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documentation:standard_operators:coulomb_repulsion [2017/02/27 12:44] – Maurits W. Haverkortdocumentation:standard_operators:coulomb_repulsion [2017/02/27 13:30] – Maurits W. Haverkort
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 ### ###
-For the case where $n_1=n_2=n_3=n_4$ and $l_1=l_2=l_3=l_4$, i.e. Coulomb repulsion within one shell one defines:+{{:documentation:standard_operators:coulomb_diagram_lll.png?nolink&200 |}}For the case where $n_1=n_2=n_3=n_4$ and $l_1=l_2=l_3=l_4$, i.e. Coulomb repulsion within one shell one defines:
 \begin{equation} \begin{equation}
 F^{(k)} = R^{(k)}[\tau_1\tau_2\tau_3\tau_4]. F^{(k)} = R^{(k)}[\tau_1\tau_2\tau_3\tau_4].
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 ### ###
-{{:documentation:standard_operators:coulomb_diagram_llll.png?nolink&200 |}}+{{:documentation:standard_operators:coulomb_diagram_llll.png?nolink&200 |}}The Coulomb repulsion in the general case allows four different principle quantum numbers and angular momenta. The radial integral is given as: 
 +\begin{equation} 
 +R^{(k)}[n_1l_1\:n_2l_2\:n_3l_3\:n_4l_4]=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_{n_1l_1}[r_i]R_{n_2l_2}[r_j]R_{n_3l_3}[r_i]R_{n_4l_4}[r_j]\mathrm{d}r_i\mathrm{d}r_j, 
 +\end{equation} 
 +with $\mathrm{Max}[|l_1-l_3|,|l_2-l_4|] \leq k \leq \mathrm{Min}[l_1+l_3,l_2+l_4]$ and $(l_1+l_3)$, $(l2_+l_4)$ either both even or both odd. 
 +###
  
 +###
 +In Quanty one can implement these operators as:
 +<code Quanty Example.Quanty>
 +NewOperator("U", NF, IndexUp_1, IndexDn_1, IndexUp_2, IndexDn_2, IndexUp_3, IndexDn_3, IndexUp_4, IndexDn_4, Rk)
 +</code>
 +For $l_1=3$, $l_2=0$, $l_3=2$ and $l_4=1$ one has $k=1$ and one could define:
 +<code Quanty Example.Quanty>
 +OppR1pd = NewOperator("U", NF, IndexUp_1, IndexDn_1, IndexUp_2, IndexDn_2, IndexUp_3, IndexDn_3, IndexUp_4, IndexDn_4, {1})
 +</code>
 ### ###
  
 +###
 +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. 
 +###
  
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