The Hartree–Fock idea

Intuition

The central field of the last lesson was a guess; Hartree turned it into a calculation. Give each electron its own orbital and ask the energy of the product state to be as low as possible. The answer is a set of one-electron Schrödinger equations: each electron moves in the field of the nucleus and of the charge clouds of all the others. But the clouds are made of the orbitals being sought, so the equations are solved by iteration — guess the orbitals, compute the field, solve for new orbitals, and repeat until the orbitals reproduce the field they move in: a self-consistent field. Fock replaced the product by a Slater determinant, which adds an exchange term between electrons of the same spin and removes each electron’s repulsion with itself. What remains beyond the best determinant is called correlation, about 1 eV for helium, and chemistry often needs it.

A crowd walking through a square: each person steers by where the others are, and where the others are depends on how everyone steers. Iterate the picture until nobody needs to change course, and the crowd’s pattern is self-consistent.

The charge one electron of helium sees at a distance $r$ from the nucleus, in units of $e$ with $r$ in Bohr radii, when the other electron is spread over a $1s$ cloud of charge $\bar{Z}=\tfrac{27}{16}$ : $1+(1+\bar{Z}r)e^{-2\bar{Z}r}$ . Close in it is the whole nuclear charge, 2; beyond the other electron’s cloud it is screened to 1. This is the self-consistent field in its simplest form.

Each electron moves in the field of the nucleus, $\hat{h}$ , the potential $\hat{U}$ of the charge cloud of all the electrons, and an exchange term $\hat{K}$ from the electrons of the same spin. The orbitals solve

\begin{aligned}\bigl(\hat{h}+\hat{U}-\hat{K}\bigr)\phi_{i}&=\varepsilon_{i}\phi_{i}\\ \hat{h}&=-\frac{\hbar^{2}}{2m}\nabla^{2}-\frac{Ze^{2}}{4\pi\varepsilon_{0}r}\end{aligned}

Properties

$\hat{U}$ includes each electron’s own cloud; $\hat{K}$ cancels that self-repulsion exactly and adds the exchange between electrons of the same spin.
$\hat{K}$ is not an ordinary potential: at one point depends on everywhere.

The Hartree equations from the variational principle

\begin{aligned}\bigl(\hat{h}+U_{i}\bigr)\phi_{i}&=\varepsilon_{i}\phi_{i}\\ U_{i}(\mathbf{r})&=\sum_{j\neq i}\int\frac{e^{2}|\phi_{j}(\mathbf{r}')|^{2}}{4\pi\varepsilon_{0}|\mathbf{r}-\mathbf{r}'|}\,d^{3}r'\end{aligned}

Take a product of one orbital for each electron and write its average energy: one-electron energies plus the direct repulsion of each pair of clouds. Make it stationary under any change of one orbital, keeping every orbital normalised with a Lagrange multiplier. Each pair of electrons appears twice in the sum over pairs, which cancels its half, and what is left is a Schrödinger equation for one electron in the field of the others.

Proof steps

\Psi=\phi_{1}(\mathbf{r}_{1})\cdots\phi_{N}(\mathbf{r}_{N}),\qquad \braket{\phi_{i}|\phi_{i}}=1

Hartree’s trial state: one orbital for each electron.

E=\sum_{i}\braket{\phi_{i}|\hat{h}|\phi_{i}}+\tfrac12\sum_{i\neq j}J_{ij}

$J_{ij}$ is the direct repulsion of the clouds $e|\phi_{i}|^{2}$ and , and each pair is counted twice in the sum.

\frac{\delta}{\delta\phi_{i}^{*}(\mathbf{r})}\Bigl(E-\sum_{k}\varepsilon_{k}\braket{\phi_{k}|\phi_{k}}\Bigr)=0

Make the energy stationary under every change of $\phi_{i}^{*}$ at every point, with the multipliers $\varepsilon_{k}$ keeping the orbitals normalised.

\hat{h}\phi_{i}+U_{i}\phi_{i}=\varepsilon_{i}\phi_{i}

$J_{ij}$ and $J_{ji}$ both contain $\phi_{i}$ ; their two halves make the potential of all the other clouds.

\phi^{(0)}\to U^{(0)}\to\phi^{(1)}\to U^{(1)}\to\cdots

$U_{i}$ depends on the orbitals being sought: iterate until they stop changing.

Applications

Practice

Self-Consistency

Each electron moves in the field of the others’ charge clouds, which are made of the very orbitals being sought. The equations are solved by iteration.

Try it

Why are the Hartree equations solved by iteration?