Question

When we worked on the variational treatment for the hydrogen molecular ion in class, we solved for E₊. By substituting back in to the original “secular equations,” C_A = C_B can be obtained. Confirm it.

Answer 1

f = Cafa +C_bf_b

E(C_a ,C_b) = Integral f H^ f dτ / integral h²dτ

H_aa = integral ψa H^f ψ_a dτ

Hbb = integral ψb H^f ψb dτ

Hab = Hba =integral ψa H^f ψb dτ

SAB = integral ψA ψB dτ

E(cA,cB) = C_a² H_aa + 2C_aC_b H_ab + Cb² H_bb /C_a² + 2C_aC_b SAB + C_b2

∂E/ ∂cA = 0, ∂E / ∂cB =0

det(HAA − E HAB − ESAB HAB − ESAB HBB − E) =0

E +- = α ± β /1 ± S

E + = α + β /1 + S bonding

E + = α - β /1 - S anti bonding

E − EA = (HAB − ESAB)²/ E − EB

E ≈ EA + (HAB − EASAB)² / EA − EB

E ≈ EB - (HAB − EASAB)²  / EA − EB