# Modern Quantum Chemistry

Chapter03 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33 - 34 - 35 - 36 - 37 - 38 - 39 - 40 - 41 - 42 - 43 - 44

Exercise. This is the solution to exercise 3.1 in the book.

Solution. We have that:

$f{\chi }_{j}\left(1\right)=h\left(1\right){\chi }_{j}\left(1\right)+\sum _{b}{J}_{b}\left(1\right){\chi }_{j}\left(1\right)-\sum _{b}{K}_{b}\left(1\right){\chi }_{j}\left(1\right)$

Multiply by ${\chi }_{i}^{\ast }\left(1\right)$ and integrate over ${x}_{1}$.

$\begin{array}{rcll}〈{\chi }_{i}|f|j〉& =& \int {\chi }_{i}^{\ast }\left(1\right)h\left(1\right){\chi }_{j}\left(1\right)d{x}_{1}+& \text{}\\ & & +\sum _{\begin{array}{c}b\\ b\ne j\end{array}}\int {\chi }_{i}^{\ast }\left(1\right){J}_{b}\left(1\right){\chi }_{j}\left(1\right)d{x}_{1}-\sum _{\begin{array}{c}b\\ b\ne j\end{array}}\int {\chi }_{i}^{\ast }\left(1\right){K}_{b}\left(1\right){\chi }_{j}\left(1\right)d{x}_{1}& \text{}\end{array}$

Replace definitions for

${J}_{b}\left(1\right){\chi }_{j}\left(1\right)=\int {\chi }_{b}^{\ast }\left(2\right)\frac{1}{{r}_{12}}{\chi }_{b}\left(2\right){\chi }_{j}\left(1\right)d{x}_{2}$

${K}_{b}\left(1\right){\chi }_{j}\left(1\right)=\int {\chi }_{b}^{\ast }\left(2\right)\frac{1}{{r}_{12}}{\chi }_{j}\left(2\right){\chi }_{b}\left(1\right)d{x}_{2}$

Hence,

$〈{\chi }_{i}|f|{\chi }_{j}〉=〈{\chi }_{i}|h|{\chi }_{j}〉+\sum _{\begin{array}{c}b\\ b\ne j\end{array}}\int {\chi }_{i}^{\ast }\left(1\right){\chi }_{j}\left(1\right)\frac{1}{{r}_{12}}{\chi }_{b}^{\ast }\left(2\right){\chi }_{b}\left(2\right)d{x}_{1}d{x}_{2}-\sum _{\begin{array}{c}b\\ b\ne j\end{array}}\int {\chi }_{i}^{\ast }\left(1\right){\chi }_{b}\left(1\right)\frac{1}{{r}_{12}}{\chi }_{b}^{\ast }\left(2\right){\chi }_{j}\left(2\right)d{x}_{1}d{x}_{2}$

or

$〈{\chi }_{i}|f|{\chi }_{j}〉=〈{\chi }_{i}|h|{\chi }_{j}〉+\sum _{\begin{array}{c}b\\ b\ne j\end{array}}\left[ij|bb\right]-\sum _{\begin{array}{c}b\\ b\ne j\end{array}}\left[ib|bj\right]$

Since

$\left[ib|bb\right]-\left[ib|bb\right]=0$

we can also add index $j=b$, therefore

$〈{\chi }_{i}|f|{\chi }_{j}〉=〈{\chi }_{i}|h|{\chi }_{j}〉+\sum _{b}\left[ij|bb\right]-\sum _{b}\left[ib|bj\right]$

or

$〈{\chi }_{i}|f|{\chi }_{j}〉=〈{\chi }_{i}|h|{\chi }_{j}〉+\sum _{b}\left\{〈ib|jb〉-〈ib|bj〉\right\}$

$〈{\chi }_{i}|f|{\chi }_{j}〉=〈{\chi }_{i}|h|{\chi }_{j}〉+\sum _{b}〈ib||jb〉$