# Tag Info

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The defining property of the double semion model is the nature of the ground state as a superposition of loop pattern with alternating signs, and not the form of its Hamiltonian. As you noticed, it is not clear how to count loops on a square lattice. As far as I see, this is one reason why the string-net models are defined on honeycomb lattices, since it ...

2

The vector (1,1,1) is normal to a plane x+y+z=c. The nearest lattice point is (+1,-1,0) from this. There are six of these, and it is easy to show the distance from eg +1,-1,0 to +1,0,-1 or +1,0,-1 to 0,0,0 are equal, and thus it must be a triangular lattice. The same vector in N dimensions (1,1,1,...), produces a simplex-based lattice, is usually how one ...

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The easiest way is to simply construct the (111) face and then it will be self-evident.

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Generally speaking we use the term adsorption when we're talking about a specific molecular interaction. In contrast the term condensation is really a phase transition of a bulk fluid. Consider the interaction of water with a silica surface (this is a system I studied about 30 years (!!!) ago). A clean silica surface contains OH bonds that have a high ...

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Adsorption can catalyse condensation, but it's a distinct process. Molecules adsorb to a surface and then act as a (heterogeneous) nucleus for the gas-to-liquid phase transition (i.e. condensation).

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because the index of summation only relates to G,you can forget about "k",and also k=G+k(that shows the transnational symmetry). and look here.

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Topological order can not be described in Ginzburg-Landau symmetry breaking paradigm. It is actually fair to say that topological order are more or less the properties of (gapped) quantum phases that can not be captured by GL. One way to define it is to use the notion of adiabatic continuity: if two gapped phases of matter can be connected by adiabatically ...

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An ordered SDW phase breaks both the continuous $SU(2)$ spin-rotation symmetry and the time-reversal symmetry (because the presence of either of these two symmetries would force the order parameter of SDW vanishing). It is the spontaneously broken of continuous spin-rotation symmetry that leads to the gapless Goldstone mode. Here is a related issue. The ...

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The interaction term is not invariant under the gauge transformation. $b_k\rightarrow e^{i\theta_k}b_k$ is only a symmetry for the non-interacting Hamiltonian $H=\sum_k E(k)b_k^\dagger b_k$.

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For the question itself, since in the Brillouin zone $k$ and $k+G$ are simply the same point, $\theta(k)$ has to be the same as $\theta(k+G)$. However, in the paper arxiv.org/pdf/1105.4867v2.pdf, the reason why they performed a gauge transformation $c_k\rightarrow c_k e^{i(k_x-k_y)/2}$ is that the Hamiltonian $h(k)$ they took as the starting point (see the ...

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Q1: Consider the example of a Haldane phase protected by time-reversal symmetry. There is a spin-1/2 on the edge, and to polarize it (so the system is gapped and the ground state is unique) one must break the symmetry. In this case, the ground state is always degenerate when the symmetry is not broken. Cenke Xu's statement is better understood for ...

1

Naively, we would say: Since the Bloch waves are an energy eigenbasis, and the unitary operator $\psi_k \to \mathrm{e}^{\mathrm{i}\theta(k)}\psi_k$ defined on that basis is obviously diagonal, hence commutes with the Hamiltonian, and therefore is a symmetry, without any restriction on $\theta$. Now, the "the Bloch waves" in the first sentence needs a ...

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The first step, probably you have already know, is to get the ground state. We may write the Hamiltonian on a set of orthogonal basis, e.g., \left| n_{1,\uparrow} n_{2,\uparrow} \cdots n_{N,\uparrow}, n_{1,\downarrow} n_{2,\downarrow} \cdots n_{N,\downarrow} \right\rangle =\prod_{i} (c_{i,\uparrow}^{\dagger})^{n_{i,\uparrow}} \prod_{i} ...

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I just found that the solution of $SU(2)$ matrices is really simple. When there is no hopping term, the projected spin state of the above $d+id$ mean-field Hamiltonian indeed has TR symmetry. Because there exist global $SU(2)$ matrices $G_i$ which implement the TR transformation, say $G_i=i\tau ^x$.

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All topological insulators can be classified according to their symmetry classes. There is time reversal symmetry ($T$) , charge conjugation symmetry ($C$) and the combination $S=T*C$ symmetry. The $T$ and $C$ symmetries can be either positive or negative, i.e the energy spectrum may change sign under the symmetry operation. The $p+ip$ superconductor is a ...

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I don't know what is a "p+ip insulator". There is indeed p+ip superconductor, which belongs to the class D and characterized by an integer invariant, the Chern number.

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yes you can write the green function in $G(w,k)=w-E(k)$ form if you have diagnolized your hamiltonian

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The Reason for ongoing flow is very simple: the cooper pairs do not have a choice. There is no resistance and the magnetic field of the current acts like mass in mechanics. The current in a circular super conductor just "goes on". The real question (which is not so easy) is, how do You start some current in the first place? I think marad has the ...

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This is a interesting question. I think the above two answers have already provided what you want to understand. However, here I propose a new question: is it possible for the Cooper pair to carry a supercurrent when their total momentum is exactly zero? (Note that in the above discussion, the supercurrent occurs when the Cooper pair carry finite ...

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This is an excellent question. One of the very few textbooks, that I found actually talking about it is Ibach and Lüth's "Solid-State Physics: An Introduction to Theory and Experiment", chapter 10.6. Supercurrents and Critical Currents. They show that a Galilean transformation applied to Cooper pairs only changes the phase of the superconducting order ...

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Obviously if one starts from spin-1/2 physical electrons and want to get effective spinless fermions, one has to break time-reversal symmetry. But let us imagine that we just have spinless electrons to begin with, and time-reversal symmetry is just complex conjugation(since now electrons do not have any internal degrees of freedom). Assume that ...

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The non-interacting classification was obtained in the seminal "periodic table" papers by Kitaev and Ryu/Snyder/Furusaki/Ludwig. The interacting classification of 2D fermionic SPT phases with time-reversal symmetry has been considered by several groups already, including: Gu and Wen's group super-cohomology theory, http://arxiv.org/abs/1201.2648 Kapustin ...

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You're probably misled by the usual form factor of diamonds. Let's first think about a common material that's easy to break: glass. Think about a glass window pane and a glass marble. I think you'll intuitively say that it's a lot easier to break the window. But what if you'd put the marble on a concrete floor and hit it with a hammer? You'd crush the ...

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The link you provided already had enough information. Well, unlike hardness, which denotes only resistance to scratching, diamond's toughness or tenacity is only fair to good. That is, it is easily breakable by a hammer. The toughness of diamond is about 2.0 MPa which is good compared to other gemstones, but poor compared to most engineering materials. So ...

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This may not be how people generally think about topological degeneracy, but if you define it to be any degeneracy or asymptotic degeneracy in the thermodynamic limit with the following two properties: (1) it cannot be lifted by local perturbative Hamiltonians that respect the symmetry; (2) it cannot be attributed to symmetry breaking, i.e. irreducible ...

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The Kitaev model belongs to class D of the Altland-Zirnbauer classification. Here's the periodic table of non-interacting (gapped) fermionic topological systems. The one circled in red corresponds to the 1D $p$-wave superconductor (or Kitaev chain). As you can see from the symmetry columns, it only possesses particle-hole symmetry ($\Xi$), while the ...

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I suspect this relates to the phase diagram of sugary water. When a liquid contains a solute (sugar for example) that lowers the melting point of water, then as you cool it down you get a segregation: a small ice crystal forms, and at the tip of that crystal pure water will crystallize. The sugar is pushed into the solution where it further lowers the ...

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