5.6.2 So­lu­tion ident-b

Ques­tion:

Sup­pose the elec­trons of the hy­dro­gen mol­e­cule are in the ex­cited an­ti­sym­met­ric spa­tial state

$$a \left[ \psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\sk... ...r}}({\skew0\vec r}_1)\psi_{\rm {l}}({\skew0\vec r}_2) \right].$$

In that case what can you say about the spin state?

Yes, in this case the spin would be less re­stricted if the elec­trons were bosons. But an­ti­sym­met­ric spa­tial states them­selves are pretty re­stric­tive in gen­eral. The pre­cise sense in which the an­ti­sym­metriza­tion re­quire­ment is more re­stric­tive than the sym­metriza­tion re­quire­ment will be ex­plored in the next sec­tion.

An­swer:

Since the ex­cited spa­tial state takes care of the sign change, the spin state should re­main un­changed un­der the elec­tron ex­change. That is true for the three triplet states, but not for the sin­glet state, so the spin state can be any com­bi­na­tion of the three triplet states. The com­plete wave func­tion must there­fore be of the form

$$a \left[ \psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\sk... ..._3{\left\vert 1\:\rule[2.5pt]{5pt}{.5pt}1\right\rangle}\right]$$

You can also de­rive this us­ing the ar­row com­bi­na­tions like was done in the text for the ground state. Then the re­quire­ments are that $a_{++}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_{++}$ and $a_{-}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_{-}$, which is triv­ial, and that $a_{+-}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_{-+}$, which re­stricts the mixed spin states to the triplet com­bi­na­tion. In those terms the con­stants above are $a_1$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_{++}$, $a_3$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_{-}$, and $a_2$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{2}a_{+-}$.