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documentation:standard_operators:coulomb_repulsion [2017/02/27 13:30] – Maurits W. Haverkortdocumentation:standard_operators:coulomb_repulsion [2017/02/27 14:59] – Maurits W. Haverkort
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-{{:documentation:standard_operators:coulomb_diagram_ll.png?nolink&400 |}}The Coulomb repulsion between two shells which does not change the number of electrons is given by a direct term ($l_1=l_3$ and $l_2=l_4$) and an indirect or exchange term ($l_1=l_4$ and $l_2=l_3$). The direct term is given by the Slater integrals:+{{:documentation:standard_operators:coulomb_diagram_ll.png?nolink&400 |}}The Coulomb repulsion between two shells which does not change the number of electrons is given by a direct term ($n_1l_1=n_3l_3$ and $n_2l_2=n_4l_4$) and an indirect or exchange term ($n_1l_1=n_4l_4$ and $n_2l_2=n_3l_3$). We here assume that $n_1l_1\neq n_2l_2$. The direct term is given by the Slater integrals:
 \begin{equation} \begin{equation}
 F^{(k)}=e^2\int_0^{\infty}\int_0^{\infty}\frac{\mathrm{Min}[r_i,r_j]^k}{\mathrm{Max}[r_i,r_j]^{k+1}}R_1[r_i]^2R_2[r_j]^2\mathrm{d}r_i\mathrm{d}r_j, F^{(k)}=e^2\int_0^{\infty}\int_0^{\infty}\frac{\mathrm{Min}[r_i,r_j]^k}{\mathrm{Max}[r_i,r_j]^{k+1}}R_1[r_i]^2R_2[r_j]^2\mathrm{d}r_i\mathrm{d}r_j,
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-Note that in the general case you need to sum over all possible permutations of $n_1l_1$, $n_2l_2$, $n_3l_3$ and $n_4l_4$. Permuting $n_1l_1$ with $n_2l_2$ and at the same time $n_3l_3$ with $n_4l_4$ will not change the value and if $n_1l_1$ is different from $n_2l_2$ and $n_3l_3$ is different from $n_4l_4$ one can add a factor of two. If one of them is the same a permutation will not lead to a new configuration and the factor of two disappears. If you just sum over all possible $n_il_i$ combinations things go right automatically. +Note that in the general case you need to sum over all possible permutations of $n_1l_1$, $n_2l_2$, $n_3l_3$ and $n_4l_4$. Permuting $n_1l_1$ with $n_2l_2$ and at the same time $n_3l_3$ with $n_4l_4$ will not change the value and form of the operator. If $n_1l_1$ is different from $n_2l_2$ and $n_3l_3$ is different from $n_4l_4$ one can add a factor of two in front of the operator and only add one of the permutations. If one of the $n_1l_1$ is the same as $n_2l_2$ or $n_3l_3$ is the same as $n_4l_4$ a permutation will not lead to a new configuration and the factor of two disappears. If you just sum over all possible $n_il_i$ combinations things go right automatically. 
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