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Why is it true that in nearly free electron compunds, complete neglect of the lattice potential is usually a good approximation as long as one considers crystal momenta remote from the boundaries of the Brillouin zone? or more precisely, what's the essential difference between the electronic states with crystal momenta close to or far away from the Brillouin zone boundary?

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  • $\begingroup$ I'm pretty sure it's because far from the zone edges the dispersion relation (i.e. relation between momentum and energy) is nearly that of a free particle, namely $E = (\hbar k)^2 / 2m$, but with a modified mass which comes from the curvature of the band. On the other hand, near the zone edges the dispersion relation is not like this at all so you can't approximate it as a free particle. $\endgroup$ – DanielSank Dec 29 '14 at 5:04
  • $\begingroup$ @DanielSank but why is it like that? $\endgroup$ – M. Zeng Dec 29 '14 at 5:06
  • $\begingroup$ If you're asking why the bands do funky things near the zone edges then the quick answer is "I don't remember" and the long answer involves revisiting the Bloch theorem and thinking about exactly what "zone edges" really are. This is why I'm commenting and not writing a proper answer. $\endgroup$ – DanielSank Dec 29 '14 at 5:27
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    $\begingroup$ In fact, this is not quite right: electrons in almost any symmetric point of Brillouin zone, including BZ boundary, can be described in effective mass approximation. But if the symmetric point has some non-trivial symmetry, there may be multiple parameters (like Luttinger parameters for holes in cubic crystals), and the effective mass in some directions may be negative. And only between these symmetric points (i.e. far away from them) the dispersion relation becomes highly nonparabolic, and effective mass approximation breaks down. $\endgroup$ – Ruslan Dec 29 '14 at 9:57
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    $\begingroup$ Take a look at band structures of various cubic materials like Si, GaAs, Ge and note the common feature: in all the points like $\Gamma$, $X$, $W$, $K$ there're extrema of dispersion curves.Naturally, since $E(\vec k)$ is differentiable at those points and symmetric with respect to $\vec k\to-\vec k$, the first Taylor term is $\sim k^2$ (actually $k_ik_j$, not isotropic).This is the basis of effective mass approximation for $k\approx0$.And since $E(\vec k)$ is periodic in $\vec k$,we can move the Brillouin zone origin to these points. $\endgroup$ – Ruslan Dec 30 '14 at 6:18
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For a simple crystal with more or less cubic symmetry and with low free electron density, for example sodium, Fermi surface is more or less a sphere. This is because it is small and deep inside Brillouine zone. Spherical Fermi surface resembles that of a free electrons with parabolic dispersion...in a crystal we do not have this parabolic dispersion of electrons, because crystal potential modifies it and this is more prominent for the values of crystal momentum of electrons near the values at BZ...because Bragg law gives that exactly at these values electrons interact very strongly with the lattice and here crystal potential deforms dispersion relation. So if you add more and more electrons in a crystal they fill more and more states and are getting near the value at the edge of BZ. That is why I said with low electron density, meaning of course, conduction electrons. The electrons at the bottom just experience conditions like in a parabolic dispersion, like free ones, and as you fill the band up, right around the middle, they experience the crystal potential and act accordingly. Now when you calculate conductivity, you realize that only electrons at the top of Fermi surface are ones being effected, so if a metals Fermi surface is near the edges of the zone this surface will be deformed because dispersion relation is deformed and because electrons scatter just in this narrow area around the surface, their behavior depends strongly on the shape of this surface. Why dont all the other electrons deep inside Fermi surface scatter? Because there is not enough energy available. Only electrons in a narrow thermal layer participate in this, and it is narrow compared to the Fermi energy.Another reason is, of course, Pauli exclusion principle. This is actually, now I see, very broad question, and I can only say, look it up in Zimann or Kittel, Solid state theory for more elaboration.

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