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I will try to answer in a simple and intuitive way. If you need more details I suggest you, for example, Ashcroft and Mermin, Solid State Physics. The bands in the $E$-$k$ diagram are filled with electrons from the lowest energy state (the bottom of the dispersion curve) up to the Fermi energy. This is a consequence of the Pauli exclusion principle and of ...

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It is not possible to get the dispersion in any direction using just the given dispersion. Some knowledge of the underlying crystal and electronic structure is required.

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These are two different problems. What Jon was saying is correct. However, it does not explain LO-TO splitting. Like Jon said, because you can tell when you are on a Ga or As atom, the degeneracy of the optical modes are lifted at the Gamma point. This is in regards of 3 different optical modes separating. However, the phenomena Cardona is refers to involves ...

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Electricity needs charges particles (or quasi-particles) to conduct. Heat can be conducted with almost any quasi-particle. Diamond is one of the best conductors of heat in existence, and it's because of phonons, ie quasi-particles of lattice vibrations, which are strong because the diamond lattice is strong.

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If you have a very low frequency, the material behaves as a conductor: the electrons respond instantaneously to the excitation, and therefore the metal becomes a reflector (the boundary condition of "no E field parallel to the surface" is met). If you have a very high frequency, the electrons don't have "time to react" at all - so the amplitude of their ...

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Just to be clear, the two or more atoms do not have to be of different type. Optical phonons are related to the relative vibrations of atoms within the unit cell, while acoustic phonons describe the relative vibration of different unit cells. Optical phonons arise whenever the unit cell has at least one such degree of freedom, meaning at least two atoms in ...

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Crystal lattices are classified in a way that is not necessarily the most natural one. A first classification is by their lattice class, which is determined by its associated unit cell, which is not necessarily the same as a fundamental domain for the lattice. There can be different Bravais lattices in each lattice class, determined by possible additional ...

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Under the substitution ℏk→ℏk−qA $<p|x>=<0| a_{p} a_{x}^{+}|0>=exp(-ipx/ h)$ will become $<p|x>=<0| a_{p} a_{x}^{+}|0>=exp(-i(p-qA)x/ h)$ effectively, the change in operator: $a_{p} a_{x}^{+} \rightarrow a_{p} a_{x}^{+} e^{iqAx/h}$ Then it looks as if: $a_{x}^{+} \rightarrow a_{x}^{+} e^{iqAx/h}$ $... 0 This is a second order process. The photon and the phonon are simultaneously absorbed. You may just as well draw the arrows in a different order. Now, if the first arrow extended all the way to the green line, followed by the second arrow to the conduction band minimum, then the order would be important. Then we would interpret this as absorption of a ... 1 A single photon with 3-momentum$p=\hbar k$incident on the system is scattered to an outgoing state with momentum$\hbar k'$. If the interaction operator between the photon and the sample is$U$, and the initial and final states are labelled by$|k\rangle$and$|k'\rangle$respectively (we ignore photon polarization/spin), then the amplitude for this ... 1 The lack of the$e^{-i \omega t}$term is just because we're using complex wave notation. If you've ever taken an electrical engineering course, it's the same sort of thing that is used there: We're using$A e^{i \omega t}$to stand for$A \cos (\omega t - \delta)$, with the implicit assumption that we're only interested in the real part of the quantities ... 2 I'm not quite sure what you mean by "crossing at$[E_F,0.5]$" but the Fermi level doesn't change with temperature by definition. The Fermi level is defined as the energy of the highest energy electrons at zero temperature when the system is in its ground state. It is a property of the system that is only dependent on the quantum mechanical eigenfunctions and ... 1 The$kT$comes from the Bose-Einstein statistics. The photons are governed by this statistic, nothing suspicious is here. The power three appears when we go form the variable$E$to the variable$E/kT$in the integral. 0 I'm don't have high enough reputation to comment yet, so here's a half answer - half comment... It is indeed possible to separate the problem, as you say, into$\psi(x,y,z) = \chi(x,y)\zeta(z)$and the total energy of the state is then $$E = E_z + \frac{\hbar^2 k_{\parallel}^2}{2m^*_{\parallel}}$$ Now, read on if you would ... 1 There are essentially two ways of generating coherent radiation from solid-state devices: Classical electronic oscillators, in which charge is made to oscillate back and forth within a device... the frequency of radiation corresponds to the frequency of charge oscillation. Solid-state lasers, in which charge-carriers undergo a transition between two ... 0 Current density is defined as electrical charge per unit time for a certain cross-section. Since a cross-section is a two-dimensional entity, it has to be$ A / m^2 $. In some cases it can be simplified to$ A / m $. 2 Question: The above analysis seems to treat the oscillators as distinguishable. But aren't they indistinguishable? In Einstein's model, the oscillators are supposed to sit at (oscillate around) definite place in space. So you could say they are distinguishable. For example, by their cartesian coordinates with respect to lab frame. 1)That the Gibbs ... 0 In this particular example the photoluminescence is of quantum wells. The reason for the asymmetry is because the density of states is not symmetric. At the low energy side the density of states has a excition Lorentzian line shape to the absorptivity. At higher energies the density of states becomes step like. The photoluminescence intensity,$$I(E) ... 0 On the very theoretical side of solid state physics is the holographic AdS/CFT correspondence which links strongly coupled condensed matter systems to gravitational theories. Recent work has been done on describing things like phase transitions in this theory. For example models of superconductivity in the gravity dual are promising in describing difficult ... 0 There is theory of dispersion in crystals. One can say that the differential geometry is used there. I think it is Group theory + differential geometry. 3 Yes, para hydrogen, in a limited manner. Recent work at Göttingen has revealed convincing evidence for superfluidity in liquid hydrogen, the only liquid other than helium to exhibit this quantum behaviour. From a spectroscopic experiment on droplets of parahydrogen, it has been discovered that properties of superfluid are observed in a system of ... 1 Sorry I was too lazy to retype everything, but here is a set of notes I made for myself a little while ago. Note that below$\epsilon(q,0) = 1+V(q)\Pi(q)$, where$V(q) = e^2/\epsilon_0q^2\$. If my notes aren't clear, let me know.

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