I know of several definitions of phases of matter:

  • The first is the "old" one, Landau theory and symmetry breaking. In this definition we pick a local order parameter $m$ (as far as I can tell this is quite vaguely defined). If the state had the same symmetry of the Hamiltonian, we would have $\langle m\rangle=0$, but as some parameter is varied and crosses a critical value, we observe spontaneous symmetry breaking and $\langle m \rangle\neq 0$. We know this definition does not describe all phases as it doesn't include topological phases.
  • Let $H_0$ and $H_1$ be two Hamiltonians. The two Hamiltonians (or their ground states) are in the same phase if there exists a continuous gapless path $\gamma \mapsto H_\gamma$ interpolating them. If we include the requirement that $H_\gamma$ must have some symmetry for all $\gamma$, we encounter the notion of symmetry protected topological order.
  • One of the keywords in Landau's definition is local order parameter. We can find new phase transitions by considering non-local order parameters, such as the string order parameter in the Haldane phase.

Now: the Haldane phase is the prototypical example of both spontaneous symmetry breaking signalled by a non-local order parameter and symmetry protected topological order. Since in SPTO we require the whole phase to have a symmetry, by definition the symmetry must be broken to exit the phase. Since we call this "topological" order, I guess this symmetry breaking cannot be detected by a local order parameter, so, my questions are the following:

  • Is the symmetry breaking between SPTO phases always signalled by a non-local order parameter?

  • Conversely, is symmetry breaking with a non-local order parameter always SPTO (there exists a gapless path...)

  • In general, what is the relation between non-local order parameters and topological phases?


2 Answers 2


When you refer to SPTO transitions as being 'nonlocal symmetry-breaking transitions', I guess you are referring to the idea of 'hidden symmetry-breaking'? This indeed the old way of looking at SPTOs (i.e., 80s and 90s), before the more modern understanding in terms of topological invariants was developed. For clarity, let me briefly review this with an illustrative example before then addressing your questions:

Minimal SPT example in 1D

The minimal SPT phase in one dimension is the Haldane phase, e.g., protected by $\mathbb Z_2 \times \mathbb Z_2$ symmetry. A minimal exactly-solvable spin chain in this phase is the cluster model: $$ H_\textrm{cluster} = -\sum_n X_{n-1} Z_n X_{n+1},$$ where I denote the spin-$1/2$ Pauli matrices as $X,Y,Z$. The protecting symmetries are $P_1 = \prod_n Z_{2n-1} $ and $P_2 = \prod_n Z_{2n}$. It turns out that the ground state has long-range order in a string order parameter: $$ \lim_{|n-m|\to \infty} \langle X_{2m} Z_{2m+1} Z_{2m+3} \cdots Z_{2n-3} Z_{2n-1} X_{2n} \rangle \neq 0. $$ Note that the endpoint operator of the string (e.g., $X_{2n}$ on the right) is charged under the $P_2$ symmetry. Hence, it looks a bit like a symmetry-breaking order parameter, except for the nonlocal ($P_1$) string connecting the two endpoints.

''Hidden'' symmetry-breaking

Now, there is in fact a nonlocal mapping that makes this into a symmetry-breaking phase: define $$ \tilde X_n = \cdots Z_{n-5} Z_{n-3} Z_{n-1} X_n \qquad \textrm{and} \qquad \tilde Z_n = Z_n.$$ Ignoring boundary issues, this gives a well-defined Pauli algebra. In these new variables, we have $$ H_\textrm{cluster} = - \sum_n \tilde X_{n-1} \tilde X_{n+1}. $$ It is just an Ising chain on the even and odd sites! Moreover, the above string order parameter now just becomes $$ \lim_{|n-m|\to \infty} \langle \tilde X_{2m} \tilde X_{2n} \rangle \neq 0, $$ i.e., it is just the conventional symmetry-breaking order parameter. If you wish, you can see this as a derivation of the earlier claim that the original cluster model has long-range order in that string order parameter (since in these effective Ising variables it is obvious that we have the local order parameter).

Pros and cons

A positive feature of this mapping is that it makes readily apparent that, e.g., the cluster model $H_\textrm{cluster}$ cannot be adiabatically connected to the trivial Hamiltonian $H_\textrm{triv} = - \sum_n Z_n$ (at least if we preserve $\mathbb Z_2 \times \mathbb Z_2$ symmetry). Indeed, under the above nonlocal mapping, the latter maps to $H_\textrm{triv} = - \sum_n \tilde Z_n$, which is symmetry-preserving and must thus be separated from the $-\sum_n \tilde X_{n-1} \tilde X_{n+1}$ phase by a thermodynamic phase transition.

The downside is that the nonlocal mapping hides the physics of the original model. After all, the cluster model has no symmetry-breaking, in particular, its ground state is unique for periodic boundary conditions. (This is possible due to the nonlocal mapping we used, which inevitably screws up when attempting to deal with periodic boundary conditions.) A more modern understanding of SPT phases is built on the concept of symmetry fractionalization (which I don't really want to go into here, if not just for space constraints, but for further reading I can point you to the intro of my paper here: p2 and p3 of arxiv:1707.05787; disclaimer: I did not invent the concept of symmetry fractionalization, and all the appropriate references can be found there).

Answering your questions

Is the symmetry breaking between SPTO phases always signalled by a non-local order parameter?

Roughly speaking: yes. More precisely, for 1D SPT phases this general construction was done in 2012 by Pollmann and Turner ( arxiv:1204.0704 ). This works for the vast majority of SPT phases, although there are funky exceptions discussed in the paper (which have thus far not appeared in any concrete model in the literature). Relatedly, Else, Barlett and Doherty ( arxiv/1304.0783 ) generalized the above nonlocal mapping from SPT to symmetry-breaking phases. Moreover, in higher dimensions the notion of symmetry fractionalization also allows to prove that one has long-range order for a symmetry membrane operator dressed with an appropriate unitary on its boundary (i.e., the higher-dimensional analogue of what we saw above for the cluster model, where the string had some endpoint operators). However, in that case I am not aware of a dual 'hidden symmetry breaking' picture (which is yet another downside of this way of understanding SPT phases).

Conversely, is symmetry breaking with a non-local order parameter always SPTO (there exists a gapless path...)

Sticking to the simpler one-dimensional case for convenience: basically, yes. More precisely, if you have an on-site symmetry $\prod_n U_n$ which has long-range order with some endpoint operator $\mathcal O_n$, and this endpoint operator is non-trivially charged under some (other) symmetry of your model, then you have a non-trivial SPT phase on your hands (the one caveat is that you have to make sure that this charge is discrete and cannot be smoothly connected to zero, e.g., $-1$ is not a discrete charge for $U(1)$, but it is a discrete charge for $\mathbb Z_2$).

In general, what is the relation between non-local order parameters and topological phases?

I hope the above comments already address this point, at least for the case of SPT phases. Even intrinsic topological order can be related to the long-range order of certain string order parameters, known as the Bricmont-Frohlich-Fredenhagen-Marcu order parameter, which is conceptually completely unrelated to the above SPT case (for a condensed matter perspective, see this paper by Gregor, Huse, Moessner and Sondhi: arxiv:1011.4187 ).

  • 1
    $\begingroup$ Thank you very much for this comprehensive answer! $\endgroup$ Commented Feb 11, 2021 at 16:36

Topological means large scale. If you can detect a property by looking at a small patch of your space, then that property is not topological, by definition. Something is topological when you need to look at points separated by a finite distance.

A local order parameter is an operator situated at a point. As such, it can only probe that point, and a small neighborhood around it. You can never detect a topological transition by looking at a specific point.

A non-local order parameter is an extended operator. As such, it can probe points separated by a finite distance. Topological properties can only be detected by such operators.

That being said, non-local operators can also detect local properties. They are sensitive to large-scale structure, but (as a matter of principle) they can also be sensitive to short distances. For example, given one non-local operator you can construct another non-local operator by multiplying the first by a local operator.The resulting object is sensitive to the physics around the position the local operator.

So, in summary: topological order requires extended operators, but the converse is not necessarily true – symmetry breaking detected by a non-local operator need not be topological.

  • 1
    $\begingroup$ Thanks! So, if I understand correctly, topological phase transitions are always signalled by a non-local order parameter, but the converse is not necessarily true. The first statement looks very much not trivial to me, do you happen to have a reference? $\endgroup$ Commented Nov 11, 2020 at 14:11
  • $\begingroup$ Which statement? "Topological means large scale"? how familiar are you with topology, in general? Have you checked the wikipedia entry? $\endgroup$ Commented Nov 12, 2020 at 18:27
  • 1
    $\begingroup$ I meant the statement "topological phase transitions always have a non-local order parameter", which seems to imply that if there exists no gapless continuous path between two Hamiltonian, then there exists a non-local observable that is 0 in a phase and non-zero in the other. This does not look trivial at all to me, and I'm pretty sure it's not in the Wikipedia entry for topology. Apologies if I'm wrong. $\endgroup$ Commented Nov 13, 2020 at 8:38
  • $\begingroup$ I see, I thought that by "first statement" you meant the first statement in my answer, not in your comment. Anyway, I am not claiming that every non-local operator will detect a topological phase transition; what I am claiming is that if an operator detects it, then the operator will be non-local. The reason is the one I give in the answer: local operators are insensitive to global properties: they only see what is immediately next to them. So if the operator is to detect a topological phenomenon, the operator cannot be local, i.e., it will be non-local. $\endgroup$ Commented Nov 14, 2020 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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