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I don't think that is the case. A useful reference is: http://journals.aps.org/rmp/abstract/10.1103/RevModPhys.80.1083. One way to approach a theory of anyons is to start by writing down a list of particle types along with their fusion rules. Once doing this one may obtain consistency equations from solving the hexagon and pentagon equations arising from ...


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This is a very a general question, I think I could provide some insight but it will certainly need to be elaborated on by someone with this specific expertise. The $\mathbb{Z}_k$ para fermions arise in several statistical mechanics models. They are both interesting and subtle because their exchange statistics depend on their positions (in one-dimension). ...


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I think I can answer my question now. The Lieb's theorem here is a statement about which flux configuration (or flux configurations) minimizes the ground state energy of the quadratic Hamiltonians with each corresponding flux configuration/sector. In the case that $J_x=J_y=0$ (or more generic cases that $J_xJ_yJ_z=0$), the corresponding $x,y$ link terms are ...


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TylerHG: Yes it is easy to calculate the density of states. But what I'm really asking here is "why." Note that a thin circular ring in $\mathbf{k}$-space of thickness $dk$ has area $dA=2\pi k\,dk$ (by elementary geometry). In $E$-space, since $E\propto k^2$, that ring corresponds to a patch of width $dE=2k\,dk$. Thus $$\frac{dA}{dE}=\pi.$$ But ...


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It is easy to show that the total number of electrons in a 3D fermi sphere is : $$N(e)=\frac{V}{3\pi^2}*k_F^3$$ Where $k_F$ is the Fermi wave vector and $V$ is the real space volume of your sphere. Now if you rearrange for $k_F$ in terms of the total number of electrons you'll get a particular equation. It is know that ...


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The classical situation with no symmetry breaking is the case of the, so-called, isostructural transitions. The word "isostructural" is misleading, since what is meant is "isosymetric". However, historically the term emerged. There is a number of examples of such transiotions. One is the alpha-alpha' transitions in the hydrogen-metal systems, another is ...


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Here we go : Lets considere the lagrangian density of our free membrane. For convenience, we will use a cartesian set of coordinates, so that : $$\mathcal{L}=\frac{\rho}{2}\dot{h}-\frac{\kappa}{2}\left[{\partial_x}^2 h+{\partial_y}^2 h\right]^2$$ Apparently, this lagrangian depends on a quite large number of variables : ...


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In Mahan may-particle physics P15: (for bilinear Hamiltonian)It is only necessary to find the eigenvalues of the Hamiltonian matrix. Usually the matrix is of infinite dimensionality. But one may often diagonalize it exactly for many problems. Computers allow very accurate solutions for any case of interest. If all Hamiltonians had only bilinear ...


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Why do you say for N particles you have N! distinguishable state? Lets just make sure you got it right. Lets say you have N particles on a lattice having M sites. In the ideal case, no excluded volume therefore each particles can access M sites. The total number of states for distinguishable particles in M^N. Now, if you considere that the particles are ...


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It's all about the electronic structure. "Electronic structure" is the term for the available energy states and transitions of electrons in the crystal. Largely, the nuclei in a crystal are much of a muchness; big positive things that don't move a lot. The electrons, however, can move around to different degrees depending on how many of them there are, the ...


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Fermi pockets (or Fermi surfaces) are contours of Fermi energy in the Brillouin zone. Depending on the effective mass $m^*$ of quasi-particles, the Fermi pockets can be divided into electron pockets (if $m^*>0$) and hole pockets (if $m^*<0$). For weakly interacting Fermion systems, according to the Fermi liquid theory, all the low-energy physics ...


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Because in the bulk, the energy between two dislocations is proportional to the distance between them. This means the dislocations are confined in the bulk. So they can not appear in side the crystal.


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The surface of the film is a [001] plane through the silicon (or diamond) lattice, space group Fd-3m (#227),


1

Indeed there is no relation between the sign of a potential and the sign of its Fourier transformation. But why should we care about this? In the field theory, the criterion is very simple, an interaction is attractive if its coefficient (in the Hamiltonian) is negative, and is repulsive if its coefficient is positive. According to this criterion, Altland ...


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One can notice that: $$(n_{i\uparrow}-1/2)(n_{i\downarrow}-1/2) = n_{i\uparrow}n_{i\downarrow} -\frac{1}{2}(n_{i\uparrow}+n_{i\downarrow}) +\frac{1}{4} $$ To show the equivalence you can absorb the $(n_{i\uparrow}+n_{i\downarrow})$ term in the chemical potential. We don't care about the kinetic term, and have: $$ ...


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I do not think this is a real space statement. The precise statement is: when you have a particle-hole pair with momentum $\vec q$ and another particle-hole pair with the opposite momentum (look back at the definition of $\rho$, you will see it represents a particle-hole pair), when the exchange phonon has a frequency smaller than that momentum, the ...


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These two models are exactly the same. The first model has the additional advantage that particle-hole symmetry takes place at $\mu = 0$. The second model takes place at $\mu = U/2$.


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1) Gauge theory is a theory where we use more than one label to label the same quantum state. 2) Gauge “symmetry” is not a symmetry and can never be broken. This notion of gauge theory is quite unconventional, but true. When two different quantum states $|a\rangle$ and $|b\rangle$ (i.e. $\langle a|b\rangle=0$) have the same properties, we say that there ...


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Let me answer your first question: Phase transition do not necessarily imply a symmetry breaking. This is clear in the example your are mentioning : The liquid-gas transition is characterized by a first order phase transition but there is no symmetry breaking. Indeed, liquid and gas share the same symmetry (translation and rotation invariance) and may be ...


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The energies of electrons in tightly bound states (some of which might be referred to as "core states", depending on circumstances) are very nearly the same as the energies of the free molecule. These electrons do not interact very much with their neighbors, so their properties are nearly as if they are isolated. In such cases, with little or no overlap ...


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The negative sign comes from the anti-commutation of the fermion operators $\psi$ and $\psi^\dagger$. That is $\psi^\dagger(r)\psi(r') = - \psi(r')\psi^\dagger(r) + \delta(r-r')$. The delta function just gives the some constant that will be absorbed into the $const$ term. Maybe it is bothering you that this term is formally divergent but this is removed by ...



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