Formulas

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 $\ket*{_{ij}^{\alpha \beta}}$,
+i) the pure doubles $\hT_{ij}^{\alpha \beta}\ket*{I}$,
-ii) the mixed doubles $\ket*{_{ij}^{a \beta}}$, and
+ii) the mixed doubles $\hT_{ij}^{\alpha b}\ket*{I}$, and
-iii) the pure singles $\ket*{_{i}^{\alpha}}$.
+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}