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?


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? $\endgroup$ – user2723984 Nov 11 '20 at 14:11
  • $\begingroup$ Which statement? "Topological means large scale"? how familiar are you with topology, in general? Have you checked the wikipedia entry? $\endgroup$ – AccidentalFourierTransform Nov 12 '20 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. $\endgroup$ – user2723984 Nov 13 '20 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. $\endgroup$ – AccidentalFourierTransform Nov 14 '20 at 14:45

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.