For the sake of practical simplicity (the use of Slater rules and a two-slope space geometry [2]) one may employ the larger, just antisymmetric but non-spin-adapted space spanned by ( ) Slater determinants. [Pg.73]

To solve Eq. (7.11), we need to know how to evaluate matrix elements of the type defined by Eq. (7.12). To simplify matters, we may note that the Hamiltonian operator is composed only of one- and two-electron operators. Thus, if two CSFs differ in their occupied orbitals by 3 or more orbitals, every possible integral over electronic coordinates hiding in the r.h.s. of Eq. (7.12) will include a simple overlap between at least one pair of different, and hence orthogonal, HF orbitals, and the matrix element will necessarily be zero. For the remaining cases of CSFs differing by two, one, and zero orbitals, the so-called Condon-Slater rules, which can be found in most quantum chemistry textbooks, detail how to evaluate Eq. (7.12) in terms of integrals over the one- and two-electron operators in the Hamiltonian and the HF MOs. [Pg.212]

A somewhat special case is the matrix element between the HF determinant and a singly excited CSF. The Condon-Slater rules applied to this situation dictate that... [Pg.212]

matrix elements r o are quite straightforward to evaluate. Before leaving them, however, it is worthwhile to make some qualitative observations about them. First, the Condon-Slater rules dictate that for the one-electron operator r, the only matrix elements that survive are those between determinants differing by at most two electronic orbitals. Thus, only absorptions generating singly or doubly excited states are allowed. [Pg.510]

Condon-Slater rules, 212-213, 221, 510 Configuration interaction, see also CIS),... [Pg.583]

If we transform the MO s such that condition (5 11) is fulfilled, the resulting transition density matrix will be obtained in a mixed basis, and can subsequently be transformed to any preferred basis The generators Epq of course have to be redefined in terms of the bi-orthonormal basis, but this is a technical detail which we do not have to worry about as long as we understand the relation between (5 9) and the Slater rules. How can a transformation to a bi-orthonormal basis be carried out We assume that the two sets of MO s are expanded in the same AO basis set. We also assume that the two CASSCF wave functions have been obtained with the same number of inactive and active orbitals, that is, the same configurational space is used. Let us call the two matrices that transform the original non-orthonormal MO s

Now, by its very definition, the global hardness rj is a measure of the HOMO-LUMO gap of a compound. Consequently, it seems reasonable to assume that the AEE AE should scale with the global hardness. The <(ag/r)3> term describes the p-orbital expansion or contraction as electrons are added or removed from the shielded nucleus. Using effective atomic numbers Znp and Slater rules ... [Pg.292]

The original Slater rules for selecting the values of C conform to the idea of being related to the atomic ionization potentials (see above). However other schemes are also in use. [Pg.43]

Assume that we have computed CASSCF wave function for two different electronic states. Now we want to compute transition properties, for example, the transition dipole moment. How can we do that. The two states will in general be described by two non-orthonormal sets of MOs, so the normal Slater rules cannot be applied. Let us start by considering the case where two electronic states fi and v are described by the same set of MOs. The transition matrix element for a one-electron operator A is then given by the simple expression ... [Pg.140]

The second approach is to truncate the expansion at some level of excitation. By Brillouin s theorem, the single excited configurations will not mix with the HF reference. By the Condon-Slater rules, this leaves the doubles configurations as the most important for including in the Cl expansion. Thus, the smallest reasonable truncated Cl wavefunction includes the reference and all doubles configurations (CID) ... [Pg.15]

Minimum STO basis with Slater rules expo-. nents,... [Pg.30]

For an explicit calculation of the energy order. Slater rules yield the Cl matrix elements... [Pg.234]

Problem 5.9 Using the Slater rules, calculate Z for the 4s orbital for a valence configuration Sd, 48, and conclude that a 4s—>3d transition leads to a decrease of Z for both 4s and 3d. [Pg.104]

In Cizek s original paper [24], which derives from an even earlier dissertation, he reports results for semi-empirical Hamiltonians like PPP, but also even a partly ab initio result for Ni- However, his use of second-quantized based, diagrammatic techniques to derive the CC equations was unfamiliar to most quantum chemists (see [34]), likely delaying the appreciation of the CC method, although for simple cases like CCD, conventional Slater rule matrix evaluation can be applied [35] Also, explicit mles for diagrams were given and could have been used to derive more complicated CC equations. [Pg.1196]

A consequence of the antisymmetry property of the determinantal wave functions is that the matrix elements can be evaluated according to the Slater rules. Before the application of the Slater rules the determinantal functions should be ordered in such a way that maximum coincidence among the positions of the spinorbitals is secured. Notice that each transposition in the position of spinorbitals alters the sign of the determinantal function. [Pg.19]

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