D.50 Spin degeneracy

To see that generally speaking the basic form of the Hamiltonian produces energy degeneracy with respect to spin, but that it is not important for using the Born-Oppenheimer approximation, consider the example of three electrons.

Any three-electron energy eigenfunction $\psi^{\rm E}$ with $H\psi^{\rm E}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E^{\rm E}\psi^{\rm E}$ can be split into separate spatial functions for the distinct combinations of electron spin values as

\psi^{\rm E}& = &
\psi_{+++}^{\rm E}{\uparrow}{\uparrow}{\...
\psi_{++-}^{\rm E}{\uparrow}{\uparrow}{\downarrow}.

Since the assumed Hamiltonian $H$ does not involve spin, each of the eight spatial functions $\psi_{\pm\pm\pm}$ above will separately have to be an eigenfunction of the Hamiltonian with eigenvalue $E^{\rm E}$ if nonzero. In addition, since the first four functions have an odd number of spin up states and the second four an even number, the antisymmetry requirements apply only within the two sets, not between them. The exchanges only affect the order of the spin states, not their number. So the two sets satisfy the antisymmetry requirements individually.

It is now seen that given a solution for the first four wave functions, there is an equally good solution for the second four wave functions that is obtained by inverting all the spins. Since the spins are not in the Hamiltonian, inverting the spins does not change the energy. They have the same energy, but are different because they have different spins.

However, they are orthogonal because their spins are, and the spatial operations in the derivation of the Born-Oppenheimer approximation in the next note do not change that fact. So they turn out to lead to nuclear wave functions that do not affect each other. More precisely, the inner products appearing in the coefficients $a_{n{\underline n}}$ are zero because the spins are orthogonal.