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Next: Slater Determinants Up: Introduction to Hartree-Fock Previous: What Problem Are We

Motivation and the Hartree Product

The basic idea of Hartree-Fock 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 ${\hat V}_{ee} = 0$). If that was true, then the Hamiltonian would be separable, and the total electronic wavefunction $\Psi({\mathbf r}_1, {\mathbf r}_2)$ describing the motions of the two electrons would just be the product of two hydrogen atom wavefunctions (orbitals), $\Psi_H({\mathbf r}_1) \Psi_H({\mathbf r}_2)$ (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

\Psi_{HP}({\mathbf r}_1, {\mathbf r}_2, \cdots, {\mathbf r}_...
...thbf r}_1) \phi_2({\mathbf r}_2) \cdots \phi_N({\mathbf r}_N),
\end{displaymath} (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 space-spin coordinates. By space-spin coordinates, we mean that fermions have not only three spatial degrees of freedom, but also an intrinsic spin coordinate, which we will call $\alpha$ or $\beta$. We call a generic (either $\alpha$ or $\beta$) spin coordinate $\omega$, and the set of space-spin coordinates ${\mathbf x} = \{ {\mathbf r}, \omega \}$. We will also change our notation for orbitals from $\phi({\bf r})$, a spatial orbital, to $\chi({\bf x})$, a spin orbital. Except in strange cases such as the so-called General Hartree Fock or Z-Averaged Perturbation Theory, usually the spin orbital is just the product of a spatial orbital and either the $\alpha$ or $\beta$ spin function, i.e., $\chi({\bf x}) =
\phi({\bf r}) \alpha$. [Note: some textbooks write the spin function formally as a function of $\omega$, i.e., $\alpha(\omega)$].

More properly, then, with the full set of coordinates, the Hartree Product becomes

\Psi_{HP}({\mathbf x}_1, {\mathbf x}_2, \cdots, {\mathbf x}_...
...thbf x}_1) \chi_2({\mathbf x}_2) \cdots \chi_N({\mathbf x}_N).
\end{displaymath} (4)

This wavefunction does not satisfy the antisymmetry principle! To see why, consider the case for only two electrons:
\Psi_{HP}({\mathbf x}_1, {\mathbf x}_2) =
\chi_1({\mathbf x}_1) \chi_2({\mathbf x}_2).
\end{displaymath} (5)

What happens when we swap the coordinates of electron 1 with those of electron 2?
\Psi_{HP}({\mathbf x}_2, {\mathbf x}_1) =
\chi_1({\mathbf x}_2) \chi_2({\mathbf x}_1).
\end{displaymath} (6)

The only way that we get the negative of the original wavefunction is if
\chi_1({\mathbf x}_2) \chi_2({\mathbf x}_1) =
- \chi_1({\mathbf x}_1) \chi_2({\mathbf x}_2),
\end{displaymath} (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 up previous
Next: Slater Determinants Up: Introduction to Hartree-Fock Previous: What Problem Are We
David Sherrill 2002-05-30