Modern Quantum Chemistry

Chapter05 - Home
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Exercise. This is the solution to exercise 5.1 in the book.

Solution. a. There is only one electron pair.

1E corr(FO) = 11̄||22̄2 2(ϵ1 ϵ2) = K122 2(ϵ1 ϵ2)

The latter equality holds because

11̄||22̄ = 11̄|22̄11̄|2̄2 = 12 1̄2̄ 12̄ 1̄2 = 12 12 0 = 12 21 = K12

12 12 = 12 21 because molecular orbitals for the minimal H2 basis set are real, hence Kij = ij ij.

b.

1E corr = Δ Δ1 + K12 2 Δ2

If we approximate

1 + K122 Δ2 = 1 + K122 2Δ2

then

1E corr(approx)Δ Δ ΔK122 2Δ2 = K122 2Δ

If we approximate Δ = ϵ2 ϵ1 then

1E corr(approx) K122 2(ϵ1 ϵ2) = 1E corr(FO)