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

$$\begin{eqnarray*} \psi^{\rm E}& = & \psi_{+++}^{\rm E}{\uparrow}{\uparrow}{\up... ...{\uparrow}+ \psi_{++-}^{\rm E}{\uparrow}{\uparrow}{\downarrow}. \end{eqnarray*}$$

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.