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back to single weight

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Pierre-Francois Loos 2020-02-21 21:09:48 +01:00
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Manuscript/FarDFT.tex
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@ -377,7 +377,7 @@ Combining these, we build a two-state weight-dependent correlation functional:
\begin{figure}
\includegraphics[width=\linewidth]{fig/fig1}
\caption{
Reduced (i.e., per electron) correlation energy $\e{\co}{(I)}$ [see Eq.~\eqref{eq:ec}] as a function of $R = 1/(\pi \n{}{})^{1/3}$ for the ground state ($I=0$), and the first doubly-excited state ($I=1$) of the (spin-unpolarised) two-electron FUEG.
Reduced (i.e., per electron) correlation energy $\e{\co}{(I)}$ [see Eq.~\eqref{eq:ec}] as a function of $R = 1/(\pi^2 \n{}{})^{1/3}$ for the ground state ($I=0$), and the first doubly-excited state ($I=1$) of the (spin-unpolarised) two-electron FUEG.
The data gathered in Table \ref{tab:Ref} are also reported.
}
\label{fig:Ec}
@ -388,7 +388,7 @@ Combining these, we build a two-state weight-dependent correlation functional:
\begin{table}
\caption{
\label{tab:Ref}
$-\e{\co}{(I)}$ as a function of the radius of the glome $R = 1/(\pi \n{}{})^{1/3}$ for the ground state ($I=0$), and the first doubly-excited state ($I=1$) of the (spin-unpolarised) two-electron FUEG.
$-\e{\co}{(I)}$ as a function of the radius of the glome $R = 1/(\pi^2 \n{}{})^{1/3}$ for the ground state ($I=0$), and the first doubly-excited state ($I=1$) of the (spin-unpolarised) two-electron FUEG.
}
\begin{ruledtabular}
\begin{tabular}{lcc}
@ -511,10 +511,10 @@ Total energies (in hartree) and excitation energies (in \titou{hartree}) of \ce{
LDA & -1.12120 & 0.379745 & 1.50095 & -1.12120 & 1.49536 & -0.370725 & 1.50565 \\
eLDA & -1.12120 & 0.175337 & 1.29654 & -1.12120 & 1.31995 & -0.462421 & 1.30839 \\
CID & -1.13728 & 0.481138 & 1.61841 & \\
Exact\fnm[1] & & & & \\
accurate\fnm[1] & & & & \\
\end{tabular}
\end{ruledtabular}
\fnt[1]{Reference \onlinecite{}.}
\fnt[1]{FCI/aug-cc-pV5Z excitation energies computer with QUANTUM PACKAGE. \cite{QP2}}
\end{table*}
%%% %%% %%% %%%
@ -523,7 +523,7 @@ Total energies (in hartree) and excitation energies (in \titou{hartree}) of \ce{
\includegraphics[width=\linewidth]{fig/GSetDES_exact_HF_LDA_eLDA}
\caption{
Total energies (in hartree) of \ce{H2} as a function of $\RHH$ (in bohr) for various methods with the STO-3G minimal basis.
\label{tab:Energies}
%\label{fig:Energies}
}
\end{figure}
%%% %%% %%% %%%
@ -533,7 +533,7 @@ Total energies (in hartree) of \ce{H2} as a function of $\RHH$ (in bohr) for var
\includegraphics[width=\linewidth]{fig/ExcitationEnergyExact_wHF_wLDA_weLDA_w=0etw=0.5}
\caption{
Excitation energies (in hartree) of \ce{H2} as a function of $\RHH$ (in bohr) for various methods with the STO-3G minimal basis.
\label{tab:Energies}
%\label{fig:Energies}
}
\end{figure}
%%% %%% %%% %%%
@ -543,7 +543,7 @@ Excitation energies (in hartree) of \ce{H2} as a function of $\RHH$ (in bohr) fo
\includegraphics[width=\linewidth]{fig/EnsembleEnergy_wHF_wLDA_weLDA_wHFbarre_wLDAbarre_weLDAbarre_R=1.4}
\caption{
Ensemble energies (in hartree) of \ce{H2} with $\RHH = 1.4$ bohr as a function of the weight $\ew{}$ for various methods with the STO-3G minimal basis.
\label{tab:Energies}
%\label{tab:Energies}
}
\end{figure}
%%% %%% %%% %%%
@ -558,7 +558,7 @@ We select STO-3G as minimal basis, and study the behaviour of the total energy o
This minimal-basis example is quite pedagogical as the molecular orbitals are fixed by symmetry.
We have then access to the individual densities of the ground and doubly-excited states (which is not usually possible in practice).
Therefore, thanks to the spatial symmetry and the minimal basis, the individual densities extracted from the ensemble density are equal to the \textit{exact} individual densities.
In other words, there is no density-driven error and the only error that we are going to see is the functional-driven error (and this is what we want to study).
In other words, there is no density-driven error and the only error that we are going to observe is the functional-driven error (and this is what we want to study).
The bonding and antibonding orbitals of the \ce{H2} molecule are given by
\begin{subequations}
@ -712,7 +712,8 @@ For HF, we have
\label{eq:bEwHF}
\begin{split}
\tE{\HF}{\ew{}}
& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
& = \Ts{\ew{}}[\n{}{\ew{}}(\br{})]
+ \int \vext(\br{}) \n{}{\ew{}}(\br{}) d\br{}
+ \frac{1}{2} \iint \frac{\n{}{\ew{}}(\br{})\n{}{\ew{}}(\br{}')}{\abs{\br{} - \br{}'}} d\br{} d\br{}'
\\
& = 2 (1-\ew{}) \eHc{1} + 2 \ew{} \eHc{2}
@ -725,7 +726,8 @@ In the case of the LDA, it reads
\label{eq:bEwLDA}
\begin{split}
\tE{\LDA}{\ew{}}
& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
& = \Ts{\ew{}}[\n{}{\ew{}}(\br{})]
+ \int \vext(\br{}) \n{}{\ew{}}(\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{}
\\
@ -742,7 +744,8 @@ For eLDA, the ensemble energy can be decomposed as
\label{eq:bEweLDA}
\begin{split}
\tE{\eLDA}{\ew{}}
& = \titou{\int \hHc(\br{}) \n{}{\ew{}}(\br{}) d\br{}}
& = \Ts{\ew{}}[\n{}{\ew{}}(\br{})]
+ \int \vext(\br{}) \n{}{\ew{}}(\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{}
\\