`Direct' and `Exchange' Operators

We defining these one-particle operators by their matrix elements (excessive use of Dirac notation, hurrah!)

$$\langle \mu \vert\hat{J_i} \vert\nu\rangle \equiv \langle \mu \nu_i \vert U \vert\nu_i \nu\rangle \leadsto\langle \... ...\nu_{j}\rangle = \langle \delta \nu_{j}\nu_{i}\vert U\vert\nu_{i}\nu_{j}\rangle$$ (3.22)

$$\langle \mu \vert\hat{K_i} \vert\nu\rangle \equiv \langle \mu \nu_i \vert U \vert\nu \nu_i\rangle \leadsto \langle ... ...nu_{j}\rangle = \langle \delta \nu_{j}\nu_{i}\vert U\vert\nu_{j}\nu_{i}\rangle.$$ (3.23)

Note that both these operators depend on the still to be determined single particle states $\vert\nu_i\rangle$!.

We can now write the functional derivate in a very elegant manner,

$$\frac{\delta}{\delta \Psi} \frac{1}{2}\sum_{ij}\left[\langle\nu_{... ...{j=1}^{N}\langle \delta \nu_{j}\vert \hat{J}-\hat{K}\vert \nu_{j}\rangle+(H.c.)$$

$$\hat{J} \equiv \sum_i\hat{J}_i,\quad \hat{K}\equiv \sum_i\hat{K}_i,$$ (3.24)

and the total functional derivative becomes

$$\frac{\delta F[\Psi]}{\delta \Psi} = \sum_{j=1}^{N}\langle \delta \nu_{j}\vert \hat{H_0}+\lambda_j + \hat{J}-\hat{K}\vert \nu_{j}\rangle + (H.c.).$$ (3.25)

As we set the functional derivative to zero

$$\frac{\delta F[\Psi]}{\delta \Psi} = 0\leadsto \left(\hat{H_0}+\lambda_j + \hat{J}-\hat{K}\right)\vert \nu_{j}\rangle=0,$$ (3.26)

as all the deviations $\delta \nu_{j}$ are independent.