Why does the density of states in a solid scale as $\sqrt{E}$? In three dimensions, the density of states of a free electron is the square root of the energy of the electron. Can somebody explain the relationship between this dependence and the shape/formation of sub-bands in k-space? In other words, how does the shape and quantity of bands always lead to this same dependence across various materials?
 A: The density of states is always uniform in k, because in a box of length L, you have a uniform lattice in k space, and you imagine making the box bigger and bigger. So given $E\propto k^\alpha$, then you ask how much k volume there is in a band of size dE around E. The answer is always proportional to $|k|^{d-1}$ in $d$ dimensions, just from the area of a sphere of radius $|k|$, times the width of the spherical annulus, $d|k|$, which is
$$d|k| = {dE\over(dE/dk)} \propto {dE\over k^{\alpha-1}} \propto {dE\over E^{\alpha-1\over\alpha}}$$
So the density of states is
$$ \rho(E) = E^{(d-\alpha)\over\alpha}$$
which for d=3 and $\alpha=2$, gives $\rho(E)\propto \sqrt E$.
A: Because the density of states is proportion to the k, while we know for free electrons the energy is $E=\frac{\hbar^2k^2}{2m}$, so $k \propto \sqrt{E}$, therefore, the density states scales of $\sqrt{E}$.
A: The relation  $\rho \sim \sqrt{E}$ is still pretty universal near the bottom of a band becasue for  electrons with low $k$ (large wave-length) lattice details are irrelevant and their dispersion is like that of free electrons, albeit with a renormalized (effective)  mass. See effective mass theory.
A: it is some kind of WEYL LAW ..
in a potential well in d-dimension the number of states goes as $ N(E)=AE^{d/2} $ with -d- being the dimension of the lattice
