# 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 ${\displaystyle Z}$. 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: ${\displaystyle {\hat {V}}_{ne}=-{\frac {Ze^{2}}{4\pi \epsilon _{0}r}}}$
• 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 ${\displaystyle \psi _{i}\left({\vec {r}}\right)\,i=1,\dots ,Z}$ , which constitute a starting ground state ${\displaystyle \left|\psi _{1},\dots ,\psi _{Z}\right\rangle _{A}}$ (a Slater determinant), we can define a charge distribution

${\displaystyle \rho \left({\vec {r}}\right)=-e_{0}\sum _{i}\left|\psi _{i}\left({\vec {r}}\right)\right|^{2}}$
We can now introduce the Hartree potential ${\displaystyle {\hat {V}}_{H}\left({\vec {r}}\right)}$ as
${\displaystyle {\hat {V}}_{H}\left({\vec {r}}\right)=\int _{0}^{+\infty }d^{3}{\vec {r}}{\frac {\rho \left({\vec {r}}'\right)}{\left|{\vec {r}}-{\vec {r}}'\right|}}}$
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
${\displaystyle {\hat {V}}_{F}={\frac {e_{0}^{2}}{4\pi \epsilon _{0}}}\sum _{j}-\int _{0}^{+\infty }d^{3}{\vec {r'}}{\frac {\psi _{j}^{*}\left({\vec {r}}'\right)\psi _{j}\left({\vec {r}}\right)\psi _{i}\left({\vec {r}}'\right)}{\left|{\vec {r}}-{\vec {r}}'\right|}}}$
The sum ${\displaystyle {\hat {V}}_{H.F}={\hat {V}}_{H}+{\hat {V}}_{F}}$ is the Hartree-Fock potential

The Hartree-Fock hamiltonian for a single electron is then:[1]

${\displaystyle {\hat {H}}_{H.F}={\frac {-\hbar ^{2}}{2m}}\nabla ^{2}-{\frac {Ze^{2}}{4\pi \epsilon _{0}r}}+{\hat {V}}_{H}\left({\vec {r}}\right)+{\hat {V}}_{F}\left({\vec {r}}\right)}$
The equation we'd like to solve is then
${\displaystyle {\hat {H}}_{H.F}\psi _{i}\left({\vec {r}}\right)=\epsilon _{i}\psi _{i}\left({\vec {r}}\right)}$

It's important to note that both ${\displaystyle \rho \left({\vec {r}}\right)}$ and ${\displaystyle \psi _{j}}$, which enter in the definition of ${\displaystyle {\hat {V}}_{H}}$ and ${\displaystyle {\hat {V}}_{F}}$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 ${\displaystyle \psi _{i}}$ until we fix in some way ${\displaystyle \rho \left({\vec {r}}\right)}$ and ${\displaystyle \psi _{j}}$

Hartree-Fock method consist in a self-consistent procedure:

• Chose an initial charge density, which means a starting ground state ${\displaystyle \left|\psi _{1},\dots ,\psi _{Z}\right\rangle _{A}}$ and compute the corresponding ${\displaystyle \rho \left({\vec {r}}\right)}$
• solve the hartree equation with that ${\displaystyle \rho }$ and ${\displaystyle \psi _{j}}$, find the energy corresponding to it ${\displaystyle E_{\rho }}$ and find the new wavefunctions ${\displaystyle \psi '_{j}}$
• compute ${\displaystyle \rho '\left({\vec {r}}\right)=-e_{0}\sum _{i}\psi '_{j}\left({\vec {r}}\right)}$
• put ${\displaystyle \rho '}$ and ${\displaystyle \psi '_{j}}$ in the hartree-fock potential and solve again the equation. If ${\displaystyle E_{\rho '} ${\displaystyle \rho '}$ and ${\displaystyle \psi '_{j}}$ 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 ${\displaystyle {\hat {V}}_{H.F}}$ as a central potential, since we are thus allowed to separate variables in the Hartree-Fock equation. Solutions are then of the form

{\displaystyle {\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 ${\displaystyle R_{n,l}}$ is different from the radial function of hydrogenlike solutions. Furthermore, since the ${\displaystyle {\frac {1}{r}}}$ dependence of Coulomb potential has been broken, energy eigenvalues depend not only on ${\displaystyle n}$ but also on ${\displaystyle l}$: Hartree-Fock model thus provides an explanation for the l-ordering observed in atomic spectra. Indeed, because of the centrifugal term ${\displaystyle {\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}}$ which grows with ${\displaystyle l}$ , lower ${\displaystyle l}$ 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 ${\displaystyle 1s^{2}2s^{2}2p^{4}}$ means that 2 electrons are in the state ${\displaystyle n=1,l=0}$, 2 electrons in the state ${\displaystyle n=2,l=0}$ and 4 electrons in ${\displaystyle n=2,l=1}$. In Hartree-Fock approximation there is no need for specifying ${\displaystyle m_{l}}$ and ${\displaystyle m_{s}}$ values since the energy does not depend on them.

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