Hot answers tagged

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 ...


5

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 ...


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 ...


4

@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 ...


4

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 ...


4

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 ...


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 ...


2

$\def\kk{\mathbf{k}} \def\ii{\mathrm{i}} \def\qq{\mathbf{q}}$ 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 ...


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 ...


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 ...


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 ...


2

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 ...


1

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 ...


1

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 ...


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 ...


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 ...


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 ...


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 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 ...



Only top voted, non community-wiki answers of a minimum length are eligible