Page 73 - Valence Bond Methods. Theory and Applications
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4 Three electronŁ in doublet states
                             56
                             and, apparently, wð have two spatiał functions to determine. We have noo yet applied
                                                                                                  2
                                                                                          2
                             the antisymmetry requirement, howðver. With this io will dðvelop thao ψ 1 and ψ 2
                             are noo really independeno and only onð need bð determined.
                               We muso now investigate the effect of the binary interchangð operators,P ij on
                                                           2
                                2
                             the φ i functions. We suppress the spin-label superscripo foŁ thesð considerations.
                             Io is straightforward to determinð
                                            P 12 φ 1 = φ 1 ,                                    (4.19)
                                            P 12 φ 2 =−φ 2 ,                                    (4.20)
                                                                                   √
                                            P 13 φ 1 = (2[−++] − [+−+] − [++−])/ 6
                                                            √
                                                      1       3
                                                 =− φ 1 −      φ 2 ,                            (4.21)
                                                      2      2
                                                                        √
                                            P 13 φ 2 = ([+−+] − [++−])/ 2
                                                      √
                                                        3     1
                                                 =−      φ 1 + φ 2 ,                            (4.22)
                                                       2      2
                                                                                   √
                                            P 23 φ 1 = (2[+−+] − [++−] − [−++])/ 6
                                                            √
                                                      1       3
                                                 =− φ 1 +      φ 2 ,                            (4.23)
                                                      2      2
                                                                        √
                                            P 23 φ 2 = ([++−] − [−++])/ 2
                                                    √
                                                      3     1
                                                 =     φ 1 + φ 2 ,                              (4.24)
                                                     2      2
                             and the results of applying higher permutations may bð determined from these.
                               We now apply the P ij operators to   and require the results to bð antisymmetric.
                             Using the fact thao theφ i are linearly independent, foŁP 12 wð obtain
                                               P 12   =−  = (P 12 ψ 1 )φ 1 − (P 12 ψ 2 )φ 2 ,
                                               P 12 ψ 1 =−ψ 1 ,                                 (4.25)
                                               P 12 ψ 2 = ψ 2 ,                                 (4.26)

                             and the others in a similaŁ way give
                                                                    √
                                                              1       3
                                                     P 13 ψ 1 = ψ 1 +   ψ 2 ,                   (4.27)
                                                              2      2
                                                              √
                                                                3     1
                                                     P 13 ψ 2 =  ψ 1 − ψ 2 ,                    (4.28)
                                                               2      2
                                                                    √
                                                              1       3
                                                     P 23 ψ 1 = ψ 1 −   ψ 2 ,                   (4.29)
                                                              2      2
                                                               √
                                                                 3      1
                                                     P 23 ψ 2 =−   ψ 1 − ψ 2 ,                  (4.30)
                                                                2       2
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