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?
