A continuous degeneracy

The classical Heisenberg model

We examine the columnar antiferromagnetic phase of the classical Heisenberg model on a square lattice given by

\[{\cal H}=\sum_{\langle ij\rangle}\sum_{\alpha\beta} S_i^\alpha J_{ij}^{\alpha\beta}S_j^\beta.\]

with exchange $J_{ij}^{\alpha\beta}=J_1\delta_{\alpha\beta}$ between nearest and $J_{ij}^{\alpha\beta}=J_2\delta_{\alpha\beta}$ between next-nearest neighbors. We assume $\left\vert{\bf S}_i\right\vert^2=1$ and $J_2/\vert J_1\vert\ge1/2$ (CAF phase) here.

The common way to find the ground state of this model is to write

\[\begin{align} {\bf S}_i&={\bf s}_0{\rm e}^{ {\rm i}{\bf kR}_i}\\ \Rightarrow{\cal H} &= \frac N2\sum_nJ_n{\rm e}^{ {\rm i}{\bf kR}_n} \end{align}\]

where the sum runs over all neighbors of an arbitrary site. We minimize $\cal H$ with respect to $\bf k$. The vector $\bf Q$ for which $\cal H$ is minimal is called the ordering vector and the spin pattern in real space is given by \({\bf S}_i={\bf s}_0{\rm e}^{ {\rm i}{\bf QR}_i}\). Here ${\bf s}_0$ denotes an arbitrary unit vector.

Columnar phase

In the CAF phase, we have two ordering vectors ${\bf Q}_0=(\pi,0)$ and $ {\bf Q}_1=(0,\pi)$ which together form the star of ${\bf Q}_0$ (or ${\bf Q}_1$). Both yield the same ground-state energy per spin

\[E= \frac12 \sum_nJ_n {\rm e}^{ {\rm i}{\bf Q}_0{\bf R}_n} = \frac12 \sum_nJ_n {\rm e}^{ {\rm i}{\bf Q}_1{\bf R}_n} =-2J_2\]

independent of $J_1$.

Double-Q structure

Physically that means the lattice decouples into two interpenetrating sublattices with exchange $J_2$ only. On each sublattice, the spins are Néel ordered, and we can rotate the two sublattices by an arbitrary angle $\phi$ relative to each other without energy cost. But why do we find only two special ground states out of the continuous manifold?

We cannot always find all states minimizing $\cal H$ by writing ${\bf S}_i$ with a single $\bf Q$ vector as above. If ${\bf Q}_1\ne{\bf Q}_0+\bf\tau$ with $\bf\tau$ a reciprocal lattice vector (like here), we can as well add two states and write

\[{\bf S}_i={\bf s}_0\cos\phi{\rm e}^{ {\rm i}{\bf Q}_0{\bf R}_i} +{\bf s}_1\sin\phi{\rm e}^{ {\rm i}{\bf Q}_1{\bf R}_i}\]

for the ground state with an arbitrary global relative angle $\phi$ and two orthonormal vectors ${\bf s}_0$ and ${\bf s}_1$. With this form the two-spin exchange is

\[{\bf S}_i{\bf S}_j = \cos^2\phi{\rm e}^{-{\rm i}{\bf Q}_0({\bf R}_i-{\bf R}_j)} +\sin^2\phi{\rm e}^{-{\rm i}{\bf Q}_1({\bf R}_i-{\bf R}_j)}\]

and the ground-state energy

\[E= \frac12\sum_nJ_n\left(\cos^2\phi{\rm e}^{ {\rm i}{\bf Q}_0{\bf R}_n} +\sin^2\phi{\rm e}^{ {\rm i}{\bf Q}_1{\bf R}_n}\right)\]

remains independent of $\phi$.

The following animated spin pattern with ${\bf s}_0=(1/\sqrt2)(1,1,0)$ and ${\bf s}_2=(1/\sqrt2)(-1,1,0)$ illustrates the degeneracy of $\cal H$:

As a guide to the eye, the sites of one of the two sublattices are connected by dashed lines.

Order from disorder

We stress that this continuous degeneracy is found in the classical case only. Quantum fluctuations lift the degeneracy and select exactly one of the two states with $\phi=0$ or $\phi=\pi/2$, a phenomenon coined as order from disorder. This appears somewhat counter-intuitive because quantum fluctuations typically destabilize classical order and even lead to the appearance of new, disordered phases.

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[1] J. Villain: A magnetic analogue of stereoisomerism : application to helimagnetism in two dimensions. Journal de Physique 38, 385 (1977).

[2] B. Schmidt and P. Thalmeier: Frustrated two dimensional quantum magnets. Physics Reports 703, 1 (2017).