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Let us understand Brownian motion in liquids before we look at the motion in solids. If you observe a glass of water at rest on a table, it "appears" to be motionless. However, all we need is a magnifying glass to observe the random, incessant motion of water on the surface. This random motion is a manifestation of heat. The same thing happens in a solid. A ...


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Since there is no interaction between the parts of your lattice, the response of the total system is sum of responses of the individual parts. These parts are harmonic oscillators, whose response function is well-known. If these have many different frequencies, the total system will have response function with peaks at these frequencies. No need for boson ...


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There are two unknowns A and $\rho/R_0$. So you'll need two equations to solve for two unknowns. $V$ at equilibrium separation is the dissociation energy. After you find equilibrium separation, substitute in $V$ to get dissociation energy.


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You miss first of all that Cooper pairs do not exist as some physical quantities. If you prefer, they are not particles as electrons. They are just correlations. The current is a collective response to a gradient of phase. You can generate such a gradient by a magnetic field, a voltage, a break of the condensate (like in Josephson system), with different ...


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The conduction and valence band electron are only differed by their Wannier orbital (wave function within each unit cell), so exciting a Bloch electron from one band to another changes the Wannier orbital of the electron, for example, from a bounding orbital to an anti-bounding orbital. In the simplest 1D example of a periodic potential, exciting the ...


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The statement is not true. Counter example: quantum spin ice or U(1) spin liquid. In gapless spin liquid phase, the boson (spin excitations) are emergent U(1) photons in the deconfined phase, which are gapless but not condensed.


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When you are exciting an electron from VB to CB, you normally talk in terms of transition from a state $|k\rangle$ to $|k'\rangle$ and $E(k')>E(k)$. Specifying $|k\rangle$ automatically rules out an precise determination of $|x\rangle$ as the state $|k\rangle$ is spread over the entire real space with specific weight at each point.


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I met the author of that paper today. He said it's more of an empirical observation rather than a statement based on solid arguments. He asked for a counterexample, which I don't have. Please post it here once you got one.


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DFT is exact concerning ground state properties. However, the bandgap is not a ground state property. Not sure, if this simple explanation is correct, but I find it somehpw intuitive: in order to speak about a bandgap, you either need a (at least fictitious) electron in the conduction band, which therefore is in an excited state, or you need a ...


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In a system with non-interacting particles you may know that the right quantity to look at is $\rho \Lambda^3 \sim \left(\Lambda/l \right)^3$ where $\Lambda$ is the (thermal) de Broglie wavelength and $l = \rho^{-1/3}$ the typical inter-particle distance in the system. Basically if $\Lambda/l \ll 1$, then your system behaves classically i.e. you do not ...


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Also what about the other 7/8th portion of the sphere..Is it just unoccupied by electrons? Looking at an older version of Griffiths' QM online, on page 232 and page 233 he is treating a free electron in a three dimensional box. In this case he applies boundary conditions such that the wave function is zero on the sides of the box. Therefore (just like ...


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You probably (you should really give more details when posting questions, so that those without your book can still help you) came across this in the Einstein model of the harmonic oscillators for the heat capacity estimation, in which case the energy eigenvalues in 1D (say for x-component) are given by: $$ E_{nx} =\hbar \omega (1/2+n_x) $$ With $n_x\ge 0$ ...


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Two programs that can do the job are: XCrySDen (free) and Material Studio (commercial). I think the MS visualizer is free. For XCrySDen and the other software you might need to convert from one format to another. For instance from .cif to .struct or .xyz, you can find scripts online to help you do the format conversion and then use the software to visualize ...


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+1 for python, scipy, numpy. The other advantage is that it does scripting so if you have to do this for many experiments, you'll be able to do the whole directory, rather than just one at a time.


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What you observe as mechanical deformation of a steel spring is an actual displacement (motion) of the atoms constituting the spring. In places, atoms will be slightly closer to their neighbors (compression) and in some other places actually futher apart (tension). The combination of compression on one side and compression on the other side of a beam or a ...


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Exact statements The Hohenberg-Kohn theorems, which are the theoretical foundation of DFT, essentially say that the ground state properties of a many-electron system are only a function of the electron density. Any quantity you want to calculate can be re-expressed in terms of the electron density $n(r)$, including the many-body ground state wavefunction, ...


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In principle it's an exact theory, except you don't know the expression of the exchange-correlation functional. Also exact theory is not opposite to mean field theory, in terms of which DFT can be understood.


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we see plots of the energy band structure from DFT simulation. How these eigen-energies are obtained as function of k within the DFT framework? It depends on what specific software program was used to do the calculation. Are they the physical quasiparticle energy of the system or just the eigen-energies of the Kohn-Sham equation? Well, they ...


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After Fourier transform you get an equation $\displaystyle (E-\frac{k^2}{m})\alpha_{\vec{k}}=\sum_{\vec{k}'}V_{\vec{k}\vec{k}'}\alpha_{\vec{k}'}$ Here $V_{\vec{k}\vec{k}'}\sim \int \mathrm{d} \vec{r} e^{i(\vec{k}-\vec{k}')r}V(\vec{r})$. Now we need to make some simplifying assumption about $V$. For $s$-wave, usually we take ...


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Ferromagnetism is a collective behavior so the answer to your 1st question is no. In a ferromagnetic material the moments of the magnetic atoms not only align themselves with an external magnetic field but they also align spontaneously in the same direction even without any external field. The regions in the material where the magnetic moments have the same ...


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Starting with some background information from Wikipedia, we have that under time reversal the position is unchanged while the momentum changes sign. In quantum mechanics we can express the action of time reversal on these operators as $\Theta\,\mathbf{x}\,\Theta^\dagger = \mathbf{x}$ and $\Theta\,\mathbf{p}\,\Theta^\dagger = -\mathbf{p}$. It is worth ...



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