The Hartree-Fock equations

The full Hartree-Fock equations are given by

$$H_{HF} \psi_k = \epsilon_k \psi_k,$$ (97)

with

$$H_{HF} \psi_k = \left[-\frac{1}{2}\nabla^2-\sum_n \frac{Z_n}{\ver... ...rt\psi_l({\bf x}')\vert^2\frac{1}{\vert{\bf r} - {\bf r}'\vert} \psi_k({\bf x})$$ (98)

$$- \sum_{l=1}^N \int dx' \psi_l^*({\bf x}')\frac{1}{\vert{\bf r} - {\bf r}'\vert} \psi_k({\bf x}')\psi_l({\bf x}).$$ (99)

Notice that the wavefunctions deppend on the generalize coordinate ${\bf x}$, which includes the orbital and spin parts. The right hand side of the equations consists of four terms. The first and second give rise are the kinetic energy contribution and the electron-ion potential. The third term, or Hartree term, is the simply electrostatic potential arising from the charge distribution of $N$ electrons. As written, the term includes an unphysical self-interaction of electrons when $l=k$. This term is cancelled in the fourth, or exchange term. The exchange term results from our inclusion of the Pauli principle and the assumed determinantal form of the wavefunction. Notice that thsi term is non-local, its value at ${\bf r}$ is determined by the value assumed by $\psi_k$ at all possible positions ${\bf r}'$.

We can rewrite $H_{HF}$ as the sum of different terms:

$$H_{HF} = h +J - K,$$ (100)

with

$$h = \sum_i \left[-\frac{1}{2}\nabla_i^2-\sum_n \frac{Z_n}{\vert{\bf r_i}-{\bf R_n}\vert}\right]$$ (101)

$$J({\bf x})\psi_k({\bf x}) = \sum_{l=1}^N \int dx' \vert\psi_l({\bf x}')\vert^2\frac{1}{\vert{\bf r} - {\bf r}'\vert}\psi_k({\bf x})$$ (102)

$$K({\bf x})\psi_k({\bf x}) = \sum_{l=1}^N \int dx' \psi_l^*({\bf x}')\psi_k({\bf x}')\frac{1}{\vert{\bf r} - {\bf r}'\vert} \psi_l({\bf x}).$$ (103)

We can eliminate the sum over the spin indices by summing over them, to find the operators acting on the orbital part, only. The $h$ operators remains the same since it does not contain any spin dependence. The new operators $J$ and $K$ acting on the orbital parts then read:

$$J({\bf r})\phi_k({\bf r}) = 2 \sum_{l=1}^N \int d^3r' \vert\phi_l({\bf r}')\vert^2\frac{1}{\vert{\bf r} - {\bf r}'\vert} \phi_k({\bf r})$$ (104)

$$K({\bf r})\phi_k({\bf r}) = \sum_{l=1}^N \int d^3r' \phi_l^*({\bf r}')\phi_k({\bf r'})\frac{1}{\vert{\bf r} - {\bf r}'\vert} \phi_l({\bf r}).$$ (105)

Multiplying by $\phi_k^*({\bf r})$ and integrating over ${\bf r}$ we obtain the expression for the energy:

$$E = 2 \sum_k \langle\phi_k\vert h\vert\phi_k\rangle + \sum_k \langle \phi_k \vert 2J - K\vert\phi_k \rangle$$ (106)

where the factors 2 arise from the sum over the spin indices.