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documentation:standard_operators:coulomb_repulsion [2017/02/27 12:44] – Maurits W. Haverkort | documentation:standard_operators:coulomb_repulsion [2017/05/23 16:43] (current) – Maurits W. Haverkort | ||
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The radial part of the operator ($\frac{\mathrm{Min}[r_i, | The radial part of the operator ($\frac{\mathrm{Min}[r_i, | ||
\begin{equation} | \begin{equation} | ||
- | R^{(k)}[\tau_1\tau_2\tau_3\tau_4]=e^2\int_0^{\infty}\int_0^{\infty}\frac{\mathrm{Min}[r_i, | + | R^{(k)}[\tau_1\tau_2\tau_3\tau_4]=e^2\int_0^{\infty}\int_0^{\infty}\frac{\mathrm{Min}[r_i, |
\end{equation} | \end{equation} | ||
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### | ### | ||
- | For the case where $n_1=n_2=n_3=n_4$ and $l_1=l_2=l_3=l_4$, | + | {{: |
\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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### | ### | ||
- | {{: | + | {{: |
\begin{equation} | \begin{equation} | ||
F^{(k)}=e^2\int_0^{\infty}\int_0^{\infty}\frac{\mathrm{Min}[r_i, | F^{(k)}=e^2\int_0^{\infty}\int_0^{\infty}\frac{\mathrm{Min}[r_i, | ||
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### | ### | ||
- | {{: | + | {{: |
+ | \begin{equation} | ||
+ | R^{(k)}[n_1l_1\: | ||
+ | \end{equation} | ||
+ | with $\mathrm{Max}[|l_1-l_3|, | ||
+ | ### | ||
+ | ### | ||
+ | In Quanty one can implement these operators as: | ||
+ | <code Quanty Example.Quanty> | ||
+ | NewOperator(" | ||
+ | </ | ||
+ | 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(" | ||
+ | </ | ||
### | ### | ||
+ | ### | ||
+ | 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. | ||
+ | ### | ||
===== Table of contents ===== | ===== Table of contents ===== | ||
{{indexmenu> | {{indexmenu> |