Suppose we have some Hamiltonian $H$ which is known to exhibit spontaneous symmetry breaking (SSB), at least in some parameter regime. For simplicity, we might consider the 2D Ising model below its critical temperature.

Now suppose we add some weak perturbation such that the new Hamiltonian is

$$ H^{\prime} = H + \lambda W, $$

where we assume that $W$ is bounded above by $H$, $\lVert W \rVert \leq \lVert H \rVert$, and $\lambda \ll 1$ is a dimensionless parameter controlling the strength of the perturbation. Note that I do not assume that $W$ preserves the symmetries of $H$.


Can we say anything about whether or not $H^{\prime}$ will also exhibit SSB? Furthermore, if it does exhibit SSB, how are the new symmetry breaking states related to the old ones? Are they the same, at least up to some $\mathcal{O}(\lambda)$ error? If there is a phase transition, are the critical parameters the same?

I believe the folklore answer to this question is "$H^{\prime}$ will still exhibit SSB for $\lambda \ll 1$, and the new symmetry breaking states are the same as the old ones", at least in the simple case of the Ising model. However, I would be interested to hear if any of this has been rigorously proven, again even if only for some specific models.


I will attempt to clarify my question. I am mainly interested in the equilibrium states of the perturbed Hamiltonian, and whether they break the symmetry of the original Hamiltonian, while still forming a closed space under the action of the original symmetry group.

Copying from my earlier comments, I believe one example of where this happens is with discrete time crystals. For concreteness let us consider a many-body localized time crystal, following this paper as a reference. There one drives a system of qubits periodically with a two-pulse protocol. This can be described by the Floquet operator

$$ U_{f} = \exp(-i t_{0} H_{\mathrm{MBL}}) \exp\left(i t_{1} \sum_{i} \sigma_{i}^{x}\right), $$

where $H_{\mathrm{MBL}}$ is given by

$$ H_{\mathrm{MBL}} = \sum_{i} (J_{i} \sigma_{i}^{z}\sigma_{i+1}^{z} + h_{i}^{z} \sigma_{i}^{z} + h_{i}^{x} \sigma_{i}^{x}), $$

with the parameters $J_{i}$, $h_{i}^{z}$ and $h_{i}^{x}$ chosen uniformly from $J_{i} \in [(J/2), (3J/2)]$, $h_{i}^{z} \in [0, h^{z}]$ and $h_{i}^{x} \in [0, h]$. If $t_{1} = \pi/2$ then the spins are exactly flipped during the first pulse. In this case, if one takes some observable, say $\sigma^{z}$ for one of the spins, and calculates the Fourier spectrum of its expectation value, one will see a peak at $\nu_{\mathrm{drive}}/2$, where $\nu_{\mathrm{drive}} = (t_{0} + t_{1})^{-1}$ is the drive frequency. In this way, we say that the discrete time translation symmetry is spontaneously broken.

The relevance of this example to my question is that, even if $t_{1} = \pi/2 + \epsilon$ for some small $\epsilon > 0$, this peak in the Fourier spectrum remains pinned at $\nu_{\mathrm{drive}}/2$, even though naively one would expect it to move if the spins aren't perfectly flipped. This is observed in experiments such as this one.

What I am wondering is how common this "robustness to error" is in conventional examples of SSB.

  • 1
    $\begingroup$ Can you provide a reference for the "Folklore answer"? Would be interesting to learn about the details $\endgroup$
    – ohneVal
    Aug 9, 2018 at 12:41
  • $\begingroup$ I'm afraid I don't have a single good reference for the "folklore answer". It's mainly motivated from examples. As an example, in discrete time crystals, one drives a system of qubits periodically with a two-pulse protocol. The first pulse consists of some interactions, and the second pulse flips all the spins. This Hamiltonian has a $\mathbb{Z}$ time translation symmetry by construction, but this symmetry is spontaneously broken because the spins need to be flipped twice to get back to their original states. What is non-trivial, and related to this question, is that even if... (cont.) $\endgroup$
    – anon1802
    Aug 9, 2018 at 13:27
  • $\begingroup$ one doesn't perfectly flip all the spins, the system still remains in the original symmetry breaking state (one talks of a "subharmonic peak at 1/2 the drive frequency" which is robust to pulse errors). See this paper for a reference. $\endgroup$
    – anon1802
    Aug 9, 2018 at 13:28
  • $\begingroup$ Your example doesn't match your question. Nowhere you talk about Floquet physics, it all sounds like ground state or (thermal) equilibrium physics. $\endgroup$ Aug 9, 2018 at 14:13
  • $\begingroup$ The common theme is "spontaneous symmetry breaking". In the discrete time crystal literature, the robustness of the peak at $\nu_{\mathrm{drive}}/2$ to weak perturbations is highlighted as a crucial criterion for this to be considered SSB. I don't have much familiarity with the literature discussing conventional SSB, so I am wondering whether this criterion of "robustness to error" is motivated by comparison with conventional SSB. See e.g. this paper for an example of this discussion. It even has "rigidity" in the title! $\endgroup$
    – anon1802
    Aug 9, 2018 at 14:26

1 Answer 1


No, even infinitesimal perturbations will remove the SSB if they explicitly break the symmetry. A symmetry can't be spontaneously broken if it's already explicitly broken.

  • $\begingroup$ Thanks for your answer! I think I should have been more specific about what I was interested in. Of course, your statement that "A symmetry can't be spontaneously broken if it's already explicitly broken" is true. However, might it not still be the case that the equilibrium states of the perturbed Hamiltonian individually break the symmetry of the original Hamiltonian, but remain transitive under the original symmetry group? (cont.) $\endgroup$
    – anon1802
    Aug 9, 2018 at 12:48
  • $\begingroup$ For example, if we perturb the classical Ising model by a symmetry breaking term $\epsilon \sum_{\langle i j k \rangle} \sigma_{i} \sigma_{j} \sigma_{k}$, then I believe that for $\epsilon$ sufficiently small the equilibrium states are still those with the spins pointing all up or all down. See the comments to this question and the book referenced therein for details. $\endgroup$
    – anon1802
    Aug 9, 2018 at 12:51
  • $\begingroup$ Thus, even if the new system doesn't technically exhibit spontaneous symmetry breaking because there is no symmetry to break, from a macroscopic point of view it will still look like it does. $\endgroup$
    – anon1802
    Aug 9, 2018 at 12:56
  • $\begingroup$ @OliverLunt (1) Edit this into your question. (2) Give an example where this happens. The example you quote seems incorrect. $\endgroup$ Aug 9, 2018 at 13:33
  • $\begingroup$ @NorbertSchuch I have performed the edits you asked for. Please let me know if you would like me to clarify anything further. $\endgroup$
    – anon1802
    Aug 9, 2018 at 13:56

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