From a6af34ee9decd9268c38c331d0ac137103936d93 Mon Sep 17 00:00:00 2001 From: Anthony Scemama Date: Wed, 16 Oct 2019 18:35:08 +0200 Subject: [PATCH] Formulas --- Manuscript/CI-F12.tex | 35 +++++++++++++++++++++++++++++++---- 1 file changed, 31 insertions(+), 4 deletions(-) diff --git a/Manuscript/CI-F12.tex b/Manuscript/CI-F12.tex index 3f9f2c3..f8b37fb 100644 --- a/Manuscript/CI-F12.tex +++ b/Manuscript/CI-F12.tex @@ -229,7 +229,7 @@ Assuming, witout loss of generality that $\cD{0}$ is the largest coefficient $\c \section{Matrix elements} %---------------------------------------------------------------- Compared to a conventional CI calculation, new matrix elements are required. -The simplest of them $f_{IJ}$ --- required in Eq.~\eqref{eq:IHF} --- can be easily computed by applying Condon-Slater rules. \cite{SzaboBook} +The simplest of them $f_{IJ}$ --- required in Eq.~\eqref{eq:IHF} --- can be easily computed by applying Slater-Condon rules. \cite{SzaboBook} They involve two-electron integrals over the correlation factor $f_{12}$. Their computation has been thoroughly studied in the literature in the last thirty years. \cite{Kutzelnigg91, Klopper92, Persson97, Klopper02, Manby03, Werner03, Klopper04, Tenno04a, Tenno04b, May05, Manby06, Tenno07, Komornicki11, Reine12, GG16} These can be more or less expensive to compute depending on the choice of the correlation factor. @@ -265,9 +265,36 @@ The set $\mC$ is defined by two simple rules. First, because $f$ is a two-electron operator [and thanks to the matrix element $f_{AJ}$ in \eqref{eq:IHF-RI}], we know that the sum over $A$ is restricted to the singly- or doubly-excited determinants with respect to the determinant $\kJ$. Second, to ensure that $H_{IA} \neq 0$, $A$ must be connected to $\kI$, i.e.~differs from $\kI$ by no more than two spin orbitals. Three types of determinants have these two properties (see Fig.~\ref{fig:CBS}).: -i) the pure doubles $\ket*{_{ij}^{\alpha \beta}}$, -ii) the mixed doubles $\ket*{_{ij}^{a \beta}}$, and -iii) the pure singles $\ket*{_{i}^{\alpha}}$. +i) the pure doubles $\hT_{ij}^{\alpha \beta}\ket*{I}$, +ii) the mixed doubles $\hT_{ij}^{\alpha b}\ket*{I}$, and +iii) the pure singles $\hT_{i}^{\alpha}\ket*{I}$. + +\alert{ +The matrix element between two determinants differing by a double excitation $\hT_{ij}^{kl}$ is given by +\begin{equation} + \mel{I}{\hH f}{J} = \{ ij || kl \} - \sum_m \{ ijm || mkl \} \Delta_{mI} \Delta_{mJ} +\end{equation} +where +\begin{equation} +\Delta_{mI} = \mel{I}{a_m^\dagger a_m}{I}, +\end{equation} +\begin{equation} + \{ ijm || mkl \} = \sum_{\alpha} \langle i j || \alpha m \rangle [ \alpha m || k l ] + + \langle i j || m \alpha \rangle [ m \alpha || k l ], +\end{equation} +\begin{equation} + \{ ij || kl \} = \sum_{\alpha \beta} \langle i j || \alpha \beta \rangle [ \alpha \beta || k l ] + \sum_m \{ ijm || mkl \} +\end{equation} +The matrix element between two determinants differing by a single excitation $\hT_{i}^{k}$ is given by +\begin{equation} + \mel{I}{\hH f}{J} = \sum_j \Delta_{jI} \Delta_{jJ} \qty( \{ ij || kj \} - \sum_m \{ ijm || mkj \} \Delta_{mI} \Delta_{mJ} ) +\end{equation} +and the diagonal terms are +\begin{equation} + \mel{I}{\hH f}{I} = \sum_{ij} \Delta_{iI} \Delta_{jI} \qty( \{ ij || ij \} - \sum_m \{ ijm || mij \} \Delta_{mI} ) +\end{equation} +} + Although $\mel{0}{\hH}{_{i}^{a}} = 0$, note that the Brillouin theorem does not hold in the CABS, i.e.~$\mel{0}{\hH}{_{i}^{\alpha}} \neq 0$. Here, we will eschew the generalized Brillouin condition (GBC) which set these to zero. \cite{Kutzelnigg91}