Frustration and Order by Disorder

Frustration occurs when local energetic preferences cannot all be satisfied. It can create an extensively or subextensively degenerate classical ground-state set, but degeneracy alone does not imply a spin liquid. Thermal, quantum, or quenched fluctuations often lift an accidental degeneracy and select an ordered state—a mechanism called order by disorder.

Required background. Exchange Interactions and Effective Spin Hamiltonians supplies competing bonds and symmetry constraints; Spin Waves and Magnons supplies harmonic fluctuation spectra. Helpful background. Vector Models, Auxiliary Fields, and Large-N Saddles supplies a complementary controlled treatment of constrained spins.

Constraint competition and classical manifolds

For antiferromagnetic Heisenberg spins on one triangle,

$$H_\triangle=J(\mathbf S_1\cdot\mathbf S_2+ \mathbf S_2\cdot\mathbf S_3+ \mathbf S_3\cdot\mathbf S_1) =\frac J2\left(\mathbf S_1+\mathbf S_2+\mathbf S_3\right)^2 -\frac{3JS^2}{2}.$$

The minimum satisfies $\mathbf S_1+\mathbf S_2+\mathbf S_3=0$, realized by coplanar $120^\circ$ order. Every bond cannot be antiparallel. On a lattice of corner-sharing triangles or tetrahedra, the local zero-sum constraints may leave many global configurations. Whether their number is extensive and whether they are connected by local moves are separate questions.

Another route is competing exchanges, such as a square-lattice $J_1$–$J_2$ model. At special ratios, several ordering wave vectors or relative sublattice angles can share the same classical energy. Some degeneracy follows from an exact symmetry; some is accidental. Fluctuations cannot lift symmetry-required degeneracy, but they can lift accidental degeneracy without explicitly breaking the Hamiltonian symmetry.

Thermal and quantum selection

Expand about a classical ground state labeled by $\lambda$. If the nonzero harmonic stiffnesses are $\kappa_\nu(\lambda)$, the low-temperature classical contribution is

$$F_{\mathrm{th}}(\lambda)=E_{\mathrm{cl}} +\frac T2\sum_\nu\ln\!\left(\frac{\beta\kappa_\nu(\lambda)}{2\pi}\right)+\cdots .$$

States with more low-stiffness fluctuations have greater entropy and lower free energy. Quantum zero-point motion instead gives

$$E_{\mathrm{q}}(\lambda)=E_{\mathrm{cl}} +\frac12\sum_{\nu,\mathbf k}\hbar\omega_{\nu\mathbf k}(\lambda)+O(S^0).$$

The state with the smallest zero-point correction is selected. Thermal and quantum selection need not agree because one depends logarithmically on stiffness while the other sums frequencies. Villain and collaborators gave the canonical thermal construction Villain et al. 1980, pp. 1263–1272; fluctuation selection among frustrated spin states was systematized by Henley 1989, pp. 2056–2059.

Accidental zero modes

If a harmonic mode is exactly soft for every $\lambda$, the Gaussian determinant is singular and quartic terms set its scale. Simply deleting the zero eigenvalue can reverse a selection. One must separate symmetry Goldstone modes, which occur for all states, from accidental soft modes whose phase-space volume differs. Finite size, boundary conditions, and the order in which $T\to0$ and $N\to\infty$ are then consequential.

What selection does and does not establish

Order by disorder is demonstrated by calculating a fluctuation-dependent free-energy difference within a controlled expansion and showing that the selected state is stable after interactions. Observing an ordered state that belongs to a classically degenerate set is not enough: weak further-neighbor exchange, dipolar interactions, strain, or quenched disorder may select it at the classical level.

Conversely, failure of harmonic selection does not establish a quantum spin liquid. Anharmonic terms, nonperturbative tunneling, or small symmetry-allowed couplings may act at lower scales. A credible hierarchy compares the predicted selection energy with omitted exchanges, disorder broadening, finite-size splitting, and experimental transition scales.

The semiclassical parameter is usually $1/S$, improved by high coordination but degraded by a dense set of soft modes. For $S=1/2$ frustrated magnets, spin-wave calculations remain informative diagnostics, yet unbiased numerics or exact constraints are needed to determine the phase.

Exercises

1. Derive the $120^\circ$ condition for three equal classical spins on an antiferromagnetic triangle.

Solution

The rewritten Hamiltonian is minimized when the vector sum vanishes. Three equal vectors summing to zero form a closed equilateral triangle in spin space, so each pair has dot product $S^2\cos120^\circ=-S^2/2$.

2. Two classically degenerate states have harmonic frequencies $(\omega,2\omega)$ and $(\sqrt2\omega,\sqrt2\omega)$. Are they selected differently at harmonic order?

Solution

The quantum frequency sums are $3\omega$ and $2\sqrt2\,\omega$, respectively. Since $2\sqrt2<3$, quantum fluctuations favor the second state. Their thermal products are both $2\omega^2$, so the logarithmic harmonic free energies are equal. This illustrates that thermal and quantum criteria differ.

References

• Henley, C. L. “Ordering Due to Disorder in a Frustrated Vector Antiferromagnet.” Physical Review Letters 62 (1989): 2056–2059. DOI.
• Villain, J., R. Bidaux, J.-P. Carton, and R. Conte. “Order as an Effect of Disorder.” Journal de Physique 41 (1980): 1263–1272. DOI.