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Next: About this document ... Up: Introduction to Hartree-Fock Previous: Energy Expression

The Hartree-Fock Equations

Again, the Hartree-Fock method seeks to approximately solve the electronic Schrödinger equation, and it assumes that the wavefunction can be approximated by a single Slater determinant made up of one spin orbital per electron. Since the energy expression is symmetric, the variational theorem holds, and so we know that the Slater determinant with the lowest energy is as close as we can get to the true wavefunction for the assumed functional form of a single Slater determinant. The Hartree-Fock method determines the set of spin orbitals which minimize the energy and give us this ``best single determinant.''

So, we need to minimize the Hartree-Fock energy expression with respect to changes in the orbitals $\chi_i \rightarrow \chi_i + \delta \chi_i$. We have also been assuming that the orbitals $\chi$ are orthonormal, and we want to ensure that our variational procedure leaves them orthonormal. We can accomplish this by Lagrange's method of undetermined multipliers, where we employ a functional ${\cal L}$ defined as

{\cal L}[ \{ \chi_i \} ] = E_{HF} [ \{ \chi_i \} ]
- \sum_{ij} \epsilon_{ij} ( <i\vert j> - \delta_{ij} )
\end{displaymath} (18)

where $\epsilon_{ij}$ are the undetermined Lagrange multipliers and $<i\vert j>$ is the overlap between spin orbitals $i$ and $j$, i.e.,
<i\vert j> = \int \chi_i^*({\mathbf x}) \chi_j({\mathbf x}) d{\bf x}.
\end{displaymath} (19)

Setting the first variation $\delta {\cal L} = 0$, and working through some algebra, we eventually arrive at the Hartree-Fock equations defining the orbitals:

h({\mathbf x}_1) \chi_i({\mathbf x}_1)
+ \sum_{j \neq i} \...
...ght] \chi_j({\mathbf x}_1) = \epsilon_i \chi_i({\mathbf x}_1),
\end{displaymath} (20)

where $\epsilon_i$ is the energy eigenvalue associated with orbital $\chi_i$.

The Hartree-Fock equations can be solved numerically (exact Hartree-Fock), or they can be solved in the space spanned by a set of basis functions (Hartree-Fock-Roothan equations). In either case, note that the solutions depend on the orbitals. Hence, we need to guess some initial orbitals and then refine our guesses iteratively. For this reason, Hartree-Fock is called a self-consistent-field (SCF) approach.

The first term above in square brackets,

\sum_{j \neq i} \left[ \int d{\mathbf x}_2 \vert \chi_j({\mathbf x}_2) \vert^2
r_{12}^{-1} \right] \chi_i({\mathbf x}_1),
\end{displaymath} (21)

gives the Coulomb interaction of an electron in spin orbital $\chi_i$ with the average charge distribution of the other electrons. Here we see in what sense Hartree-Fock is a ``mean field'' theory. This is called the Coulomb term, and it is convenient to define a Coulomb operator as
{\cal J}_j({\bf x}_1) = \int d{\mathbf x}_2 \vert \chi_j({\mathbf x}_2) \vert^2
\end{displaymath} (22)

which gives the average local potential at point ${\mathbf x}_1$ due to the charge distribution from the electron in orbital $\chi_j$.

The other term in brackets in eq. (20) is harder to explain and does not have a simple classical analog. It arises from the antisymmetry requirement of the wavefunction. It looks much like the Coulomb term, except that it switches or exchanges spin orbitals $\chi_i$ and $\chi_j$. Hence, it is called the exchange term:

\sum_{j \neq i} \left[ \int d{\mathbf x}_2 \chi_j^*({\mathb...
...i_i({\mathbf x}_2) r_{12}^{-1}
\right] \chi_j({\mathbf x}_1).
\end{displaymath} (23)

We can define an exchange operator in terms of its action on an arbitrary spin orbital $\chi_i$:
{\cal K}_j({\mathbf x}_1) \chi_i({\mathbf x}_1) =
\left[ \...
\chi_i({\mathbf x}_2) \right] \chi_j({\mathbf x}_1).
\end{displaymath} (24)

In terms of these Coulomb and exchange operators, the Hartree-Fock equations become considerably more compact.

\left[ h({\mathbf x}_1) + \sum_{j \neq i} {\cal J}_j({\mathb...] \chi_i({\mathbf x}_1)
= \epsilon_i \chi_i({\mathbf x}_1).
\end{displaymath} (25)

Perhaps now it is more clear that the Hartree-Fock equations are eigenvalue equations. If we realize that
\left[ {\cal J}_i({\mathbf x}_1) - {\cal K}_i({\mathbf x}_1) \right]
\chi_i({\mathbf x}_1) = 0,
\end{displaymath} (26)

then it becomes clear that we can remove the restrictions $j \neq i$ in the summations, and we can introduce a new operator, the Fock operator, as
f({\mathbf x}_1) = h({\mathbf x}_1) + \sum_{j} {\cal J}_j({\mathbf x}_1)
- {\cal K}_j({\mathbf x}_1).
\end{displaymath} (27)

And now the Hartree-Fock equations are just
f({\mathbf x}_1) \chi_i({\mathbf x}_1) = \epsilon_i \chi_i({\mathbf x}_1).
\end{displaymath} (28)

Introducing a basis set transforms the Hartree-Fock equations into the Roothaan equations. Denoting the atomic orbital basis functions as ${\tilde \chi}$, we have the expansion

\chi_i = \sum_{\mu=1}^K C_{\mu i} {\tilde \chi}_{\mu}
\end{displaymath} (29)

for each spin orbital $i$. This leads to
f({\mathbf x}_1) \sum_{\nu} C_{\nu i} {\tilde \chi}_{\nu}({\...
...lon_i \sum_{\nu} C_{\nu i} {\tilde \chi}_{\nu}({\mathbf x}_1).
\end{displaymath} (30)

Left multiplying by ${\tilde \chi}_{\mu}^*({\mathbf x}_1)$ and integrating yields a matrix equation
\sum_{\nu} C_{\nu i}
\int d{\mathbf x}_1 {\tilde \chi}_{\mu...
...i}_{\mu}^*({\mathbf x}_1)
{\tilde \chi}_{\nu}({\mathbf x}_1).
\end{displaymath} (31)

This can be simplified by introducing the matrix element notation
$\displaystyle S_{\mu \nu}$ $\textstyle =$ $\displaystyle \int d{\mathbf x}_1 {\tilde \chi}_{\mu}^*({\mathbf x}_1)
{\tilde \chi}_{\nu}({\mathbf x}_1),$ (32)
$\displaystyle F_{\mu \nu}$ $\textstyle =$ $\displaystyle \int d{\mathbf x}_1 {\tilde \chi}_{\mu}^*({\mathbf x}_1)
f({\mathbf x}_1)
{\tilde \chi}_{\nu}({\mathbf x}_1).$ (33)

Now the Hartree-Fock-Roothaan equations can be written in matrix form as
\sum_{\nu} F_{\mu \nu} C_{\nu i} = \epsilon_i \sum_{\nu} S_{\mu \nu}
C_{\nu i}
\end{displaymath} (34)

or even more simply as matrices
{\mathbf F}{\mathbf C} = {\mathbf S} {\mathbf C}{\bf\epsilon}
\end{displaymath} (35)

where ${\bf\epsilon}$ is a diagonal matrix of the orbital energies $\epsilon_i$. This is like an eigenvalue equation except for the overlap matrix ${\mathbf S}$. One performs a transformation of basis to go to an orthogonal basis to make ${\mathbf S}$ vanish. Then it's just a matter of solving an eigenvalue equation (or, equivalently, diagonalizing ${\mathbf F}$!). Well, not quite. Since ${\mathbf F}$ depends on it's own solution (through the orbitals), the process must be done iteratively. This is why the solution of the Hartree-Fock-Roothaan equations are often called the self-consistent-field procedure.

next up previous
Next: About this document ... Up: Introduction to Hartree-Fock Previous: Energy Expression
David Sherrill 2002-05-30