# Tag Info

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The material may become opaque for radiation that has frequency close to resonance frequencies of the material. The electrons in matter are sensitive to radiation in certain ranges of frequencies (absorption peaks or bands) and can get excited. This is accompanied by stronger absorption of the radiation. The electrons are sensitive to certain resonance ...

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The general two-particle state will look like $\displaystyle \int dp_1 dp_2 \psi(p_1,p_2) a^\dagger_{p_1} a^\dagger_{p_2}| 0\rangle$ Here $\psi(p_1,p_2)$ is the momentum-space wavefunction. Since the creation operators commute, only the symmetric part matters, so we may as well take $\psi(p_1,p_2)=\psi(p_2,p_1)$ (there would be a minus sign if they were ...

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"Total spin conservation" means global $SU(2)$ spin-rotation symmetry (a continuous symmetry) of the Heisenberg model, and "spin wave" indicates an ordered ground state that spontaneously breaks the spin-rotation symmetry. Thus, according to Goldstone theorem, there must be a gapless mode for spin wave.

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In general, both IQHE and FQHE are rigid quantum states, whose rigidness is protected by the finite energy gap ($h\omega$ for IQHE) between the ground state(s) and the exited states. Finite temperature can support excitations to overcome the gap, which destroys the rigidness of the state. Under finite temperature, the quantization of the Hall conductivity is ...

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Yes, there is a structural reason for the existence of flat band on the Kagome lattice. This is related to the wave function localization due to the destructive interference on the lattice. The flat band has many physical interpretations. In the momentum space, looking at the dispersion relation, a flat band means the effective mass of the particle is ...

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The short answer is yes. One can convince oneself this is indeed the case by doing the dimensional counting as it was done by Everett You. However, it is by no means a proof. The problem is that the valence bond states are not linearly independent. Even though there are much more valence bond states than the number of singlets made from $N$ spin-one-half ...

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The short answer is, yes, the chiral edge state is determined by bulk topological property. It is known as bulk-edge correspondence. The paper you should read is: Protected edge modes without symmetry - 1301.7355. Generically, to determine whether the edge states are robust is by determining whether the edge states are protected'' by any of the three ...

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A rule of thumb in electrons systems is that only the states close to the Fermi surface are perturbed. If you apply a perturbation of typical frequency $\omega$, then the states of energy between $\epsilon_F-\omega$ and $\epsilon_F+\omega$ will be perturbed. Same thing with the temperature.

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What comes into my mind is the three-magnon process, in which an optical magnon with $k=0$ and $\omega=\omega_0$ splits into two acoustic magnons, one $k=k_1, \omega=\frac{\omega_0}{2}$, the other $k=-k_1, \omega=\frac{\omega_0}{2}$. Note that in this process energy and momentum are conserved. This 3-magnon process can lead to a finite life time. Not sure.

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Yes, a spin-singlet state is also an RVB state. The valence bound states (singlet-product states) over-complete the Hilbert space of spin-singlet states.

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You might find Franchini's lecture notes on the Bethe Ansatz techniques useful. Chapter four is relevant for your question I would say.

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So a "pure" paramagnetic system has a positive magnetic susceptibility and for which the dipoles don't interact. Usually the effect of the "thermal bath" outweighs an applied magnetic field so the net magnetization is 0. I would say that in general the answer is yes seeing as it takes a Squid magnetometer to detect a paramagnetic system, but it is certainly ...

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The best way is to use the inverse value, the compressibility, this one is more easilly found.

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The expression for $\epsilon$ above written as real + imaginary part is $$\epsilon(\omega) = 1- \frac{\omega_p^2}{\omega^2 + \Gamma^2}+\frac{i\omega_p^2\Gamma}{\omega^3 + \omega\Gamma^2},$$ so in fact it is the imaginary part that goes to infinity as $\omega$ is lowered down to 0. What does this mean? $\epsilon(\omega)$ for metals is defined by the ...

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The law of mass actions says that in steady-state or equilibrium the product for electron concentration $n$ and hole concentration $p$ is a constant at all locations in a semiconductor, $$np=n_i^2$$ It's true that the intrinsic carrier concentration $n_i$ is a function of temperature but the law does not break down as this would break charge neutrally. ...

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A Google books search for "bulk modulus of liquid helium" turned up this result: Helium, edited by Paul Muljadi. On page 7, you will find the value of the bulk modulus as on the order 50 MPa. There is a reference linked to this value, but it is not part of the free preview, so I cannot tell you what it is.

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It is the atoms (more precisely, the electrons of the atoms) that contribute to X-ray diffraction. The aperture comparison could work, but you should see the atoms themselves as the apertures. You can invoke the separation between structure factor (given by the lattice) and form factor (given by the shape of the repeating unit, in your case the atoms). The ...

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Yes, there is such a point. The precise formula varies as a function of the scattering geometry, but if we consider a special case: normal incidence on a flat sample and small scattering/diffraction angle it is quite simple: the scattered intensity is proportional with the sample thickness $d$ but it gets attenuated as $\exp(-\mu d)$ (the Beer-Lambert ...

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With solids atoms are mostly locked in place so it makes sense there can be lots of different crystal structures and atomic packings. For liquids and gases though their defining characteristic is that their atoms are mobile enough to flow and fill a container. You can't both have structure and mobility.

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Yes, it is called a black hole. Some are not stable due to too little mass, or other problems. Example:Photon Sphere

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They don't relate. The primitive unitcell is a property of the lattice. The lattice has nothing to do with the basis. You can have a single atomic basis or a thousand atomic basis but both have the same lattice, and therefore the same primitive unit cell. One of the most important facts you have to wrap your head around is that a lattice point has nothing ...

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Again, thanks to the $SU(2)$ PSG proposed by prof.Wen, I can answer my question now, $THT^{-1}$ is in fact $SU(2)$ gauge equivalent to $H$, and the statement "$H$ is also not SU(2) gauge equivalent to the time-reversal transformed Hamiltonian $THT^{-1}$" in my question is wrong. Let's rewrite the Hamiltonian as ...

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I don't know the article you refer to, but I believe the Hamiltonian you discuss should get a $\pi$-phase shift after one turn around a (2D) lattice cell. So I guess it should read $H=F^{\dagger}\cdot H_{\pi}\cdot F$ with H_{\pi}=t\left(\begin{array}{cccc} 0 & e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4}\\ e^{-\mathbf{i}\pi/4} & 0 & ...

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Temperature in the classical model is mapped to imaginary time in the quantum model. By analytic continuation, one can obtain the real-time evolution. The matrix elements of the time-evolution operator of the quantum model at zero temperature will get mapped to the matrix elements of the transfer matrix of the classical model at an appropriate temperature ...

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Simple, combine both real- and $\mathbf{k}$-space pictures! The basic idea is to split up your $n$-dimensional system into multiple $(n-1)$-dimensional systems. For example, say you have a 2D square lattice and you define your edges along the $x$-direction. Then you need to break the 2D lattice into 1D lattices pointing in the $x$-direction. In other words, ...

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First of all, the Pauli matrices are not space-time dependent, so of course you can pass the derivative right through them. Second of all, $\operatorname{Tr} [\partial(\vec{\pi}\cdot\vec{\tau})]^2 = \operatorname{Tr}\partial_\mu \pi^i \partial^\mu \pi^j \tau^i \tau^j$ Now remember $\tau^i \tau^j = i \epsilon_{ijk} \tau^k + \delta^{ij} I_{2x2}$ So ...

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Topological degeneracy is only defined in the thermodynamic limit on a closed manifold. The ground state degeneracy of a finite-sized system or on an open manifold is not "topological", and can not be called topological degeneracy. Considering your examples. (1) The ground state degeneracy is ill-defined with open boundary condition. Because there might be ...

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