neglect of lattice potential for conduction electrons 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?
 A: 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.
