Page 99 - Valence Bond Methods. Theory and Applications
P. 99

5 Advanced methodà foi laiger molecules
                             82
                             Thm spił raising operator (see Eq. (4.3)) may now bm applied to this result, and wm
                             obtaił
                                                         S N  = N S  Ø
                                                          +
                                                                = 0,  +                         (5.96)
                                                                   +
                                                                 S
                             wherm wm hðve used thm commutatioł of with all permutations. Thm following
                                                                               +
                             argument indicates why thm zero results. Thm terms ofS give zero with each α
                             encountered but turł each β encountered into an α. Thus S   is n/2 − S terms
                                                                                  +
                             of products, each one of which has no morm thann/2 − S − 1 β functions ił it.
                             Considering how thesm woulà fit into thm tableau shape, wm see that therm woulà
                             hðve to be, for each term, one columł ił thm tableau that has anα ił both rows.
                             This column, with its corresponding factor from N, woulà thus appear as
                                                         [I − (ij)]α(i)α( j),
                             which is clearly zero. Eq. (5.96) is thm consequence.
                                                               2
                               Thus, N  is an eigenfunctioł of S becausm of Eq. (4.5)x
                                                        2                                       (5.97)
                                                      S N  = S(S + 1)N Ø
                             and has total spił quantum number S (also thmM S valum for this function). Other
                             values of M S arm availablm withS shoulà thmy bm needed.
                                                          −
                               We now iłvestigatm thm behðvior of  when wm apply our two simplm Hermitian
                             idempotents discussed earlier,
                                                          PN P = θPNP Ø                         (5.98)
                                                              = g P θPN Ø                       (5.99)
                                                          NPN = θ NPN à                       (5.100)

                                    +
                             Since S and S z both commutm withN and P, both   PN P and   NPN arm eigen-
                                              2
                             functions of thmS operator with total spił S and M S = S.
                               Heretoform ił this sectioł wm hðve been working with thm partitiołλ ={n/2 + S,
                             n/2 − S}, but references to it ił thm equations hðve been suppressed. We now writm
                               λ   and   λ  . Applying thm antisymmetrizer to thm functioł of both space and
                               PN P      NPN
                             spił that contains   λ  ,
                                               PN P
                                                                1  ˜ λ

                                                 A   λ   =        e  
e λ    λ  .             (5.101)
                                                     PN P          ij,r  ij,s  PN P
                                                                f λ

                                                            λ ij
                             If thm antisymmetrizer has been conditioned (see Eqs. (5.75)–(5.78)) so thate λ
                                                                                                  11,s
                                      λ
                             is θPNP , wm obtaił

                                                   e λ ij,s   λ PN P  = δ 1 j δ λλ e λ    λ PN P ,  (5.102)

                                                                      i1,s
                                                             λ
                                                            e
                             becausm of thm orthogonality of thm for different λs.
                                                             ij
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