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In atomic units, the displayed integral is the direct electrostatic contribution between two normalized atomic orbitals $a,b$. It enters a two-electron energy expectation together with the exchange integral.

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]

Spin and spatial exchange symmetry. Write the normalized spin singlet and triplet states asThe singlet is antisymmetric under exchange and has total spin zero; the triplet is symmetric and has total spin one. For orthogonal orbitals defineIdentical spin one-half particles are fermions, so the complete state must be antisymmetric. Thus $C{\chi }_{00}$ is allowed, and $A{\chi }_{1M}$ is allowed for each triplet component. The spatially symmetric same-orbital states $|aa⟩,|bb⟩$ also require the singlet. A same-orbital triplet vanishes by the Pauli exclusion principle.

These products give a spin-adapted basis. More generally the physical state can be writtenA superposition of the singlet and triplet sectors need not factor into one spatial function times one spin function. Factorization holds for a definite-spin sector, as used in the subsequent model. If the original normalized orbitals are not orthogonal, the corresponding $ab\pm ba$ normalization is $\sqrt{2(1\pm |⟨a|b⟩{|}^2)}$; the remainder of this calculation uses the stipulated orthogonal orbitals.

Integral matrix elements and positivity. Let $a(\mathbf{r})$ and $b(\mathbf{r})$ denote the localized spatial orbitals. With fixed proton positions, the one-electron electronic operator isThe proton-proton repulsion is a separate common nuclear energy. The requested one-electron and direct two-electron elements areSince $H_1=h\otimes I$,Likewise,The integrand is nonnegative and is positive on a set of nonzero measure. This is the repulsion cost of double occupancy; the coincidence singularity is integrable for ordinary smooth localized orbitals.

One electron on each center. Suppress the normalized singlet spin factor and use $|\psi ⟩=C$. For each of the spatial products $ab$ and $ba$, the one-body sum has expectation $2E_0$. The cross terms vanish because a one-body hop leaves an orthogonal spectator orbital. The interaction contribution is the direct integral plus the exchange integral, both neglected in the model. ConsequentlyThis state has no same-center double occupancy, but remains an exchange-symmetric spatial superposition rather than assigning a distinguishable electron permanently to each proton.

Two electrons in the bonding orbital. The normalized one-electron bonding orbital is $g=(a+b)/\sqrt{2}$. The trial spatial state is $|\varphi ⟩=|gg⟩$, whose one-body expectation isExpanding the product gives $|\varphi ⟩=(|aa⟩+|ab⟩+|ba⟩+|bb⟩)/2$. The total probability of double occupancy is therefore $1/2$. Keeping only on-site Coulomb repulsion gives the intended interaction truncation,and the interaction expectation is $U/2$. HenceThis truncation must be stated. In particular, mixed Coulomb elements such as $X=⟨aa|H_{1,2}|ab⟩$ do not vanish merely by orbital orthogonality or inversion symmetry. The printed list does not explicitly name them. If they are retained and are equal and real by site symmetry, their contribution to $E_{\varphi }$ is $2X$; the requested expression uses their neglect along with the other off-site interaction terms. No numerical value of $X$ is supplied or invented.

Energy comparison and the bond. Put $x=|t|/U$ with $U>0$ and $t=-|t|$. ThenThe straight line starts at $1/2$ and crosses zero at $x=1/4$. The purely bonding-orbital trial state is better for $x>1/4$, while the singly occupied covalent trial state is better for $x<1/4$.

For localized orbitals of approximately fixed atomic size, the hopping magnitude $|t|$ decreases rapidly, typically exponentially times a polynomial, as the proton separation grows. The on-site integral $U$ remains of atomic Coulomb scale and tends to the isolated-orbital value; it does not decay as the intersite repulsion $1/R$. If the orbitals themselves are allowed to relax, $U$ also changes, so no universal monotonic law for $U(R)$ follows from the data.

At small $|t|/U$, electron correlation suppresses the ionic configurations $aa$ and $bb$, favoring a singlet with roughly one electron on each center. A single doubly occupied bonding orbital has an artificial finite ionic weight even as the centers separate. At large $|t|/U$, delocalization gains enough kinetic energy that the bonding-orbital description becomes useful and charge fluctuations are substantial.

The crossing of two restricted trial energies is not a sharp transition of the actual two-site ground state. To see this without adding empirical data, include both $C$ and the symmetric ionic state $I_{+}=(|aa⟩+|bb⟩)/\sqrt{2}$. This two-site Hubbard dimer reduces toIts ground state is $\cos \theta C+\sin \theta I_{+}$, with $\tan 2\theta =4|t|/U$ and $0\le \theta \le \pi /4$. The ionic probability isIt changes smoothly from zero to $1/2$. At large separation, the electronic lowering is $-4t^2/U$ to leading order and vanishes as hopping vanishes. An absolute molecular binding curve would additionally require $E_0(R)$, the nuclear repulsion and the neglected intersite terms; this simplified comparison alone does not determine an equilibrium bond length.

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.

The $3d$ and $4s$ atomic orbitals of chromium are close enough in energy that electron-electron terms can overturn the simple filling order. The configuration $[\mathrm{Ar}]3d^{5}4s^1$ permits five singly occupied $3d$ orbitals and a parallel-spin $4s$ electron. Relative to $3d^{4}4s^2$, it removes the $4s$ pair and permits more parallel-spin exchange interactions, lowering the Coulomb-and-exchange contribution enough to offset the orbital promotion cost.

As qualitative bookkeeping, six parallel electron spins give $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ same-spin pairs, compared with $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ in high-spin $3d^{4}4s^2$. The individual exchange integrals are unequal, so this is an explanation of the tendency, rather than a quantitative atomic energy calculation. The observed ground state configuration is $\boxed{[\mathrm{Ar}]3d^{5}4s^1}$.^[ref]

For fixed orthonormal orbitals $a,b$, symmetrizing or antisymmetrizing $a(1)b(2)$ gives electron-repulsion expectations $J_{ab}+K_{ab}$ and $J_{ab}-K_{ab}$. Antisymmetry of the complete electronic state therefore places the singlet above the triplet by $2K_{ab}$, with $K_{ab}$ the positive Coulomb exchange integral. Orbital relaxation and further correlation refine this comparison. Exchange splitting here is a consequence of Fermi statistics and the electrostatic interaction, not a separate classical spin-spin force.