Editing Hartree–Fock method (section)

=== Derivation ===
According to the [[Slater–Condon rules]], the energy expectation value of the [[Molecular Hamiltonian#Clamped nucleus Hamiltonian|molecular electronic Hamiltonian]] <math>\hat{H}^e</math> for a [[Slater determinant]] is

: <math display="inline">\begin{aligned} E[\psi^{HF}] &= \left\langle\psi^{HF}|\hat{H}^e|\psi^{HF}\right\rangle \\
&=  \sum_{i=1}^N \int\text{d}\mathbf{x}_i \, \phi_i^*(\mathbf{x}_i) \hat{h}(\mathbf{x}_i) \phi_i(\mathbf{x}_i) \\ 
&+ \frac{1}{2} \sum_{i=1}^N\sum_{j=1}^N \int \mathrm{d}\mathbf{x}_i \int \text{d}\mathbf{x}_j \phi_i^*(\mathbf{x}_i)\phi_j^*(\mathbf{x}_j) \frac{1}{|\mathbf{x}_i-\mathbf{x}_j|}\phi_i(\mathbf{x}_i)\phi_j(\mathbf{x}_j) \\ 
&- \frac{1}{2} \sum_{i=1}^N\sum_{j=1}^N \int \text{d}\mathbf{x}_i \int \text{d}\mathbf{x}_j\phi_i^*(\mathbf{x}_i)\phi_j^*(\mathbf{x}_j) \frac{1}{|\mathbf{x}_i-\mathbf{x}_j|}\phi_i(\mathbf{x}_j)\phi_j(\mathbf{x}_i)  \end{aligned}
 </math>

where <math>\hat{h}</math> is the one electron operator including electronic kinetic energy and electron-nucleus Coulombic interaction, and
: <math>\begin{aligned}
 \psi^{HF} = \psi(\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_N) =
  \frac{1}{\sqrt{N!}}
  \begin{vmatrix} \phi_1(\mathbf{x}_1) & \phi_2(\mathbf{x}_1) & \cdots & \phi_N(\mathbf{x}_1) \\
                      \phi_1(\mathbf{x}_2) & \phi_2(\mathbf{x}_2) & \cdots & \phi_N(\mathbf{x}_2) \\
                      \vdots & \vdots & \ddots & \vdots \\
                      \phi_1(\mathbf{x}_N) & \phi_2(\mathbf{x}_N) & \cdots & \phi_N(\mathbf{x}_N)
  \end{vmatrix}.
\end{aligned}</math>

To derive the Hartree-Fock equation we minimize the energy functional for N electrons with orthonormal constraints.

: <math>\delta E[\phi_k^*(x_k)] = \delta \left\langle\psi^{HF}|\hat{H}^e|\psi^{HF}\right\rangle - \delta\left[\sum_{i=1}^N \sum_{j=1}^N \lambda_{ij} \left( \left\langle\phi_i, \phi_j\right\rangle - \delta_{ij}\right)\right] \stackrel{!}{=}\, 0,</math>

We choose a basis set <math>\phi_i(x_i)</math> in which the [[Lagrange multiplier]] matrix <math>\lambda_{ij}</math> becomes diagonal, i.e. <math>\lambda_{ij} = \epsilon_i \delta_{ij}</math>. Performing the [[Functional derivative|variation]], we obtain

: <math>\begin{aligned}
\delta E[\phi_k^*(x_k)] &= \sum_{i=1}^N \int\text{d}\mathbf{x}_i \, \hat{h}(\mathbf{x}_i) \phi_i(\mathbf{x}_i) \delta(\mathbf{x}_i -\mathbf{x}_k) \delta_{ik}\\ &+ \sum_{i=1}^N\sum_{j=1}^N \int \mathrm{d}\mathbf{x}_i \int \text{d}\mathbf{x}_j\phi_j^*(\mathbf{x}_j) \frac{1}{|\mathbf{x}_i-\mathbf{x}_j|}\phi_i(\mathbf{x}_i)\phi_j(\mathbf{x}_j) \delta(\mathbf{x}_i-\mathbf{x}_k)\delta_{ik}\\ 
&- \sum_{i=1}^N\sum_{j=1}^N \int \text{d}\mathbf{x}_i \int \text{d}\mathbf{x}_j\phi_j^*(\mathbf{x}_j) \frac{1}{|\mathbf{x}_i-\mathbf{x}_j|}\phi_i(\mathbf{x}_j)\phi_j(\mathbf{x}_i) \delta(\mathbf{x}_i-\mathbf{x}_k)\delta_{ik}\\
 &- \sum_{i=1}^N \epsilon_i \int \text{d}\mathbf{x}_i \, \phi_i(\mathbf{x}_i)  \delta(\mathbf{x}_i-\mathbf{x}_k)\delta_{ik}\\
&= \hat{h}(\mathbf{x}_k) \phi_k(\mathbf{x}_k)\\ 
&+ \sum_{j=1}^N  \int \text{d}\mathbf{x}_j\phi_j^*(\mathbf{x}_j) \frac{1}{|\mathbf{x}_k-\mathbf{x}_j|}\phi_k(\mathbf{x}_k)\phi_j(\mathbf{x}_j)\\ 
&- \sum_{j=1}^N  \int \text{d}\mathbf{x}_j\phi_j^*(\mathbf{x}_j) \frac{1}{|\mathbf{x}_k-\mathbf{x}_j|}\phi_k(\mathbf{x}_j)\phi_j(\mathbf{x}_k)\\ 
&- \epsilon_k \phi_k(\mathbf{x}_k)=0. \\
\end{aligned}</math>

The factor 1/2 before the double integrals in the molecular Hamiltonian drops out  due to symmetry and the product rule. We may define the [[Fock matrix|Fock operator]] to rewrite the equation

: <math>\hat{F}(\mathbf{x}_k)\phi_k(\mathbf{x}_k) \equiv \left[ \hat{h}(\mathbf{x}_k) + \hat{J}(\mathbf{x}_k) - \hat{K}(\mathbf{x}_k) \right]\phi_k(\mathbf{x}_k) = \epsilon_k \phi_k(\mathbf{x}_k),</math>

where the [[Coulomb operator]] <math>\hat{J}(\mathbf{x}_k)</math> and the [[exchange operator]] <math>\hat{K}(\mathbf{x}_k)</math> are defined as follows

: <math>\begin{aligned}
\hat{J}(\mathbf{x_k}) &\equiv \sum_{j=1}^N \int \mathrm{d}\mathbf{x}_j \frac{\phi_j^*(\mathbf{x}_j) \phi_j(\mathbf{x}_j)}{|\mathbf{x}_k-\mathbf{x}_j|}= \sum_{j=1}^N \int \mathrm{d}\mathbf{x}_j \frac{\rho(\mathbf{x}_j)}{|\mathbf{x}_k-\mathbf{x}_j|},\\
\hat{K}(\mathbf{x_k})\phi_{k}(\mathbf{x}_k) &\equiv  \sum_{j=1}^N  \phi_{j}(\mathbf{x}_k) \int \text{d}\mathbf{x}_j \frac{\phi_j^*(\mathbf{x}_j) \phi_k(\mathbf{x}_j)}{|\mathbf{x}_k-\mathbf{x}_j|}.\\
\end{aligned}</math>

The exchange operator has no classical analogue and can only be defined as an integral operator.

The solution <math>\phi_k</math> and <math>\epsilon_k</math> are called molecular orbital and orbital energy respectively.

Although Hartree-Fock equation appears in the form of a eigenvalue problem, the Fock operator itself depends on <math>\phi</math> and must be solved by a different technique.