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Hartree-Fock Equations

We write out Eq. (IV.3.26) in detail, setting \bgroup\color{col1}$ \lambda_j=-\varepsilon_j$\egroup,
$\displaystyle \left(\hat{H_0} + \hat{J}-\hat{K}\right)\vert \nu_{j}\rangle$ $\displaystyle =$ $\displaystyle \varepsilon_j\vert \nu_{j}\rangle$ (3.27)
$\displaystyle \langle \mu \vert\hat{J} \vert\nu\rangle$ $\displaystyle \equiv$ $\displaystyle \sum_i\langle \mu \nu_i \vert U \vert\nu_i \nu\rangle,\quad
\lang...
...} \vert\nu\rangle \equiv \sum_i\langle \mu \nu_i \vert U \vert\nu \nu_i\rangle,$  

where we again stated the definition of the two operators \bgroup\color{col1}$ \hat{J}$\egroup and \bgroup\color{col1}$ \hat{K}$\egroup. How do these equations look in the coordinate representation? Let's write
$\displaystyle \left(\hat{H_0} + \hat{J}-\hat{K}\right)\vert \nu_{j}\rangle$ $\displaystyle =$ $\displaystyle \varepsilon_j\vert \nu_{j}\rangle \leadsto$  
$\displaystyle \langle {\bf r}\vert \left(\hat{H_0} + \hat{J}-\hat{K}
\right)\vert \nu_{j}\rangle$ $\displaystyle =$ $\displaystyle \varepsilon_j \langle {\bf r}\vert \nu_{j}\rangle \leadsto$  
$\displaystyle \langle {\bf r}\vert\hat{H_0}\vert\nu_{j}\rangle
+ \sum_i\langle ...
..._i \nu_j\rangle
- \sum_i\langle {\bf r} \nu_i \vert U \vert\nu_j \nu_i\rangle $ $\displaystyle =$ $\displaystyle \varepsilon_j \langle {\bf r}\vert \nu_{j}\rangle \leadsto$  
$\displaystyle \hat{H_0}\psi_{\nu_{j}}({\bf r}) + \sum_i \int d{\bf r'} \psi_{\n...
...({\bf r'}) U(\vert{r-r'}\vert) \psi_{\nu_{i}}({\bf r'})
\psi_{\nu_{j}}({\bf r})$      
$\displaystyle - \sum_i \int d{\bf r'} \psi_{\nu_{i}}^*({\bf r'}) U(\vert{r-r'}\vert) \psi_{\nu_{j}}({\bf r'}) \psi_{\nu_{i}}({\bf r}) \delta_{\sigma_i
\sigma_j}$ $\displaystyle =$ $\displaystyle \varepsilon_j \psi_{\nu_{j}}({\bf r}) .$ (3.28)

These are the Hartree-Fock equations in the position representation; we write them out again,
$\displaystyle \left[ \hat{H_0} + {\sum_i \int d{\bf r'} \vert\psi_{\nu_{i}}({\bf r'})\vert^2 U(\vert{r-r'}\vert)} \right]
\psi_{\nu_{j}}({\bf r})$      
$\displaystyle - {\sum_i \int d{\bf r'} \psi_{\nu_{i}}^*({\bf r'}) U(\vert{r-r'}...
...t) \psi_{\nu_{j}}({\bf r'}) \psi_{\nu_{i}}({\bf r}) \delta_{\sigma_i
\sigma_j}}$ $\displaystyle =$ $\displaystyle \varepsilon_j \psi_{\nu_{j}}({\bf r}) .$ (3.29)

This looks like a set of \bgroup\color{col1}$ j=1,...,N$\egroup stationary Schrödinger equations, but things are actually more complicated as the equations are non-linear.



Subsections
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Next: Direct Term Up: Hartree-Fock Equations Previous: `Direct' and `Exchange' Operators   Contents   Index
Tobias Brandes 2005-04-26