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documentation:standard_operators:coulomb_repulsion [2017/02/27 11:34] – 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 (Min[ri,rj]kMax[ri,rj]k+1) is more difficult and will be cast into parameters: | The radial part of the operator (Min[ri,rj]kMax[ri,rj]k+1) is more difficult and will be cast into parameters: | ||
\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 n1=n2=n3=n4 and l1=l2=l3=l4, | + | {{: |
\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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===== Two shells, shell occupation conserving ===== | ===== Two shells, shell occupation conserving ===== | ||
+ | |||
### | ### | ||
- | The Coulomb repulsion between two shells which does not change the number of electrons is given by a direct term ($l_1=l_3andl_2=l_4)andanindirectorexchangeterm(l_1=l_4andl_2=l_3$). 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, | F^{(k)}=e^2\int_0^{\infty}\int_0^{\infty}\frac{\mathrm{Min}[r_i, | ||
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</ | </ | ||
### | ### | ||
- | |||
===== General case of 4 different shells ===== | ===== General case of 4 different shells ===== | ||
+ | |||
+ | ### | ||
+ | {{: | ||
+ | \begin{equation} | ||
+ | R^{(k)}[n_1l_1\: | ||
+ | \end{equation} | ||
+ | with Max[|l1−l3|,|l2−l4|]≤k≤Min[l1+l3,l2+l4] and (l1+l3), | ||
### | ### | ||
+ | ### | ||
+ | In Quanty one can implement these operators as: | ||
+ | <code Quanty Example.Quanty> | ||
+ | NewOperator(" | ||
+ | </ | ||
+ | For l1=3, l2=0, l3=2 and l4=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 n1l1, n2l2, n3l3 and n4l4. Permuting n1l1 with n2l2 and at the same time n3l3 with n4l4 will not change the value and form of the operator. If n1l1 is different from n2l2 and n3l3 is different from n4l4 one can add a factor of two in front of the operator and only add one of the permutations. If one of the n1l1 is the same as n2l2 or n3l3 is the same as n4l4 a permutation will not lead to a new configuration and the factor of two disappears. If you just sum over all possible nili combinations things go right automatically. | ||
+ | ### | ||
===== Table of contents ===== | ===== Table of contents ===== | ||
{{indexmenu> | {{indexmenu> |