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Use a three-dimensional electron gas with a uniform positive background, the jellium model. The Hartree-Fock method minimizes the expectation of the interacting Hamiltonian over normalized Slater determinants. Antisymmetry is exact within that trial class, but correlations requiring a superposition of determinants are omitted. For orthonormal occupied spin orbitals ${\varphi }_i$, the energy iswhere $J_{ij}$ is the direct Coulomb integral and $K_{ij}$ the exchange integral; the spin overlap makes exchange vanish between opposite-spin orbitals. The self-interaction terms $J_{ii}$ and $K_{ii}$ cancel. Variation gives a self-consistent Fock operator $\hat{f}=h_0+{\sum }_j({\hat{J}}_j-{\hat{K}}_j)$; exchange is nonlocal. Its eigenfunctions define occupied orbitals of a stationary determinant, and the determinant must be compared with competing self-consistent solutions to select a variational ground state.

The eigenvalues are orbital Lagrange multipliers. Their sum counts each interaction twice:Frozen-orbital removal energies may be estimated using Koopmans' theorem, but measured quasiparticle energies can contain relaxation and correlations absent from this approximation. Unoccupied orbital energies are not automatically an exact excitation spectrum. In particular, unscreened exchange in a uniform three-dimensional gas produces an unphysical logarithmic slope singularity at its Fermi surface; a one-determinant mean field does not reproduce the interacting Fermi liquid's screened response.

For a spatially uniform plane-wave determinant, electron-background and background-background terms cancel the uniform direct Coulomb contribution. Retain the background when making this cancellation: an isolated infinite electron charge distribution would have a divergent direct energy. The surviving contributions per particle are kinetic energy and negative same-spin exchange energy. Let $(3/(4\pi n){)}^{1/3}=r_s{a}_B^{\ast }$ define the Dimensionless Wigner-Seitz radius, with $a_B^{\ast }=4\pi \epsilon {\hslash }^2/(m^{\ast }{e}^2)$, $E_H^{\ast }={\hslash }^2/(m^{\ast }{a}_B^{\ast 2})$ and $\zeta =(n_{\uparrow }-n_{\downarrow })/n$. ThenFor the unpolarized state this is approximately $1.105/r_s^2-0.4582/r_s$. The kinetic contribution penalizes spin polarization because it enlarges an occupied Fermi sphere; exchange favors polarization because equal-spin occupation gains more exchange energy. As density falls, the $r_s^{-1}$ exchange scale overtakes the $r_s^{-2}$ kinetic scale. The uniform fully polarized and unpolarized energies cross atThis is a variational comparison within the homogeneous Hartree-Fock electron gas, not an exact physical ferromagnetic phase boundary. It demonstrates how Pauli kinetic energy and exchange compete and why the allowed trial states matter.

Allowing spatial symmetry breaking changes the result: the Overhauser instability lowers the paramagnetic determinant's Hartree-Fock energy through a spin density wave, even in the high-density limit. A calculation restricted to uniform Fermi spheres therefore does not give the unrestricted Hartree-Fock ground state.^[ref] At still lower density, localization and a Wigner crystal compete with fluid states. electron-gas correlation energy beyond a single determinant is important in assessing their competition. Quantum Monte Carlo provides a concrete route to comparing correlated fluid and crystal energies; the original stochastic calculation demonstrates why exchange-only energy differences are insufficient.^[ref]

Hartree-Fock is a useful variational baseline with exact exchange, not a controlled complete low-density phase diagram. At high density the kinetic scale dominates, but even a small symmetry-breaking energy gain can invalidate a restricted determinant as the exact Hartree-Fock minimizer. At low density neglected correlations, screening and localization can change both transition locations and the ordering of candidate phases. A numerical phase boundary should therefore state dimension, spin constraints, background model and whether correlation and spatially modulated states were included.

$\beta =-2.4\mathrm{eV}$ and $\alpha =-6.8\mathrm{eV}$. The Hückel method gives benzene's occupied frontier level as ${ϵ}_H=\alpha +\beta$ and its unoccupied frontier level as ${ϵ}_L=\alpha -\beta$. ThereforeUsing the Koopmans' theorem approximation for the ionization energy, $I\simeq -{ϵ}_H=9.2$, gives $\alpha +\beta =-9.2$ and the stated parameters. This calibration neglects orbital relaxation, electron correlation and any distinction between a measured optical excitation and an independent-particle orbital gap.