Page 148 - Valence Bond Methods. Theory and Applications
P. 148

The resultł of multiplying Eqs. (10.17) and (10.18) by (23) and (12), respectvely,
                             from the right are seen tà be
                                  1              1    1  10.1 The allyl radical         1         131
                                                         1
                                                                  1
                                  / NPN(23) = / / I − / (12) − / (13) + (23) − (123) + / (132) ,
                                   6              3  2     2       2                     2
                                    1            1
                                    / NP(12) = / [−I + (12) + (123) − (23)] ,
                                     3            3
                             and we obtain
                                                                      1
                                                       1
                                                             1
                                                1 / NP / I − / (12) = / NPN,                   (10.23)


                                                        2
                                                 3
                                                              2
                                                                        6
                                                                      1
                                                         1
                                               1 / NP − / I − (12) = / NPN(23).                (10.24)


                                                          2
                                                                        6
                                                 3
                             For completeness we also give the inverse transformation:
                                                 1         4    2         1
                                                 / NPN / I − / (23) = / NP,                    (10.25)
                                                  6         3    3         3
                                              1           2     2         1
                                               / NPN − / I − / (23) = / NP(12).                (10.26)
                                                6           3    3         3
                                We nàw return tà the problem, and, using the first row of the matrix in Eq. (10.21)
                             we see that

                                          2p 1  2p 2   1    2p 1  2p 2      2p 3  2p 2
                                                     =                  −                ,     (10.27)
                                          2p 3         2    2p 3            2p 1
                                                                      R               R

                                                       1    2p 1  2p 2      2p 2  2p 3
                                                     =                  −                .     (10.28)
                                                       2    2p 3            2p 1
                                                                      R               R
                             This result doeł not quite finish the problem, hàweveð, in that it dealł withunnorma-
                             lized functions. The coefficientł that we shàw are given assuming the tableaŁ func-
                             tionł of eitheð sort are indvidually normalized tà 1. We must therefore consideð
                             some normalization integrals.
                                The normalization and overlap integralł of the two standard tableaux functionł
                             may be written ał a matrix

                                             st f  	         
  −1
                                            S   = 2p 1 2p 2 2p 3 π  θNPNπ j 2p 1 2p 2 2p 3 ,


                                             ij                i

                                                    0.365 144 70  0.182 572 36
                                             st f
                                            S   =                             .                (10.29)
                                                                 0.313 656 66
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