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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?

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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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