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  Topological order vs. Symmetry breaking: what does (non-)local order parameter mean?

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Topological order are sometimes defined in opposition with the order parameter originating from a symmetry breaking. The latter one being possibly described by a Landau theory, with an order parameter.

Then, one of the distinctions would be to say that topological order can not be described by a local order parameter, as e.g. in this answer by Prof. Wen. I thus suppose that a Landau theory describes a local order parameter.

I have deep difficulties to understand what does local mean ? Does it mean that the order parameter can be inhomogeneous (explicit position dependency), as $\Delta\left(x\right)$ ? Does it mean that one can measure $\Delta$ at any point of the system (say using a STM tip for instance) ? (The two previous proposals are intimately related to each other.) Something else ?

A subsidiary question (going along the previous one of course) -> What is the opposite of local: is it non-local or global ? (Something else ?) What would non-local mean ?

Any suggestion to improve the question is warm welcome.

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user FraSchelle
asked Jul 15, 2013 in Theoretical Physics by FraSchelle (390 points) [ no revision ]
I'm no expert on topological order. But I found this article very enlightening: dx.doi.org/10.1063/PT.3.1641 This article, at least for me, represents the big moment of epiphany when topological order first made sense to me; it would be a bold (probably arrogant) claim to say it made me "understand" it. I actually find it very convenient to think of topological order from a disorder perspective: a system with topological order can look disordered from a Landau perspective, i.e. no measurable local order parameter but a finite topological order parameter.

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user NanoPhys
@NanoPhys Thanks so much for this nice pedagogical review by Read. It definitely helps, even if I need more time to (start to) appreciate some details about the topological order. Particularly the point [Quote: There remains the possibility of degenerate ground-state that cannot be mapped onto one another by application of any local operator and that cannot be distinguished by the expectation value of any local operator. ] I think I'll need several lives to understand

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user FraSchelle

1 Answer

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In theories with spontaneous symmetry breaking, the phase transition can usually be characterized by a local order parameter $\Delta(x)$, which is not invariant under the relevant symmetry group $G$ of the Hamiltonian. The expectation value of this field has to be zero outside the ordered phase $\langle\Delta(x)\rangle = 0$, but non-zero in the phase $\langle\Delta(x)\rangle \neq 0$. This shows that there has been a spontaneous breaking of $G$ to a subgroup $H\subset G$ (where $H$ is the subgroup that leaves $\Delta(x)$ invariant).

What local means in this context, is usually that $\Delta(x)$ at point $x$, can be constructed by looking at a small neighborhood around the point $x$. Here $\Delta(x)$ can be dependent on $x$ and need not be homogeneous. This happens for example when you have topological defects, such as vortices or hedgehogs. One powerful feature of these Landau-type phases, is that there will generically be gapless excitations in the system corresponding to fluctuations of $\Delta(x)$ around its expectation value $\langle\Delta(x)\rangle$ in the direction where the symmetry is not broken (unless there is a Higgs mechanism). These are called Goldstone modes and their dynamics are described by a non-linear $\sigma$-model with target manifold $G/H$.

An example is the order parameter for s-wave superconductors $\langle\Delta(x)\rangle = \langle c_{\uparrow}(x)c_{\downarrow}(x)\rangle$, which breaks a $U(1)$ symmetry down to $\mathbb Z_2$. But there are no Goldstone modes due to the Higgs mechanism, the massive amplitude fluctuations are however there (the "Higgs boson"). [Edit: see EDIT2 for correction.]

A non-local order parameter does not depend on $x$ (which is local), but on something non-local. For example, a non-local (gauge-invariant) object in gauge theories are the Wilson loops $W_R[\mathcal C] = \text{Tr}_R{\left(\mathcal Pe^{i\oint_{\mathcal C}A_\mu\text dx^\mu}\right)},$ where $\mathcal C$ is some closed curve. The Wilson loop thus depends on the whole loop $\mathcal C$ (and a representation $R$ of the gauge group) and cannot be constructed locally. It can also contain global information if $\mathcal C$ is a non-trivial cycle (non-contractible).

It is true that topological order cannot be described by a local order parameter, as in superconductors or magnets, but conversely a system described by a non-local order parameter does not mean it has topological order (I think). The above mentioned Wilson loops (and similar order parameters, such a the Polyakov and 't Hooft loop), is actually a order parameter in gauge theories which probe the spontaneous breaking of a certain center-symmetry. This characterizes the deconfinement/confinement transition of quarks in QCD: in the deconfined phase $W_R[\mathcal C]$ satisfies a perimeter law and quarks interact with a massive/Yukawa type potential $V(R)\sim \frac{e^{-mR}}R$, while in the confined phase it satisfy an area law and the potential is linear $V(R)\sim \sigma R$ ($\sigma$ is some string tension). There might be other examples of spontaneous symmetry breaking phases with non-local order parameter. [Edit: see EDIT2.]

Let me just make a few comments about topological order. In theories with with spontanous symmetry breaking, long-range correlations are very important. In topological order the systems are gapped by definition, and there is only short-range correlation. The main point is that in topological order, entanglement plays the important role not correlations. One can define the notion of long-range entanglement (LRE) and short-range entanglement (SRE). Given a state $\psi$ in the Hilbert space, loosely speaking $\psi$ is SRE if it can de deformed to a product state (zero entanglement entropy) by LOCALLY removing entanglement, if this is not possible then $\psi$ is LRE. A system which has a ground state with LRE is called topological order, otherwise its called the trivial phase. These phases have many characteristic features which are generally non-local/global in nature such as, anyonic excitations/non-zero entanglement entropy, low-energy TQFT's, and are characterized by so-called modular $S$ and $T$ matrices (projective representations of the modular group $SL(2,\mathbb Z)$).

Note that, unlike popular belief, topological insulators and superconductors are SRE and are NOT examples of topological order!

If one requires that the system must preserve some symmetry $G$, then not all SRE states can be deformed to the product state while respecting $G$. This means that SRE states can have non-trivial topological phases which are protected by the symmetry $G$. These are called symmetry protected topological states (SPT). Topological insulators/superconductors are a very small subset of SPT states, corresponding to restricting to free fermionic systems. Unlike systems with LRE and thus intrinsic topological order, SPT states are only protected as long as the symmetry is not broken. These systems typically have interesting boundary physics, such as gapless modes or gapped topological order on the boundary. Characterizing them usually requires global quantities too and cannot be done by local order parameters.


EDIT: This is a response to the question in the comment section.

I am not sure whether there are any reference which discuss this point explicitly. But the point is that you can continuously deform/perturb the Hamiltonian of a topological insulator (while preserving the gap) into the trivial insulator by breaking the symmetry along the way (they are only protected if the symmetry is respected). This is equivalent to locally deforming the ground state into the product state, which is the definition of short range entanglement. You can find the statement in many papers and talks. See for example the first few slides here. Or even better, see this (slide with title "Compare topological order and topological insulator" + the final slide).

Let me make another comment regarding the distinction between intrinsic topological order and topological superconductors, which at first seems puzzling and contrary to what I just said. As was shown by Levin-Wen and Kitaev-Preskill, the entanglement entropy of ground state for a gapped system in 2+1D has the form $S = \alpha A - \gamma + \mathcal O(\tfrac 1A)$, where $A$ is the boundary area (this is called the area law, not the same area law I mentioned in the case of confinement), $\alpha$ is a non-universal number and $\gamma$ is universal and called the topological entanglement entropy (TEE). What was shown in the above papers is that the TEE is equal to $\gamma = \log\mathcal D$, where $\mathcal D\geq 1$ is the total quantum dimension and is only strictly $\mathcal D>1$ ($\gamma\neq 0$) if the system supports anyonic excitations.

Modulo some subtleties, LRE states always have $\gamma\neq 0$, which in turn means that they have anyonic excitations. Conversely for SRE states $\gamma = 0$ and there are no anyons present.

This seems to be at odds with the existence of 'Majorana fermions' (non-abelian anyons) in topological superconductors. The difference is that, in the case of topological order you have intrinsic finite-energy excitations which are anyonic and the anyons correspond to linear representations of the Braid group. While in the case of topological superconductors, you only have non-abelian anyons if there is an extrinsic defect (vortex, domain wall etc.) which the zero-modes can bind to, and they correspond to projective representation of the Braid group. The latter type anyons from extrinsic defects can also exist in topological order, but intrinsic finite-energy ones only exist in topological order. For more details, see the recent set of papers from Barkeshli, Jian and Qi.


EDIT2: Please see my comments below for some corrections and subtleties. Such as, it is in a sense not correct that superconductors are described by a local order parameter. It only appears local in a particular gauge. Superconductors are actually examples of topological order, which is rather surprising.

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user Heidar
answered Jul 15, 2013 by Heidar (855 points) [ no revision ]
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Now, the fact zero modes bound to extrinsic defects in topological superconductors (for example, certain p-wave superconductors) lead to non-abelian anyons is not due to the above mentioned topological order and originate from different physics. This is because, at that point superconductors are described Bogoliubov-de Gennes Hamiltonian which mathematically can be though of as an insulator with particle-hole symmetry (in the Nambu notation). When I said that topological superconductors are SRE, its from this starting point essentially. (continued)

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user Heidar
Thus the fact that Wen say superconductors are LRE, it has nothing to do with the existence of topological superconductors. It governs different kind of physics and is valid for all superconductors (no matter gap symmetry). But be warned: the above is just my current understanding, and as stated I don't understand all the details as clearly as I would like.

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user Heidar
Thanks so much for the clarification. I've tried reading the Hansson, Oganesyan & Sondhi paper few months ago without big success. When I said superconductor can be wherever you want due to gap symmetry I was indeed thinking to Elitzur's theorem which forbids a local order parameter to exist in superconductor (even for $s$-wave), whereas anyonic excitations only exist for $p$-wave. At some point Wen defines topological phase has having fractional statistic/charge at finite energy defect. I think only $p$-wave exhibits this. Thank again for the discussion.

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user FraSchelle
@Oaoa I think the point is that even an old and boring s-wave superconductor has (abelian) anyonic excitations, in the sense that it has fermions and boson with non-trivial mutual statistics. A s-wave superconductor is actually in the same topological phase as the (3 million dollar) Kitaev toric code model, the so-called $\mathbb Z_2$ topological order. However, I am becoming less certain about p-wave superconductors. I now think the correct characterization is symmetry enriched topological order (SET) (LRE + symmetry) and not SPT (SRE + symmetry) as I originally thought.

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user Heidar
Thanks a lot for the discussion, it has given me some stuff to think more deeply about that I thought I had understood.

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user Heidar
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@Oaoa: Superconductivity is much more subtle than usually presented. The slightly wrong comment of mine above is that I claimed superconductors can be described by a local order parameter $\langle \Delta(x)\rangle=\langle c_{\uparrow}(x)c_{\downarrow}(x)\rangle$, this is however not correct in a sense. According to Elitzurs theorem [prd.aps.org/abstract/PRD/v12/i12/p3978_1 ], it is impossible to find a gauge invariant and local order parameter in the presence of the electromagnetic gauge field. (continued)

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user Heidar
@Oaoa One can construct a non-local gauge invariant order parameter however, which in the Coulomb gauge reduces to the above and looks as if it was local in this gauge. Thus superconductivity cannot be described by a local order parameter, actually it turns out that superconductors are topologically ordered (in the sense of LRE, not in the sense of topological insulators/superconductors). I think one heuristic way to see this is that in the superconducting phase there is still a $\mathbb Z_2$ gauge symmetry, and discrete gauge theories are topological. (continued)

This post imported from StackExchange Physics at 2014-04-05 03:28 (UCT), posted by SE-user Heidar

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