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It is more difficult to think about the free particle case, because it is unnatural to put it on a lattice and not gap the Brillioun Zone edges. However, if you do want to do such a thing, the dispersion would look like (b) in the figure below for an arbitrary choice of lattice spacing $a$: Why is this so? It is because you have made something in real ...

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In a metal the Fermi energy is somewhere in an unfilled band. At any temperature above absolute zero (which you can never reach) there are states available for electrons to get to and result in conduction at the Fermi surface. This will occur in any metal. Superconductivity is a separate phenomena that I won't touch on here.

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At present, there is a belief (though obviously not verifiable) by solid-state physicists that a metal cannot exist at absolute zero. The Fermi surface of the metal will be unstable to order of some sort such as superconductivity, charge density waves, magnetic ordering, etc. With that said, let us concentrate on your scenario though. If there are no ...

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(1) Since $u(\textbf{r}) = u(\textbf{r}+\textbf{R})$, we can expand this part in terms of reciprocal lattice vectors, $u_k(\textbf{r}) = \sum_\textbf{G}{e^{i\textbf{G}\cdot \textbf{r}}u_\textbf{k-G}}$. We can therefore write: \psi_{\textbf k+\textbf K} = e^{i(\textbf k + \textbf K)\cdot \textbf r}\sum_\textbf{G'}{e^{i\textbf{G'}\cdot ...

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Because that is the obvious place to start - the mental experiment of bringing two "bulk" materials together and seeing what happens. That is the straightforward, simple calculation to explore. I'm sure Schottky and Mott knew that it did not capture all possible issues, but you do the simple first and see how bad it is. As it turned out, the theory was ...

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This Question is Beyond the Scope of This Site This subject takes much, much more room to precisely answer than a single post can do here. People get degrees to answer this question. A single post here will not answer your question with exactness. A few years studying physics and then going on to study material science- that will answer your question as ...

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Do you understand that the electric field within a conductor is zero? The charge is mobile, so the internal charge rearranges itself until there is no longer any force to move them: there is no field in the interior. If you understand that, then you will realize that a test particle within the conductor will feel no force, so no work will be done in ...

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If atoms have well define energy levels and the differences correspond to the frequencies of light that can be absorbed, This is correct how is it that opaque objects absorb all or most visible light frequencies get absorbed and you basically don't have any visible light coming out on the other end? The crux of the matter is the word "objects". ...

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min Zhang, The Hubbard model (offen reffered to as the U and J terms in ab-initio DFT or tight-binding models) is a little bit more complicated than it looked like. It is indeed an additional energy that you add locally to some states (d or f bands usually) to locallized them. Usually you want to do this because DFT tends to spouriously delocallize ...

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Well, when did you want it to be written down? As it is, it was pretty much simultaneously and independently arrived at by Hubbard, Kanamori, and Gutzwiller an awful long time ago. Probably others too. The point is, it was written down when there were experimental phenomena that justified including interactions in the model. It wasn't some great conceptual ...

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It seems like a very arbitrary question. It would be much easier if you could just elaborate on one given "specific" and well defined scenario, that you wish to solve, then one can help. From which then you would analogously try to solve other similar scenarios. From what you've asked thus far, your starting point should be Conservation of momentum. Since ...

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The formula $$E = \frac{\hbar^{2}}{2m} |k|^{2}$$ is valid in the vicinity of $k = 0$, this is called parabolic band approximation, which is - as the name suggests - only an approximated formula. Regarding the actual calculation of energy, you need to calculate the eigenvalues of the Hamiltionian of the periodic system, then the theorem is trying to say that ...

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Have you heard of superconductivity? This is a phenomenon where a material exhibits zero resistivity near absolute zero: it clearly contradicts your assertion that thermal excitation is needed for conductivity near absolute zero. For a semiconductor, it is true that electrons need to be kicked into the conduction band by thermal fluctuations - but for a ...

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In the image above, you can see a series of Bragg planes drawn in the crystal. This is called one "set of planes". Another "set of planes" would be if one would just draw a series of horizontal lines through the atoms. (Of course by lines I mean planes, but they are projected here onto a 2D image). The planes are those formed by the atoms, so in that ...

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The problem is that light has almost no momentum to give to the electrons in the solid (one usually makes the assupmtion of $k=0$). The dispersion relation for light is extremely steep because of the light speed ($E=ck$). So the electron has to get (or give) some momentum to make the indirect transition. This is typically achieved through interactions with ...

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Yes, this understanding is correct. The Born-Oppenheimer is assumed in the derivation of the Jahn-Teller effect. Also, the scenario you describe in the last paragraph is qualitatively similar to the Dynamic Jahn-Teller effect.

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