Ground State Energy

The ground state energy in Hartree-Fock can be expressed using our equations Eq. (IV.2.9) and Eq. (IV.2.5),

$$\langle \Psi \vert\hat{H}\vert \Psi\rangle = \langle \Psi \vert\hat{\mathcal H}_0\vert \Psi\rangle + \langle \Psi \vert\hat{U}\vert \Psi\rangle$$ (3.35)

$$= \sum_{i=1}^{N}\langle\nu_{i}\vert\hat{H}_0\vert\nu_{i}\rangle + \... ...t\nu_{i}\nu_{j}\rangle -\langle\nu_{i}\nu_{j}\vert U \vert\nu_{i}\nu_{j}\rangle$$

$$= \sum_{i=1}^{N}\langle\nu_{i}\vert\hat{H}_0+\frac{1}{2}\left(\hat{J}-\hat{K}\right)\vert\nu_{i}\rangle,$$ (3.36)

where again we used the direct and exchange operators $\hat{J}$ and $\hat{K}$, Eq. (IV.3.27). Since the $\vert\nu_i\rangle$ are the solutions of the HF equations Eq. (IV.3.27),

$$\left(\hat{H_0} + \hat{J}-\hat{K}\right)\vert \nu_{j}\rangle = \varepsilon_j\vert \nu_{j}\rangle,$$ (3.37)

we obtain

$$E_{\Psi} \equiv \langle \Psi \vert\hat{H}\vert \Psi\rangle = \sum_{i=1}^{N} \left[ \varepsilon_i- \frac{1}{2}\langle\nu_{i}\vert\left(\hat{J}-\hat{K}\right)\vert\nu_{i}\rangle\right]$$ (3.38)

$$= \frac{1}{2}\sum_{i=1}^{N} \left[ \varepsilon_i+ \langle\nu_{i}\vert\hat{H}_0 \vert \nu_{i}\rangle\right].$$ (3.39)