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In insulators, the valence band and lower bands comprise just enough quantum states for the number of electrons in uncharged material. If there's no thermal energy to promote electrons into higher bands, the valence band will, at equilibrium, be full. And since it is an insulator, there's no electrons in the next higher (conduction) band. If the material ...


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I try to give you an intuitive reason for this: As has already been said in the comments, the time-frequency DFT of a signal is also bounded in therms of the maximum frequency that can accurately be measured/reconstructed. This limitation stems from the sampling frequency of your hardware. It is therefore not a fundamental limitation, but merely imposed ...


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For a transition like electron-hole recombination, its probability is linear with time with some characteristic time scale that depends on the system: the probability of having a transition between $0$ and $dt$ is $\frac{dt}{\tau}$. So there is a non-vanishing transition probability at arbitrarily small times. In momentum space, the evolution of a ...


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The answer lies in the band structure of the two materials. The band structure describes how the electrons in a solid are bound, and what other energy states are available to them. Very simply, the band gap for transparent diamonds is very wide (see this link): Normally, diamond is not a conductor: all the electrons live in the "valence band", and you ...


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Diamonds are unstable compared to coal (or more exactly, graphite) so high temperature and pressure are required for diamonds to form from graphite. The reason that coal (graphite) is black and diamonds are clear has to do with how the carbon atoms are connected together in the two different forms of carbon. In diamond each carbon atom is bonded to ...


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$|\psi(+)|^2$ has its peaks at the ions, while $|\psi(-)|^2$ has its maxima between the ions. Since those represent the chance of finding the electron at a certain place we expect $|\psi(+)|^2$ to have a lower potential energy as $|\psi(-)|^2$, lower energy because the coulomb potential between the electron and the ions is attractive. What he is calculating ...


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If you're talking about a solid, the integral is bounded to the first Brillouin zone. This gives you a finite number of states. Otherwise, you have for each energy an infinite set of possible $k_z$ so the DOS diverges.



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