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Chiral $p$-wave superconductor and He A phase can be considered being equivalent phases of matter, for the following reason: The fact that the superconductor is charged does not make too much difference in this regard, because it only affects the electromagnetic response (one has Meissner effect, the other does not), but electromagnetic field is an external ...


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Let us consider the S matrix of Ising anyons as an example: $S_{\sigma\sigma}=0$. To be more precise, suppose we have four $\sigma$'s, labeled as $1,2,3,4$. We assume $1$ and $2$ are in a definite fusion channel. Then we braid $3$ around $1$ (or $2$, no difference). Then the state of $1$ and $2$ must be orthogonal to the initial one: they must be in the ...


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The general picture is the following: All Levin-Wen string nets models are conjectured to be derivable from representation categories of weak-Hopf algebras. The most well-known case is the quantum double of finite groups, the Kitaev model. Here the mapping between the ground states are worked out (using Fourier transformation on the group) in the paper ...


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The most general statement one can make about the string operators is that, for a Levin-Wen model constructed from some fusion category $\mathcal{C}$, the irreducible string operators correspond to the simple objects in the braided fusion category $Z(\mathcal{C})$, where $Z$ denotes the Drinfeld center. If one lets $\mathcal{C}$ be the category of ...


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I'm just going to put a slightly more general remark, essentially the same as the answer by Norbert. The condensation of Lagrangian subalgebra (it is not just a set, really an algebra satisfying several conditions) in a MTC always leads to a trivial phase, namely, a state that can be continuously connected to a product state (I'm ignoring a subtlety of ...


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The argument works fine for PBC, but as you observe it might give problems for OBC -- this is, the phase is the same in the bulk, but not at the boundary. This is not surprising, since depending on how you choose the boundary, it can either condense e or m particles (smooth/rough boundaries), and thus breaks e/m duality. Nevertheless, I would argue that this ...


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I am new to this and just happened to have similar questions. Here are some of my thoughts -- not sure they are correct and I would certainly welcome some discussion. I think it is important to remember what is drawn in your attached figure are Wannier orbitals rather than band states. To answer your first question -- it is not shown. The figure you ...



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