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Classification of Basis Functions by Excitation Level

Now we will discuss the importance of various excitation classes to the CI wavefunction. As noted in equation (2.9), the CI expansion is typically truncated according to excitation level; in the vast majority of CI studies, the expansion is truncated (for computational tractability) at doubly-excited configurations. Since the Hamiltonian contains only two-body terms, only singles and doubles can interact directly with the reference (for the sake of simplicity, we are assuming only a single reference for now). This is a direct result of Slater's Rules (cf. section 2.4).  The structure of the CI matrix with respect to excitation level is given below (adapted from Szabo and Ostlund [1], p. 235), where $\vert S \rangle, \vert D \rangle, \vert T \rangle,$ and $\vert Q \rangle$ represent blocks of singly, doubly, triply, and quadruply excited determinants, respectively. The Hamiltonian matrix H is Hermitian; if only real orbitals are used, as is usually the case, then the Hamiltonian is also symmetric. Thus only the lower triangle of H is shown below.

\begin{displaymath}{\bf H} =
\begin{array}{c}
\langle \Phi_0 \vert \\
\langl...
...\vdots & \vdots & \vdots & \vdots & \vdots
\end{array}\right]
\end{displaymath} (tex2html_deferredtex2html_deferred4.tex2html_deferred14H =
\begin{array}{c}
\langle \Phi_0 \vert \\
\langle S \vert \\
\langle D \vert \\
\langle T \vert \\
\langle Q \vert \\
\vdots
\end{array}
[
\begin{array}{cccccc}
\langle \Phi_0 \vert H \vert \Phi_0 \rangle & & & & & \cd...
...e & \cdots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots
\end{array}
] )

Note that the matrix elements $\langle S \vert H \vert \Phi_0 \rangle$ are given as 0. This is due to Brillouin's theorem, which is valid when our reference function $\vert \Phi_0 \rangle$ is obtained by the Hartree-Fock method (Hartree-Fock guarantees that off-diagonal elements of the Fock matrix are zero, and it turns out that the matrix element between two Slater determinants which differ by one spin orbital is equal to an off-diagonal element of the Fock matrix). Furthermore, the blocks $\langle X \vert H \vert Y \rangle$ which are not necessarily zero may still be sparse; for example, the matrix element $\langle \Phi_{ab}^{rs} \vert H \vert \Phi_{cdef}^{tuvw} \rangle$, which belongs to the block $\langle D \vert H \vert Q \rangle$, will be nonzero only if a and b are contained in the set $\{ c, d, e, f \}$ and if r and s are contained in the set $\{ t, u, v, w \}$.

Since only the doubles interact directly with the Hartree-Fock reference, we expect double excitations to make the largest contributions to the CI wavefunction, after the reference state. Indeed, this is what is observed. Even though singles, triples, etc. do not interact directly with the reference, they can still become part of the CI wavefunction (i.e. have non-zero coefficients) because they mix with the doubles, directly or indirectly. Although singles are much less important to the energy than doubles, they are generally included in CI treatments because of their relatively small number and because of their greater importance in describing one-electron properties (dipole moment, etc.)


next up previous contents index
Next: Energy Contributions of the Up: Reducing the Size of Previous: Symmetry Restrictions on the
C. David Sherrill
2000-04-18