Hartree-Fock method

The Hartree-Fock method is an independent particle model which uses variational methods to find the best individual electron spin-orbitals.

Let us consider a many-electron atom with atomic number $Z$ . The task is to find an equation describing the effective potential felt by a single electron.

Several effects have to be taken into account:

• the nucleus attraction: ${\hat {V}}_{ne}=-{\frac {Ze^{2}}{4\pi \epsilon _{0}r}}$ • the interaction with the other electrons
• the necessity for the system wavefunction to be a Slater determinant (antisymmetric)

Given the set of the electrons' wavefunctions $\psi _{i}\left({\vec {r}}\right)\,i=1,\dots ,Z$ , which constitute a starting ground state $\left|\psi _{1},\dots ,\psi _{Z}\right\rangle _{A}$ (a Slater determinant), we can define a charge distribution

$\rho \left({\vec {r}}\right)=-e_{0}\sum _{i}\left|\psi _{i}\left({\vec {r}}\right)\right|^{2}$ We can now introduce the Hartree potential ${\hat {V}}_{H}\left({\vec {r}}\right)$ as
${\hat {V}}_{H}\left({\vec {r}}\right)=\int _{0}^{+\infty }d^{3}{\vec {r}}{\frac {\rho \left({\vec {r'}}\right)}{\left|{\vec {r}}-{\vec {r'}}\right|}}$ This potential represents the coulomb electron-electron repulsion. However it does not take into account spin effects and consider also the interaction of the electron with itself. These effects are taken into account by Fock-potential, which hasn't a classical interpretation
${\hat {V}}_{F}={\frac {e_{0}^{2}}{4\pi \epsilon _{0}}}\sum _{j}-\int _{0}^{+\infty }d^{3}{\vec {r'}}{\frac {\psi _{j}^{*}\left({\vec {r'}}\right)\psi _{j}\left({\vec {r}}\right)\psi _{i}\left({\vec {r'}}\right)}{\left|{\vec {r}}-{\vec {r'}}\right|}}$ The sum ${\hat {V}}_{H.F}={\hat {V}}_{H}+{\hat {V}}_{F}$ is the Hartree-Fock potential

The Hartree-Fock hamiltonian for a single electron is then:

${\hat {H}}_{H.F}={\frac {-\hbar ^{2}}{2m}}\nabla ^{2}-{\frac {Ze^{2}}{4\pi \epsilon _{0}r}}+{\hat {V}}_{H}\left({\vec {r}}\right)+{\hat {V}}_{F}\left({\vec {r}}\right)$ The equation we'd like to solve is then
${\hat {H}}_{H.F}\psi _{i}\left({\vec {r}}\right)=\epsilon _{i}\psi _{i}\left({\vec {r}}\right)$ It's important to note that both $\rho \left({\vec {r}}\right)$ and $\psi _{j}$ , which enter in the definition of ${\hat {V}}_{H}$ and ${\hat {V}}_{F}$ depend on the solutions of the hamiltonian equation and are not given "a priori". This is the reason why Hartree Fock equation is not, in fact, a Schrödinger equation for $\psi _{i}$ until we fix in some way $\rho \left({\vec {r}}\right)$ and $\psi _{j}$ Hartree-Fock method consist in a self-consistent procedure:

• Chose an initial charge density, which means a starting ground state $\left|\psi _{1},\dots ,\psi _{Z}\right\rangle _{A}$ and compute the corresponding $\rho \left({\vec {r}}\right)$ • solve the hartree equation with that $\rho$ and $\psi _{j}$ , find the energy corresponding to it $E_{\rho }$ and find the new wavefunctions $\psi '_{j}$ • compute $\rho '\left({\vec {r}}\right)=-e_{0}\sum _{i}\psi '_{j}\left({\vec {r}}\right)$ • put $\rho '$ and $\psi '_{j}$ in the hartree-fock potential and solve again the equation. If $E_{\rho '} $\rho '$ and $\psi '_{j}$ are better approximations for the ground state according to the variational principle.
• repeat the procedure until you reach a satisfactory agreement between the previous solutions and the latest.

Central field approximation

We can introduce further simplifications if we build ${\hat {V}}_{H.F}$ as a central potential, since we are thus allowed to separate variables in the Hartree-Fock equation. Solutions are then of the form

{\begin{aligned}&\psi _{i}\left({\vec {r}}\right)=R_{n,l}\left(r\right)Y_{l,m_{l}}\left(\theta ,\phi \right)\chi _{m_{s}}\\&{\hat {H}}\left|n,l,m_{l},m_{s}\right\rangle =\epsilon _{n,l}\left|n,l,m_{l},m_{s}\right\rangle \end{aligned}} This decomposition is a feature shared by central potentials. However $R_{n,l}$ is different from the radial function of hydrogenlike solutions. Furthermore, since the ${\frac {1}{r}}$ dependence of Coulomb potential has been broken, energy eigenvalues depend not only on $n$ but also on $l$ : Hartree-Fock model thus provides an explanation for the l-ordering observed in atomic spectra. Indeed, because of the centrifugal term ${\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}$ which grows with $l$ , lower $l$ implies lower energy (other things being equal). Electronic configuration of atoms is expressed specifying the number of electrons in each eigenstate: for example the expression $1s^{2}2s^{2}2p^{4}$ means that 2 electrons are in the state $n=1,l=0$ , 2 electrons in the state $n=2,l=0$ and 4 electrons in $n=2,l=1$ . In Hartree-Fock approximation there is no need for specifying $m_{l}$ and $m_{s}$ values since the energy does not depend on them.

1. same hamiltonian without ${\hat {V}}_{F}$ is the Hartree hamiltonian, less correct but much easier to solve