The force between two atoms takes the generic shape of a Lennard-Jones potential. It has an attractive part caused by dipole-dipole attraction and a short-range repulsive part which is said to be caused by Pauli's exclusion principle.

Can some one explain rigorously how the repulsive part of the potential arises from Pauli exclusion principle?


2 Answers 2


The Lennard-Jones potential is $$U(r) = A r^{-12} - B r^{-6}.$$The function of the first term is to make the energy go to $+\infty$ as $r\to 0$ as a way of modeling Pauli exclusion with a potential barrier that stops two particles from being in the same location. In fact this term is purely heuristic, and any power $n>6$ would work, but higher powers have the advantage that we expect the actual long-range Pauli effect to decay like $e^{-kr}$ as $r$ gets larger, since when we solve for the wavefunction of the Hydrogen atom we find that the wavefunction of the electron cloud decays radially like $e^{-r/2r_\text{Bohr}}$ and the overlap of the two electron clouds has to be the source of the Pauli exclusion principle; this must decay faster than $r^{-n}$ for any $n$. So we prefer higher powers as they decay to 0 faster and therefore introduce less long-range weirdness, while still allowing for a repulsive effect that keeps the atoms from falling into each other.

The reason $r^{-12}$ is used in practice is that it is a very high $n$ which can be gotten from the $r^{-6}$ that one already has to calculate (which can be explained as the London forces) by one single multiplication instruction; $r^{-6}\cdot r^{-6} = r^{-12}.$

The 1924 paper by John Lennard-Jones actually considered a wide range of $n$ and $m$ for potentials of the form $U(r) = A r^{-n} - B r^{-m},$ trying to match his results against the measured viscosity of liquid argon. He found that good fitting of this potential required $m=6$ (the paper actually discusses forces so it says $m=5$ but this integrates to an $m=6$ potential), but that a lot of different $n$ seemed to be valid choices, and any $n > 10$ worked quite well, with probably his best results for this particular experiment of $n = 15 + 1/3.$ So going up another power to $r^{-18}$ or $r^{-24}$ does not seem to "buy us much" over $r^{-12}$ and we just use that in practice.

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    $\begingroup$ If the spatial part of the wavefunction is symmetric two electrons can share the same location $r_1=r_2$. It wouldn't make the antisymmetric wavefunction to vanish i.e., Pauli principle doesn't forbid two electrons to share the same location as long as other quantum numbers are different. In the ground state of the $H_2$ molecule, the spatial part of the wavefunction is symmetric and electrons are in the spin songlet state. So what do you mean by the fact that they cannot share the same location? @CR Drost $\endgroup$
    – SRS
    Aug 30, 2017 at 18:16
  • $\begingroup$ @SRS yeah I think you may be right; as I recall the Pauli exclusion indeed usually has a form of adding a term $J \vec\sigma_1 \cdot \vec\sigma_2$ to the Hamiltonian which creates a lower-energy state for the spin-antisymmetric case and a higher-energy state for the spin-symmetric case. You cannot describe all of the physics without saying that these big massive protons are repelling each other and then coming to some mean distance, and that's the dominant cause of the repulsion term. Calling it "Pauli exclusion" is probably a vast oversimplification, if that's your question. $\endgroup$
    – CR Drost
    Aug 30, 2017 at 21:15

Really nice question, wich is rarely arise.

In fact the answer is I guess :

If you determine the true wave function of your molecule by solving the true hamiltonian of the two well potential + electronic potential, you get the true result which is very close of the results you get using a Lennard-Jones potential.

The Pauli exclusion principle is include in the quantum mechanic !

Where ?

It is not a fundamental principe of quantum mechanic, it is derived from the commutation of the hamiltonien with the exchange operator $P$ . If the hamiltonian $H$ commute with $P$ then you have two familly of solution : symmetric and anti-symmetric. When solving the hamilotnian for electron you explicitly keep only anti-symetric part. And you can easly demonstrate that anti-symmetric functions follow Pauli exclusion principle !

So it is by explicitly keeping only anti-symmetric solutions that you insure the Pauli Exclusion Principle.

But you can't show ( i'm not sure) that the Lennard-Jones potential describe well this phenomenone. It is totally empiric way but which work


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