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=== Derivation ===
According to [[Slater–Condon rules]], the expectation value of energy of the [[Molecular Hamiltonian#Clamped nucleus Hamiltonian|molecular electronic Hamiltonian]] <math>\hat{H}^e</math> for a [[Slater determinant]] is
: <math>\begin{aligned} E[\psi^{HF}] &= \left\langle\psi^{HF}|\hat{H}^e|\psi^{HF}\right\rangle \\
To derive Hartree-Fock we minimize the energy functional for N electrons▼
&= \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)\mathbf{x}_j\phi_j^*(\mathbf{x}_j) \frac{1}{|\mathbf{r}_i-\mathbf{r}_j|}\phi_i(\mathbf{x}_i)\phi_j(\mathbf{x}_j) \\
: <math>\delta E[\phi_k^*(x_k)] = \delta \left\langle\psi^{HF}|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>▼
&- \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{r}_i-\mathbf{r}_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 operators 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) \\
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\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 Hartree-Fock equation we minimize the energy functional for N electrons with orthonormal constraints.
are orthonormal. Since we can choose the basis of <math>\phi_i(x_i)</math>, we choose a basis 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>\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>
▲
: <math>\begin{aligned}
\delta E[\phi_k^*(x_k)] &= \sum_{i=1}^N \int\text{d}\mathbf{x}_i \, \hat{h
&- \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{r}_i-\mathbf{r}_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
&+ \sum_{j=1}^N \int \text{d}\mathbf{x}_j\phi_j^*(\mathbf{x}_j) \frac{1}{|\mathbf{r}_k-\mathbf{r}_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{r}_k-\mathbf{r}_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 in the molecular Hamiltonian drops out before the double integrals due to symmetry and the product rule. We find the Fock equation
: <math>\hat{F}(\mathbf{x}_k)\phi_k
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{r}_k-\mathbf{r}_j|}= \sum_{j=1}^N \int \mathrm{d}\mathbf{x}_j \frac{\rho(\mathbf{x}_j)}{|\mathbf{r}_k-\mathbf{r}_j|},\\
\hat{K}(\mathbf{x_k})\phi_{k}(\mathbf{x}_k) &
\end{aligned}</math>
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