Slater determinantal

The normalisation factor is assumed. It is often convenient to indicate the spin of each electron in the determinant this is done by writing a bar when the spin part is P (spin down) a function without a bar indicates an a spin (spin up). Thus, the following are all commonly used ways to write the Slater determinantal wavefunction for the beryllium atom (which has the electronic configuration ls 2s ) ... 

VVc can now see why the normalisation factor of the Slater determinantal wavefunction is I v/N . If each determinant contains N terms then the product of two Slater determinants, ldeU rminant][determinant], contains (N ) terms. However, if the spin orbitals form an oi lhonormal set then oidy products of identical terms from the determinant will be nonzero when integrated over all space. We Ccm illustrate this with the three-electron example, k ljiiiidering just the first two terms in the expansion we obtain the following ... 

In short, the Slater determinantal moleculai orbital and only the Slater determinantal moleculai orbital satisfies the two great generalizations of quantum chemistry, uncertainty (indistinguishability) and fermion exchange antisymmetry. 

Unfortunately, the A-representability constraints from the orbital representation are not readily generalized to the spatial representation. A first clue that the A-representability problem is more complicated for the spatial basis is that while every A-representable Q-density can be written as a weighted average of Slater determinantal Q-densities in the orbital resolution (cf. Eq. (54)), this is clearly not true in the spatially resolved formulation. For example, the pair density (Q = 2) of any real electronic system will have a cusp where electrons of opposite spin coincide but a weighted average of Slater determinantal pair densities,... 

The Kohn-Sham scheme then provides a mapping from the true interacting system to a Slater determinantal approximation. 

The variational procedure in Eq. (100) is in the spirit of the Kohn-Sham ansatz. Since satisfies the (g, K) conditions, it is A-representable. In general, Pij ig corresponds to many different A-electron ensembles and one of them,, corresponds to the ground state of interest. However, for computational expediency in computing the energy, a Slater determinantal density matrix,... 

Note that this property is a correlation effect unique to electrons of the same spin. If we consider the contrasting Slater determinantal wave function formed from different spins... 

If we carry out a restricted HF calculation, one or other of the degenerate frontier pair will be chosen to be occupied, the calculation will optimize the shapes of all of the occupied orbitals, and we will end up with a best possible single-Slater-determinantal wave function formed from those MOs. But it should be fairly obvious that an equally good wave function... 

The wavefunction F° then follows as an antisymmetrized product built from the single-particle functions q>i(r, ms) for the Z electrons (Slater determinantal wave-function, see below and Section 7.2), where r is the spatial vector and ms the spin magnetic quantum number. 

Due to the properties of determinants, a Slater determinantal wavefunction ° automatically fulfils the Pauli principle and takes care of the antisymmetric character of fermions. If written explicitly in terms of the single-particle orbitals,... 

For example, the ground state f 0 of the magnesium atom for which 12 electrons must be placed in the spin-orbitals lsOm% 2sOm% 2prn"s,3sOms is represented by the Slater determinantal wavefunction... 

In the independent particle picture, the ground state of helium is given by Is2 xSo. For this two-electron system it is always possible to write the Slater determinantal wavefunction as a product of space- and spin-functions with certain symmetries. In the present case of a singlet state, the spin function has to be... 

Table 3.1. Slater determinantal wavefunctions for the 2p2-electron configuration (see Sla60J).

In order to calculate the matrix elements with the Coulomb operator Vc, one again uses Slater determinantal wavefunctions, for the intermediate state xp(Mp, t) as well as for the complete final state which contains the doubly charged ion, f, and the two ejected electrons, x<, (Ka, Kb). Assuming that there is no correlation between the two escaping electrons and that their common boundary condition applies separately to each single-particle function, the directional emission property is included in the factors f( ka) and f( kb), and one gets for this Coulomb matrix element C... 

The essence of the hole-particle formalism lies in a new meaning given the vacuum state. Consider now the closed shell ground state Slater determinante I d>0 > expressed by means of Eq. (25) as... 

Methods based upon Slater determinantal functions (SDF). When we take this approach, we are, in effect, applying the antisymmetrization requirement first. Only if the orbitals are all doubly occupied among the spin orbitals is the SDF automatically, at the outset, an eigenfunction of the total spin. In all other cases further manipulations are necessary to obtain an eigenfunction of the spin, and these are written as sums of SDFs. 

RMS root-mean-square SCVB spin coupled valence bond SDF Slater determinantal functions STF standard tableau function VB valence bond... 

It is well-known that the electron repulsion perturbation gives rise to LS terms or multiplets (also known as Russell-Saunders terms) which in turn are split into LSJ spin-orbital levels by spin-orbit interaction. These spin-orbital levels are further split into what are known as Stark levels by the crystalline field. The energies of the terms, the spin-orbital levels and the crystalline field levels can be calculated by one of two methods, (1) the Slater determinantal method [310-313], (2) the Racah tensor operator method [314-316]. 

Slater determinantal method Part of the microstate table for f2 is given in Table 8.32. 

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