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In Hartree-Fock theory the time-independent electronic energy of a single (restricted) determinant electronic wavefunction consists of one electron terms, $h_{ii}$, Coulomb interaction energies, $J_{ij}$, and exchange interaction energies $K_{ij}$.

Exchange interaction energy terms result from integration over the 2-electron repulsion operator $r_{12}^{-1}$ where the electrons in the 2-electron integral have parallel spin.

I have read that the exchange interaction is not a true physical interaction, but arises because of the single determinant description of the system. It is a partial description of the correlated motion of electrons, which is better described in many-determinant methods.

My question is this: does spin alone have any effect on the physical interactions of particles?

For example, consider 2 uncharged particles with spin 1/2. They may either have parallel or antiparallel spins. In physical reality, would these 2 particles repel one another if they have parallel spin?

If I was to model this system with a Hartree-Fock calculation, I believe there would be no exchange interaction, because the Hamiltonian would feature no two particle operator, so no Coulomb or exchange integrals would arise. However, perhaps if I used a higher level of theory, I might see an interaction?

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Interactions have to conserve angular momentum and spin is angular momentum, so trivially "Yes".

In your hypothetical with two uncharged (I assuming you mean the electric charge here) particles there will be no electromagnetic repulsion because there will be no electromagnetic interaction. But Hartree-Fock is used for interacting systems...

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