We now find the coefficients in (7.18)for each of the roots of the secular equation.From the linear homogeneous equations in Chapter 5, we have (H-W)c+(Hh-S)c=0 (7.27 Substituting in W from(7.26),we get Ca=l Co (7.28) =Ca(Osa +oisb) (7.29) We fix ca by normalization: <g,9,>cP∫(2++244h)d=1 Ico (7.30) 2+28 The normalized trial function corresponding to the energy W is thus: 1 01= =(9a+中b） V2+2S (7.31) For the root W2,we find c=-co and 1 0=2-2 =(4La-1b) (7.32) Let us evaluate Haa,Hab,and Sab Overlap integral Sat Sw=e1+R+片2R] (7.34) Coulomb integral Haa (7.35) Resonance integral HabWe now find the coefficients in (7.18) for each of the roots of the secular equation. From the linear homogeneous equations in Chapter 5, we have ( − ) + ( − ) = 0 aa a ab ab b H W c H S c (7.27) Substituting in W1 from (7.26), we get = 1 b a c c (7.28) ( ) 1 a 1sa 1sb ϕ = c φ +φ (7.29) We fix ca by normalization: ab a a sa sb sa sb S c c dv 2 2 1 | | | | | ( 2 ) 1 1 1 2 1 2 1 2 1 1 + = < >= + + ⋅ = ∫ ϕ ϕ φ φ φ φ (7.30) The normalized trial function corresponding to the energy W1 is thus: ( ) 2 2 1 1 1sa 1sb ab S ϕ φ + φ + = (7.31) For the root W2, we find cb = -ca and ( ) 2 2 1 2 1sa 1sb ab S ϕ φ −φ − = (7.32) Let us evaluate Haa, Hab, and Sab. Overlap integral Sab ] 3 1 [1 2 2 S e kR k R kR ab = + + − (7.34) Coulomb integral Haa ) 1 ( 1 2 1 2 2 R e k R H k k kR aa = − − + + − (7.35) Resonance integral Hab