2+1D lattice gauge theory can emerge in a spin system through fractionalization. Usually if the gauge structure is broken down to $\mathbb{Z}_N$, it is believed that the fractionalized spinons are deconfined. However in general, $\mathbb{Z}_N$ gauge theory also have a confined phase. The question is how to determine if the discrete emergent gauge theory is really deconfined or not?

For example, I am considering a $\mathbb{Z}_3$ gauge-Higgs model defined on the Kagome lattice with the Hamiltonian $H=J\sum_{\langle i j\rangle}\cos(\theta_i-\theta_j-A_{ij})$, where $\theta_i=0,\pm2\pi/3$ is the matter field and $A_{ij}=0,\pm2\pi/3$ is the gauge field. If the matter field is in a ferromagnetic phase, then I can understand that the gauge field will be Higgs out. But the matter field here is a Kagome antiferromagnet, which is strongly frustrated and may not order at low temperature. So in this case, I would suspect that the effective $\mathbb{Z}_3$ gauge theory will be driven into a confined phase. Is my conjecture right? How to prove or disprove that?

Thanks in advance.

  • $\begingroup$ Hope I'm not raising the dead here: but naively thinking, couldn't you try and compute the $\beta$-function and find out its sign? Like you do in QFTs normally? $\endgroup$ Apr 18, 2013 at 12:25
  • $\begingroup$ @Afriendlyhelper Thanks, but I am not sure what is the RG scheme for a lattice gauge theory. The lattice geometry is very important. Like the Kagome lattice I considered here is highly frustrated. Shouldn't that make a difference with the usual QFT RG? $\endgroup$ Apr 19, 2013 at 4:03
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    $\begingroup$ The only way I know to "prove" or "disprove" confinement is simulating the system on a computer. Some other techniques do exist, but every time I attend some confinement-related conference, there's some people fighting each other about the validity of these methods. BTW, computing the $\beta$-function won't work, as (if I'm not mistaken) a Higgs-phase gauge theory may still have negative $\beta$-function while being completely and utterly deconfined. $\endgroup$ Aug 14, 2013 at 15:00
  • $\begingroup$ @Afriendlyhelper: Figuring out the sign of the beta function is an indication of asymptotic freedom, but confinement requires non-peturbative methods. $\endgroup$
    – JamalS
    Apr 7, 2014 at 15:12
  • $\begingroup$ What are the examples of confined ZN gauge theory? Is that within Landau Ginzburg symmetry breaking or not? $\endgroup$
    – user32229
    Nov 8, 2016 at 16:26

1 Answer 1


I have to admit that I do not know anything about the model you are working on, but the standard way to determine whether a gauge theory is confining or not is to calculate the vacuum expectation value of Wilson loops. The latter are gauge invariant operators that describe parallel transport around a closed loop in spacetime. If the vacuum expectation of a Wilson loop decreases exponentially with the area it encloses, the theory is confining. It is also possible to formulate such loops within the framework of lattice gauge theory, which seems to be of interest for your application. For a nice and accessible introduction see chapter 82 of Srednicki's book on QFT.

  • $\begingroup$ +1 I can also vouch for the recommend text, Srednicki's 82nd chapter on wilson loops is straightforward, and applies to your problem. $\endgroup$
    – JamalS
    Apr 7, 2014 at 15:16
  • $\begingroup$ I have a question on the Wilson loop approach, do we have to choose the loop to have the shape like a rectangle whose one edge is time $T$ and the other edge is a distance $R$ and require $T\gg R$? In a theory with Lorentz invariance, this choice may not be too important because time and space are not really different. But in a discrete lattice gauge theory we may not have Lorentz invariance, so it seems this choice of the loop is relevant? Another question: can we say confinement comes if the gauge fluctuation is so strong that electric flux can penetrate the system and charges can interact? $\endgroup$ Aug 24, 2014 at 3:02
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    $\begingroup$ The wilson loops method doesn't work for a Higgs-gauge model. The reason is when you have the Higgs filed or matter field, the Wilson loop will always behave as the perimeter law. $\endgroup$
    – user21090
    Mar 10, 2015 at 9:47
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    $\begingroup$ @ hongchaniyi, could you give a Ref on what you said? " have the Higgs filed or matter field, the Wilson loop will always behave as the perimeter law?" $\endgroup$
    – user32229
    Nov 10, 2016 at 16:44

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