3
$\begingroup$

The Heisenberg model$^1$ has a symmetry $O(3)=\Bbb{Z}_2 \times SO(3)$. The Heisenberg model in $d\ge 3$ has a second order para-ferromagnetic phase transition corresponding to the spontaneous symmetry breaking (ssb) of this $O(3)$ symmetry. Mermin-Wagner prevents the ssb of a continuous symmetry in $d\le 2$, for discrete symmetries the corresponding 'lower critical dimension' is $d=1$.

If everything I have said in the preceding paragraph is correct (please correct me if not). Then this begs the question in $d=2$ do we get spontaneous symmetry breaking of the $\Bbb{Z}_2$ part of $O(3)$? If so - what does it look like physically and if not why not?

$^1$ Why does a Heisenberg magnet break the O(3) symmetry in stead of SU(2)?

$\endgroup$
  • $\begingroup$ I am pretty sure that the two-dimensional ferromagnetic Heisenberg model undergoes a BKT-transition at finite temperature. Would that answer your question? $\endgroup$ – Ruben Verresen May 23 '18 at 18:48
  • $\begingroup$ I think it's important to stress that you speak of nonzero temperature. Otherwise, you can spontaneously break continuous symmetry in any dimension. Just take the Heisenberg model with a nearest-neighbor ferromagnetic coupling on a square lattice. You can easily find its exact ground state. It definitely breaks continuous spin symmetry, even in one dimension (spin chain). $\endgroup$ – Tomáš Brauner May 25 '18 at 19:15
  • $\begingroup$ @TomášBrauner That's not quite right. If you treat the system as quantum-mechanical, then the lowest dimension in which you can spontaneously break continuous symmetries at zero temperature actually depends on the dynamical exponent $z$. A ferromagnetic Heisenberg spin chain can indeed spontaneously order at zero temperature, but an antiferromagnetic one can't, because it has a lower dynamical exponent. If you treat the system as classical even at zero temperature, you always need at least three dimensions. $\endgroup$ – tparker May 25 '18 at 19:36
  • $\begingroup$ @tparker Yes, I did mean quantum-mechanical, not classical, and ferromagnetic, not antiferromagnetic :) Thanks anyway for making my comment more precise! $\endgroup$ – Tomáš Brauner May 25 '18 at 21:39
  • $\begingroup$ @tparker "The "2D Heisenberg model" [...] orders ferromagnetically below the critical temperature in a conventional second-order transtition." Do you have any reference for this statement? While it is true that Mermin-Wagner does not apply to ferromagnets in 1D at $T=0$, I believe it does apply to ferromagnets in 2D at finite temperature. (Indeed, I think this was one of the original settings for their theorem.) And it's not just me saying that: see, for example, the second paragraph in PRL 88, 047203 (2002). $\endgroup$ – Ruben Verresen May 25 '18 at 22:21
5
+50
$\begingroup$

Great question. I assume that you mean the classical Heisenberg model (whose spins are just arrows). Jump down the last sentence if you only want the actual answer.

The Heisenberg model does not have symmetry group $\mathrm{O}(3)$ - that's just the spin part of the symmetry group. The full symmetry group is $\mathrm{O}(3) \times S$, where $S$ is the space group of the lattice on which the model is defined. (The most common choice is a $d$-dimensional hypercubic lattice, for which $S$ is a hyperoctahedral group.) This is important, because while the spin and spatial symmetries act independently on the Hamiltonian, it's possible for the system's symmetry to be spontaneously broken down to a smaller symmetry group that combines them. This is sometimes called "spontaneously induced spin-orbit coupling". I'll give an example below.

In the decomposition $\mathrm{O}(3) \cong \mathbb{Z}_2 \times \mathrm{SO}(3)$ of the spin-space symmetry, the $\mathrm{SO}(3)$ corresponds to a rigid rotation of all the spins in spin space, and the $\mathbb{Z}_2$ corresponds to time reversal (TR) - i.e. simultaneously flipping all the spins. It's impossible to spontaneously break the $\mathbb{Z}_2$ TR symmetry while leaving the $\mathrm{SO}(3)$ (and, implicitly, the space group $S$) unbroken. That's because an individual spin (which is the fundamental atomic unit as long as the space symmetry remains unbroken) is just an arrow, and transforms in the dipolar representation of $\mathrm{SO}(3)$: rotating a spin about its own axis doesn't do anything. Therefore, at the level of a single spin, the $\mathrm{Z}_2$ time-reversal transformation that flips the spin doesn't actually add anything, because you can do the same thing just using the $\mathrm{SO}(3)$. (E.g. if the spin starts out pointing in the $+\hat{z}$ direction, you can reverse it to $-\hat{z}$ via the time-reversal operator, but also by rotating by $180^\circ$ about the $x$- or $y$-axes. Mathematically, we say that the dipolar representations of $\mathrm{O}(3)$ and $\mathrm{SO}(3)$ are isomorphic.) Since the $\mathbb{Z}_2$ time-reversal symmetry doesn't do anything independently of the $\mathrm{SO}(3)$ in this case, it doesn't make any sense to break it individually.

But once you factor in the space symmetry group, things get much more interesting. It can spontaneously break along with magnetic symmetry in a way that, roughly speaking, spontaneously enlarges the magnetic unit cell. And if the enlarged magnetic unit cell contains multiple spins, then it transforms under spin-space rotations as a multipolar (rather than dipolar) representation of $\mathrm{SO}(3)$. In these higher representations, the $\mathbb{Z}_2$ and $\mathrm{SO}(3)$ operators are not equivalent, because the enlarged magnetic unit cell can be chiral. For example, if you have three labeled spins, then any $\mathrm{O}(3)$ transformation of all three spins together will either be chiral and require a time-reversal spin-space inversion, or not, and the two cases can't be related by a simple $\mathrm{SO}(3)$ rotation.

In this case you can indeed just break the $\mathbb{Z}_2$ while preserving the $\mathrm{SO}(3)$. But you can't see the effects by just considering a single spin, which will remain $\mathrm{SO}(3)$ symmetric and therefore "not point anywhere in particular". You need to look at the correlated behavior of several spins simultaneously to observe the TR-symmetry breaking. The most natural context in which this arises is when the lattice of spins contains triangles (i.e. either the triangular or kagome lattice). In this case we can consider each triangle's "scalar chirality" $\langle S_A \cdot (S_B \times S_C) \rangle$, where the $S_i$ are the spins at the three corners of the triangle. This collective order parameter is a pseudoscalar product which is invariant under $\mathrm{SO}(3)$ but changes sign under TR, so it becomes nonzero if TR is spontaneously broken, even if the $\mathrm{SO}(3)$ symmetry is preserved. (If the Hamiltonian only has $\mathrm{O}(2) \cong \mathbb{Z}_2 \times \mathrm{U}(1)$ symmetry, like the $XY$ model, then a simpler order parameter is the signed angle $\theta_i - \theta_j$ between two adjacent spins.)

Such phases do indeed occur in the Heisenberg model on the Kagome lattice with both first- and second-nearest-neighbor couplings, as described in https://journals.aps.org/prb/abstract/10.1103/PhysRevB.72.024433.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.