I have read the following paper, "Bypassing the Kohn-Sham equations with machine learning", published from Nature Communications in 2017.


The Hohenberg-Kohn theorem states that there is one-to-one correspondence between the external potential $V_{ext}$ and the electron density $\rho$. I believe that the form of $V_{ext}$ is the Coulomb as

$V_{ext}(r) = \sum_{i=1}^M \frac{Z_i}{||r-R_i||}$,

where $r$ is the electron position, $R$ is the atomic position, and $Z$ is the nuclear charge. Instead of the Coulomb, however, this paper uses the Gaussian external potential as

$V_{ext}(r) = \sum_{i=1}^M Z_i e^{-||r-R_i||^2}$,

because "The external (Coulomb) potential diverges for 3D molecules and is, therefore, not a good feature to measure the distance in machine learning (ML). Instead, we use an artificial Gaussian potential...", and here I have a question about the selection of $V_{ext}$.

In the Hohenberg-Kohn map between the input $V_{ext}$ and the output $\rho$, can $V_{ext}$ be anything form such as the Gaussian, not the Coulomb? For example, the following figure shows the electron density values (calculated by Gaussian09 software) on the C=C bond in the benzene molecule, and the Coulomb potential has a very extreme form; in contrast, the Gaussian potential seems to be closer to the calculated density.

enter image description here

When considering the map from $V_{ext}$ to $\rho$, the Gaussian seems to be more reasonable in a molecule (also may be reasonable for other molecules) than the Coulomb from a practical point of view. Nevertheless, is the actual external potential Coulomb?


1 Answer 1


The actual external potential of molecules is indeed a sum of the Coulomb potentials of the nuclei (except when one uses a pseudopotential, where the potential will be Gaussian-like near the nuclei; but pseudopotentials are approximations). The point is that, there is also a one-to-one correspondence between the sum of nuclear-centered Gaussians and the sum of nuclear-centered Coulomb potentials (you can easily verify this). Thus, the fact that there is a one-to-one correspondence between the external potential and the density (when the ground state is not degenerate), means that there is also a one-to-one correspondence between the artificial sum-of-Gaussian potential and the density.

However, there are some potentially misleading points that I'd like to clarify:

  1. It seems that the ML algorithm proposed in the paper only used the fact that $v(r)$ uniquely determines $\rho(r)$ (which is only true if the ground state is non-degenerate). This is a much weaker, in fact trivial, statement than the Hohenberg-Kohn theorem, which also states that $\rho(r)$ uniquely determines $v(r)$.

  2. The Hohenberg-Kohn theorem does not only apply to nuclear Coulomb potentials, but also to non-nuclear and/or non-Coulomb potentials. For example, if you have a molecule in a possibly non-uniform external electric field, then one can reveal the exact distribution of the electric field simply by examining the ground state density of the molecule. Or if the sizes of the nuclei are taken into account, so that the nuclear potentials do not have a cusp. Or if we used pseudopotentials. Or if we lived in a weird universe where the nuclear Coulomb interaction is $V(r) = -Z \exp(-\alpha r)/r$ instead of $V(r)=-Z/r$. This is what makes the Hohenberg-Kohn theorem a very deep result. In fact if we already know that the potential is the sum of the Coulomb potentials of a finite number of nuclei, we can find the potential simply by looking at the positions and strengths of the cusps of the density (owing to Kato's conditions), which would make the one-to-one correspondence between $v(r)$ and $\rho(r)$ a trivial result.

  3. The similarity between the Gaussian potential and the density does not guarantee a better mapping. Besides, sufficiently near the nuclei, the Gaussian potential does not look like the density. If you draw your A and B points at the centers of the respective carbon atoms, and if you properly include the contributions of the core electrons, you will find very localized and very strong cusps in the density, but the Gaussian potential will be much more slow-varying and rather featureless.

  4. One can view the Gaussian potentials as yet another method to convert the nuclear coordinated into ML descriptors (besides the Coulomb matrix, rotationally symmetric functions, etc.). In this sense, this work represents simply a ML method that predicts the electronic density and energy from nuclear coordinates.

  5. It's much harder to predict the energy than the density, at least when the required accuracy of the density is not very high. For example, you can get a not-too-bad density by simply superimposing the neutral atomic densities, which is already sufficient for some applications, e.g. solving XRD structures. What is difficult is to calculate the energy given the density. Traditional non-ML approaches typically given atomization energies that are in error of > 200 kcal/mol, which is useless for any application. Thus, what should be really highlighted in this paper is the good energies.

  • $\begingroup$ Thank you for the very detailed and insightful answer! I haven't understood all the answers yet, but I will continue to study them. $\endgroup$
    – neco
    Commented Jul 2, 2021 at 3:09

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