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Pierre-Francois Loos 2019-11-25 17:02:17 +01:00
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Manuscript/FarDFT.tex
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@ -241,9 +241,9 @@ From the GOK-DFT ensemble energy expression in Eq.~\eqref{eq:Ew-GOK}, we obtain
\end{equation} \end{equation}
where $\Eps{I}{\bw} = \sum_{p}^{\Norb} \ON{p}{(I)} \eps{p}{\bw}$, $\eps{p}{\bw}$ is the $p$th KS orbital energy given by the ensemble KS equation where $\Eps{I}{\bw} = \sum_{p}^{\Norb} \ON{p}{(I)} \eps{p}{\bw}$, $\eps{p}{\bw}$ is the $p$th KS orbital energy given by the ensemble KS equation
\begin{equation} \begin{equation}
\qty( -\frac{\nabla^2}{2} + \vext(\br{}) + \fdv{\E{\Hxc}{\bw}[\n{}{}]}{\n{}{}(\br{})}) \MO{p}{\bw}(\br{}) = \eps{p}{\bw} \MO{p}{\bw}(\br{}), \qty( \hHc(\br{}) + \fdv{\E{\Hxc}{\bw}[\n{}{}]}{\n{}{}(\br{})}) \MO{p}{\bw}(\br{}) = \eps{p}{\bw} \MO{p}{\bw}(\br{}),
\end{equation} \end{equation}
(where $\MO{p}{\bw}(\br{})$ is a KS orbital), $\ON{p}{(I)}$ its occupancy for the state $I$, and $\n{}{\bw} = \sum_{I=0}^{\Nens-1} \ew{I} \n{}{(I)}$ is the ensemble density. where $\hHc(\br{}) = -\frac{\nabla^2}{2} + \vext(\br{})$, $\MO{p}{\bw}(\br{})$ is a KS orbital, $\ON{p}{(I)}$ its occupancy for the state $I$, and $\n{}{\bw} = \sum_{I=0}^{\Nens-1} \ew{I} \n{}{(I)}$ is the ensemble density.
Equation \eqref{eq:dEdw} is our working equation for computing excitation energies. Equation \eqref{eq:dEdw} is our working equation for computing excitation energies.
%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%
@ -587,15 +587,16 @@ which are all, by construction, linear with respect to $\ew{}$.
These energies given in Eqs.~\eqref{eq:EwHF}, \eqref{eq:EwLDA} and \eqref{eq:EweLDA} can also be obtained directly from the ensemble density $\n{}{\ew{}} = (1-\ew{}) \n{}{(0)} + \ew{} \n{}{(1)}$. These energies given in Eqs.~\eqref{eq:EwHF}, \eqref{eq:EwLDA} and \eqref{eq:EweLDA} can also be obtained directly from the ensemble density $\n{}{\ew{}} = (1-\ew{}) \n{}{(0)} + \ew{} \n{}{(1)}$.
(This is what one would do in practice, \ie, by performing a KS ensemble calculation.) (This is what one would do in practice, \ie, by performing a KS ensemble calculation.)
We will label these energies as $\bE{}{\ew{}}$. We will label these energies as $\bE{}{\ew{}}$.
\begin{widetext}
For HF, we have For HF, we have
\begin{equation} \begin{equation}
\begin{split} \begin{split}
\bE{\HF}{\ew{}} \bE{\HF}{\ew{}}
& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} & = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
+ \frac{1}{2} \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}' + \frac{1}{2} \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
\\ \\
& = \ldots & = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2}
+ (1-\ew{})^2 (2\eJ{11}- \eK{11}) + \ew{}^2 (2\eJ{22}- \eK{22}) + 2 (1-\ew{})\ew{} (2 \eJ{12} - \eK{12}),
\end{split} \end{split}
\end{equation} \end{equation}
which is clearly quadratic with respect to $\ew{}$ due to the ghost interaction error in the Hartree term. which is clearly quadratic with respect to $\ew{}$ due to the ghost interaction error in the Hartree term.
@ -604,12 +605,15 @@ For LDA, we have
\begin{equation} \begin{equation}
\begin{split} \begin{split}
\bE{\LDA}{\ew{}} \bE{\LDA}{\ew{}}
& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} & = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}' + \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
+ \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
\\ \\
& + \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{} & = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2}
+ 2(1-\ew{})^2 \eJ{11} + 2\ew{}^2 \eJ{22} + 4 (1-\ew{})\ew{} \eJ{12}
\\ \\
& = \ldots , & + (1-\ew{}) \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{(0)}(\br{}) d\br{}
+ \ew{} \int \e{\xc}{\LDA}[\n{}{\ew{}}(\br{})] \n{}{(1)}(\br{}) d\br{}
\end{split} \end{split}
\end{equation} \end{equation}
which is also clearly quadratic with respect to $\ew{}$ because the (weight-independent) LDA functional cannot compensate the ``quadraticity'' of the Hartree term. which is also clearly quadratic with respect to $\ew{}$ because the (weight-independent) LDA functional cannot compensate the ``quadraticity'' of the Hartree term.
@ -618,16 +622,23 @@ For eLDA, we have
\begin{equation} \begin{equation}
\begin{split} \begin{split}
\bE{\eLDA}{\ew{}} \bE{\eLDA}{\ew{}}
& = (1-\ew{}) 2 \eHc{1} + \ew{} 2 \eHc{2} & = \int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}
+ \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}' + \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
+ \int \be{\xc}{\ew{}}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{}
\\ \\
& + \int \be{\xc}{\ew{}}[\n{}{\ew{}}(\br{})] \n{}{\ew{}}(\br{}) d\br{} & = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2}
+ 2(1-\ew{})^2 \eJ{11} + 2\ew{}^2 \eJ{22} + 4 (1-\ew{})\ew{} \eJ{12}
\\ \\
& = \ldots , & + (1-\ew{})^2 \int \be{\xc}{(0)}[\n{}{\ew{}}(\br{})] \n{}{(0)}(\br{}) d\br{}
+ \ew{}^2 \int \be{\xc}{(1)}[\n{}{\ew{}}(\br{})] \n{}{(1)}(\br{}) d\br{}
\\
& + (1-\ew{})\ew{} \int \be{\xc}{(0)}[\n{}{\ew{}}(\br{})] \n{}{(1)}(\br{}) d\br{}
+ \ew{}(1-\ew{}) \int \be{\xc}{(1)}[\n{}{\ew{}}(\br{})] \n{}{(0)}(\br{}) d\br{}
\end{split} \end{split}
\end{equation} \end{equation}
which *could* be linear with respect to the weight if the weight-dependent xc functional is very well constructed. which \textit{could} be linear with respect to the weight if the weight-dependent xc functional is very well constructed.
This would be, for example, the case with the exact xc functional. This would be, for example, the case with the exact xc functional.
\end{widetext}
%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%
%%% CONCLUSION %%% %%% CONCLUSION %%%