# Tag Info

## New answers tagged condensed-matter

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$\def\kk{\mathbf{k}} \def\ii{\mathrm{i}} \def\qq{\mathbf{q}}$ You can prove this using translation invariance, without any other assumptions on the nature of the interacting ground state. I am going to give the argument in $D$ dimensions, which obviously holds in $D=1$ as a special case. Spatial translations of the system as a whole are generated by the ...

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@GerryHarp's answer contains the gist of the main idea, but there is a point that doesn't quite make sense: There is no such thing as bare ions crystal in solid state physics. To answer your last question: The "bare phonon frequencies" definitely do not refer to phonon frequencies in the absence of electrons. In fact, an ion crystal without electrons is ...

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Eq. 2.A.30 is a somewhat non-trivial identity for the ground-state $|\psi_0\rangle$ which only uses the ground-state property that $\eta_q|\psi_0\rangle =0$. (Of course, as the OP has noted, $c_i|\psi_0\rangle \neq 0$.) What we need to show is that $$I=\langle \psi_0 | (c_j+c_j^\dagger)(c_i+c_i^\dagger)|\psi_0\rangle =\delta_{ij}.$$ Using Eq. 2.A.37a, we ...

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Lets start with the bare ions. You set up a linear lattice of protons where the boundaries are held in place by some means. The protons want to get as far apart from each other as possible. So the minimum energy state has the protons spaced on the line with equal distances between them. So yes, the ions form a regular lattice without the electrons. Now ...

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Because in the FIR, $\omega \rightarrow 0$ and therefore the $4\pi i \sigma/\omega$ term dominates the $\epsilon_\infty$ term. It can therefore be safely neglected. You are right that the $\epsilon_\infty$ represents the contribution to $\epsilon(\omega)$ of the bound (or dipole-like) electrons.

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The mechanism for increasing the thermal conductivity is phonon assisted hopping. For a disordered system, one which do not preserve the long range order, the electronic wave function becomes localized. The wave function extent is typically much smaller than the system size and is characterized by the localization length $\xi$, a parameter in theory. In this ...

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The higher the dimension, the more phase space you have for your fluctuations to spread. Imagine you make a small local disturbance in your system (caused by e.g. thermal fluctuations). In 1D this disturbance can freely propagate in the system without decaying : it destroys long-range order. In 3D, the disturbance also propagates but it quickly dies out ...

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You can introduce the would-be'' bosonic mean field exactly, using the Hubbard-Stratonich (a.k.a partial bosonization) method, see wikipedia and Interacting fermions on a lattice and Hubbard-Stratonovich transformation and mean-field approximation . The mean field approximation correspond to performing the integral over the bosonic field using the ...

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Mean-field theory is exact (in the thermodynamic limit) in the case of long-range interaction (which is not the case for the nearest-neighbor Ising model). Therefore, mean-field theory is exact for BCS, where you have an effective long-range interaction. As for rigorous results, Bogoliubov rigorously proved that in the ground state (zero temperature) the ...

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Mean field theory is only good when fluctuations are small, which means that the free energy of a fluctuation must be much smaller than the total free energy. The free energy of the typical fluctuation is of order $kT$ and its size is determined by the correlation length $\xi$, and is of order $\xi^d$, with $d=$ dimension: $$F_{fluct}\sim ... 0 The justification of interpreting the energy band gradient with the velocity comes from semiclassics, and it turns out that the formula you give is correct only in certain circumstances and only to leading order. To attribute these terms to multiparticle effects is misleading, they are due to neglecting transitions to other bands, and they occur even if you ... 0 E_F is the highest occupied energy level, however in the context of semi-conductors E_F should be taken as the chemical potential of the semiconductor. The chemical potential is in the band gap of a semiconductor in most cases and thus it is incorrect to say that there are occupied energy levels at the Fermi level. I understand this goes against the ... 0 These two points of view are not so different in fact. To see that, let's work in the grand-canonical ensemble (which is the most natural to talk about the chemical potential in the Mott phase, since it is not well defined in the canonical ensemble). At a given (and small enough) t/U, there is a range of chemical potential [\mu_-,\mu_+] where the ... 5 Phonons follow a wave equation, which is at least in first approximation simply a standard wave equation, the only difference to relativistic particles is that the speed of the waves is not c but the speed of sound c_s. But this does not change the mathematics of the equation, so that in general there may be phonons which follow a massless wave equations ... 19 Phonons are indeed massless, as you can see from their dispersion relation or from the fact that they are Goldstone bosons. The phonon dispersion relation that you wrote down tells us that we can excite a phonon mode, with some finite momentum, using an arbitrarily small amount of energy, hence they have no rest mass (in condensed matter language, they are ... 1 To simplify things--it's just what you do with probabilities. Consider 2 non interacting particles uniformly distributed in a box of length 1. Their wave functions are:  \psi_1(x) = \Pi(x_1)   \psi_2(x) = \Pi(x_2)  where \Pi is the unit box function equal to 1 on [0, 1] and 0 elsewhere. If we were to add probabilities (or amplitudes), the ... 1 In general your relation is$$ \vec{B}(\omega) = (1 + \chi_m(\omega))\vec{B}_0(\omega) $$or in the time domain$$ \vec{B}(t) =\vec{B}_0(t) + \int\limits_{-\infty}^\infty \chi_m(t,t') \vec{B}_0(t') \;\rm{d}t' $$Only in the case of instantanous material response, i.e. \chi_m(t,t') = \chi_{m,0} \cdot \delta(t-t') , your equation is correct. This already ... 0 Consider this example: Imagine that the state vectors are represented by real-space wave functions. For any particular value of the possible location of particle A, the entire space of values for B are available for the position of B. For another location of A, again, the entire space is available for B. In order to account for all possible ... 2 The wave function for any composite system consisting of two uncorrelated (unentangled) subsystems is the (tensor) product$$ |\psi_A\rangle |\psi_B\rangle \equiv |\psi_A\rangle \otimes |\psi_B\rangle $$This multiplicative behavior of states isn't a specific feature of identical particles – or any particles. It holds for any physical system that may be ... 0 The reason is an n doped material has additional atoms (donor atoms) added near the conduction band. They provide electrons which can easily go to the conduction band, as the energy gap between the donor level and the conduction band is low. There are now more levels above the Fermi level, thus skewing the distribution of electrons higher. This is seen by ... 0 Negative sign just indicates the attractive potential. The factor of 1/2 is to average the double counting of the same term. It is the average of two terms for every value of i and j. 2 I asked my advisor this exact same question a couple years ago. He said that there's no sense of anyonic statistics in momentum space (or in any basis other than real space). The reason for this is that anyons typically emerge from a microscopic Hamiltonian that is spatially local, and so strictly speaking, anyons are only well-defined when they stay far ... 2 The anomalous Hall effect should be present in the Gapped case as well (Although I don't know if it has been experimentally observed in a gapped system). The reason is that for massive Dirac Fermions, the effective Bloch dynamics is also governed by a Berry gauge field, this time a non-Abelian gauge field. Please see Chen, Pang, Pu and Wang equations ... 1 You are correct that Goldstone's theorem does not apply to the 1-D Heisenberg chain, or indeed to any local 1-D system, because there is no continuous symmetry breaking in 1-D for local Hamiltonians. But Goldstone modes are not the only possible kinds of gapless excitations - you don't need continuous SSB for a system to be gapless. There are plenty of ... 1 The definition of a spin liquid as a spin system "with no spontaneously broken symmetries" is out of date and no longer used, partially for the reason you describe. If you perturb as spin-liquid Hamiltonian by adding small terms that break all the symmetries, then the ground state will still be a spin liquid even though there are no longer any symmetries ... 0 It's not necessary to use QFT to predict the lowest potential energy conformations of e.g. ethane and butane. Quantum chemistry, in particular the Molecular Orbital and VSEPR models, allow chemists to determine the electron configurations of such molecules. The conformational isomer with minimum internal electrostatic repulsion, that is minimum potential ... 3 The FRG can be thought of as a modern version of Wilson RG, although the technical details are of course very different. But all in all, if one could do all calculations exactly, these different versions would all be the same. Now, about these technical differences. In Wilson RG (and in Polchinski's functional version) one work with a low energy action for ... 0 The Kondo effect is a phenomenon that occurs when we have a magnetic impurity located in one place of a non-magnetic metal. The magnetic impurity has an residual spin due your electronic configuration. The electrons of the conduction band would interact with this electron via exchange interaction. We can see in equation 10 of the wiki page that the ... 0 The coherent state path integral is basically a recipe for converting a Hamiltonian into a Lagrangian. In condensed matter, we often start with a "microscopic" Hamiltonian description of a material at the level of individual atoms/electrons, and want to convert that into a Lagrangian so that we can more easily do QFT. In high energy, it's usually easier to ... 0 A very late answer, but for symmetry-protected topological (SPT) phases, I believe it is true (certainly, no counterexamples are known) that the boundary is "non-trivial" if and only if the bulk is a non-trivial SPT phase. Here "non-trivial" boundary has a very specific meaning. A boundary is "non-trivial" if there is NO symmetry-respecting terms that we can ... 2 The model of Levin and Gu is built out of products of the spin-1/2 operators S_i^x, S_i^y and S_i^z at each site i. These operators commute with each other at different sites ([S_i^\alpha, S_j^\beta] = 0 for i \neq j), which is the reason why we say this model is bosonic, and the SPT phases in this model are bosonic SPT's. By constrast, a ... 1 The presence of a ductile to brittle transition temperature implies there are insufficient (ductile) deformation modes at low temperatures to support plastic deformation and therefore fracture occurs to release energy/load. In FCC materials, dislocation slip of both edge and screw dislocations is relatively athermal and due to the number of active slip ... 1 I think what is observed is the two phases difference, not each one separately. for example take look at this one: Direct measurement of the Zak phase in topological Bloch bands As you said, the Zak phase depends on the unit cell you choose, so it can not be physical. But the difference is physical in the sense that if you fix your unit cell and start with ... 0 The quantity of electrons traveling across this interesting capacitor is necessarily hindered by the formula you used. Although there would be this consideration for electron emission, the potential through the dielectric would be chiefly concerned with the polarization and I would therefore consider that. Polarization is the net shift in dipole moments of ... 6 Wood ash is most mineral. Opinions seem to differ about the composition, but it's things like calcium and potassium carbonates, phosphates and oxides. The minerals in wood are distributed throughout the wood, so as the organic material burns away you are left with a very fine network of aggregated particles of the minerals. Googling will find you lots of ... 1 Consider the Ising antiferromagnet$$ H=\sum Z_i Z_{i+1}  on a ring. It even has product ground states $|0101\cdots\rangle$ and $|1010\cdots\rangle$ (and only those product ground states). Thus, the lowest energy product states are not translational invariant.

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There are no easy to verify criteria which will work in general. One way to see it is by noting that for every classical StatMech model, we can define a PEPS with the same correlation functions (for which the tensors can be easily constructed from the StatMech model), see http://arxiv.org/abs/quant-ph/0601075. On the other hand, for StatMech models it is ...

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