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Example: \bgroup\color{col1}$ N=2$\egroup, `closed shell'

In the case \bgroup\color{col1}$ N=2$\egroup, we are back to the Helium atom ( \bgroup\color{col1}$ N=2$\egroup electrons). We assume
$\displaystyle \psi_{\nu_{1}}({\bf r},\sigma) = \psi ({\bf r})\vert\uparrow \rangle,\quad
\psi_{\nu_{2}}({\bf r},\sigma) = \psi ({\bf r})\vert\downarrow \rangle,$     (3.32)

i.e. we only have two spin orbitals with opposite spin. From the Hartree-Fock equations Eq. (E.4.2), we obtain
$\displaystyle \left[ \hat{H_0} + {\sum_{i=1}^2 \int d{\bf r'} \vert\psi({\bf r'})\vert^2 U(\vert{r-r'}\vert)} \right]
\psi({\bf r})$      
$\displaystyle - {\sum_{i=1}^2 \int d{\bf r'} \psi^*({\bf r'}) U(\vert{r-r'}\vert) \psi({\bf r'}) \psi({\bf r}) \delta_{\sigma_i
\sigma_j}}$ $\displaystyle =$ $\displaystyle \varepsilon_j \psi({\bf r}) .$ (3.33)

Formally, these are still two equations due to the label \bgroup\color{col1}$ j$\egroup ( \bgroup\color{col1}$ =1,2$\egroup), but the two equations are the same and we may set \bgroup\color{col1}$ \varepsilon_1=\varepsilon_2=\varepsilon$\egroup. The sum in the exchange part has only one term which is half the direct part, and therefore (we re-insert the explicit expression for \bgroup\color{col1}$ \hat{H}_0$\egroup)
$\displaystyle \left[ -\frac{\hbar^2}{2m}\Delta + V({\bf r}) + \int d{\bf r'} \v...
...})\vert^2 U(\vert{r-r'}\vert) \right]
\psi({\bf r})= \varepsilon \psi({\bf r}).$     (3.34)

Since we have only one orbital wave function, we only have one equation.


next up previous contents index
Next: Ground State Energy Up: Hartree-Fock Equations Previous: Exchange Term   Contents   Index
Tobias Brandes 2005-04-26