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In the Toric code, the excitations are e, m, fermion $\epsilon$ and vacuum. Thus, the total quantum dimension is $D= \sqrt{\sum{d_{a}^{2}}} = 2$. It seems one takes into account all sorts of possible anyons in the system and not just the minimum possible set of "most basic" ones- because fermion $\epsilon$ is just the fused anyon out of $e$ and $m$.

But for the Toric code with dislocations, we get the dislocation behave as an Ising anyon and thus we get the 'Ising anyon model'. Now, I have the Ising anyon besides the other anyons e, m, fermion $\epsilon$ and vacuum as shown here http://arxiv.org/abs/1004.1838 So, should I take all the anyons into account or not ? Here, in this paper http://arxiv.org/abs/1303.4455, to calculate $D$ for Toric code with dislocations, they take into account only the anyons of the Ising model and thus, they get $D=2$. So, why is that and what is the correct approach to calculate the quantum dimension- which anyons to take into account and which not ? Do we take only the set of anyons into account which contribute to the topological degeneracy of the system according to the definition here Quantum dimension in topological entanglement entropy ?

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The definition of $D$ requires sum over all the anyons, not just the "basic" ones -- after all, you can not really say which ones are "more basic" than others. In certain realizations, like in toric code, it is easy to write down the operators that create $e$ and $m$, but there is no sense that they are more basic than $\psi$, since you can as well take $m$ and $\psi$ as your "basis" and say $e=m\times\psi$.

Toric code with dislocations are not exactly Ising anyons. Dislocations behave like Ising anyons in many ways, but they are extrinsic defects, not true anyonic quasiparticle excitations of the system. Your question is not really about calculating the quantum dimensions (it is just a mathematical definition), because the significance of quantum dimension comes from (partly) its appearance in the topological entanglement entropy. In http://arxiv.org/abs/1303.4455 they looked at the TEE of a disk with a single twist defect. From the general TQFT argument, we expect the TEE is $\ln (D/d_\sigma)=1/2$, which is what they find.

To be more precise, one should think about the general structure of defects. These defects are symmetry defects, for example the dislocation corresponds to a symmetry of the model that exchanges $e$ and $m$, and we can consider the symmetry group to be $G=\mathbb{Z}_2$. Now for each $g\in G$, we can collect all the "defects" which are associated with the action of $g$, call them $a_g$ where $a$ labels different types of $g$ defects (not to be confused with the anyon labels). In the toric code case, there are actually two distinct types of dislocations, each of them is Ising like (see Bombin, for example). Now for each $g$ one should define a total quantum dimension $D_g=\sqrt{\sum_{a_g}d_{a_g}^2}$. However, one can actually prove that $D_g$ is always equal to the total quantum dimension of the anyon model $D$. I may be too brief, but the punch line is that now consider a region with a single $g$-defect $a_g$ in it. The TEE should be $-\ln D_g/d_{a_g}$.

For more details on the general structure you can look at http://arxiv.org/abs/1410.4540.

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  • $\begingroup$ thanks!!. I do realize the difference of defects (extrinsic anyons) w.r.t to the usual intrinsic anyons. Thanks for clarifying that for $D$ of underlying model (with defects on top of it), one takes into account just the intrinsic ones, right ? But when one has extrinsic 'Ising anyon' in Toric code, do you consider other anyon of the Ising model $\psi$ as "defect" associated with action of g. Is there an intuitive way to argue why for each $g$, one needs to define a total quantum dimension $D_{g}$ and why $D_{g}$ is same as the total quantum dimension of the anyon model $D$ ? $\endgroup$ – cleanplay Sep 24 '15 at 0:43
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    $\begingroup$ $\psi$ is the fermion in the toric code. It's not like we bring in a whole new set of Ising anyons into the theory, we only bring in the dislocation $\sigma$ and $\sigma\times\sigma =1+\psi$. When we fuse two $\sigma$, the $g$ action cancels because it is $\mathbb{Z}_2$ so we get back anyons. $\endgroup$ – Meng Cheng Sep 24 '15 at 1:53
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    $\begingroup$ As I have argued, there are actually two types of dislocations that are topologically distinct, call them $\sigma_\pm$, each with dimension $\sqrt{2}$. $D_g$ in this case should be $D_g=\sqrt{d_{\sigma_+}^2+d_{\sigma_-}^2}=2$. Heuristically, defects corresponding to different $g$ belong to completely different superselection sectors. So each sector has its own dimensions. $D_g=D$ is proven in the paper I linked. $\endgroup$ – Meng Cheng Sep 24 '15 at 1:54
  • $\begingroup$ @ Meng Cheng : thanks; I agree that $\psi$ is the fermion of Toric code to begin with. Things got clear with your second comment that there is a set of defects corresponding to each element of the symmetry group and one has to sum over them. Thanks. $\endgroup$ – cleanplay Sep 24 '15 at 3:10

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