The Hartree-Fock Method and Its Consequences

Abstract

We now return to the one-electron approximation of Chapter 4, and our first question is a technical one: How can we determine the best 9, wave functions for calculating the energy (and other properties) of the wave function? We write down the expectation value of the energy, based on Eqs. (4.14)-(4.16):

$$E = \left\langle {\Phi |\hat H\Phi } \right\rangle = \left\langle {\Phi |{{\hat H}_0}\Phi } \right\rangle + \left\langle {\Phi |\sum\limits_{i = 1}^n {{{\hat h}_1}\left( i \right)\Phi } } \right\rangle + \left\langle {\Phi |\mathop {\mathop \Sigma \limits^n \mathop \Sigma \limits^n }\limits_{i < 1} {{\hat h}_2}\left( {ij} \right)\Phi } \right\rangle $$

((6.1))

where Ф is the Slater determinant of the system under study, ĥ ₁(i) is the one-electron operator from (4.15), and ĥ ₂(ij) = (e ²/ r _ij ), the two-electron operator, comes from Eq. (4.16). We know that the determinant is simply the linear combination of product wave functions. Since the ĥ ₁(i) operators affect only one function (the i ^th) from such a product and ĥ ₂(ij) affects only two (the i^th and j ^th), a number of integrals will vanish from (6.1).