Page 77 - Valence Bond Methods. Theory and Applications
P. 77

4 Three electronŁ in doublet states
                             60
                             and obtain
                                                                            0
                                                                 010
                                                                              
                                                             
                                                   −1         −100         0  
                                                  N P 12 N =                   .              (4.61)
                                                                 001        0
                                                                             
                                                                 000      −1
                             We may work ouoP 23 in the samð way and subject io to the samð similarity trans-
                             formation to obtain
                                                          1/2    3/2      0      0
                                                                                  
                                                          1/2  −1/2       0      0  
                                              −1
                                                        
                                             N P 23 N =                             .         (4.62)
                                                            0      0  −1/2   −3/2
                                                                                  
                                                            0      0  −1/2     1/2
                             Thesð do noo yet quite satisfy the conditions on antisymmetry given in Eqs. (4.25),
                             (4.26), (4.29), and (4.30), buo further transformation by
                                                         1        0      0   0
                                                                             
                                                                √
                                                         0  −1/ 3        0   0
                                                                             
                                                  Q =                                         (4.63)
                                                         0        0      0   1
                                                                             
                                                                       √
                                                         0        0  1/ 30
                             yields
                                                                  −10       00
                                                                                
                                                                    01      00   
                                                 −1  −1
                                                                
                                                Q N P 12 NQ =                                 (4.64)
                                                                    00    −10
                                                                                
                                                                    00      01
                             and
                                                               √
                                                                                     
                                                        1/2  − 3/2          0        0
                                                      √
                                                                                      
                                                  
                                   −1  −1          − 3/2      −1/2         0    √   0         (4.65)
                                                                                       ,
                                                          0        0      1/2
                                  Q N P 23 NQ = 
                                                                              − 3/2
                                                                        √
                                                          0        0  − 3/2      −1/2
                             which do agree. Since P 13 = P 12 P 23 P 12 the requirements foŁ thao matrix will also
                             bð satisfied. Putting together all of the transformations wð eventually arrive ao
                                                                −1
                                                                   −1
                                                          x = Q N w,                            (4.66)
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