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?

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You've answered your own question -- that's the dependence for free electrons (i.e., a single band with $E(\vec{k}) \propto k^2$). Real materials have complicated densities of states that depend crucially on the non-universal microscopic details. – wsc Sep 15 '11 at 4:05

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 $\rho(E)\propto \sqrt E$.

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Does this relationship hold for atypical band k-space shapes like graphene (linear, forming cones in 3d)? In other words, would the DoS for graphene in 3 dimensions scale as $E^2$, since $\alpha=1$? – Caedar Sep 17 '11 at 18:24
@Caedar: Yes. This works for any power law. – Ron Maimon Sep 17 '11 at 19:11
Erm, except that there's no "graphene in 3 dimensions" ... but yes, graphene does have a linear density of states scaling (near the Dirac point), as Ron lays out. – wsc Sep 18 '11 at 3:29
@wsc: massless neutrinos are graphene in 3d. – Ron Maimon Sep 18 '11 at 4:44
No, they aren't. Low-energy electrons in graphene are a subset of objects with a linear dispersion - not vice-versa. – wsc Sep 18 '11 at 5:15

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 $\sqrt{E}$.

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This is kind of circular. I don't think it's at all physically obvious that DOS scales linear in $k$ (in 3D!), except by knowing that it scales as $\sqrt{E}$ and inverting your reasoning... – wsc Sep 17 '11 at 1:03

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.

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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

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