effective potential. Here, is the meaning of non-interacting electrons (1) or (2) or both?

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The non-interacting orbitals in exact KS-DFT still interact between different distances and different orbitals, so the answer is (1) and (2).

Your spatial representation is basically a projection into position eigenstates, while the real orbital has contributions from all over the system. To evaluate the one-orbital contributions correctly, you need to integrate over the system, giving you (1).

Next, already the Coulomb contribution, which is included in the exact Kohn-Sham functional, includes interactions between orbitals (between orbital contributions to the total density) which gives (2).

The "non-interacting" just means that the Kohn-Sham wave function is a single Slater determinant, while the true wave function consists of exponentially many determinants.

Addendum: what you attempted to do with the grid, i.e. a real-space solution of the wave function can actually be achieved with e.g. finite differences or finite elements, see my open access review in Int J Quantum Chem 119, e25968 (2019).

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  • $\begingroup$ +1 for another very timely and helpful answer indeed! And congratulations on reaching 12000 points here :) Also, on reaching 400/400 score on the DFT tag! $\endgroup$
    – Nike Dattani
    Aug 25, 2021 at 19:12
  • $\begingroup$ Thank you very much for the answer. I understood the meaning of the "non-interacting" in DFT. $\endgroup$
    – neco
    Aug 26, 2021 at 0:07

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