3.The solution of ForH达that is a e-Ta B 1S4= =Ψa R Wπ e-n =Ψb Vπ Note:we have as many linear combinations as we have atomic orbitals φ=CaΨa+CbΨ6 Trial functions =CaΨa+CbΨb seqular equation Hoa-ESaa Hab-ESab =0 Hpa-ESpa Hob-ESbb has the same form as ∴Ha=Ho, Hab=Ha (Ha-ESa)2=(Hb-ESb)月 Haa-ESaa=(Hab-ESab) 计 Ha-ESa=-(Hab-ESab） E= Haa+Hab_a+B 1+Sab 1+S Haa-ESoa=Hab-ESab E2 Hoe-Hab= -B 1-Sab 1-S3. The solution of H2 + A B e- r r b a R θ For H2 + that is : Note : we have as many linear combinations as we have atomic orbitals a r A a e s ψ π = = − 1 b 1 ψ π = = − br B e s a a b b φ = c ψ + c ψ Trial functions ( ) ( ) ( ) has the same form as 2 2 , aa aa ab ab aa aa ab ab a b aa bb ab ba H ES H ES H ES H ES H H H H − = ± − − = − Qψ ψ ∴ = = = 0 − − − − = + ba ba bb bb aa aa ab ab a a b b H ES H ES H ES H ES seqular equation φ c ψ c ψ S S H H E if H ES H ES S S H H E if H ES H ES ab aa ab aa aa ab ab ab aa ab aa aa ab ab − − = − − = − = − + + = + + = − = − − 1 1 1 1 ( ) 2 1 α β α β