Hartree-Fock Theory

The lack of exchange in Hartree theory is overcome using Hartree-Fock theory, where the many body electronic wavefunction is rewritten [14]. Instead of using a simple product wavefunction, (Equation 2.1.5 above), a Slater determinant is used,

$$\Psi({\bf r}_1 s_1, {\bf r}_2 s_2, \ldots ,{\bf r}_N s_N) =\... ...\ \ .\ \ .\ \ & \psi_N({\bf r}_N s_N) \end{array} \right\vert.$$

Incorporating electron spin through a spin function, $\chi_\alpha(s)$,this can be rewritten as

$$\Psi(r) = \frac{1}{\sqrt{N!}} {\rm det} \vert\psi_\lambda(r_\mu)\vert$$

$$\psi_\lambda(r) = \psi_i({\bf r}) \chi_\alpha(s)\\$$

$$\sum_s{\chi^*_\alpha(s)\chi_\beta(s)} = \delta_{\alpha\beta}.$$

We can define the electron density in terms of these orbitals,

$$n({\bf r}) = \sum_{\lambda s}\vert\psi_\lambda({\bf r},s)\vert^2.$$

In order to find the ground state it is necessary to minimise the energy while keeping the orbitals, $\psi_\lambda$, orthonormal. This is done by applying the variational principle, adding Lagrange multipliers, $E_{\lambda\mu}$ to Equation 2.1.7,

$$\left(- \frac{1}{2}\nabla^2 + V_{eff}(r) - E_\lambda \right)... ...lambda(r) - \sum_{\mu\neq\lambda}E_{\lambda\mu}\psi_\mu(r) = 0.$$

These set of equations are known as the Hartree-Fock equations. Writing them more completely, for a given orbital, $\lambda$,

 

$$\left\{ -\frac{1}{2}\nabla^2 + V_{e-i}({\bf r}) + V_H({\bf r... ..._\lambda(r) = \sum_{\mu\neq\lambda} E_{\lambda\mu} \psi_\mu(r),$$ (4)

$$V_H({\bf r})\psi_\lambda(r) = \frac{\delta E_H}{\delta \psi^... ...}_1)\psi_\lambda(r)}{\vert{\bf r} - {\bf r}_1\vert}d{\bf r}_1,$$

$$V_\lambda^x({\bf r}) \psi_\lambda(r) = \frac{\delta E_x}{\de... ...\frac{1}{\vert{\bf r}_1 - {\bf r}\vert} \psi_\mu(r)d{\bf r}_1,$$

with the orthonormality constraint

$$\sum_s{\int{\psi_\mu^*\psi_\lambda d{\bf r}}} = \delta_{\lambda\mu}.$$

Diagonalising $E_{\lambda\mu}$ removes the Lagrange multiplier term on the right hand side, and the equation can then be solved self-consistently. To find the total energy, E_TOT, Equation 2.2.10 is multiplied by $\psi^*_\lambda(r)$,integrated over r and summed over s and $\lambda$ to give

$$E_{TOT} = \sum_\lambda E_\lambda - E_H - E_x + E_{i-i}.$$

$E_\lambda$ is the ionisation energy for the $\lambda$ electron excluding any redistribution in the remaining orbitals once an electron is removed.

The orbitals are normally expanded in terms of plane waves, and for a homogeneous electron gas the exact plane wave solution is given by

$$\psi_\lambda({\bf r},s) = \frac{1}{\sqrt{\Omega}}e^{i{\bf k}.{\bf r}}\chi_\alpha(s).$$

In the spin polarised case these are split into the spin `up' and spin `down' populations, each with its own Fermi wavevector; the total energy is a sum of these two populations (n_s is the electron density of spin state s). For the homogeneous electron gas this is given by

$$E = \Omega \sum_s \left\{ \frac{3n_s}{10}(6\pi^2n_s)^\frac{2... ...ft(\frac{3}{4\pi}\right)}^\frac{1}{3} n_s^\frac{4}{3} \right\}.$$