research35 min studyLesson 552 in Physics

Superexchange from a strong-coupling expansion

Derive antiferromagnetic exchange from virtual double occupancy and benchmark its coefficient against an exact dimer.

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What you’ll understand

  • Project a half-filled repulsive Hubbard model into its singly occupied sector.
  • Derive the sign and coefficient of the leading exchange.
  • State the limits of a spin-only effective Hamiltonian.

Suppressing real charge motion does not make an interaction-driven insulator magnetically inert. A pair of electrons can virtually visit a doubly occupied configuration and return, lowering some spin states more than others. This superexchange process survives at energies far below the atomic charge gap. Its derivation is a controlled strong-coupling expansion for a specified one-orbital repulsive model; its positive exchange is not a universal sign rule for every magnetic material.

Separate low and high charge sectors

At exactly one electron per site and U much larger than the hopping magnitude, define a projector onto the subspace with no empty or doubly occupied sites. A single hop leaves that subspace by creating a neighboring hole and doublon. Consequently the projected hopping vanishes to first order. The intermediate configuration costs U in the simple uniform model, so a second hop can return to the low-energy subspace with an energy shift of order hopping squared over U.

Eliminating the higher-energy sector gives a negative second-order operator because its energy denominators are positive relative to the low-energy reference. That negative sign represents energy lowering through virtual admixture. The operators still retain fermionic signs and spin exchange; treating the process as classical random jumps loses the relative signs that distinguish the singlet from the triplet. The expansion applies to low energies and temperatures compared with the charge gap, with corrections organized in powers of hopping over interaction.

Heff(2)=−PHtQ (QHUQ)−1QHtP,Q=1−P,∣t∣/U≪1H_{\rm eff}^{(2)}=-P H_tQ\,(QH_UQ)^{-1}QH_tP,\qquad Q=1-P,\qquad |t|/U\ll1
P projects onto one electron per site, Q onto eliminated charge excitations, and Ht and HU are the hopping and interaction operators. This form assumes zero low-sector interaction energy and positive charge denominators.↔ Scroll wide equations horizontally.

Read the coefficient from spin symmetry

For two sites and two electrons, the spin triplet cannot couple to an on-site pair because one orbital with two electrons supports only a spin singlet. The low-energy singlet does mix with doubly occupied states and gains energy. Rotational invariance therefore fixes the projected bond interaction to an antiferromagnetic Heisenberg form plus a constant. Use dimensionless spin operators with eigenvalues one half, and count each undirected bond once; these choices fix the numerical coefficient.

The resulting exchange is four times squared hopping magnitude over U. Keeping the density offset makes the triplet bond energy zero and the singlet bond energy minus the exchange, directly matching the Hubbard reference energies. Dropping that constant changes absolute energies but not their separation. For complex spin-independent hopping the leading magnitude remains squared modulus, while spin-orbit-dependent hopping can produce anisotropic exchange. Multiple orbitals, Hund coupling and direct exchange can change the sign or structure and are absent here.

Hspin=∑⟨ij⟩J(Si⋅Sj−14),J=4∣t∣2U>0,Es=−J,Et=0H_{\rm spin}=\sum_{\langle ij\rangle}J\left(\mathbf S_i\cdot\mathbf S_j-\frac14\right),\qquad J=\frac{4|t|^2}{U}>0,\qquad E_s=-J,\quad E_t=0
The spin operators are dimensionless Si = σi/2 within the singly occupied sector; J has energy units. The one-bond singlet-triplet splitting is J for the once-counted bond convention.↔ Scroll wide equations horizontally.

Benchmark the approximation before using a lattice

The Hubbard dimer can be diagonalized exactly in its singlet sector. Its lower singlet energy contains a square root that expands to the second-order result plus higher even powers of hopping. This exact benchmark reveals the leading error without invoking a thermodynamic phase transition. Agreement for a small hopping ratio establishes the effective bond scale, but does not solve a frustrated quantum spin lattice or establish its ordered state.

At finite doping the low-energy subspace also contains mobile holes, so a spin-only Hamiltonian is insufficient and projected hopping must remain. Even at half filling, longer virtual loops generate additional interactions at higher order. Exchange sets a useful temperature scale after division by Boltzmann constant, but an ordering temperature also depends on dimension, anisotropy and fluctuations. The two-level lab can illustrate singlet-state superposition algebra only after a basis is declared; it cannot perform the strong-coupling projection or represent an interacting electron lattice.

Esexact=U−U2+16t22=−4t2U+16t4U3+O(t6/U5)E_s^{\rm exact}=\frac{U-\sqrt{U^2+16t^2}}2=-\frac{4t^2}{U}+\frac{16t^4}{U^3}+O(t^6/U^5)
This exact energy is for an isolated two-site, two-electron Hubbard dimer with real hopping, zero on-site energies and no chemical-potential offset. Its triplets have energy zero in that same convention.↔ Scroll wide equations horizontally.

Check the exchange against the dimer

A Hubbard dimer has t=0.10 eVt=0.10\,\mathrm{eV} and U=2.0 eVU=2.0\,\mathrm{eV}. Compute the leading exchange, approximate singlet energy and exact singlet energy. Estimate the relative error in the predicted splitting.

  1. The small ratio is t/U=0.05t/U=0.05. The exchange is J=4(0.10)2/2.0=0.020 eVJ=4(0.10)^2/2.0=0.020\,\mathrm{eV}, so the projected triplet is at zero and the approximate singlet is at −0.020 eV-0.020\,\mathrm{eV}.

  2. The exact energy is [2.0−4.0+0.16]/2=−0.019804 eV[2.0-\sqrt{4.0+0.16}]/2=-0.019804\,\mathrm{eV}. It is less negative than the second-order estimate, consistent with the positive fourth-order correction.

  3. The exact splitting is 0.019804 eV0.019804\,\mathrm{eV}. The excess predicted splitting is approximately 0.000196 eV0.000196\,\mathrm{eV}, or about 0.99%0.99\% of the exact value. This checks a bond-scale expansion, not a bulk ordering temperature.

The result

The leading exchange is 20 meV; the exact singlet-triplet splitting is 19.8 meV, giving about a 1% second-order error.

Explore the model

Interactive lab

Interactive lab · quantum

Two spin configurations as a limited basis analogy

Reset and set BOTH Mixing angle θ = 90° and Relative phase φ = 0°, then Reset and set θ = 90° and φ = 180°. Interpret the basis states as |↑↓⟩ and |↓↑⟩ only for comparing symmetric and antisymmetric superpositions; predict equal basis probabilities in both. This pure two-state lab does not include double occupancy, compute exchange energies or solve a many-electron Hubbard model.

Two spin configurations as a limited basis analogyA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRelative phase distinguishes spin superpositions even when their basis probabilities agree.

Check your understanding

For the once-counted bond Hamiltonian J(Si·Sj − 1/4), with dimensionless spin-one-half operators and positive J, what is the singlet-triplet energy difference?

Try it yourself

Take t=0.20 eVt=0.20\,\mathrm{eV} and U=4.0 eVU=4.0\,\mathrm{eV} in the same half-filled strong-coupling convention. Compute J and the scale J/kB using kB=8.62×10−5 eV/Kk_B=8.62\times10^{-5}\,\mathrm{eV/K}. Does the result determine a Néel temperature?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The exchange is an energy. Converting it to kelvin gives a microscopic scale, not an automatic transition temperature.

Compare with the solution

The exchange is 4(0.20)2/4.0=0.040 eV4(0.20)^2/4.0=0.040\,\mathrm{eV}. Its temperature scale is 0.040/(8.62×10−5)≈464 K0.040/(8.62\times10^{-5})\approx464\,\mathrm K. This does not determine a Néel temperature: ordering depends on lattice dimension, frustration, anisotropy and fluctuations, and an ideal two-dimensional isotropic short-range Heisenberg model does not develop conventional finite-temperature continuous-spin order. The calculation only identifies the leading low-energy exchange scale under the stipulated single-orbital assumptions.

Take it with you

  • Virtual charge excitations produce low-energy spin interactions at second order.
  • The positive coefficient 4|t|²/U requires the stated orbital, filling and bond conventions.
  • An exchange scale does not solve the spin lattice or determine its ordering temperature.

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