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I read different things, some textbooks say that the repulsive part of the lennard jones potential is due to the repulsion of the nuclei when they are too close, others say it is due to the pauli repulsion between the electrons.

Or can I imagine that the pauli repulsion creates an antibonding sigma orbital, which means that the electron density between the nuclei is no longer so high and therefore the nuclei repel each other?

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    $\begingroup$ Some say that new kids are found in cabbage patches, and some say they are brought by a stork. Then you grow up and have your first course in Quantum Mechanics, and things come out in a very, very different light. $\endgroup$ Dec 23, 2021 at 21:13
  • $\begingroup$ @IvanNeretin So both approaches are permissible? Electron repulsion by pauli repolsion or/(and?) repulsion between the nuclei? $\endgroup$
    – iwab
    Dec 23, 2021 at 21:34
  • $\begingroup$ Electrons repel each other by their charge, too . $\endgroup$
    – Alchimista
    Dec 24, 2021 at 9:13
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    $\begingroup$ If you learn nothing else from this post and ensuing discussion, learn this: it's one guy - John Lennard-Jones - hence the hyphen. $\endgroup$ Dec 24, 2021 at 16:51
  • $\begingroup$ it isn't even clear clear that the best way to think about Pauli "repulsion" is as a force. $\endgroup$
    – matt_black
    Dec 18, 2023 at 21:27

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The short answer is: the Lennard-Jones potential is a completely heuristic model, and the short-range part of the potential is not physically meaningful except in the sense that it becomes very repulsive very quickly. Pauli repulsion is known, also heuristically, to decay as an exponential, so using a very steep polynomial works reasonably well. As to whether it is modelling the nuclear repulsion or the Pauli repulsion, the answer is both. Also, these two effects are closely related when discussing intermolecular interactions because exchange has the effect of de-shielding electrons and hence enhancing nuclear repulsion.


To be more explicit, the Lennard-Jones potential, $$ V(r)=4\epsilon\left[\left(\frac{\sigma}{r}\right)^{12}-\left(\frac{\sigma}{r}\right)^{6}\right] $$ aimed to provide a simple functional form that reproduced the interactions between neutral gas molecules. In short, any potential that is repulsive at short-range, has a minimum at some intermediate distance, and goes to a constant as $r\rightarrow\infty$ is a heuristic model for the medium-range to long-range interactions between nonpolar molecules. What I mean by medium-range is something like twice the diameter of the molecule in question.

The special thing that the Lennard-Jones potential does is get the correct asymptotic form of the attractive interaction. Namely, it is known from the London Dispersion formula (and other ways) that the dispersion interaction decays as $r^{-6}$. Not only that, but the induced dipole interaction between two nonpolar molecules also decays as $r^{-6}$.

So, this brings us to the short-range part of the potential. As you mention, there are two effects which result in short-range repulsion: nuclear-nuclear repulsion and so-called Pauli repulsion due to the electron exchange. Note that classically, the reason exchange results in repulsion is that it de-shields electrons in the internuclear region and hence results in increased nuclear repulsion.[1]

In fact, many theories predict that exchange repulsion decreases exponentially with distance, so using a polynomial distance dependence is somewhat arbitrary. Due to this, there are other potentials that choose to use an even steeper polynomial which goes as $r^{-14}$, which tends to give somewhat better results in simulations.

Now the question is why does the Lennard-Jones potential choose the specific polynomial of $r^{-12}$? The answer is rather funny. Back in the day when computers were very slow and floating point operations were very expensive, one needed to be very mindful of how expensive calculations were. Now, calculating exponents can be rather expensive. To calculate $(\sigma/r)^6$ requires a division and six multiplies. It would be a shame to have to do even more multiplication operations for the short-range part of the potential, which is the least physically meaningful part anyways. Hence, the exponent 12 was chosen because it is simply the product $(\sigma/r)^6\cdot (\sigma/r)^6$. Hence, one can save many floating point operations by doing this.

Note that this story about choosing the exponent 12 for computational speed is mentioned on the wikipedia page and I have been told this story by multiple scientists I trust, but I haven't looked closely enough at the original paper to know if this is really what Lennard-Jones had in mind.


References:

[1]: Rackers, J. A., & Ponder, J. W. (2019). Classical Pauli repulsion: An anisotropic, atomic multipole model. The Journal of chemical physics, 150(8), 084104.

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    $\begingroup$ "To calculate (σ/r)6 requires a division and six multiplies". No it doesn't: rsq = r * r; r4 = rsq * rsq; r6 = r4 * rsq; V = s/r6. In general any integer power takes approximately log to the base 2 of that power .multiplies. But your general point standsl $\endgroup$
    – Ian Bush
    Dec 19, 2023 at 7:42

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