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Time indepedendent Schrödinger equation for a system (atom or molecule) consisting of N electrons can be written as (with applying Born - Oppenheimer approximation): $$ \left[\left(\sum_{i=1}^N - \frac {h^2} {2m} \nabla _i ^2\right) + \sum_{i=1}^N V(r_i) + \sum_{i < j}^N U(r_i,r_j)\right] \Psi = E \Psi $$

Terms in Hamiltonian are as follows:

  1. Kinetic energy of electrons
  2. Potential energy of electron - nuclei interaction
  3. Potential energy of electron - electron interaction

It is said that for N electron system, kinetic energy of electrons and potential energy of electron - electron interaction are system independent which means that their value depends only on number of electrons $N$ (Because of that they are called universal operators). Potential energy of electron - nuclei interaction depends on specific system and isn't determined only by $N$.

Source: DFT wikipedia, section: Derivation and Formalism, 2nd paragraph.

https://en.wikipedia.org/wiki/Density_functional_theory.

Second source is this page where Hohenberg and Kohn theorems are proved; statements are made after equation 1.31.

http://cmt.dur.ac.uk/sjc/thesis_ppr/node12.html

Why is this? This is usually mentioned in DFT materials, but I didn't find any source which explains it.

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  • $\begingroup$ Each electron in the universe exactly equals another (they are "identical particles"). What else do you expect from an explanation? $\endgroup$
    – oliver
    Commented Feb 19, 2022 at 10:12
  • $\begingroup$ They are, yeah. However, what does that have to do with my question? I am not talking about exchange symmetry, I am talking about kinetic energy and electron - electron interaction energy and their universality regarding number of electrons. $\endgroup$ Commented Feb 19, 2022 at 10:16
  • $\begingroup$ Exchange symmetry is just one consequence of all electrons being the same thing. If N electrons in Los Angeles are the same as N electrons in Berlin, how could their operators of kinetic or interaction potential be different? Shouldn't you ask the question "why not" instead of "why"? Maybe I am missing your point. $\endgroup$
    – oliver
    Commented Feb 19, 2022 at 10:25
  • $\begingroup$ Yes, I've got this in the meantime after discussions on PhysicsForums. Born - Oppenheimer approximation fixes the nuclei and this actually makes operator non - universal since $R_i$ is than system specific since every substance (in which Born - Oppenheimer approximation is valid) has unique distribution of nuclei positions. $\endgroup$ Commented Feb 19, 2022 at 11:16

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The Born-Oppenheimer approximation introduces a dependency on the nuclear coordinates. The operator $\hat V = \hat V(R)$ can be seen as function of these parameters. We solve the electronic time independent Schrödinger equation only for one particular choice of $R$ and in that sense, the solution is not universal, since the operator itself isn't universal. The universal solution would be solving the Schrödinger equation without the Born-Oppenheimer approximation, i.e. treating the nuclei not as fixed but just like the electrons. In that case we wouldn't have to specify any nuclear coordinates and could use an universal potential operator. The electron interaction operator on the other hand is universal. $$ \hat U = \sum^N _{i=1}\sum^N_{j>i} \frac{e^2}{4\pi \varepsilon_0|\hat r_i- \hat r_j|} $$ It only depends on $N$ the number of electrons but besides that, it always has exactly this form.

The interaction with the nuclei however is not a "pure" operator within the Born-Oppenheimer approximation, $$ \hat V(R) = \sum^N _{i=1}\sum^N_{j>i} \frac{-Z_ie^2}{4\pi \varepsilon_0|R_i- \hat r_j|} $$ This is different to the universal interaction $$ \hat V = \sum^N _{i=1}\sum^N_{j>i} \frac{-Z_ie^2}{4\pi \varepsilon_0|\hat R_i- \hat r_j|}. $$

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  • $\begingroup$ Yes, thank you. After some discussions on PhysicsForums, I think I've got it. As you said, Born - Oppenheimer approximation fixes the nuclei and this actually makes operator non - universal since $R_i$ is than system specific since every substance (in which Born - Oppenheimer approximation is valid) has unique distribution of nuclei positions. $\endgroup$ Commented Feb 19, 2022 at 11:14

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