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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.

A helium atom has two electrons bound to nuclear charge $+2e$. In the nonrelativistic fixed-nucleus approximation, using atomic units,The last term couples the electrons, so a product of independent hydrogenic orbitals is not an exact solution. The dominant ground configuration is $1s^2$. Its spatial wavefunction is symmetric, so Fermi antisymmetry requires the spin singletA triplet cannot put both electrons in the same $1s$ spatial orbital. Correlation changes the exact spatial function, including its dependence on $r_{12}$, while preserving this ground-state symmetry.

A useful helium variational effective charge takes normalized orbitals ${\varphi }_{\zeta }(\mathbf{r})=({\zeta }^3/\pi {)}^{1/2}{e}^{-\zeta r}$ for both electrons and multiplies their product by ${\chi }_{00}$. The energy in Hartree units isThe reduced effective charge expresses screening. Neglecting repulsion gives $-4E_h$, while its expectation in the unscreened $\zeta =2$ product gives $-2.75E_h$; optimizing the orbital improves that trial estimate. The exact ground energy is lower than the optimized product energy, since the product omits explicit electron correlation. This calculation keeps nuclear charge two in $H$; only the trial orbital parameter is varied.

For a singly excited helium state with distinct orthonormal orbitals $a=1s$ and $b=n\ell$, the spatial combinations areThe symmetric ${\Phi }_{+}$ pairs with the antisymmetric spin singlet, while antisymmetric ${\Phi }_{-}$ pairs with the symmetric spin triplet. With the same orbitals, their leading energies areHere $J_{ab}$ is the direct Coulomb integral and $K_{ab}$ is the exchange integral. The latter is positive for the Coulomb interaction, so the triplet is lower for this fixed-orbital comparison. singlet-triplet exchange splitting in helium arises from antisymmetry and spatial correlation in the Coulomb interaction, not from inserting a separate classical spin-spin force. Spin-singlet helium states are parahelium; spin-triplet helium states are orthohelium. The two spectra approximately form separate families because the leading electric dipole operator does not change total spin.^[ref]

The lowest $1s2s$ terms are ${}^3{S}_1$ and ${}^1{S}_0$, with the triplet lower. The $1s2p$ configuration gives ${}^3{P}_{0,1,2}^{\mathrm{o}}$ and ${}^1{P}_1^{\mathrm{o}}$, where the odd parity allows singlet $P$ to connect to the even-parity ground singlet by an electric dipole transition. Both $1s2s$ levels lack an allowed direct electric dipole decay to $1s^2$: the triplet is spin forbidden, and singlet $S\to S$ is forbidden by orbital angular momentum and parity. This explains helium metastability without implying an infinite lifetime once weaker radiative processes or collisions are allowed.

The observed first $1s2s$ excitation energies are approximately $19.82$ and $20.62\mathrm{eV}$ for the triplet and singlet, respectively. Higher singly excited Rydberg series approach the He II ground-state ionization threshold; doubly excited states above that threshold can autoionize and are not all stable bound excited levels.^[ref]

Let $P_{12}$ exchange the complete labels of the two electrons. The spatial factors obey $P_{12}{\Phi }_{\pm }=\pm {\Phi }_{\pm }$. The singlet state spin factor is antisymmetric, while each triplet state spin factor is symmetric. Thus in all four casesas required by the Pauli exclusion principle. In particular, the $M_S=0$ triplet state remains a triplet despite having opposite measured spin projections in each product term.

For a spin-independent Hamiltonian and the same fixed spatial atomic orbitals, direct evaluation giveswhere $J_{ab}$ is the direct Coulomb integral and $K_{ab}>0$ is the exchange integral for the helium $1s,2s$ pair. ThereforeThe antisymmetric spatial wavefunction vanishes when the two positions coincide and reduces their Coulomb repulsion. The lower energy of the maximum-spin member of this fixed electron configuration is the first of Hund's rules.

In atomic units, define the Coulomb integral and exchange integral byExpanding the four products in the expectation of $r_{12}^{-1}$ yieldsThe one-electron terms have the same value for these two states. Within this common-orbital approximation,so $\boxed{E_S-E_T=2K}$. The exchange integral is nonnegative: it is the positive Coulomb quadratic form of the transition density $a^{\ast }b$, or equivalently has a nonnegative Fourier representation. The triplet is lower. Actual orbital relaxation and correlation make the common-orbital $2K$ an approximate physical splitting.