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Slater Determinants

An electronic wavefunction for $N$ particles must be a function of $4N$ coordinates: for each electron, we have $x$, $y$, and $z$ Cartesian coordinates plus a spin coordinate (sometimes designated $\omega$, which can have values $\alpha$ and $beta$). The Cartesian coordinates for electron $i$ are usually denoted by a collective index ${\bf r}_i$, and the set of Cartesian plus spin coordinates is often denoted ${\bf x}_i$.

What is an appropriate form for an $N$-electron wavefunction? The simplest solution would be a product of one-particle functions (``orbitals''):

\begin{displaymath}
\Psi({\mathbf x}_1, {\mathbf x}_2, \cdots, {\mathbf x}_N) = ...
...thbf x}_1) \chi_2({\mathbf x}_2) \cdots \chi_N({\mathbf x}_N).
\end{displaymath} (28)

This is referred to as a Hartree Product. Since the orbitals $\chi_i({\mathbf x}_i)$ depend on spatial and spin coordinates, they are called spin orbitals. These spin orbitals are simply a spatial orbital times a spin function, i.e., $\chi_i({\mathbf x}_i) =
\phi_i({\mathbf r}_i) \vert \alpha \rangle $ or $\chi_i({\mathbf x}_i) =
\phi_i({\mathbf r}_i) \vert \beta \rangle $.

Unfortunately, the Hartree product is not a suitable wavefunction because it ignores the antisymmetry principle (quantum mechanics postulate #6). Since electrons are fermions, the electronic wavefunction must be antisymmetric with respect to the interchange of coordinates of any pair of electrons. This is not the case for the Hartree Product.

If we simplify for a moment to the case of two electrons, we can see how to make the wavefunction antisymmetric:

\begin{displaymath}
\Psi({\mathbf x}_1, {\mathbf x}_2) = \frac{1}{\sqrt{2}}
\le...
...x}_2)
- \chi_1({\mathbf x}_2) \chi_1({\mathbf x}_1) \right].
\end{displaymath} (29)

The factor $1/sqrt{2}$ is just to make the wavefunction normalized (we're assuming our individual orbitals are orthonormal). The expression above can be rewritten as a determinant as
$\displaystyle \Psi({\mathbf x}_1, {\mathbf x}_2)$ $\textstyle =$ $\displaystyle \frac{1}{\sqrt{2}}
\left\vert
\begin{array}{cc} \chi_1({\bf x}_1)...
...i_2({\bf x}_1) \\
\chi_1({\bf x}_2) & \chi_2({\bf x}_2)
\end{array}\right\vert$ (30)

Note a nice feature of this; if we try to put two electrons in the same orbital at the same time (i.e., set $\chi_1 = \chi_2$), then $\Psi({\mathbf x}_1, {\mathbf x}_2) = 0$. This is just a more sophisticated statement of the Pauli exclusion principle, which is a consequence of the antisymmetry principle!

This strategy can be generalized to $N$ electrons using determinants.

\begin{displaymath}
\Psi = \frac{1}{\sqrt{N!}}
\left\vert \begin{array}{cccc}
...
...f x}_N) & \cdots & \chi_N({\bf x}_N)
\end{array} \right\vert.
\end{displaymath} (31)

A determinant of spin orbitals is called a Slater determinant after John Slater. By expanding the determinant, we obtain $N!$ Hartree Products, each with a different sign; the electrons $N$ electrons are arranged in all $N!$ possible ways among the $N$ spin orbitals. This ensures that the electrons are indistinguishable as required by the antisymmetry principle.

Since we can always construct a determinant (within a sign) if we just know the list of the occupied orbitals $\{ \chi_i({\bf x}),
\chi_j({\bf x}), \cdots \chi_k({\bf x}) \}$, we can write it in shorthand in a ket symbol as $\vert \chi_i \chi_j \cdots \chi_k \rangle $ or even more simply as $\vert i j \cdots k \rangle $. Note that we have dropped the normalization factor. It's still there, but now it's just implied!

How do we get the orbitals which make up the Slater determinant? This is the role of Hartree-Fock theory, which shows how to use the Variational Theorem to use those orbitals which minimize the total electronic energy. Typically, the spatial orbitals are expanded as a linear combination of contracted Gaussian-type functions centered on the various atoms (the linear combination of atomic orbitals molecular orbital or LCAO method). This allows one to transform the integro-differential equations of Hartree-Fock theory into linear algebra equations by the so-called Hartree-Fock-Roothan procedure.

How could the wavefunction be made more flexible? There are two ways: (1) use a larger atomic orbital basis set, so that even better molecular orbitals can be obtained; (2) write the wavefunction as a linear combination of different Slater determinants with different orbitals. The latter approach is used in the post-Hartree-Fock electron correlation methods such as configuration interaction, many-body perturbation theory, and the coupled-cluster method.


next up previous
Next: Simplified Notation for the Up: intro_estruc Previous: Dirac Notation
David Sherrill 2003-08-07