In the seminal work by Levin and Gu in 2012 ( Braiding statistics approach to symmetry-protected topological phases ) they give a concrete prescription for how to gauge a global symmetry to a local one. As is well-known, this involves introducing degrees of freedom which live on the links of the original lattice. To avoid speaking in too general a case, let me specify to the case of the toric code. Their procedure basically ends up with the following form of the toric code:

$$ H'_\textrm{TC} = - \sum (A_v + B_p \; P_p) $$

where $A_v$ and $B_p$ are the usual vertex and plaquette terms, respectively, and $P_p$ is a projector which is zero whenever one (or more) of the vertices $v$ neighboring the plaquette does not satisfy $A_v = +1$ (for those who'd like to check my statement, see Eq.19 of that paper). Contrast this to the `usual' toric code model:

$$ H_\textrm{TC} = - \sum (A_v + B_p) .$$

Is this $H'_\textrm{TC}$ really what we want? Suppose we start in the ground state ($A_v = B_p = 1$ for all $v,p$). If we introduce a single $e$ excitation (i.e. there is one vertex with $A_v = -1$), then now in the Hamiltonian all $B_p$ terms disappear for all plaquettes neighbouring this vertex. This means that there is zero energy cost to then create a flux $m$. This seems undesirable? Or should I just see this as a perturbation of the TC where the $e$ and $m$ particles have a local attraction (but they remain non-interacting at long-distances, which is the crucial thing)?

(To be clear: they are not working in the limit of $H_\textrm{TC}$ where the coupling in front of $A_v$ goes to infinity, which would end up in an effective Hilbert space where $A_v = +1$. They address that case as well, which they call the `restricted toric code', a special limit of $H'_\textrm{TC}$.)

  • $\begingroup$ I think your understanding is correct that there is effectively an attraction between $e$ and $m$ in this case. $\endgroup$ – Everett You Feb 4 '18 at 1:09

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