Density of States in NOT Free Electron Gas I think that I understand how the density of states works for a free electron gas. It is effectively just a conversion factor between summing over values of k and integrating over values of E. If you want, you can look at it as the density of points along the k-axis of a discrete plot of the electronic dispersion.
From there, you can compute most thermodynamic quantities, like internal energy and heat capacity, along with their temperature dependencies.
However, once we move beyond the simple model, I don't really know what to do. (i.e. we can calculate the dispersion in the Kronig-Penney model. Is it useful to re-calculate the density of states for this? Would it even look different than the free electron gas density of states, given the potential has no k-dependence? Do we gain any physical insight by calculating the density of states when we have band theory--beyond that higher bands have more states in 3D, the same number of states in 2D, and fewer states in 1D--assuming the free electron gas density of states holds?)
Thanks!
 A: This is all pretty standard stuff that you can find elsewhere on this site, but it is not always clearly presented and I feel like working through it again:
Intuitively, anything that changes the dispersion relation $E(\vec{p})$ also changes the density of states. This is just because we typically label the states by their momentum and assume there are two states (spin-up and spin-down) for each momentum. This means that for continuous $p$:
$n=2\int d^Dp $
which is an integral in $D$ dimensions.
So anything that changes the number of momenta at a given energy will affect this by:
$DOS=\frac{\partial n}{\partial E}=2\frac{\partial}{\partial E}[\int d^Dp]$
You can write the integral in spherical coordinates in p-space:
$d^Dp=A_Dp^{D-1}dp$, so
$DOS=\frac{\partial}{\partial E}[2A_D\int p^{D-1}dp]$
$A_D$ is a constant (it is the surface area of a $D$-dimensional sphere with radius 1). Now you use the inverse of the dispersion relation $p(E)$ to make a substitution:
$dp=(\frac{dp(E)}{dE})dE$
$DOS(E)=\frac{\partial}{\partial E}[2A_D\int_0^{E} p(E')^{D-1}(\frac{dp(E')}{dE'})dE']=2A_Dp(E)^{D-1}(\frac{dp(E)}{dE})$
So the dispersion, in its inverted form, is right there. All you've really done is a change of variables, but the way those variables change depends on the dispersion. This exact procedure is only possible if E is a function of the magnitude of p. If it isn't, like in a higher-dimensional band structure, you have to do the integration over the components of $p$ individually but it doesn't really change the situation.

As an example, let's go to D=1 and compare a free particle and band structure. The free particle has $E(p)=p^2/2$, and the band structure has a tight-binding form $E(p)=1-\cos(\pi p)$, where $p$ are now actually quasimomenta that are limited between 0 and 1, which means $E$ is limited between 0 and 2.
For the free particle, you get
$DOS=\frac{\partial}{\partial E}[2A_1\int_0^E p(E')^{1-1}(\frac{dp(E')}{dE'})dE']$
$=\frac{\partial}{\partial E}[2\int_0^E (\sqrt{E'})^{-1}dE']$
$=2(\sqrt{E})^{-1}$
For the band structure, it is:
$DOS=\frac{\partial}{\partial E}[2\int_0^E \frac{1}{\pi\sqrt{1-(1-E')^2}} dE']=\frac{2}{\pi}\left(\frac{1}{\sqrt{1-(1-E)^2}}\right)$
As you can see, these two expressions are clearly not the same. In particular, both go to infinity at $E=0$, but the band structure expression also does at the top of the band, $E=2$. This is a Van Hove singularity, and it is important for determining various properties of the band structure material.
The Kroenig-Penny dispersion would look more complicated and have higher bands, but in the lowest band it would be qualitatively similar.
