Formulas
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@ -229,7 +229,7 @@ Assuming, witout loss of generality that $\cD{0}$ is the largest coefficient $\c
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\section{Matrix elements}
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%----------------------------------------------------------------
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Compared to a conventional CI calculation, new matrix elements are required.
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The simplest of them $f_{IJ}$ --- required in Eq.~\eqref{eq:IHF} --- can be easily computed by applying Condon-Slater rules. \cite{SzaboBook}
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The simplest of them $f_{IJ}$ --- required in Eq.~\eqref{eq:IHF} --- can be easily computed by applying Slater-Condon rules. \cite{SzaboBook}
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They involve two-electron integrals over the correlation factor $f_{12}$.
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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}
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These can be more or less expensive to compute depending on the choice of the correlation factor.
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@ -265,9 +265,36 @@ The set $\mC$ is defined by two simple rules.
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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 $\kJ$.
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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.
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Three types of determinants have these two properties (see Fig.~\ref{fig:CBS}).:
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i) the pure doubles $\ket*{_{ij}^{\alpha \beta}}$,
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ii) the mixed doubles $\ket*{_{ij}^{a \beta}}$, and
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iii) the pure singles $\ket*{_{i}^{\alpha}}$.
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i) the pure doubles $\hT_{ij}^{\alpha \beta}\ket*{I}$,
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ii) the mixed doubles $\hT_{ij}^{\alpha b}\ket*{I}$, and
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iii) the pure singles $\hT_{i}^{\alpha}\ket*{I}$.
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\alert{
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The matrix element between two determinants differing by a double excitation $\hT_{ij}^{kl}$ is given by
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\begin{equation}
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\mel{I}{\hH f}{J} = \{ ij || kl \} - \sum_m \{ ijm || mkl \} \Delta_{mI} \Delta_{mJ}
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\end{equation}
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where
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\begin{equation}
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\Delta_{mI} = \mel{I}{a_m^\dagger a_m}{I},
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\end{equation}
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\begin{equation}
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\{ ijm || mkl \} = \sum_{\alpha} \langle i j || \alpha m \rangle [ \alpha m || k l ]
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+ \langle i j || m \alpha \rangle [ m \alpha || k l ],
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\end{equation}
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\begin{equation}
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\{ ij || kl \} = \sum_{\alpha \beta} \langle i j || \alpha \beta \rangle [ \alpha \beta || k l ] + \sum_m \{ ijm || mkl \}
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\end{equation}
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The matrix element between two determinants differing by a single excitation $\hT_{i}^{k}$ is given by
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\begin{equation}
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\mel{I}{\hH f}{J} = \sum_j \Delta_{jI} \Delta_{jJ} \qty( \{ ij || kj \} - \sum_m \{ ijm || mkj \} \Delta_{mI} \Delta_{mJ} )
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\end{equation}
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and the diagonal terms are
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\begin{equation}
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\mel{I}{\hH f}{I} = \sum_{ij} \Delta_{iI} \Delta_{jI} \qty( \{ ij || ij \} - \sum_m \{ ijm || mij \} \Delta_{mI} )
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\end{equation}
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}
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Although $\mel{0}{\hH}{_{i}^{a}} = 0$, note that the Brillouin theorem does not hold in the CABS, i.e.~$\mel{0}{\hH}{_{i}^{\alpha}} \neq 0$.
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Here, we will eschew the generalized Brillouin condition (GBC) which set these to zero. \cite{Kutzelnigg91}
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