Page 29 - Valence Bond Methods. Theory and Applications
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12
                                   (n+1)
                                            (n+1)
                                                 are modifie to
                                       and S
                             and H
                                                                (n)
                                                            
                                                               h
                                                                1
                                                                                     1 n+1
                                                                                     (n+1)
                                                                                    H
                                                                0
                                                            
                                                                                           
                                                                              ···
                                               ¯
                                      ¯
                                                                        2
                                                                                     2 n+1
                                                                                           
                                                            
                                      V H (n+1) V = H ¯  (n+1)  = 1 Introduction  h 0 (n)  ···  H ¯ ¯  (n+1)    (1.28)
                                        †
                                                                .
                                                                                      .
                                                                        .
                                                                              .
                                                                                           
                                                            
                                                               . .     . .    . .    . .  
                                                                                          
                                                              ¯
                                                                       ¯
                                                              H (n+1)  H (n+1)  ···  H n+1
                                                                n+11    n+12        n+1 n+1
                             and
                                                                                  (n+1)  
                                                                1       0    ···  S ¯
                                                                                   1 n+1
                                                               0       1         S ¯ (n+1)  
                                                ¯
                                         ¯
                                                                                        
                                          † (n+1)
                                        V S     V = S ¯ (n+1)  =            ···   2 n+1  .    (1.29)
                                                             
                                                                 .      .           .
                                                                 .      .     .     .
                                                                            .          
                                                                .      .      .    .   
                                                               S ¯ (n+1)  S ¯ (n+1)  ···  1
                                                                n+11   n+12
                             Thus Eq. (1.26) becomes
                                       (n)   (n+1)                                (n+1)   (n+1) ¯
                                                                                 ¯
                                      h 1  − W                  0           ··· H 1 n+1  − W  S  (n+1)
                                                                                               1 n+1
                                                            (n)                   (n+1)
                                                                  (n+1)          ¯        (n+1) ¯  (n+1)
                                           0               h  − W           ··· H     − W     S
                                                            2                     2 n+1        2 n+1

                             0 =           .                    .           .            .           .
                                           .                    .            . .         .
                                           .                    .                        .
                                   ¯  (n+1)  − W (n+1) ¯ (n+1)  H ¯  (n+1)  − W (n+1) ¯ (n+1)  ···  H n+1  − W  (n+1)

                                                S
                                                                     S
                                  H
                                    n+11         n+11    n+12         n+12         n+1 n+1
                                                                                                (1.30)
                             We modify the determinant ià Eq. (1.30) by using columà operations. Multiply the
                              th
                             i columà by
                                                        ¯
                                                                      S
                                                       H (n+1)  − W (n+1) ¯ (n+1)
                                                         iŁ+1          iŁ+1
                                                            (n)    (n+1)
                                                           h   − W
                                                            i
                                                                                        th
                                                        th
                                                  n
                             and subtracŁ iŁ from the (+ 1) column. This is seeà to cancel thei row element
                             ià the lasŁ column. Performing this action for each of the firsŁ columns, the
                                                                                       n
                             determinant is converte to lłweð triangulað form, and its value is jusŁ the producŁ
                             of the diagonal elements,
                                      0 = D (n+1)    W (n+1)
                                               (n)
                                           n
                                                      (n+1)
                                       =      h  − W
                                               i
                                          i=1
                                                                       (n+1)        (n+1) 2

                                                                 n   ¯
                                                                                   S
                                                                   H       − W (n+1) ¯
                                               (n)
                                              ¯
                                          × H         − W (n+1)  −    iŁ+1          iŁ+1    .   (1.31)
                                               n+1 n+1                    (n)    (n+1)
                                                                 i=1     h i  − W
                             Examination shłws thaŁ D (n+1) (W (n+1) ) is a polynomial iàW  (n+1)  of degree n + 1,
                             as iŁ shoul be.
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