From this question and answer, I understood the Hohenberg-Kohn theorem and found that there is a one-to-one correspondence between the external potential $V_\text{ext}$ and the electron density $\rho$ as entire functions, not pointwise.

Now I would like to know the relationship between $\rho$ and the effective potential $V_\text{eff}$ (= $V_\text{ext}$ + $V_{H}$ + $V_{xc}$, where $V_{H}$ is the Hatree term and $V_{xc}$ is the exchange-correlation term). Is there any kind of pointwise correspondence between $V_\text{eff}$ and $\rho$ in systems? I understand that the DFT can be described as the electron cloud moving in $V_\text{eff}$, that is, the electrons do not interact with each other and are spread on $V_\text{eff}$ like the following figure of Schrödinger's and DFT's views (https://www.semanticscholar.org/paper/TOPICAL-REVIEW%3A-Designing-meaningful-density-theory-Mattsson-Schultz/9e57d4b4d5eec0f1b65fe774638b6d97a99f5f53/figure/0).

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Here, is there any pointwise correspondence between $V_\text{eff}$ and $\rho$? The reason for thinking this way is that, I was wondering if there is such a mapping between the effective potential and the non-interacting electron cloud, I believe that we may be able to learn the mapping by machine learning method described in the following paper.


Postscript: the word of 'point-wise' mapping means the mapping between the value of $V_\text{eff}(r)$ on a point $r$ and the value of $\rho(r)$ on $r$, not the the mapping between their entire functions, $V_\text{eff}$ and $\rho$ in the system. The (non-interacting) electrons are moving on the effective potential, so if we have or can consider a mapping between the scalars, $V_\text{eff}(r)$ and $\rho(r)$, such situation is easy for machine learning to learn the mapping from data.


Roughly speaking and provided that the Hohenberg-Kohn theorem applies, if the $v$-representable ground state density of an interacting system $\rho$ is additionally non-interacting $v$-representable, then by definition there exists a non-interacting system with potential $V$ and $\rho$ as its ground state density. By the virtue of the Hohenberg-Kohn theorem, $V$ uniquely determines $\rho$ and vice versa. In the Kohn-Sham approach, then an explicit form of $V=V_\mathrm{eff}=V_\mathrm{eff}[\rho]$ is constructed, such that solving the Kohn-Sham equations self-consistently yields $\rho$.

I don't know what exactly you mean with 'point-wise'; the mapping between $V_\mathrm{eff}$ and $\rho$ is similar to the mapping between the external potential in the interacting system $V_\mathrm{int}$ and $\rho$, i.e. is a mapping between the 'complete' functions, since it is also governed by the Hohenberg-Kohn theorem.

  • $\begingroup$ Thank you very much for the quick reply. The word of 'point-wise' mapping means the mapping between the value of $V_{eff}(r)$ on a point $r$ and the value of $\rho(r)$ on $r$, not the the mapping as entire functions, $V_{eff}$ and $\rho$ in the system. I have added the postscript. $\endgroup$
    – neco
    Aug 25 at 11:10
  • $\begingroup$ @neco I see. But as the answer to your linked question also writes, the Hohenberg-Kohn theorem (if it applies) just states that the entire functions determine each other. There is no statement about the 'point-wise' behaviour. So there can or can not be a unique point-wise map; from the HK theorem alone we don't know. But from my naive point of view, I'd say that this does not hold in general, as also your example in the linked question shows. $\endgroup$
    – Jakob
    Aug 25 at 11:24
  • 1
    $\begingroup$ I see. Thank you very much! I will continue to think about it. $\endgroup$
    – neco
    Aug 25 at 11:27

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