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So what people mean by 'non-local' varies from context to context and person to person. Wen has a very particular meaning to this. 1) In fermionization in $D=1+1$ the Jordan-Wigner fermions are, in the bosonic language, operators supported over many sites. The emergent (mutual)-fermions in the toric code are also supported at the ends of strings. 2) ...

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Metals are good conductors of electricity because the outer (valence) electrons of the metal atoms are only loosely bound to the nucleus and form molecular orbitals known as the conduction band. Electrons can move more or less freely through the conduction band and so metals conduct electricity generally well. When a metal is chemically oxidised its outer ...

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Friction causes the chalk to stay on the chalkboard. While the chalkboard appears smooth, under a microscope its surface is rough. Chalk is a much weaker material than the chalk board. When it is forced across the chalk board, small parts of chalk ('dust') are broken and remain trapped by friction in the surface asperities of the chalk board. The rougher ...

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Yes, there is a similar mean-field approach to toric code. It was introduced in Xiao-Gang Wen's book Section 10.2 and 10.3, known as the Wen plaquette model. In the toric code model, the qubits are defined on the links, which are not very convenient in terms of the spin liquid language. So the first step is to redefine a smaller square lattice, such that ...

2

So: I assume you want to diagonalize this problem by rewriting the Hamiltonian as $H=\sum E_id_i^\dagger d_i$, where $d_i$ are quasiparticle operators which obey the Fermionic commutation relations. If we only had $c^\dagger c$ terms, we would be able to write H as $$H=H_{ij}c_i^\dagger c_j$$ We could then prove that if $\{c_i\}$ obey the Fermion ...

2

Are fermions non-local objects, in a sense in which gauge bosons are not? As far as I understand, the answer is definitely NO. Fermionic particles are local objects as bosonic ones. Based merely on the non-local form of the bosonized Jordan-Wigner fermions, one cannot conclude that fermions are non-local. Jordan-Wigner transformation, like any ...

2

After x-rays hit a substance they will be scattered in all directions; if the material is a crystal then you will obtain a diffraction pattern where each point is created by the constructive interference of the scattered rays. The connection between the diffraction pattern and the reciprocal space is readily found: take a crystal and consider an atom ...

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The surface plasmons are bosons. The bosonic nature of photons is preserved. Plasmons are hybridizations of photons and excitons. Although electrons are fermions, their particle-hole excitations (excitons) are bosonic. Because to create a exciton, one needs to move an electron from one state to another, which is implemented by a fermion bilinear operator ...

2

Magnetic fields certainly can influence thermal conductivity. This shows up, not surprisingly, when there is a strong influence of the magnetic field on other properties, particularly electronic ones. One (non-metal) example is 'Thermal conductivity tensor in YBa$_{2}$Cu$_{3}$O$_{7-x}$: Effects of a planar magnetic field' by R. Ocana and P. Esquinazi, Phys ...

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I don't believe that the thermal conductivity of most metals is very sensitive to magnetic fields. Yes, there will be some field-induced band shifting in the case of an itinerant ferromagnet which, in principle, leads to a change in the density of states at the Fermi level, but that will typically be a very small effect. If the magnetic field induced ...

1

(1) For anyons to be created locally in a physical model they must be created in groups such that the local excitation is a boson or a fermion. However, the local excitation can fractionalize into anyonic parts which can propagate independently. In terms of second quantized operators the expectation is that the the local fermionic/bosonic degree of freedom ...

1

No. The elasticity of the solids is a liquid kind of character. The non newtonian fluids provide a solid kind of character. As everything is just electromagnetism, you can split your thoughts down the an single atom level, and play it with magnets on real scale. You can push your hands between magnets, and you can make them flow. In the wikipedia ...

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First of all, the SSH model does not have particle-hole symmetry. Particle-hole symmetry is an exact symmetry (at mean field level) reserved for superconductors and is an anti-unitary symmetry. A symmetric spectrum does not mean particle-hole symmetry. SSH model and end states Let me first explain a simple way to understand the topological end states. The ...

1

When a liquid or solid evaporates, it turns into a gas. In a closed container, pressure builds as gas accumulates. There are two competing processes. In the solid or liquid, the higher energy atoms at the surface fly off. In the gas, the slower atoms stick to the surface and condense. The number of atoms available to condense is proportional to the gas ...

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Since the question is rather vague, I will just give you some key points: Debye's model treats oscillation modes of a solid as sound waves (phonons) with frequency $\omega(\mathbf{k})=v|\mathbf{k}|$ ($v$ the sound velocity). As a result, with this model, Debye shows how the heat capacity is directly related to the rate of change of the energy expectation ...

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These lectures on the QHE and FQHE given at the Les Houches summer school are a great start imho.

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TL;DR In general, no. A longer but possibly irrelevant discussion follows. Consulting the classic review RevModPhys.58.323 by Rammer and Smith, the quantities you are considering are defined as (Eq. 2.5): $$G^{<}(\boldsymbol x_1,t_1,\boldsymbol x_{1'},t_{1'})=\mp i\langle \psi^\dagger_{\mathcal H}(\boldsymbol x_1,t_1) \psi_{\mathcal H}(\boldsymbol ... 1 First you need to bring it into the following form: H=\Psi^\dagger h \Psi Here \Psi is a big column vector: \Psi=(\dots, c_{m,n}, \dots, c_{m,n}^\dagger, \dots)^T Basically, the first half of \Psi are all annihilation operators, and the second half are all creation ones. If the number of sites is N, the size of \Psi is 2N. So h is a ... 1 Fermi surface is not necessarily a sphere. It can have an arbitrary shape in the reciprocal space (momentum space) that respects the symmetry of the crystal. Because crystals are periodic arrangements of atoms in direct space, in the reciprocal space it implies that the Fermi surface must repeat itself periodically in every direction. Every such "period" is ... 1 1) \exp(-1/g) is not necessarily related to bound states. In the standard QM double well problem it is the splitting, not the binding energy, that is O(\exp(-1/g)). In conformal field theories instantons can give \exp(-1/g) effects even though there are no bound states at all. 2) Instantons are one source of \exp(-1/g) effects, but there are others. ... 1 Yes, there is an bulk invariant for 3D topological insulators known as the second Chern parity P_3 [1-3], as the integral of the Chern-Simons 3-form of the (presumably non-Abelian) Berry connection \mathcal{A} (in the momentum space) over the Brillouin zone (BZ). Note that now the Brillouin zone is a 3 dimensional manifold (as a 3D torus). ... 1 The A operators and the Bs all commute with each other because they always share an even number of sites and therefore an even number of Pauli matrices. Therefore, these are all conserved quantities and can be replaced by their expectation value. The ground state is the state with eigenvalues A = 1 = B in units where \sigma is a Pauli matrix. In ... 1 The electron configuration of iron (Z=26), at least using a rather outdated notation is:$$K=2, L=8, M=14, N=2. Your question is 'how can we rearrange this to obtain $K=2, L=8, M=8, N=8$?' because you believe that's the electron configuration of some noble gas (a Main Group VIII element). There are two things fundamentally wrong with your reasoning. ...

1

You don't need the biquadratic interaction to realize the Haldane phase. The Heisenberg spin-1 chain already does the job. Adding the biquadratic term allows one to have a parent Hamiltonian for the exact AKLT wavefunction. Here is a phase diagram for the bilinear-biquadratic Hamiltonian taken from http://arxiv.org/abs/0806.1839, with the following ...

1

Clearly the "TMI" and the slave-rotor mean-field state are very different, because the TMI, as you assume, has no topological degeneracy while the other state is topologically ordered. However, I feel this answer is not very meaningful without seeing more details of the slave-rotor mean-field state. I'm afraid this is not a very well-known (or even ...

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