Manu: gave the final expression for the individual energies

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Emmanuel Fromager 2019-09-11 18:26:57 +02:00
parent 7c1eacafa5
commit 420a464a6f

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@ -458,10 +458,34 @@ Hxc}(n^{\bw}(\br))\,n^{(I)}(\br)
Hxc}(n)}{\partial n}\right|_{n=n^{\bw}(\br)}n^{\bw}(\br)\left(n^{(I)}(\br)-n^{\bw}(\br)\right) Hxc}(n)}{\partial n}\right|_{n=n^{\bw}(\br)}n^{\bw}(\br)\left(n^{(I)}(\br)-n^{\bw}(\br)\right)
\nonumber\\ \nonumber\\
&& &&
+\sum_{K>0}\left(\delta_{IK}-w^{(K)}\right)\left. \dfrac{\partial \overline{E}^{{\bw}}_{\rm +\int d\br\,\sum_{K>0}\left(\delta_{IK}-w^{(K)}\right)n^{{\bw}}(\br)\left.
Hxc}\left[n\right]}{\partial w^{(K)}}\right|_{n=n^{{\bw}}}. \dfrac{\partial \overline{\epsilon}^{{\bw}}_{\rm
Hxc}(n)}{\partial w^{(K)}}\right|_{n=n^{{\bw}}(\br)}.
\eeq \eeq
\alert{
or, equivalently,
\beq
&&E^{(I)}\rightarrow
{\rm
Tr}\left[{\bmg}^{(I)}{\bm h}\right]+
\Tr\left[\left(\bmg^{(I)}-\dfrac{1}{2}\bmg^{\bw}\right) \, \bG \, \bmg^{\bw}\right]
\nonumber\\
&&
+\int d\br\,
\dfrac{\delta \overline{E}^{{\bw}}_{\rm
Hxc}\left[n^{\bw}\right]}{\delta
n({\br})}\,n^{(I)}(\br)
\nonumber\\
&&
-\int d\br\,\left.\dfrac{\partial \overline{\epsilon}^{{\bw}}_{\rm
Hxc}(n)}{\partial n}\right|_{n=n^{\bw}(\br)}\Big(n^{\bw}(\br)\Big)^2
\nonumber\\
&&
+\int d\br\,\sum_{K>0}\left(\delta_{IK}-w^{(K)}\right)n^{{\bw}}(\br)\left.
\dfrac{\partial \overline{\epsilon}^{{\bw}}_{\rm
Hxc}(n)}{\partial w^{(K)}}\right|_{n=n^{{\bw}}(\br)}.
\eeq
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{KS-eDFT for excited states} \subsection{KS-eDFT for excited states}