For the simplest form of Happrox. with no electron-electron
interaction, the exact eigenfunction will be a product function
where the subscripts 1, 2, 3, ..., N specify which unique quantum state the single-electron wave function represents. Any permutation of the single-electron functions, which we can call spin-orbitals, will also lead to an eigenfunction of Happrox..
In the full Hartree method, electron-electron interactions are included in the
Hamiltonian by an additional potential energy term. This extra term is
like a perturbation of the simple solution for the non-interacting electron
situation. This more complete Hamiltonian is also the starting point for the
The Hartree-Fock method introduces full quantum mechanical considerations into
the analysis of the many-electron problem. Symmetry constraints are applied to
the eigenfunction. As electrons are fermions, the total wave function must be
antisymmetric with respect to the interchange of any two of the electrons. This
is equivalent to claiming the electrons must satisfy the Pauli exclusion
principle. We can form an eigenfunction with this property by antisymmetrizing
the product function above. This can be represented in matrix notation as
and this is the Slater Determinant representation of the total wave function. If you consider matrix properties, the exclusion principle can be seen to be enforced as the determinant will be zero if any orbital i = orbital j.
As a simple example consider a helium atom with its two electrons in their
ground state configuration. The antisymmetrized wave function will then be
where the generic spin-orbital subscripts have been replaced with the definite quantum levels of the electrons. Note, in principle the set of all possible Slater determinants of order N forms a complete set for describing a system. For the helium case, we could also have contributions from the singly excited states with 1s-3s or 2s-3s; with diminishing contribution from more highly excited situations.
If we want the total wave function to consist of one Slater determinant,
how do we choose the "best" one? The expectation value of the total
energy for a state represented by a Slater determinant is given by the
expectation value of the total Hamiltonian. The "best" determinant
for the ground state should be the one which minimizes the expectation
value of H; shown here in detail (and dropping the subscript
We minimize the energy with respect to changes in the Slater determinant
eigenfunction through a variational derivative with respect to the spin-orbitals
using a lagrange multiplier which turns out to be the energy eigenvalue for the
spin-orbital. This leads to the Hartree-Fock equations.
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Created: November 12, 1998 --- Last Updated: November 28, 2001
By Mark D. Pauli