Brian C. Hoffman: SCF Theory

Basic Principles and
Hartree-Fock Theory

The Hartree-Fock Equations

Having chosen Slater determinants as trial functions and having worked out the rules needed to evaluate the Average Value Theorem, the next step in a variational method (Table 1.2) is to minimize the expectation value of the Hamiltonian with respect to the variational parameters in the Slater determinant, the set of LCAO-MO coefficients in the spin orbitals, and thereby find the best single Slater determinant description of the molecular system. Collectively, this process is called the Hartree-Fock method. [5,13,14] From case 1 of Slater's rules, one finds that the expectation value of the total energy for a molecule in the state given by the determinant $\vert S \rangle$ is

$\begin{displaymath}f = E_0 = \langle S \vert \hat{H} \vert S \rangle = \sum_{a}^......+ \frac{1}{2} \sum_{ab}^{N} \langle ab \vert \vert ab \rangle.\end{displaymath}$

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In order to use Slater's rules, it is assumed that the set of spin orbitals are orthonormal and that they satisfy

$\begin{displaymath}g = \langle a \vert b \rangle - {\delta_a}_b = 0.\end{displaymath}$

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This constraint, g, together with the energy functional, f, can be used to minimize f using the method of Lagrange's undetermined multipliers. In order to employ this method one combines the constraint and energy functionals by introducing a set of undetermined multipliers, denoted by ${\epsilon_a}_b$ , with the foreknowledge that in the end they will represent orbital energies:

$\begin{displaymath}L = f - \sum_{a}^{N} \sum_{b}^{N} ({\epsilon_a}_b \langle a \vert b \rangle-{\delta_a}_b).\end{displaymath}$

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The next step in Lagrange's method of undetermined multipliers is to set the first variation of L to zero:

$\begin{displaymath}\delta L = 0 = \delta f - \sum_{a}^{N} \sum_{b}^{N}{E_a}_b \delta \langle a\vert b \rangle.\end{displaymath}$

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Upon expansion, followed by simplification in which a and b are replaced by the respective spin orbitals, $\chi_a$ and $\chi_b$ , one eventually arrives at a sum of complex conjugates:

$\displaystyle \delta L$	=	0
	=	$\displaystyle \sum_{a}^{N} \langle \delta \chi_a \vert \hat{h} \vert \chi_a \ra......_{a}^{N} \sum_{b}^{N} \langle ab \vert ab \rangle - \langleab \vert ba \rangle$
		$\displaystyle -\sum_{a}^{N} \sum_{b}^{N} {E_a}_b \langle \delta a \vert b \rangle+ \;\; complex \;\; conjugate.$	(37)

At this juncture it is convenient to introduce two operators, the Coulomb and exchange operators, $\hat{J}$ and $\hat{K}$ , which will help reduce the equations:

$\displaystyle \hat{J}$	=	$\displaystyle \int d{x_2} {\mid \chi_b (X_2) \mid}^2 \frac{1}{{r_1}_2}$	(38)
$\displaystyle \hat{K}$	=	$\displaystyle \int d{x_2} {\chi_b}^{*} (X_2) \frac{1}{{r_1}_2} \chi_b(X_2).$	(39)

These two particular operators are chosen because they have the following expectation values:

$\displaystyle \langle \chi_a \vert \hat{J} \vert \chi_b \rangle$	=	$\displaystyle \langle ab \vert ab \rangle$	(40)
$\displaystyle {\rm and} \; \langle \chi_a \vert \hat{K} \vert \chi_b \rangle$	=	$\displaystyle \langle ab \vert ba \rangle,$	(41)

allowing the minimization condition to be rewritten as

$\displaystyle \delta L$	=	$\displaystyle \sum_{a}^{N} \int dX_1 \delta \chi_a^* (1) \left[ h(1)\chi_a (1)......hat{J}_b(1) - \hat{K}_b(1))\chi_a (1) - \sum_{b}^{N} {E_a}_b \chi_b(1) \right]$
		$\displaystyle + \; \; complex \;\; conjugate.$	(42)

Moreover, the quantity $\delta \chi_a^*(1)$ is an arbitrary variation in $\chi_a^*(1)$ , and therefore, if the right side of the above equation is to equal zero for all possible cases, the quantity in brackets must also equal zero:

$\begin{displaymath}\left[ h(1) \chi_a (1) + \sum_{b}^{N} (\hat{J}_b(1) - \hat{K}_b(1))\chi_a(1) - \sum_{b}^{N} {E_a}_b \chi_b(1) \right] = 0 .\end{displaymath}$

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Moving the term involving $\chi_b$ to the other side of the equal sign yields

$\begin{displaymath}\left[ h(1) + \sum_{b}^{N} (\hat{J}_b(1) - \hat{K}_b(1)) \right] \chi_a(1)= \sum_{b}^{N} {E_a}_b \chi_b(1).\end{displaymath}$

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The quantity in brackets on the left-hand side of the equal sign is known as the Fock operator, denoted by $\hat{f}$ . Thus, the minimization condition is reduced to the convenient operator equation

$\begin{displaymath}\hat{f} \vert \chi_a \rangle = \sum_{b}^{N} {E_a}_b \vert \chi_b \rangle.\end{displaymath}$

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The minimization condition given above is not in the standard or canonical form of a eigenvalue equation. In quantum mechanics, when an operator acts upon a state function that is compatible with it, one fully expects to get an eigenvalue equation. In the case of the Fock operator, one can expect the Slater determinant and the Fock operator to be compatible given the method of their construction; hence, an eigenvalue equation should result. The minimization criteria takes a non-canonical form because any single Slater determinant wave function $\vert \Psi_0 \rangle$ formed from a set of spin orbitals $\chi_a$ retains a certain degree of flexibility in the spin orbitals: the spin orbitals can be mixed among themselves without changing the expectation value $E_0 = \langle \Psi_0 \vert \hat{h} \vert \Psi_0 \rangle$ . Fortunately, it is possible to cast the above minimization criterion into an eigenvalue equation by transforming the spin orbitals into a set of canonical spin orbitals through a unitary transformation of the general form

$\begin{displaymath}U \{ \chi_a \} U^\dagger\end{displaymath}$

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where U is a unitary matrix and $U^\dagger$ is its inverse. The unitary matrix U is chosen such that the matrix of undetermined multipliers Ewith entries E_ab is diagonalized after the unitary transformation

$\begin{displaymath}U E U^\dagger.\end{displaymath}$

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The exact mathematical details behind the transformation are not important; what is important is that upon completion of some mathematically allowable unitary transformation of the form described, the minimization criterion is finally cast into the standard form known as the canonical Hartree-Fock equation:

$\begin{displaymath}\hat{f} \vert \chi_a \rangle = E_a \vert \chi_a \rangle.\end{displaymath}$

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