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For orthonormal orbitals $a,b$, the symmetric spatial state $C=(|ab⟩+|ba⟩)/\sqrt{2}$ multiplied by the antisymmetric spin singlet has one electron on each center and no double occupancy. The full state remains antisymmetric under fermion exchange.

The energy cost $U=\iint |a({\mathbf{r}}_1){|}^2|a({\mathbf{r}}_2){|}^2{e}^2/[4\pi {ϵ}_0|{\mathbf{r}}_1-{\mathbf{r}}_2|]{\mathrm{d}}^3{r}_1{\mathrm{d}}^3{r}_2$ for double occupancy of a localized orbital. It is positive for an ordinary nonzero orbital. A model retaining only this integral also neglects intersite and mixed Coulomb terms; orbital orthogonality alone does not make those terms vanish.

Fix canonical single-particle order $(1\uparrow ,1\downarrow ,2\uparrow ,2\downarrow )$ and write the occupation ket as $|n_1{n}_2{n}_3{n}_4⟩$. The six states in the printed order are $|1010⟩$, $|1001⟩$, $|0110⟩$, $|0101⟩$, $|1100⟩$, $|0011⟩$. The last state is double occupancy on site 2.

For a fermionic raising and lowering operator acting on mode $j$, the sign is $(-1{)}^{{\sum }_{k<j}{n}_k}$, and annihilation requires that mode occupied while creation requires it empty. Starting from the second basis state, the hop that reaches the last state isIt therefore contributes $H_{6,2}=-t$. Starting from the third state instead,because the occupied $2\uparrow$ mode precedes $2\downarrow$. The Hamiltonian's leading $-t$ makes $H_{6,3}=+t$. Finally $n_{2\uparrow }{n}_{2\downarrow }|0011⟩=|0011⟩$, so $H_{6,6}=U$. No other displayed basis state reaches $|0011⟩$ in one allowed hop. These signs verify the three nonzero entries without assigning distinguishable-particle hopping signs to fermions.

Use $|j⟩$ for the six printed basis vectors. The three singly occupied triplets areEach has energy zero: either Pauli exclusion prevents hopping to double occupancy or the two contributions cancel. DefineThe state $D_{-}$ has energy $U$. In the $S,D_{+}$ subspace the Hamiltonian is $\left(\begin{array}{cc}0& -2t\\ -2t& U\end{array}\right)$, withFor $U\gg |t|>0$, the fourth low-energy eigenstate is the normalized singletwhere $\mathcal{N}=[1+(E_{-}/(2t){)}^2{]}^{-1/2}$. Its energy isThe four lowest energies are $-4t^2/U,0,0,0$ to leading order, with the singlet and triplets just given. The upper pair lie near $U$ and $U+4t^2/U$.

The low-energy Hamiltonian can be writtenusing dimensionless spin-half operators. It places the singlet below the triplet by $J$, hence favors antiferromagnetic alignment. A lattice of such bonds can support antiferromagnetism, although its ordering pattern also depends on geometry and competing couplings. The physical energy gain comes from virtual charge fluctuations, not from a permanent double occupancy in the large-$U$ limit.

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.

The Hubbard model combines hopping with a local repulsion,Hopping favours a broad itinerant band; the interaction penalizes opposite-spin double occupancy. For uniform Stoner decoupling of the Hubbard interaction, let $s=n_{\uparrow }-n_{\downarrow }$ be the spin imbalance per site. The interaction is $U(n^2-s^2)/4$, so polarization lowers it by $U{s}^2/4$, at the expense of kinetic energy.

If $D(E_F)$ is the noninteracting density of states per spin per site, small polarization costs $s^2/[4D(E_F)]$. Thereforeis the mean-field ferromagnetic instability criterion. The exchange splitting is $\Delta =Us$, and the paramagnetic response is enhanced as $\chi ={\chi }_0/[1-UD(E_F)]$. A density of states including both spins changes the written factor in the criterion.

The model explains how itinerant bands can polarize without imposing pre-existing localized atomic spins. It is not a claim that every large-$U$ metal becomes ferromagnetic: near half filling the Hubbard model can instead form a Mott insulator with antiferromagnetic superexchange of scale $4t^2/U$. Band filling, bandwidth and fluctuations matter.^[ref] The interaction-energy and kinetic-cost expansion above is the stated mean-field derivation of the enhancement.^[ref]

In a minimal hole description of a metal–ligand–metal path, two holes initially occupy separate metal orbitals and the ligand has no hole. Creating a virtual ligand hole costs a charge-transfer energy $\Delta$. Two hops generate an effective metal–metal hopping of order $t_{pd}^2/\Delta$; two more hops through metal double occupancy generate the antiferromagnetic scale in the title. Ligand double occupancy adds a channel with denominator $2\Delta +U_p$. Orbital geometry and intra-atomic exchange determine when the antiferromagnetic channel dominates.

Two localized orbitals coupled by hopping, with an on-site penalty for double occupancy. At two-electron filling the competition between hopping and local repulsion can be solved exactly within the finite basis. It gives a compact example of a molecular bonding crossover and electron correlation; it is not by itself a complete real-molecule binding curve.