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Or is it irrelevant, as orbitals are QM while Coulomb interactions are classical physics? What I think I understood of orbitals is that particles with the same quantum numbers cannot occupy the same space (pauli exclusion principle), while with different numbers such as spin they can. Does that mean the two electrons are invisible to each other or just that they form a third entity that occupies an orbital? By third entity, I mean a collaboration or synchronization between the two electrons.

I found this document, but I am unsure of the conclusion to draw: http://magnetism.eu/esm/2013/slides/lacroix-slides.pdf

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Yes, there is Coulomb interaction, which also leads to correlation in position.

As an example, you could look at helium. The binding energy of one electron is 4 Rydberg = 54.4 eV. But the ionization energy of neutral helium is 24.6 eV.

Calculating this number is not so easy because it is a three-body problem. One way of taking into account electron-electron correlation is by "configuration interaction" with higher orbitals. Or one can use density-functional theory.

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  • $\begingroup$ This is enlightening, thank you. Position correlation is what I suspected. Is it similar to Cooper pairs? $\endgroup$ – Exocytosis Sep 28 '19 at 19:41
  • $\begingroup$ Also are the spin axis correlated too or is the quantum spin without actual orientation? $\endgroup$ – Exocytosis Sep 28 '19 at 19:44
  • $\begingroup$ @Exocytosis No, Cooper pairing is in momentum space. Yes, the total spin of the two electrons is zero in the ground state of helium. $\endgroup$ – Pieter Sep 28 '19 at 21:13
  • $\begingroup$ Total spin is zero ok. Does that mean there is a magnetic cohesion of the two electrons compensating for the electrical repulsion? $\endgroup$ – Exocytosis Sep 29 '19 at 13:27
  • $\begingroup$ @Exocytosis The magnetic dipole-dipole interaction is negligibly small in such cases. It is always the Coulomb terms that are behind the spin-spin interaction (the "exchange term"). And that will cause the triplet states (parallel spins, orthohelium) to be lower in energy because close distances are suppressed in such configurations. $\endgroup$ – Pieter Sep 29 '19 at 14:45
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Loosely speaking, yes. By 'feel each other's Coulomb potential' you mean that the behaviour of one electron is influenced by the presence of the other owing to electrodynamic effects. That is undoubtedly the case. If you were to model the behaviour of an electron in an atom by considering only the potential due to the nucleus and electrons in other orbitals you would calculate imprecise answers. To help you visualise the reasons, imagine two atoms, identical except that one is ionised by missing an electron that would normally fill an orbital; clearly the two atoms appear different to an electron passing by, and the difference must be attributed to charge effects.

To try to illustrate the point from another direction. Suppose that an electron in a given orbit did not 'feel' the Coulomb repulsion from the other electrons in that orbit. In that case the electron would feel only the Coulomb potential from the nucleus. If that were true, then a nucleus would attract an infinite number of electrons.

Another illustration is provided by the Hartree-Fock method for calculating energy levels, in atoms for example. In that method, the Schrodinger equation is solved for a single electron by considering a Hamiltonian that models the presence of, including the Coulomb interaction from, the other electrons orbiting the nucleus.

In truth the Pauli exclusion principle is a post-hoc rule to reflect the observed population of electron orbitals.

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  • $\begingroup$ Sorry I don't see how your example is connected to the question of orbitals populated by electrons with opposite spin. I know from classical electromagnetism that two atoms with different valences/charges will interact differently. But my question is about what is going on inside a single atomic orbital populated by two electrons with opposite spin. Is there anything comparable to Coulomb repulsion between them? $\endgroup$ – Exocytosis Sep 28 '19 at 19:36
  • $\begingroup$ Yes, there is. I will update my answer to try to clarity why. All the best. $\endgroup$ – Marco Ocram Sep 28 '19 at 19:42
  • $\begingroup$ I just read the 3 paragraphs you added to your answer and it is quite interesting. But first, I still cannot see where you address my main query about the behavior of two electrons sharing the same space because they have opposite spin. My concern is not about the existence of charge surrounding the atom, nor between electrons in different orbitals, nor between the nucleus and electrons, but only about these two opposite spin electrons in the same compartment. (second part in the next comment) $\endgroup$ – Exocytosis Sep 29 '19 at 13:21
  • $\begingroup$ Pieter directly addressed this aspect by saying their positions are correlated. But this probably also means part of the repulsion is cancelled by the opposite spins. Is that a magnetic attraction counteracting electrical repulsion? Secondly, you wrote something absolutely fascinating that I have never heard of before. You write that Pauli principle is a post-hoc rule invented to fit the observation of orbitals. However this principle applies to all fermions, no? $\endgroup$ – Exocytosis Sep 29 '19 at 13:24
  • $\begingroup$ @Exocytosis Yes the exclusion principle applies to Fermions, but when Pauli first proposed it he was trying to explain the behaviour of electrons in atoms. He did not then appreciate its more general implications. $\endgroup$ – Marco Ocram Sep 29 '19 at 17:43
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...particles with the same quantum numbers cannot occupy the same space (pauli exclusion principle), while with different numbers such as spin they can. Does that mean the two electrons are invisible to each other<...>?

No, this doesn't mean that they are invisible. It's just that Coulomb potential is a "soft" potential: due to Heisenberg uncertainty principle the electrons have nonzero probability density at the point of collision, despite having infinite potential energy at that point.

Had the potential had a higher power in the denominator, e.g. $r^2$ instead of $r$, like in centrifugal effective potential, the electrons would then never be able to come arbitrarily close to each other, regardless of their spins.

See also my answer to the question "Can two electrons never touch each other?".

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  • $\begingroup$ What about the spin? Do opposite spins change this probability density distribution? $\endgroup$ – Exocytosis Sep 29 '19 at 14:54
  • $\begingroup$ @Exocytosis opposite spins is an underspecification of spin state. It can be $|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle$ or $|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle$. In the former case the spatial wavefunction must be antisymmetric, i.e. $\psi(\vec r_1,\vec r_2)=-\psi(\vec r_2, \vec r_1)$, and thus the collision is impossible due to Pauli. In the latter case the collision is possible because the wavefunction must be symmetric, i.e. $\psi(\vec r_1,\vec r_2)=+\psi(\vec r_2, \vec r_1)$. So the probability distribution does indeed change depending on spin state. $\endgroup$ – Ruslan Sep 29 '19 at 16:19
  • $\begingroup$ Fantastic. And I appreciated the link to your other answer too, it is instructive. Finally, because I cannot get a proper answer to this one, are quantum spin states (up or down) here refering to electrons spatial orientation (be it relative to each other in a pair) like a macroscopic rotating body or is it more abstract? I know QM spin is not a classical rotation but it has an angular momentum so I wonder if there is an axis of the spin with polar coordinates. $\endgroup$ – Exocytosis Sep 29 '19 at 17:09
  • $\begingroup$ @Exocytosis (Note: appreciation on StackExchange network is normally expressed by upvotes (and depreciation by downvotes). And if an answer solves your problem, you can accept it by clicking the tick mark to the left of it.) The spin states refer to the values of $z$ projection of the intrinsic angular momentum of the electron. There's no axis of spinning, because in an eigenstate of $\hat s_z$ the projections of spin on $x$ and $y$ axes don't have definite values (spin projection operators don't commute). $\endgroup$ – Ruslan Sep 29 '19 at 17:49
  • $\begingroup$ I usually let sufficient time for people to answer before picking an answer. This is SE recommended way. $\endgroup$ – Exocytosis Sep 29 '19 at 18:20

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