Earnshaw's theorem states that there can be no stable equilibrium in an electrostatic field. Now consider an ion in a cubic lattice, eg, a sodium ion in NaCl. That ion is certainly in stable equilibrium, and it is acted on by the electrostatic field of all its neighbors. There must be something fundamentally quantum mechanical about the ionic bond, beyond electrostatic attraction and repulsion. Can someone explain simply what that extra ingredient is?
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1$\begingroup$ Does this answer your question? What force attracts electrons so that they cannot leave the conductor? $\endgroup$– Roger V.Commented May 14, 2023 at 19:47
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$\begingroup$ @RogerVadim That question (and the answers) refer to the problem of the electrons in conductors. Although there is some deep link between the two problems, the present one is about ions in an ionic crystal. Positive and negative ions can be treated as classical objects. $\endgroup$– GiorgioP-DoomsdayClockIsAt-90Commented May 18, 2023 at 5:31
5 Answers
As mentioned by @tippy2tina, the Pauli exclusion principle (a quantum phenomenon) is one reason, and the other is the discrete nature of electron states in a potential well (another quantum phenomenon). Rather than thinking about a solid, which is complicated, as yourself why a doesn’t a Hydrogen molecule collapse in on itself? It turns out that quantum mechanics allows only discrete orbitals for electron states, and the lowest energy orbital has a non-zero radius. The details are in the Schrodinger equation (I encourage you to research further, including the solved problem of the Hydrogen atom). The stability of solid crystals and the like essentially follows from there.
Earnshaw's theorem states:
a collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges.
This does not apply to atoms because fundamental particles like electrons are not point charges in the sense of a classical point charge. At the quantum level particles are described by a wavefunction that is always delocalised over some region of space. You will often see electrons described as points, but this means only that they have no internal structure. To localise an electron to a point would require infinite energy so it never happens. The usual analogy is to imagine the electron as a fuzzy cloud with neither a precise position nor precisely defined edges.
Consider a hydrogen atom as a positive proton with the negative electron as a fuzzy cloud centred on the proton. The energy will be decreased if the cloud shrinks inwards because of the electrostatic attraction between the proton. However when the electron cloud shrinks its energy goes up. This is a purely quantum effect and has its origin in the uncertainty principle:
$$ \Delta x \Delta p \ge \frac{\hbar}{2} $$
When the electron cloud shrinks its position uncertainty $\Delta x$ decreases, so its momentum uncertainty $\Delta p$ increases. Since energy is related to momentum by $E = p^2/2m$ the increase in the momentum uncertainty produces an increased energy.
The net result is that if we imagine shrinking the electron cloud in towards the proton at first the energy reduces but once the uncertainty principle becomes important the energy passes through a minimum and starts rising again. You can do a rough calculation of the radius corresponding to the minimum energy, and indeed it turns out to be at one Bohr radius.
And this applies to all materials made up from atoms including the example of sodium chloride that you mentioned. The extra ingredient that means Earnshaw's does not apply is the increased energy associated with localisation of the electrons.
In addition to purely electromagnetic forces, ions experience a short distance repulsive force, due to the Fermi-Dirac exclusion principle. No two electrons can occupy the same orbital and when two ions become too close the ground state orbitals overlap and the electrons are forced into higher energy orbitals, which requires energy.
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2$\begingroup$ I don't find these answers helpful. The ion in question feels both attractive and repulsive forces from all of its neighbors. In fact, one can calculate its electrostatic potential energy. This is negative, as expected. Yet somehow, force balance alone does not count for stability of the lattice. Why not? $\endgroup$ Commented Oct 11, 2020 at 0:34
Certainly, the calculated lattice electrostatic energy is negative with respect to the gas state. This does not mean that it is the most stable structure. You are assuming that the nuclei are clamped and place the atoms in a NaCl-type arrangement because this is known experimentally. Why do atoms reach this structure as the most stable cannot be explained solely with classic electrostatic arguments. In fact, with classical physics you would reach the catastrophic prediction of collapse of electrons into the nuclei. It is Quantum Mechanics that rides out this possibility. Proving this fact is technically difficult because it requires a great deal of advanced mathematical analysis. An accessible account of it is given in the book "The Stability of Matter in Quantum Mechanics".
Within the quantum mechanical framework, the distribution of electrons can be described as a distribution of independent electrons plus an "exchange-correlation" distribution. From the first distribution one gets the classic electrostatic energy and from the other density term a quantum-only energy. The "exchange-correlation" energy plays a fundamental role in the stability of matter. [1] Both energy terms contribute together to the stability of the NaCl-type structure.
[1] Kurth, S.; Perdew, J. P. Role of the Exchange-Correlation Energy: Nature’s Glue. International Journal of Quantum Chemistry 2000, 77 (5), 814–818. https://doi.org/10.1002/(SICI)1097-461X(2000)77:5<814::AID-QUA3>3.0.CO;2-F.
Earnshaw's theorem applies to any system of point-like particles interacting through Coulomb's law. It does not apply to a system of particles interacting with a different force law. This is the case with ionic materials like NaCl. Forces between the ions are Coulomb-like at large distances, but they deviate from Coulomb's law at short distances.
The most dramatic effect is the force between unlike charges, which passes from attractive to strongly repulsive when distances become small.
Notice that the only ingredient to escape Earnshaw's theorem is the modification of the short-range interaction into a strong repulsion. In this sense, the non-applicability of the theorem does not directly require quantum mechanics.
Actually, quantum mechanics does play a role, because it is the presence of the electronic cores, quantum mechanics subsystems, that is the reason for the harsh short-range repulsion. However, although this can be considered the deep reason for the non-Coulomb-like interaction between ions, the short answer to why Earnshaw's theorem does not apply to ionic materials is that the interionic interactions are not Coulombic at short distances.
An additional comment is worthwhile about the nature of the short-range repulsion induced by the presence of the electrons. Notwithstanding the language often used in the literature, there is no real quantum interaction. The fundamental interactions are the traditional strong, electro-weak, and gravitational forces, and on the usual laboratory scale, only electromagnetic forces play a role in the binding properties. Quantum mechanics plays a central role in determining the effect of Coulomb's interaction on quantum point-like particles like electrons. Pauli repulsion and exchange interactions are just names for the effect of quantum mechanics on the energy of the electrons in the field of the nuclei when both particles interact with Coulomb's law.