advanced32 min studyLesson 523 in Physics

Heisenberg and Ising spin Hamiltonians

Translate spin conventions, diagonalize a bond, and distinguish anisotropy from an approximation that discards fluctuations.

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What you’ll understand

  • Translate Pauli matrices into dimensionless spin-half operators.
  • Identify the transverse exchange terms missing from an Ising interaction.
  • Distinguish mean-field ordering estimates from exact model conclusions.

A short spin Hamiltonian carries several choices that affect every numerical prediction: operator normalization, bond counting, exchange sign, and the direction of an external field. Stating those choices is part of defining the physical model. The isotropic Heisenberg interaction allows angular momentum to move between sites, while a longitudinal Ising interaction assigns energies to fixed projection configurations. Comparing their smallest nontrivial examples makes clear why dropping transverse terms changes both eigenstates and symmetry.

Normalize operators before interpreting an exchange

Let each unordered pair occur once in H=∑⟨ij⟩Jijsi⋅sjH=\sum_{\langle ij\rangle}J_{ij}\mathbf s_i\cdot\mathbf s_j. The operators are dimensionless, so J is an energy and a physical angular momentum is ħ times s. Positive J favors antiferromagnetic correlations. An ordered sum over both i,j and j,i needs a factor one half. A paper that instead puts a minus sign before the entire sum reverses which sign of its coupling is called ferromagnetic.

For spin one half, s=σ/2\mathbf s=\boldsymbol\sigma/2, not the full vector of Pauli matrices. Consequently a coefficient multiplying a Pauli-matrix product differs by a factor of four from the same coefficient multiplying spin operators. A classical Ising variable taking values plus and minus one is another normalization. Its coupling must be translated before comparing a transition temperature or bond energy with a quantum spin-half Hamiltonian. Energy differences provide a quick invariant check on those translations.

HijH=J4σi⋅σj,HijI=Jz4σizσjz,[siα,sjβ]=iδijϵαβγsiγH_{ij}^{\mathrm H}=\frac J4\boldsymbol\sigma_i\cdot\boldsymbol\sigma_j,\qquad H_{ij}^{\mathrm I}=\frac{J_z}{4}\sigma_i^z\sigma_j^z,\qquad [s_i^\alpha,s_j^\beta]=i\delta_{ij}\epsilon_{\alpha\beta\gamma}s_i^\gamma
The first two identities apply to spin one half. The sigma matrices have eigenvalues ±1, while sz has eigenvalues ±1/2. J and Jz have energy units, and operators on different sites commute.↔ Scroll wide equations horizontally.

Transverse exchange creates different eigenstates

Resolve the scalar product into a longitudinal term and two ladder-operator terms. The longitudinal part measures the projections already present. A transverse term raises one spin while lowering its neighbor and conserves total z projection. It therefore connects the product states with opposite projections. The symmetric and antisymmetric combinations of those states become the triplet-zero and singlet eigenstates of the isotropic bond; a lone antiparallel product state is not an energy eigenstate.

For a longitudinal Ising bond, both antiparallel product states have the same energy and no such mixing occurs. The spin-only Ising Hamiltonian is diagonal in the common z basis. Adding a transverse field makes an Ising model quantum mechanically noncommuting again, so describing every Ising model as classical would be too broad. Exchange anisotropy is a physical selection of couplings, whereas mean-field factorization is a later approximation to the chosen interacting model.

si⋅sj=sizsjz+12(si+sj−+si−sj+),Es=−3J4,Et=J4\mathbf s_i\cdot\mathbf s_j=s_i^zs_j^z+\frac12(s_i^+s_j^-+s_i^-s_j^+),\qquad E_s=-\frac{3J}{4},\quad E_t=\frac J4
s±=sx±isy changes the dimensionless z projection by one. The two energy eigenvalues are for one isotropic spin-half bond with no external field; the triplet has three projection states.↔ Scroll wide equations horizontally.

Symmetry and dimensionality constrain ordering claims

The isotropic Heisenberg model has continuous global spin-rotation symmetry, whereas a longitudinal easy-axis interaction distinguishes an axis and can retain a discrete reversal symmetry. Those differences affect collective modes and thermal ordering. In an infinite one- or two-dimensional isotropic short-range Heisenberg system at positive temperature, the Mermin–Wagner result forbids spontaneous breaking of continuous spin rotation. A finite sample may have strong correlations without defining a thermodynamic singularity or evading the theorem.

A simple uniform molecular-field treatment gives the high-temperature Curie–Weiss intercept Θ=−zJS(S+1)/(3kB)\Theta=-zJS(S+1)/(3k_B) for equal nearest-neighbor couplings and coordination z in our convention. Negative J produces a positive intercept; positive J produces a negative one. This formula is an approximation, and its associated ordering scale can remain nonzero in dimensions where exact thermal order is forbidden. Frustration, anisotropy, longer-range interactions, and quantum fluctuations require further analysis; an intercept alone does not identify a Néel temperature.

Compare an isotropic and a longitudinal bond

Take two spin-half sites with J=4 meV for the isotropic Hamiltonian and Jz=4 meV for a separate longitudinal Ising Hamiltonian. Find the isotropic singlet and triplet energies and the Ising parallel and antiparallel energies. Explain what changes beyond the numerical ground-state energy.

  1. The isotropic singlet has Es=−3(4)/4=−3 meVE_s=-3(4)/4=-3\,\mathrm{meV}, and each triplet projection has Et=4/4=1 meVE_t=4/4=1\,\mathrm{meV}. The singlet–triplet gap is 4 meV.

  2. For the Ising interaction, the products of projections are ±1/4. Parallel products cost +1 meV and antiparallel products have energy −1 meV; each level has two product states.

  3. The isotropic ground state is the antisymmetric superposition of antiparallel products, with additional lowering from transverse mixing. The Ising model lacks those matrix elements and leaves the two antiparallel products degenerate. Matching coupling numbers has not made the two spectra equivalent.

The result

The isotropic energies are −3 and +1 meV; the Ising energies are −1 and +1 meV. Their ground-state structure, degeneracies, and gaps differ.

Explore the model

Interactive lab

Interactive lab · quantum

Projection alone does not specify a pure spin state

For each fresh comparison, Reset and set BOTH Mixing angle θ=90° and Relative phase φ. Compare φ=0° and φ=180° while paused. Predict equal basis probabilities one half in both cases, but Bloch x=+1 and −1. This shows how a longitudinal projection distribution can miss transverse information. The lab represents one pure two-level state, not a two-spin singlet, an exchange Hamiltonian, or a thermal many-spin ensemble.

Projection alone does not specify a pure spin stateA 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 forEqual z-basis probabilities can coexist with different transverse spin directions; the single-spin display does not model intersite entanglement.

Check your understanding

For s=σ/2, what coefficient multiplies σizσjz in the interaction Jz sizsjz?

Try it yourself

An isotropic spin-half bond has J=−4 meV and a field term b(s1z+s2z)b(s_1^z+s_2^z) with b=0.20 meV. Find its lowest energy and total spin projection. Determine the singlet energy and the gap from the lowest triplet state to the singlet.

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

Give me a hint

First obtain the exchange multiplets without the field. Then add b times the total projection; the electron magnetic moment points opposite to that projection.

Compare with the solution

The exchange triplet energy is J/4=−1 meVJ/4=-1\,\mathrm{meV} and the singlet energy is −3J/4=+3 meV-3J/4=+3\,\mathrm{meV}. The triplet has total projections −1, 0, and +1, with energies −1.20, −1.00, and −0.80 meV. Its mtotal=−1 member is lowest, corresponding to a positive field-directed electron moment. The singlet has total projection zero and stays at +3 meV. Its gap above the actual field-selected ground state is 4.20 meV, not the zero-field exchange gap of 4 meV.

Take it with you

  • Spin-half operators are Pauli matrices divided by two, so pair couplings carry a quarter.
  • Transverse exchange mixes antiparallel product states and changes the bond spectrum.
  • Mean-field scales and finite-size correlations do not override dimensional restrictions on thermal order.

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