I am interested in understanding patterns of spontaneous symmetry breaking in spin chains. I want to understand what happens when I have "competing" orders, like symmetry breaking orders that are probed via $\langle \sigma^z_i \sigma^z_{j}\rangle$ and $\langle \sigma^x_i \sigma^x_{j}\rangle$. I call these competing because $\sigma^z_i$ and $\sigma^x_i$ do not commute. I am particularly interested in when I have simultaneous ordering of two competing orders.

In particular, consider a length $L$ ($L$ is even) qubit chain in periodic boundary conditions. Suppose that it explicitly has a $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry generated by $\prod_{i=1}^L \sigma^x_i$ and $\prod_{i=1}^L \sigma^z_i$.

What is an example of such a spin-$1/2$ chain that spontaneously breaks both $\prod_{i} \sigma^x_i$ and $\prod_{i} \sigma^z_i$?

Here are some of the features that I identify with spontaneous symmetry-breaking of all of $\mathbb{Z}_2 \times \mathbb{Z}_2$. I expect there to be a ground-state manifold of four states that are extremely close in energy (splitting decaying like $e^{-cL}$) separated by a gap to the rest of the spectrum. The four states will have eigenvalues $(+1, +1), (-1,+1), (+1, -1),$ and $(-1,-1)$ under the above $\mathbb{Z}_2 \times \mathbb{Z}_2$.

I know that such a symmetry-breaking happens in spin-$1$ chains of a curious flavor, like in $$-\left( \sum_i S^x_i S^x_{i+1} + a (S^y_i e^{i \pi S^z_i}) (e^{i \pi S^x_{i+1}} S^y_{i+1}) + b S^z_i S^z_{i+1} \right)$$ which is identified (after the application of a nonlocal unitary in open boundary conditions) with the string order in regimes of the anisotropic $XYZ$ Heisenberg antiferromagnet.

However, I'm having a hard time finding a spin-1/2 model that shows this sort of symmetry breaking. For spin-$1/2$ particles, $e^{i \pi S_i^z} = i \sigma^z_i$ and so on, and so the above Hamiltonian is no longer $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetric. I've tried $-\left( \sum_i (S^x_{i} S^x_{i+1} + S^y_{i} S^y_{i+1}) + \epsilon (S^x_{i} S^x_{i+1} S^x_{i+2} S^x_{i+3} + S^y_{i} S^y_{i+1} S^y_{i+2} S^y_{i+3}) \right)$, but that appears to be gapless in the spin-$1/2$ model.

  • 1
    $\begingroup$ Sorry, I misread. $\endgroup$
    – Buzz
    Apr 15 at 20:01
  • $\begingroup$ Note that in $\mathbb{Z}_2 \times \mathbb{Z}_2$ the generators of the two copies of $\mathbb{Z}_2$ commute. The group generated by two (global) spin flips is the Pauli group, with 16 elements. $\endgroup$ May 5 at 22:06
  • $\begingroup$ @JulesLamers Could you comment on this further? When $L$ is a multiple of $4$, I get only four elements in the group generated by $\prod_i X_i$ and $\prod_i Z_i$. $\endgroup$
    – user196574
    May 5 at 23:18
  • $\begingroup$ @user196574 The $L$-dependence only sits in the ('global') representation, on the Hilbert space. Abstractly, there are two generators $X,Y$ that square to the identity $I$ and obey $X\,Y=-X\,Y=i\,Z$ and $X\,Y\,Z=i\,I$. So we have to include $1,i,-1,-i$ times $I,X,Y,Z$ to get a set that is closed under multiplication. The result is a group, called the Pauli group. See also en.m.wikipedia.org/wiki/Pauli_group $\endgroup$ May 6 at 6:30

1 Answer 1


If you do not want to impose translation symmetry, there are trivial (but valid) examples. If the number of sites is a multiple of 4, $$ H = -\sum_{i \text{ even}} (\sigma^x_i\sigma^x_{i+2} + \sigma^z_{i-1}\sigma^z_{i+1}) $$ spontaneously breaks both symmetries and has four ground states $$ (\vert \uparrow \uparrow \uparrow \dots \rangle_{\text{odd}} \pm \vert \downarrow \downarrow \downarrow \dots \rangle_{\text{odd}}) \otimes (\vert \rightarrow \rightarrow \rightarrow \dots \rangle_{\text{even}} \pm \vert \leftarrow \leftarrow \leftarrow \dots \rangle_{\text{even}}) $$ in the four symmetry sectors you mentioned. The case with single-site translation symmetry seems more difficult -- it would be interesting if there is an example there.

  • $\begingroup$ +1 Thanks for this example. I'm imagining your model as a two-leg square ladder without couplings between the legs. I suppose there should be some choices of weak couplings that would keep the order above, like a weak $\sigma^x_i\sigma^x_{i+2} \sigma^z_{i-1}\sigma^z_{i+1}$, which is a point in the quantum Ashkin-Teller phase diagram. (I think technically the broken symmetries are $\prod_{i\, odd} \sigma_i^x$ and $\prod_{i\, even} \sigma_i^z$, though I need to think on it more :)) If you don't mind, I'm going to explicitly add the constraint of one-site translation invariance to my question. $\endgroup$
    – user196574
    Apr 16 at 23:19
  • $\begingroup$ @user196574 that's right. I think the order should be stable to any symmetric perturbation, but I don't have a proof. I disagree that those are the broken symmetries -- you decide the symmetries first to decide the space of Hamiltonians, and then you check whether a specific Hamiltonian spontaneously breaks any of the chosen symmetries, and whether that is stable to any symmetric perturbation. $\endgroup$ Apr 17 at 15:05
  • $\begingroup$ With one-site translation symmetry, there seems to be some LSM physics here, but not sure how illuminating that will be for your question, in particular take a look at section 5.3, theorem 5.2. $\endgroup$ Apr 17 at 15:08
  • $\begingroup$ Thanks. I'm tempted that even a weak $\sigma^z_{i} \sigma^z_{i+1}+\sigma^x_{i} \sigma^x_{i+1}$ perturbation might affect the symmetry breaking in the model above. This perturbation is symmetric under the $\mathbb{Z}_2 \times \mathbb{Z}_2$ symmetry in the question but not symmetric under $\prod_{i\, even} \sigma_i^z$ and $\prod_{i\, odd} \sigma_i^x$ which I think is what's broken in this example. This is a little bit of an aside, since a weak perturbation doesn't restore the one-site translation symmetry, but I might ask this as a separate question since now I'm curious. $\endgroup$
    – user196574
    Apr 17 at 16:16
  • $\begingroup$ I agree that LSM is probably at play. Another paper also discusses discrete-symmetry LSM, which I hadn't appreciated. It's tricky because I think LSM doesn't forbid symmetry-breaking and just forbids unique gapped ground states in PBC, but it often points to a big change in the phase diagram. $\endgroup$
    – user196574
    Apr 17 at 16:30

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