I have seen many machine learning algorithms for prediction of quantum chemistry properties that use Coulomb matrix as their input. Coulomb matrix is defined as,

$$\boldsymbol{M}_{i j}^{\mathrm{Coulomb}}=\left\{\begin{array}{cc}{0.5 Z_{i}^{2.4}} & {\text { for } i=j} \\ {\frac{Z_{i} Z_{j}}{R_{i j}}} & {\text { for } i \neq j}\end{array}\right.,$$

where the diagonal terms:

... can be seen as the interaction of an atom with itself and are essentially a polynomial fit of the atomic energies to the nuclear charge.

I was trying to read more about these diagonal terms and how exactly this equation is fitted. Essentially, something that says: "... and from there we get $0.5Z^{2.4}$." If these energies can be extended to bonds as well, as the off-diagonal term in Coulomb matrix represents repulsion forces rather than potentials, any references would be appreciated as well.

  • 4
    $\begingroup$ The model seems to stem from Rupp et al Phys. Rev. Lett.,108(5):058301, 2012, but I did not find much explanation for it there. $\endgroup$ – Buck Thorn Sep 28 '19 at 6:45
  • $\begingroup$ @Blade could you clarify what you mean by extending these energies to bonds. There are methods (e.g. bag of bonds) related to Coulomb matrices which use bond energies, rather than atom energies , to predict an overall atomization energy. $\endgroup$ – Tyberius Sep 29 '19 at 22:19
  • $\begingroup$ @Tyberius For instance, in molecular mechanics, we describe bond potentials using bond stretching interactions (either harmonic oscillator or Morse function). I was basically wondering how the empirical relation above was derived, so that I see if there are similar empirical relations for bond energies as well. I'm not sure if there is an answer to the second part of my question, just wondering if there is such a thing. $\endgroup$ – Blade Sep 29 '19 at 23:21

In terms of how they got the relation for the diagonal elements, I believe it is relatively straightforward. Given a list of atomic energies (energy of atom relative to separated nucleus and electrons), one can try to fit these values to a function of nuclear charge. If you plot a few values, it suggests trying an exponential fit (I don't know why all the papers say polynomial fit, as this could not lead to a noninteger exponent). So, you can try to fit the values to a function of the form $aZ^b$ using, for example, least-squares regression. You should wind up with something pretty close to the equation given in your question, with some variation because I don't know what atomic energies they used to do the fit. (I'm also not sure why the function isn't negative since the atomic energies are negative. My best guess is the sign winds up not being important as a descriptor in the network).

As to why they choose to use these values along the diagonal, they describe the system in terms of the energies of the individual atoms, along with the energy of interactions between atoms given by the off-diagonal terms. The original Rupp et al. paper compared two Coulomb matrices by looking at the difference between their eigenvalues, with small differences implying similar molecules. These eigenvalues can be thought of as effective atomic energies within the molecule, that is, energies of the individual atoms with the influence of the surrounding atoms included. Similarly, the eigenvectors would describe fragments of the molecule or an effective atom which account for the influence of neighboring atoms.

| improve this answer | |
  • $\begingroup$ Wouldn't the diagonalized matrix then be in a basis consisting of linear combinations of the individual atoms, so something closer to fragments? $\endgroup$ – Buck Thorn Oct 1 '19 at 12:29
  • 1
    $\begingroup$ @BuckThorn that's my understanding. I guess, based on the context of being used for atomization energies, these eigenvector/fragments can thought of as effective atoms with the influence of their environment included as part of their extent. $\endgroup$ – Tyberius Oct 1 '19 at 13:40
  • $\begingroup$ "I don't know why all the papers say polynomial fit, as this could not lead to a noninteger exponent" good catch! I would have repeated that somewhere and looked like a fool! $\endgroup$ – Blade Oct 3 '19 at 0:05
  • 1
    $\begingroup$ @Blade yeah unfortunately the original paper didn't explain things in much detail and later papers just seem to copy the limited and somewhat misleading explanation from that original paper without thinking about it. $\endgroup$ – Tyberius Oct 3 '19 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.