## Hartree-Fock-Roothaan method Although when we say '*Hartree-Fock*', we are typically referring to this kind, to better show the difference in detail, it separate it from [[Hartree-Fork method|the previous part]]. With the [[Hartree-Fork method#^b0b90c|Fock operator]], the HF problem is then $\hat{F}(r)\Psi_{SD}(r)=\epsilon_{k}\Psi_{SD}(r)$ Now expand the $\Psi_{SD}=\sum_{\nu}c_{{\nu k}}\psi_{nu}(r)$ And define Fock matrix ($F_{\mu \nu}$) and overlap matrix ($S_{\mu \nu}$) as $F_{\mu \nu}=\int \psi_{\mu}^{*}\hat{F}\psi_{\nu} \mathrm{d}r$ $S_{\mu \nu}=\int \psi_{\mu}^{*}\psi_{\nu} \mathrm{d}r$ The above Schrödinger-like equation would form a linear problem $FC=SC\epsilon$ >[!Note] >This $C$ is a coefficient matrix and $\epsilon$ is the diagonal elements (energy). This is called Roothaan equations, and if compute $\partial C$, we may get the secular equation as $\begin{vmatrix} F_{11}-ES_{11} &F_{12}-ES_{12} & \cdots & F_{1N}-ES_{1N}\\ F_{21}-ES_{21} &F_{22}-ES_{22} & \cdots & F_{2N}-ES_{2N}\\ \vdots&\vdots & \ddots &\vdots\\ F_{N1}-ES_{N1} &F_{N2}-ES_{N2} & \cdots & F_{NN}-ES_{NN} \\ \end{vmatrix}=0$ Solve above HF secular equation, we may get coefficients, and consequently may compute the density matrix, since for orbital $\nu$, we have $\rho(r)=\sum_{\mu \nu}P_{\mu \nu}\psi_{\mu}^{*}\psi_{\nu}$ Here $P_{{\mu \nu}} = 2\sum_{i}^{N/2} c_{\mu_{i}}c^{*}_{\nu_{i}}$, this $\rho$ is the density matrix, and it will get into the HF exchange energy again, makes the problem an iterating one and therefore we can solve it as a self-consistency field problem. To conclude, the general procedure for HF would be: ![[Drawing 2023-10-08 23.14.01.excalidraw.svg]]