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The d is not separation between points in reciprocal lattice. Actually, they do not even have the same units. d is the separation between lattice planes, as you said. What is related to reciprocal lattice vectors is the change (before and after scattering) in the wave vector of light: change in k = reciprocal lattice vector, which is the Laue condition that ...


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Here's a simple-minded answer: Let's just compute the momentum of a particle with a Bloch wave function $$\begin{eqnarray} \left.\langle x \right| \hat{p}\left|\Psi \rangle\right. &=& -i\hbar \left(\frac{d}{dx}\right) u_k(x) e^{i k x} \\ &=& -i \hbar \left( i k u_k(x) e^{ikx} + u_k'(x)e^{ikx}\right) \\ &=& \left( pu_k(x) - i\hbar ...


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To get a better understand of what is going on, take a look at the plot below, also linked here: http://en.wikipedia.org/wiki/File:Dispersion_Relationship.gif What the author meant by "letting the speed of light go to infinity" is that the we let the slope of the blue line become infinite. In that case, the solid red line would not curve as shown below, ...


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I think that the most appropriate way to think about the Bragg formula is in terms of a diffraction grating. In a diffraction grating one obtains sharp peaks because there are many slits with distance $d$ between them. The derivation of the intensity maximum for the diffraction grating case is similar to the Bragg case. To obtain a diffraction grating ...


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Optical conductivity and AC electric conductivity experiments are indeed quite similar. However, they operate in different frequency regimes and measure slightly different quantities. Optical conductivity refers to an experiment using light, such as a reflectivity measurement and then using a Kramers-Kronig transform to deduce the real part of the ...


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1) The Thomas-Fermi method is a useful way to estimate the "screening cloud" surrounding electrons in a metal. A quantification of the screening is the inverse dielectric function of the material: \begin{equation} \frac{\phi_{ext}(\textbf{q})}{\epsilon(\textbf{q})} = \phi_{total}(\textbf{q}) \end{equation} Slowly-varying in this context means that the ...


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To actually measure the 3D band structure is quite a difficult proposition. In materials that are quasi-2D (materials that are weakly bonded along one axis), the best probe of band structure today is angle-resolved photoemission spectroscopy. However, it only probes the band structure of the occupied states. For instance see the Fermi surface map of Sr2RhO4 ...


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Ashcroft & Mermin devotes a (short) chapter (14 in my edition) to experimental techniques to measure Fermi surfaces (note they focus mostly on metals). These include de Haas - van Alphen, magnetoacoustic measurements, etc. For semiconductor band structures there are various techniques with (angle resolved) photoelectron spectroscopy being one of the ...


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Probably all of them and more besides. The 230 does not include quasicrystals, while nature does.


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According to Fizicheskaya Entsiklopediya (Physical Encyclopedia, in Russian, http://www.femto.com.ua/articles/part_2/3634.html ), no real crystals had been found for 4 space groups (Pcc2 and three others) as of the encyclopedia's publication in 1988-1999.


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If there were only a few that had not been seen, I suspect that Ashcroft and Mermin would have gleefully pointed that out. However, that is not proof. On the other hand, I see no way to prove that a given space group definitively could not exist in nature, given the essentially infinite number of molecules that could form crystals. We still spend lots of ...


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I would be surprised if there wasn't an example of each. For instance, you can search the American Mineralogist Crystal Structure Database by space group. Click on Cell Parameters and Symmetry and select whichever you like. I'll leave it to you to go through all 230 of them.


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I can answer that yes, momentum is definitely transferred to the surrounding lattice, though not through a direct scatting of electrons with protons. This is important in increasingly small electronics. When the cross-section of the wire or metal is extremely small, the collisions can be enough to displace the metal, and wreck a circuit. Tungsten is ...


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The Drude model is fine for thinking about some things. (It is still taught.) The electrons collide with the atoms. (or a bit more precisely, the outer electrons of the atoms.) Since this is a classical picture perhaps it's OK to have a classical picture of the atoms. Imagine they are little sphere's all joined to the other atoms by little springs. ...


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Wait a minute. Surrounding the protons are a matrix of electrons. These are the electrons not "cool enough" to exist in the conduction band. I imagine these are what collide with the electrons in the Drude model approach. The closer you get to those electrons, the more they're going to push back by the Coulomb potential. Furthermore, I imagine some Pauli ...


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In drude model, Electron don't collide with protons. The best way to figure this is that the electron collides with phonons and with impurities. But the correct Drude model do not involve collision with the nucleus. The collision gives excite phonons, and can be considered as heat from a macroscopic view point. The Drude model give the right formula only ...


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It would be useful to have the exact reference in the text where "correlation" is mentioned, but I would argue that correlation is a word which characterizes the collective behavior of electrons in certain circumstances. Such circumstances are varied, but imply that their behavior (and that may be spatial dynamics, but also spin dynamics or other quantities) ...


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The potential has to vary slowly enough so that the electrons have time to respond as if the field were static. Risking a semi-educated guess: I suppose that the potential would have to change less than a few percent in a time on the order of the transit time of an electron across a unit cell. That is, the Fermi velocity divided by the lattice constant. ...


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According to my limited understanding of density functional theory. Coulomb interaction is one of the correlation effects. Besides Coulomb interaction, there are interaction due to Pauli exclusion principle and change of kinetic energy compared with that of non-interacting electron gas.


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When solving for the spectrum, you assume the solid has infinite extent, and so the electron is not really bound. Thus you can get a continuum of momentum. Consider for example the trivial case where there are no atoms, and you just have a free electron in space. Then the energy is proportional to momentum squared, and momentum can be arbitrarily big, so ...


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Ok, I am by no means and expert on solid state but I might be a little helpful. Band structure in solids arises due only to periodicity of the lattice. It all comes down to this periodicity. Periodicity of the lattice makes the potential also periodic. This periodicity has many (interesting) consequences (Bloch states and bla bla bla) but the one that ...


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You could think of making a solid by slowly bringing all the atoms closer together. As they get closer the discrete electron orbits start to bump into each other. But the electrons can't be in the same state, and you start to get new states that are a combination of the discrete states.. And these form bands.(Yeah that last is a bit of a cop out, see a ...


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Because in solid states, the electrons are not bound. A simple derivation of the continuous bands can be made if you look at particle in a box (Wikipedia). If we approach the number of an infinite box with infinite particles (and constant density), there are infinite states, and the states get closer, as the energy gap goes as $L^{-2}$. Therefore a ...


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You'll want a much bigger heatsink!! (and maybe just one TEC) If it's being cooled only by convection then maybe a heat sink area* that is 10 times that of the TEC. (maybe bigger) The classic mistake with a TEC is to make the heat sink too small. With too small a heatsink the hot side of the TEC gets hotter, more thermal leakage through the TEC, it has ...


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If efficiency is the issue, then definitely parallel TECs (or use a single unit rated for twice the power, same thing). The only reason for stacking TECs is to get a lower temperature. However that comes at great expense to efficiency and overall power consumption. Another point is that paralleling TECs is actually more efficient overall. The reason is ...


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Let us decompose the formula to better understand what it means: $\hbar \mathbf k/m$ is the velocity of the electrons $\epsilon_k - \mu$ is the energy relative to the temperature, so if particles with higher energy than the temperature move in a direction, they produce a positive thermal energy flux, particles with lower energy an negative flux. ...


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I'm not sure what goes into a film emulsion. As to small particles, it's basically the black-body principle. No material is perfectly reflective, so the more reflections a light ray undergoes, the more attenuation occurs. A pile of tiny particles essentially acts as a "light trap." One of the better absorbing materials (up until the development of ...


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The difference between Fermi energy and Fermi level is described in a previously asked question. To summarize, you're correct that the Fermi level is the energy at which (in a 0K system) all of the lower energy states are occupied while all of the higher energy states are unoccupied. For a system not at 0K, the Fermi energy is the energy state that has a 50% ...


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Let's assume, you take a one-dimensional chain of atoms and compress it. In order to investigate the bandstructure, you will need to determine the electronic wavefunctions of quantum-mechanically allowed states. If you know your wavefunctions for the initial condition, before you compress your chain of atoms, you need to also scale the solution in order to ...


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A quick google search leads one immediately to the wikipedia page on this particular theorem. The first paragraph of this page states: The Bohr–van Leeuwen theorem is a theorem in the field of statistical mechanics. The theorem states that when statistical mechanics and classical mechanics are applied consistently, the thermal average of the ...


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A quasi-crystal has no long-range translational symmetry, but it does have long range orientational symmetry. The lattice sites all occur at well defined angles, and in well-defined planes. It is reflection from these planes that causes the well-defined spots. The fact that there is no translational symmetry within those planes does not bear on the ...



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