Brian C. Hoffman: SCF Theory

Basic Principles and
Hartree-Fock Theory

Slater Determinant Wave Functions

The faults, then, of the Hartree product wave function are threefold: it fails to satisfy the Antisymmetry Principle, it distinguishes between electons, and it has a finite probability of electrons occupying the same point in space. One method of constructing a wave function which takes electron indistinguishability into account is to use a linear combination of Hartree products with all possible distributions of the electrons. For example, in a system consisting of two electrons there are two possible Hartee Products

$\begin{displaymath}{\chi}_i(x_1) {\chi}_j(x_2) \; {\rm and} \; {\chi}_i(x_2) {\chi}_j(x_1)\end{displaymath}$

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which can then be added or subtracted to give a wave function in which the electrons cannot be distinguished. However, subtracting the two products leads to a wave function which also satisfies the Antisymmetry Principle. The subtraction of the two Hartree products is equivalent to the following determinant:

$\begin{displaymath}\left\vert\begin{array}{cc}{\chi}_i (x_1) & {\chi}_j (x_1) \\{\chi}_i (x_2) & {\chi}_j (x_2)\end{array}\right\vert .\end{displaymath}$

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Considering that the wave function for a system with two electrons can be efficiently expressed as a determinant, it is logical to consider similar determinants for N-electron systems. Before jumping ahead and writing an arbitrary determinant, though, it is important to notice that the columns of the determinant in the two-electron case have the same spin orbital down each column and identical electron coordinates across each row. Carrying this trend over to a determinant describing an N spin orbital system generates a determinant of the form:

$\begin{displaymath}(N!)^{-\frac{1}{2}} \left\vert\begin{array}{ccc}{\chi}_i ......i}_i (X_N) & \cdots & {\chi}_k (X_N)\end{array} \right\vert.\end{displaymath}$

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The factor of $(N!)^{-\frac{1}{2}}$ is just a normalization constant. The reason it is important to place identical electron coordinates across each row is because exchanging the coordinates of any two electrons is equivalent to swapping two rows of the determinant, which changes the overall sign of the determinant and guarantees that the determinant meets the Antisymmetry Principle. Moreover, the set of spin orbitals must be arranged as they are in order to include all of the many possible Hartree products and, thus, meet the electron indistinguishability requirement. A many electron wave function of this form is called a Slater determinant.[9] Slater determinants are often written in a shorthand notation in which the diagonal elements are placed in a ket:

$\begin{displaymath}\vert \chi_i (x_1) \chi_j(x_2) \cdots \chi_k (x_n) \rangle.\end{displaymath}$

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This notation implies the normalization constant and is often shortened even further by writing just the number of the spin orbitals in the ket.

$\begin{displaymath}\vert i \; j \; \cdots k \rangle .\end{displaymath}$

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In this discussion two of the desired properties of a many electron wave function were met by a Slater determinant, but the third criterion which requires the wave function to predict a vanishing probability for any two electrons to occupy the same point is space is only ``half'' met. If one considers a Slater determinant describing a two-electron system with the electrons in spin orbitals with opposite spins, one finds that there does exist a non-zero probability of the two electrons occupying the same space. However, if one now considers a two-electron Slater determinant in which the spin orbitals have parallel spins, the probability of the electrons occupying the same point in space is indeed zero. Therefore, a ``Fermi hole'' is said to exist around each electron. Because a single Slater determinant does not account for Coulombic repulsions between electrons with opposite spins, it is said to be uncorrelated.

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