Order parameters in topologically ordered systems Topologically ordered phases are characterized by a non-vanishing ground-state expectation value of a non-local operator. These operators are supported on sites whose number grows with the system size. I always understood this statement as to also imply that the expectation value vanishes for local operators.
When I started thinking about those two statements I got confused. For example, consider the toric code:
$$
H_{toric} = -\sum_p A_p - \sum_v B_v
$$
where $A_p,\; B_v$ are the plaquette and star operators. A ground state is constructed by
$$
\vert \psi \rangle = \prod_v \frac{1}{2}(1+B_v)\vert00\dots00\rangle
$$
It follows $\langle \sigma_i^z\rangle=0$ for any site $i$, as direct calculation shows. Another way is using Elitzur's theorem, and then any gauge variant operators will vanish. But then, what about $A_p, \,B_v$ themselves? They do acquire a non-vanishing expectation value.
Another troublesome case I encountered is in this paper about the reduction to $1D$ of FQH $1/3$ Laughlin state. Equation (9) in the paper describes the occupancies:
$$
\langle n_{3i} \rangle = \frac{1}{\sqrt{4t^2+1}}
$$
for some parameter $t$ (and there are formulas for the other sites as well). This certainly does not vanish at all as well. The authors describe the non-local string operator as the order parameter. However, to me the pattern of $\langle n_{3i} \rangle,\; \langle n_{3i\pm1} \rangle$ also seem to be very distinctive. So how come they are not order parameters?
 A: Let's compare what happens in toric code with e.g. spontaneous breaking of a discrete symmetry. In the latter case, suppose the local order parameter is $O$, which transforms nontrivially under the symmetry group. There are a number of characteristic phenomena related to each other and happening at the same time for SSB:

*

*When a discrete symmetry is spontaneously broken, there are multiple degenerate ground states (on any closed manifold, or just an infinite system). The projection of $O$ to the ground state subspace is a nontrivial operator in that subspace.


*We can choose to work with a symmetry-breaking ground state where $\langle O\rangle\neq 0$, or choose a symmetric ground state in which $\langle O\rangle=0$. In both cases the two-point function $\langle O^\dagger(x)O(y)\rangle$ approaches a constant value when $|x-y|$ is large. The cluster decomposition $\langle O^\dagger(x)O(y)\rangle\approx \langle O^\dagger(x)\rangle \langle O(y)\rangle$ can fail in latter type of ground state (this is "long-range correlation").


*In addition, all of above will be immediately gone when the system transitions out of the ordered phase.
Now in a toric code, even though $\langle A_v\rangle = \langle B_v\rangle=1$ in the ground state, they do not break any symmetry of the Hamiltonian. There may be degenerate ground states when the underlying manifold is a torus or a higher-genus surface, but all local operators are proportional to the identity once projected to the ground state subspace ("locally indistinguishable"), the opposite to item 1 above. As a result, in any of the ground state, cluster decomposition always holds for any local observables (so no item 2).
If you turn on a magnetic field like $-h\sum_i \sigma^z_i$, when $h$ is large enough the topological order is killed, but $\langle B_p\rangle$ or $\langle A_v\rangle$ evolve continuously with $h$ and remain non-zero even in the trivial phase. So these local observables do not characterize the topological order in any reasonable sense.
