# Modern Quantum Chemistry

Chapter05 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21

Exercise. This is the solution to exercise 5.1 in the book.

Solution. a. There is only one electron pair.

${}^{1}{E}_{corr}\left(FO\right)=\frac{{〈1\stackrel{̄}{1}||2\stackrel{̄}{2}〉}^{2}}{2\left({ϵ}_{1}-{ϵ}_{2}\right)}=\frac{{K}_{12}^{2}}{2\left({ϵ}_{1}-{ϵ}_{2}\right)}$

The latter equality holds because

$\begin{array}{rcll}〈1\stackrel{̄}{1}||2\stackrel{̄}{2}〉& =& 〈1\stackrel{̄}{1}|2\stackrel{̄}{2}〉-〈1\stackrel{̄}{1}|\stackrel{̄}{2}2〉& \text{}\\ & =& \left[12\right|\stackrel{̄}{1}\stackrel{̄}{2}]-\left[1\stackrel{̄}{2}\right|\stackrel{̄}{1}2]& \text{}\\ & =& \left(12\right|12)-0& \text{}\\ & =& \left(12\right|21)& \text{}\\ & =& {K}_{12}& \text{}\end{array}$

$\left(12\right|12)=\left(12\right|21)$ because molecular orbitals for the minimal ${H}_{2}$ basis set are real, hence ${K}_{ij}=\left(ij\right|ij)$.

b.

${}^{1}{E}_{corr}=\Delta -\Delta \sqrt{1+\frac{{K}_{12}^{2}}{{\Delta }^{2}}}$

If we approximate

$\sqrt{1+\frac{{K}_{12}^{2}}{{\Delta }^{2}}}=1+\frac{{K}_{12}^{2}}{2{\Delta }^{2}}$

then

${}^{1}{E}_{corr}\left(approx\right)\cong \Delta -\Delta -\Delta \frac{{K}_{12}^{2}}{2{\Delta }^{2}}=-\frac{{K}_{12}^{2}}{2\Delta }$

If we approximate $\Delta ={ϵ}_{2}-{ϵ}_{1}$ then

${}^{1}{E}_{corr}\left(approx\right)\cong \frac{{K}_{12}^{2}}{2\left({ϵ}_{1}-{ϵ}_{2}\right)}{=}^{1}{E}_{corr}\left(FO\right)$