# 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?

• 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. – DanielSank Dec 29 '14 at 5:04
• @DanielSank but why is it like that? – M. Zeng Dec 29 '14 at 5:06
• 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. – DanielSank Dec 29 '14 at 5:27
• 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. – Ruslan Dec 29 '14 at 9:57
• 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. – Ruslan Dec 30 '14 at 6:18