VMFC

The starting point for VMFC is the assumption that we have a set of atoms, \(\mathcal{A}\), which has been partitioned in to \(m\) non-empty, disjoint sets. The partitions are called “fragments” and we define \(\mathcal{S}\) to be the set of fragments. Traditionally, the \(m\) fragments are labeled \(I,J,K,\cdots,m\). We define \(E_{X}\left(Y\right)\) as the energy of system \(X\) computed using the molecular basis set for system \(Y\) where \(X\) and \(Y\) will be subsets of the power set of \(\mathcal{S}\).

Single Fragment Case

If there is a single fragment the total energy of the system is simply the energy of the fragment. Labeling the fragment \(I\), the total energy of the system is \(E_I\left(I\right)\) and the BSSE-free energy of the system is given by:

\[ \begin{align}\begin{aligned}\newcommand{\egy}[2]{E_{#1}\left(#2\right)} \newcommand{\begy}[1]{\epsilon_{#1}}\\\begy{I} = \egy{I}{I}\end{aligned}\end{align} \]

Two Fragment Case

The dimer \(IJ\) has total energy \(E_{IJ}\left(IJ\right)\). In the dimer basis set, monomer \(I\) has energy \(E_{I}\left(IJ\right)\) and monomer \(J\) has energy \(E_{J}\left(IJ\right)\). The BSSE-free two-body interaction energy of dimer \(IJ\), computed using the dimer basis set is then given by:

\[ \begin{align}\begin{aligned}\newcommand{\dbegy}[1]{\Delta\begy{#1}}\\\dbegy{IJ} = \egy{IJ}{IJ} - \egy{I}{IJ} - \egy{J}{IJ}\end{aligned}\end{align} \]

Computing the interaction energy such that the energy of \(I\) is computed in the basis set of \(I\), and similarly for \(J\), yields:

\[ \begin{align}\begin{aligned}\newcommand{\degy}[1]{\Delta E_{#1}}\\\degy{IJ} = \egy{IJ}{IJ} - \begy{I} -\begy{J}\end{aligned}\end{align} \]

The difference in energy yields the BSSE,

\[ \begin{align}\begin{aligned}\newcommand{\bsse}[2]{\theta_{#1}\left(#2\right)} \newcommand{\dbsse}[2]{\Delta\bsse{#1}{#2}}\\\begin{split}\dbsse{IJ}{IJ} \equiv \degy{IJ} - \dbegy{IJ} &= \bsse{I}{IJ} + \bsse{J}{IJ}\\ \bsse{I}{IJ} &\equiv \egy{I}{IJ} - \begy{I}\\ \bsse{J}{IJ} &\equiv \egy{J}{IJ} - \begy{J}\end{split}\end{aligned}\end{align} \]

Trimer Case

The trimer \(IJK\) has total energy \(E_{I,J,K}\left(I,J,K\right)\). The BSSE-free three-body interaction, \(\Delta E_{I,J,K}\left(I,J,K\right)\), is obtained by:

\[\degy{I,J,K}{I,J,K} = \egy{I,J,K}{I,J,K} - \degy{I,J}{I,J,K} - \degy{I,K}{I,J,K} - \degy{J,K}{I,J,K} - \egy{I}{I,J,K} - \egy{J}{I,J,K} -\egy{K}{I,J,K}\]

That is, using the trimer basis set for all terms, we removed from the energy of the trimer, the three two-body interactions, and the energies of the three monomers. The BSSE-free two-body interactions are obtained using the dimer formula, i.e.:

Thus the total interaction energy, \(E^{\text{int}}_{I,J,K}\left(I,J,K\right)\), is: