Available literature proposes 3 equations to calculate the coupling constant during a Broken-Symmetry approach:
J(1) = (from -(E[HS]-E[BS])/Smax**2)
J(2) = (from -(E[HS]-E[BS])/(Smax*(Smax+1))
J(3) = (from -(E[HS]-E[BS])/(HS-BS))
I'm intrigued by the fact that, from the equations, the more the syystem have unpaired electrons, the minor will be Jab. Why does this happen?
After doing some research I was able to get an answer that looks correct, so, I'm letting it here for future reference in case someone needs it too.
Jab seems to decrease as the number of unpaired electrons increases because the Hamiltonian have to take account also of all other possible spin states. As an example, take a septet. Until reaching the septet state all other microstates have to exist, so, the energy is "split" in a multiplicity ladder like 1et, 3et, 5et and finally 7et. That is why the energy looks smaller for a septet.
Hi Nora.
It is related to Magnetic couplings. This is a set of equations derived from one Hamiltonian and gives the couplings between two magnetic states: ferromagnetic and antiferromagnetic.
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