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The basic idea of HartreeFock theory is as follows. We know how to solve
the electronic problem for the simplest atom, hydrogen, which has
only one electron. We imagine that perhaps if we added another
electron to hydrogen, to obtain H, then maybe it might be reasonable
to start off pretending that the electrons don't interact with each
other (i.e., that
). If that was true, then the
Hamiltonian would be separable, and the total electronic wavefunction
describing the motions of the two
electrons would just be the product of two hydrogen atom wavefunctions
(orbitals),
(you should
be able to prove this easily).
Obviously, pretending that the electrons ignore each other is a pretty serious
approximation! Nevertheless, we have to start somewhere, and it seems
plausible that it might be useful to start with a wavefunction of the
general form

(3) 
which is known as a Hartree Product.
While this functional form is fairly convenient, it has at least
one major shortcoming: it fails to satisfy the antisymmetry principle,
which states that a wavefunction describing fermions should be
antisymmetric with respect to the interchange of any set of spacespin
coordinates. By spacespin coordinates, we mean that fermions have not
only three spatial degrees of freedom, but also an intrinsic spin
coordinate, which we will call or . We call a generic
(either or ) spin coordinate , and the set of
spacespin coordinates
.
We will also change our notation for orbitals from ,
a spatial orbital, to , a spin orbital.
Except in strange cases such as the socalled General Hartree
Fock or ZAveraged Perturbation Theory, usually the spin orbital
is just the product of a spatial orbital and either
the or spin function, i.e.,
. [Note: some textbooks write the spin
function formally as a function of , i.e.,
].
More properly, then, with the full set of coordinates, the Hartree Product
becomes

(4) 
This wavefunction does not satisfy the antisymmetry principle! To
see why, consider the case for only two electrons:

(5) 
What happens when we swap the coordinates of electron 1 with those of
electron 2?

(6) 
The only way that we get the negative of the original wavefunction is if

(7) 
which will not be true in general! So we can see the Hartree Product is
actually very far from having the properties we require.
Next: Slater Determinants
Up: Introduction to HartreeFock
Previous: What Problem Are We
David Sherrill
20020530