# Coulomb Repulsion and Spin

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.

• @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 $-1\times -1 = +1$. Commented Apr 29, 2021 at 13:45
• @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. Commented Apr 30, 2021 at 0:00