Hartree-Fock Equations

We write out Eq. (IV.3.26) in detail, setting $\lambda_j=-\varepsilon_j$,

$$\left(\hat{H_0} + \hat{J}-\hat{K}\right)\vert \nu_{j}\rangle = \varepsilon_j\vert \nu_{j}\rangle$$ (3.27)

$$\langle \mu \vert\hat{J} \vert\nu\rangle \equiv \sum_i\langle \mu \nu_i \vert U \vert\nu_i \nu\rangle,\quad \lang... ...} \vert\nu\rangle \equiv \sum_i\langle \mu \nu_i \vert U \vert\nu \nu_i\rangle,$$

where we again stated the definition of the two operators $\hat{J}$ and $\hat{K}$. How do these equations look in the coordinate representation? Let's write

$$\left(\hat{H_0} + \hat{J}-\hat{K}\right)\vert \nu_{j}\rangle = \varepsilon_j\vert \nu_{j}\rangle \leadsto$$

$$\langle {\bf r}\vert \left(\hat{H_0} + \hat{J}-\hat{K} \right)\vert \nu_{j}\rangle = \varepsilon_j \langle {\bf r}\vert \nu_{j}\rangle \leadsto$$

$$\langle {\bf r}\vert\hat{H_0}\vert\nu_{j}\rangle + \sum_i\langle ... ..._i \nu_j\rangle - \sum_i\langle {\bf r} \nu_i \vert U \vert\nu_j \nu_i\rangle = \varepsilon_j \langle {\bf r}\vert \nu_{j}\rangle \leadsto$$

$$\hat{H_0}\psi_{\nu_{j}}({\bf r}) + \sum_i \int d{\bf r'} \psi_{\n... ...({\bf r'}) U(\vert{r-r'}\vert) \psi_{\nu_{i}}({\bf r'}) \psi_{\nu_{j}}({\bf r})$$

$$- \sum_i \int d{\bf r'} \psi_{\nu_{i}}^*({\bf r'}) U(\vert{r-r'}\vert) \psi_{\nu_{j}}({\bf r'}) \psi_{\nu_{i}}({\bf r}) \delta_{\sigma_i \sigma_j} = \varepsilon_j \psi_{\nu_{j}}({\bf r}) .$$ (3.28)

These are the Hartree-Fock equations in the position representation; we write them out again,

$$\left[ \hat{H_0} + {\sum_i \int d{\bf r'} \vert\psi_{\nu_{i}}({\bf r'})\vert^2 U(\vert{r-r'}\vert)} \right] \psi_{\nu_{j}}({\bf r})$$

$$- {\sum_i \int d{\bf r'} \psi_{\nu_{i}}^*({\bf r'}) U(\vert{r-r'}... ...t) \psi_{\nu_{j}}({\bf r'}) \psi_{\nu_{i}}({\bf r}) \delta_{\sigma_i \sigma_j}} = \varepsilon_j \psi_{\nu_{j}}({\bf r}) .$$ (3.29)

This looks like a set of $j=1,...,N$ stationary Schrödinger equations, but things are actually more complicated as the equations are non-linear.