
Advanced Quantum Mechanics course sequence · Interactive lesson
Hubbard dimer: superexchange and one determinant's failure
This is where Unit 10 gets graded. The Hubbard dimer is small enough to diagonalise by hand and rich enough to hold both magnetism and a breaking bond, so every tool of the unit, from operator signs to the Hartree-Fock equations, is run against an exact answer, and we see precisely where one determinant stops being enough.
UAQM-10.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Build the model
Six states, solved exactly, grade every single-determinant method this unit has built.
The key idea
The Hubbard dimer is two sites, each with one orbital, sharing electrons that hop between the sites with amplitude t and pay an energy U whenever two of them occupy the same site.
Example. With U = 6.0 eV and t = 1.5 eV the singlet ground state lies at (6.0 − √72)/2 = −1.24 eV and the triplet at 0, while restricted Hartree-Fock gives 0.00 eV and unrestricted Hartree-Fock −0.75 eV.
The complete physical model and assumptions
Keep one orbital on each of two sites and only the largest Coulomb term, the cost U of two electrons on one site, and the Hamiltonian of topic 10.04 becomes the Hubbard model; two electrons give it six states. Hopping, of amplitude t, spreads the electrons over a bonding orbital, which puts both on one site half the time, and U taxes exactly that. Solved exactly, the triplet sits at zero because the Pauli principle forbids its only hop, while the singlet mixes its covalent and ionic parts and falls to (U − √(U² + 16t²))/2. For U ≫ t that is −4t²/U: two localised spins that gain from a virtual hop only when antiparallel. That is superexchange, the antiferromagnetic J = 4t²/U behind the exchange constants Quantum Mechanics II named. The same six states then grade this unit's methods. Restricted Hartree-Fock keeps the bonding orbital and an ionic weight of one half at every U, so its energy, −2t + U/2, climbs to U/2, not 0, as the bond is pulled apart and t falls to zero. Unrestricted Hartree-Fock escapes once U exceeds 2t by letting the two spins' orbitals lean towards opposite sites, but it reaches only −2t²/U, half the true gain, and only by being half triplet. The exact state needs two determinants. The model stops where its truncation does: repulsion between sites, further orbitals, and the real Néel order of a large lattice, where broken symmetry is no longer an artefact.
1Six states, and the symmetries that split them
Units are SI with energies in eV and ħ explicit in spin; nothing here is relativistic, so no metric signature enters (the course uses (+, −, −, −) when four-vectors appear). The model is quoted, not derived: one orbital per site, hopping −t between them, and of all the Coulomb integrals of topic 10.04 only the on-site one, U. Order the spin-orbitals 1↑, 1↓, 2↑, 2↓; two electrons in four give 6 states. H conserves N, Sz and S², and commutes with the swap P of the sites, so most of the work is done before any algebra. In c†₁↑c†₂↑|0⟩ a hop would put two parallel spins in one orbital, which the Pauli principle forbids, so it is an eigenstate at exactly 0; so is c†₁↓c†₂↓|0⟩, and by S² symmetry the third triplet state. Of the remaining Sz = 0 states, the odd ionic combination (|d₁⟩ − |d₂⟩)/√2, with |dᵢ⟩ = c†ᵢ↑c†ᵢ↓|0⟩, is the only singlet odd under P, so it sits alone at U. A 2 × 2 block is all that is left.
2Signs first, then the 2 × 2 block
Write |a⟩ = c†₁↑c†₂↓|0⟩ and |b⟩ = c†₁↓c†₂↑|0⟩. The hop c†₁↓c₂↓ acting on |a⟩ passes c₂↓ through c†₁↑ for one sign, giving −c†₁↓c†₁↑|0⟩ = +|d₁⟩ after reordering; c†₂↑c₁↑ gives |d₂⟩ directly. So T|a⟩ = −t(|d₁⟩ + |d₂⟩), while on |b⟩ each matching hop differs by one swap, so T|b⟩ = +t(|d₁⟩ + |d₂⟩). The triplet (|a⟩ + |b⟩)/√2 is therefore annihilated by T, confirming its 0, and the covalent singlet |cov⟩ = (|a⟩ − |b⟩)/√2 reaches the even ionic state |ion⟩ = (|d₁⟩ + |d₂⟩)/√2 with ⟨ion|H|cov⟩ = −2t, the 2 counting either electron hopping. The block [[0, −2t], [−2t, U]] has eigenvalues (U ∓ √(U² + 16t²))/2, derived. The trace checks it, 3 × 0 + U + U = 2U, and so does U = 0, where −2t, 0 (four states) and +2t are two free electrons in bonding (−t) and antibonding (+t) orbitals. At U = 6.0 eV and t = 1.5 eV, E₀ = −1.24 eV and the ground state's ionic weight, (1 − U/√(U² + 16t²))/2, is 0.146.
3Superexchange: the hop only the singlet can make
For U ≫ t the root expands to E₀ = −4t²/U + 16t⁴/U³ − …, derived. Second-order perturbation theory, used from Quantum Mechanics II, says why. At t = 0 the singlet and triplet with one electron per site are degenerate at 0; the hop takes the singlet to the ionic state at U with element −2t, giving (2t)²/(0 − U) = −4t²/U, and takes the triplet nowhere: a doubly occupied orbital must be a spin singlet, and hopping conserves spin. So J = 4t²/U, and Heff = J(S₁⋅S₂/ħ² − 1/4) reproduces both levels: S₁⋅S₂ = −3ħ²/4 gives −J, +ħ²/4 gives 0. The sign is the point. Helium's exchange integral K favours parallel spins; this kinetic exchange favours antiparallel ones, because Pauli blocks the parallel pair's virtual hop. The ionic admixture is small, 3.6% at U = 10t, yet it carries all of J. Measured: in copper(II) acetate monohydrate two Cu²⁺ spins form a dimer whose singlet lies about 290 cm⁻¹, 0.036 eV, below its triplet (quoted). With U ≈ 5 eV that needs t = √(JU)/2 ≈ 0.21 eV through the acetate bridges, an estimate rather than a fit.
4Restricted Hartree-Fock keeps the bond half ionic
Put both electrons in b†σ = (c†₁σ + c†₂σ)/√2. Each site then carries ½ of each spin, so the Fock operator of topic 10.06 adds the same U⟨nᵢ↓⟩ = U/2 to every site for a ↑ electron, and its eigenvectors stay bonding and antibonding: self-consistent at every U, with εb = −t + U/2. Contractions give the energy: −t per electron from hopping, U⟨nᵢ↑⟩⟨nᵢ↓⟩ = U/4 on each site, and no exchange term, because U acts only between opposite spins, whose exchange contraction vanishes. So ERHF = −2t + U/2, derived, which is 2εb less the U/2 that adding orbital energies counts twice. Expanded on sites the determinant is (|cov⟩ + |ion⟩)/√2: ionic weight ½ whatever U is. At U = 0 that is exact. Pull the bond apart and t falls towards 0: the exact energy rises to 0, two neutral atoms, but ERHF rises to U/2, as half the state is still ionic. For hydrogen U is I − A = 13.598 − 0.754 = 12.84 eV (measured values, quoted), the cost of H⁺ + H⁻ over H + H, so the model leaves RHF 6.42 eV too high at dissociation. Real H₂ fails the same way: its RHF curve ends several eV above two separated atoms (quoted).
5Unrestricted Hartree-Fock breaks spin symmetry to dissociate
Give the spins different orbitals: ↑ in cos θ φ₁ + sin θ φ₂ and ↓ in the mirror image sin θ φ₁ + cos θ φ₂. Each hop gives −t sin 2θ and the double occupancy costs 2U cos²θ sin²θ, so with s = sin 2θ, E(s) = −2ts + (U/2)s². RHF is θ = 45°, stationary at every U as Brillouin's theorem requires, but once U exceeds 2t its curvature in 2θ, −(U − 2t), turns negative and the minimum moves to s = 2t/U, where EUHF = −2t²/U, derived. The Fock operator drives it: the ↑ electron sees U⟨n↓⟩, larger on site 2, so it leans to site 1 and pushes the ↓ electron further onto site 2. The price is spin. ⟨S²⟩ = ħ²(1 − 4t²/U²), the triplet weight is (1 − 4t²/U²)/2, and site 1 carries a moment ⟨n₁↑ − n₁↓⟩ = √(1 − 4t²/U²) where the exact singlet has none. On Heff this is the Néel product state, ⟨S₁⋅S₂⟩ = −ħ²/4, with energy −J/2 = −2t²/U: correct dissociation, half the superexchange, because the spin flip that makes ↑↓ a singlet is missing. In H₂ this instability, the Coulson-Fischer point, lies near 1.2 Å, 1.6 times the 0.741 Å bond (quoted).
6Two determinants are exact: configuration interaction, named
In molecular orbitals, with a†σ = (c†₁σ − c†₂σ)/√2, the even singlet space is spanned by |RHF⟩ = (|cov⟩ + |ion⟩)/√2 and the double excitation |D⟩ = a†↑a†↓|0⟩ = (|ion⟩ − |cov⟩)/√2. There H is [[−2t + U/2, U/2], [U/2, 2t + U/2]], derived: the only coupling is U/2, and the single excitations b†↑a†↓|0⟩ and a†↑b†↓|0⟩ are odd under the swap of sites, so they cannot mix, Brillouin's theorem made exact by symmetry. Its lower eigenvalue is E₀ again, so two-determinant configuration interaction, here the full CI, is exact; CI in general is only named in this course. The weight of |D⟩, (1 − 4t/√(U² + 16t²))/2, is 0.15 at U = 4t, 0.31 at U = 10t and tends to ½ as t → 0, when the exact state becomes the purely covalent Heitler-London state (|RHF⟩ − |D⟩)/√2 and no single determinant of the right spin can follow it. That is the line this topic draws: correlation that reweights determinants at order one, as a stretched bond does, defeats any single determinant, while closed-shell helium's, topic 10.08, is a 1.4% correction to one.
Key equations
Read the units, assumptions and limits before substituting.
Hubbard dimer Hamiltonian (a model, quoted)
H = −t Σσ (c†₁σ c₂σ + c†₂σ c₁σ) + U Σᵢ nᵢ↑ nᵢ↓
Topic 10.04's Hamiltonian cut to one orbital per site and the on-site integral alone. It conserves N, Sz, S² and the swap of the two sites.
t, U in eV; c†ᵢσ adds spin σ to the orbital on site i; nᵢσ = c†ᵢσ cᵢσ; N = 2 gives 4!/(2! 2!) = 6 statesSinglet ground state, exact
E₀ = (U − √(U² + 16t²))/2
Derived from the 2 × 2 even block. Limits: −2t at U = 0, both electrons bonding; it rises to 0, two neutral atoms, as t → 0.
E₀ in eV; triplet at 0 (three states), odd ionic state at U, excited singlet at (U + √(U² + 16t²))/2Superexchange coupling
J = ET − ES = 4t²/U − 16t⁴/U³ + …
Derived twice: by expanding E₀, and at second order through the ionic state, where only the singlet can hop. Heff = J(S₁⋅S₂/ħ² − 1/4) gives both levels.
J in eV, for U ≫ t; the ground state's ionic admixture is about 4t²/U²Restricted Hartree-Fock energy
ERHF = −2t + U/2
Derived by contractions. Exact at U = 0, but as t → 0 it tends to U/2, not 0: one determinant cannot shed the ionic half of a stretched bond.
both electrons in (φ₁ + φ₂)/√2; hopping gives −t each, each site gives U(½)(½); ionic weight ½ at every UUnrestricted Hartree-Fock energy
EUHF = −2t²/U for U ≥ 2t, and −2t + U/2 below
Derived. The symmetric solution s = 1 stops being the minimum at U = 2t; at large U the result is half the exact −4t²/U.
↑ in cos θ φ₁ + sin θ φ₂, ↓ in sin θ φ₁ + cos θ φ₂; E = −2ts + (U/2)s² with s = sin 2θ, least at s = 2t/USpin contamination of the UHF determinant
⟨S²⟩UHF = ħ²(1 − 4t²/U²)
Derived from the overlap 2t/U of the two orbitals. At large U the determinant is half triplet, which gains nothing, so UHF finds half the superexchange.
U ≥ 2t; 0 for the singlet, 2ħ² for the triplet; ħ = 1.055 × 10⁻³⁴ J sInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Start at U/t = 4, where RHF sits level with the triplet. Drag t down to 1.0 eV to stretch the bond: the filled exact point slides towards the faint −4t²/U curve, the RHF readout climbs to +1.0 eV, and the open UHF point sits near half the exact depth.
Live results
- EXACT SINGLET E₀
- -1.243 eV
- SECOND ORDER −4t²/U
- -1.500 eV
- RHF −2t + U/2
- 0.000 eV
- UHF ENERGY
- -0.750 eV
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. On-site repulsion U: 6.0 eV; Hopping t (smaller t, longer bond): 1.5 eV
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting conclusion is that unrestricted Hartree-Fock has fixed the problem: it equals RHF below U = 2t, lies lower above it, and goes to the right limit, two neutral atoms, as t → 0. A lower energy is not a better state. The UHF determinant is not an eigenstate of S², with ⟨S²⟩ approaching ħ² rather than 0; it is half triplet, and it puts almost a whole unpaired spin on each site, where the exact singlet has no moment on either. Its large-U energy, −2t²/U, is half the true −4t²/U, so a singlet-triplet gap read off it is wrong by a factor of 2. The variational principle promises an upper bound on the energy, not a wavefunction that resembles the ground state; a broken symmetry is the warning that one determinant is not enough.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A two-site model has t = 0.50 eV and U = 4.0 eV. Find all six two-electron energies, the singlet-triplet gap and the ionic weight of the ground state, and compare the gap with the superexchange estimate 4t²/U.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- A doubly occupied orbital must be a spin singlet and hopping conserves spin, so the three triplet states have nowhere to hop and all sit at 0. The odd ionic state (|d₁⟩ − |d₂⟩)/√2 couples to nothing and sits at U = 4.0 eV.
- The even block is [[0, −2t], [−2t, U]] = [[0, −1.0], [−1.0, 4.0]] eV, with eigenvalues (4.0 ∓ √(16 + 4))/2 = (4.0 ∓ 4.472)/2 eV: E₀ = −0.236 eV and E₊ = 4.236 eV.
- Check the trace: 0 + 0 + 0 + 4.0 − 0.236 + 4.236 = 8.0 eV = 2U, the sum of the diagonal elements. Every term is an energy, so the dimensions agree.
- The gap ET − E₀ is 0.236 eV. The estimate 4t²/U = 4(0.25)/4.0 = 0.250 eV is 6% high; adding the next term, −16t⁴/U³ = −0.0156 eV, gives 0.234 eV.
- Ionic weight: (1 − U/√(U² + 16t²))/2 = (1 − 4.0/4.472)/2 = 0.053, close to the leading estimate 4t²/U² = 0.0625.
Answer. Energies −0.236 eV (singlet ground state), 0 (triplet, three states), 4.0 eV (odd ionic) and 4.236 eV (excited singlet). Gap 0.236 eV against 4t²/U = 0.250 eV; ionic weight 0.053.
MediumExample 2
Model a stretching bond with U = 12.84 eV, the measured I − A of hydrogen, held fixed while t falls. At t = 4.0 eV and at t = 1.0 eV, find the exact, RHF and UHF energies and the correlation energy E₀ − ERHF, and find the t below which UHF breaks the spin symmetry.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- UHF leaves RHF when U = 2t, so it breaks symmetry for t below U/2 = 6.42 eV. Both values of t qualify, so EUHF = −2t²/U at each.
- t = 4.0 eV: √(U² + 16t²) = √(164.87 + 256) = 20.515 eV, so E₀ = (12.84 − 20.515)/2 = −3.84 eV. ERHF = −8.0 + 6.42 = −1.58 eV, and EUHF = −2(16)/12.84 = −2.49 eV.
- t = 1.0 eV: √(164.87 + 16) = 13.449 eV, so E₀ = −0.304 eV, close to −4t²/U = −0.312 eV. ERHF = −2.0 + 6.42 = +4.42 eV, above even the triplet at 0, and EUHF = −2/12.84 = −0.156 eV.
- Correlation energy E₀ − ERHF: −2.26 eV at t = 4.0 eV and −4.72 eV at t = 1.0 eV. As t → 0 it tends to −U/2 = −6.42 eV, the ionic half that RHF never sheds.
- UHF's error EUHF − E₀ falls from 1.35 eV to 0.149 eV, but as a fraction of E₀ it recovers 0.65 and then only 0.51: it dissociates correctly while heading for half the superexchange.
Answer. t = 4.0 eV: E₀ = −3.84 eV, ERHF = −1.58 eV, EUHF = −2.49 eV, correlation −2.26 eV. t = 1.0 eV: E₀ = −0.304 eV, ERHF = +4.42 eV, EUHF = −0.156 eV, correlation −4.72 eV. UHF breaks symmetry below t = 6.42 eV.
HardExample 3
For U = 8t, build the UHF determinant: find θ, its energy, ⟨S²⟩ and the spin moment on site 1. Then remove its triplet part, renormalise, and compare the energy of the resulting singlet with the exact ground state.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- E(s) = −2ts + (U/2)s² is least at s = sin 2θ = 2t/U = 0.25, so 2θ = 14.48° and θ = 7.24°. The ↑ orbital is 0.992φ₁ + 0.126φ₂ and the ↓ orbital its mirror image, 0.126φ₁ + 0.992φ₂.
- EUHF = −2t(0.25) + 4t(0.0625) = −0.250t, which is −2t²/U. For comparison RHF gives −2t + 4t = +2.000t and the exact singlet (8 − √80)t/2 = −0.472t.
- Expanded, with |T₀⟩ = (|a⟩ + |b⟩)/√2, |UHF⟩ = (1/√2)|cov⟩ + (cos 2θ/√2)|T₀⟩ + (s/√2)|ion⟩, whose weights 0.5 + 0.469 + 0.031 sum to 1. ⟨S²⟩ = 2ħ² × 0.469 = 0.9375ħ² = ħ²(1 − s²).
- The moment on site 1 is ⟨n₁↑ − n₁↓⟩ = cos²θ − sin²θ = cos 2θ = 0.968 of an unpaired spin; the exact singlet has 0.
- Dropping |T₀⟩ leaves |P⟩ = (|cov⟩ + s|ion⟩)/√(1 + s²), with energy (−4ts + Us²)/(1 + s²) = (−1.000t + 0.500t)/1.0625 = −0.4706t.
- Against the exact −0.4721t the singlet part recovers 99.7% of the energy, while UHF alone recovers 53%: the missing half was the triplet, not the orbitals. For large U the same projection gives −4t²/U to leading order.
Answer. θ = 7.24°, EUHF = −0.250t, ⟨S²⟩ = 0.9375ħ², moment 0.968 on site 1. The singlet part has energy −0.4706t against the exact −0.4721t (99.7%), where UHF alone reaches 53%.
Quick check
Select an answer and explain it before checking.
Review & continue
The Hubbard dimer is two sites, each with one orbital, sharing electrons that hop between the sites with amplitude t and pay an energy U whenever two of them occupy the same site.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.