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Because of the coulomb interaction the energy of electrons of the same spin is lower - the average potential energy is less positive for parallel spin than for antiparallel spin.

This is quoted from Kittel, Ch 11, under Hund Rules.

I can't understand how Coulomb repulsion is related to spin. Is there any theoretical basis to this in QED or somewhere? All that I am aware of is that Coulomb repulsion happens due to charge of particles while spin is an intrinsic property that has nothing to do with charge. What have I missed here?

Also, in the above quote if I replace "electrons" by "uncharged fermions" does this still hold?

Any insights are highly appreciated.

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1 Answer 1

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Two electrons with the same spin projections have a symmetric spin-part of their wavefunction and therefore an antisymmetric spatial wavefunction - to be totally antisymmetric. An antisymmetric spatial wavefunction goes to zero in between the electrons, and so same-spin pairs are farther apart on average. Being farther apart means that they have lower Coulomb repulsion (or interaction) energy.

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  • $\begingroup$ okay... Correct me if I am wrong. Here if we take both the electrons to be having the same spin, which would mean the total Spin is an integer. Then by the spin statistics theorem, wouldn't that system be a bosonic system - meaning the total wavefunction should be symmetric right? $\endgroup$ Commented Apr 29, 2021 at 13:32
  • $\begingroup$ @TheImperfectCrazy The wave function of two bosons is symmetric under the exchange of the bosons, but if each boson is a bound state of two fermions, then exchanging the bosons requires exchanging two pairs of fermions. So the antisymmetry mentioned in the answer is consistent with the symmetry of bosons, because $-1\times -1 = +1$. $\endgroup$ Commented Apr 29, 2021 at 13:45
  • $\begingroup$ @ChiralAnomaly, I am more confused now. Q1. If I have a single fermion system, the total wavefunction is antisymmetric. Is this right? Q2. Now if I have 2 fermion system, the total wavefunction should be symmetric right? Because its an integer spin system. What am I missing? $\endgroup$ Commented Apr 29, 2021 at 15:47
  • $\begingroup$ @TheImperfectCrazy Suppose you have a system with two bosons, each of which is made of two fermions: boson A is made of fermions $A_1$ and $A_2$, and boson B is made of $B_1$ and $B_2$. The wavefunction $\psi(A_1,A_2,B_1,B_2)$ must change sign whenever two fermions are switched, so $$ \psi(A_1,A_2,B_1,B_2)=-\psi(A_1,B_2,B_1,A_2)=\psi(B_1,B_2,A_1,A_2). $$ Thus the wavefunction stays the same when the two bosons A and B are switched, because the two sign-changes from the two fermion-switches cancel each other. $\endgroup$ Commented Apr 30, 2021 at 0:00
  • $\begingroup$ Ahh okay thank you... $\endgroup$ Commented May 3, 2021 at 11:24

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