Manu: saving work

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Emmanuel Fromager 2019-09-18 21:48:44 +02:00
parent cba8d0360a
commit 89038f94fb

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@ -545,7 +545,7 @@ and therefore
Combining the latter Eqs. with
Eqs. (\ref{eq:exact_GIC}), (\ref{eq:var_princ_Gamma_ens}) leads to
the final ensemble energy expression
\beq
\beq\label{eq:exact_Eens_EEXX}
E^{{\bw}}={\rm
Tr}\left[{\bmg}^{{\bw}}{\bm h}\right]+\frac{1}{2} \sum_{L\geq0}w_L
\Tr(\bmg^{(L)} \, \bG \, \bmg^{(L)})
@ -567,7 +567,7 @@ c}\left[n_{\bmg^{\bw}}\right]
\eeq
For $K>0$
\beq
\beq\label{eq:XE_EEXX}
&&\dfrac{\partial E^{{\bw}}}{\partial w_K}=
{\rm
Tr}\left[\left({\bmg}^{(K)}-{\bmg}^{(0)}\right){\bm h}\right]
@ -594,6 +594,40 @@ Tr}\left[\dfrac{\partial\bmg^{(L)}}{\partial w_K}{\bm h}\right]
c}\left[n_{\bmg^{\bw}}\right]}{\delta
n({\br})}n_{\frac{\partial \bmg^{(L)}}{\partial w_K}}(\br).
\eeq
If we introduce individual Fock matrices
\beq
{\bm F}^{(L)}={\bm h}+\bG \,\bmg^{(L)}+\overline{\bm v}^{{\bw}}_{\rm
c}\left[n_{\bmg^{\bw}}\right],
\eeq
the last three terms can be simply rewritten as
\beq
\sum_{L\geq0}w_L{\rm
Tr}\left[{\bm F}^{(L)}\frac{\partial \bmg^{(L)}}{\partial w_K}\right].
\eeq
According to Eqs.~(\ref{eq:indiv_ener_from_ens}),
(\ref{eq:exact_Eens_EEXX}), and (\ref{eq:XE_EEXX}),
\beq
E^{(I)}&&={\rm
Tr}\left[{\bmg}^{(I)}{\bm h}\right]
+\frac{1}{2} \Tr(\bmg^{(I)} \, \bG \,
\bmg^{(I)})
\nonumber\\
&&+\overline{E}^{{\bw}}_{\rm
c}\left[n_{\bmg^{\bw}}\right]
+\int d\br\,\dfrac{\delta \overline{E}^{{\bw}}_{\rm
c}\left[n_{\bmg^{\bw}}\right]}{\delta
n({\br})}\left(n_{\bmg^{(I)}}(\br)-n_{\bmg^{\bw}}(\br)\right)
\nonumber\\
&&+\sum_{K>0}\left(\delta_{IK}-w_K\right)\left. \dfrac{\partial \overline{E}^{{\bw}}_{\rm
c}\left[n\right]}{\partial w_K}\right|_{n=n_{\bmg^{\bw}}}
\nonumber\\
&&
+\sum_{K>0}\left(\delta_{IK}-w_K\right)\sum_{L\geq0}w_L{\rm
Tr}\left[{\bm F}^{(L)}\frac{\partial \bmg^{(L)}}{\partial w_K}\right]
.
\eeq
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{KS-eDFT for excited states}