Free electron gas in two dimensions Can someone give a qualitative description on why the density of states for a two dimensional free electron gas is independent of energy while it is not in one and three dimensions? In one dimension it goes as $E^{-1/2}$ and in three $E^{1/2}$.
 A: 
TylerHG: Yes it is easy to calculate the density of states. But what I'm really asking here is "why."

Note that a thin circular ring in $\mathbf{k}$-space of thickness $dk$ has area $dA=2\pi k\,dk$ (by elementary geometry). In $E$-space, since $E\propto k^2$, that ring corresponds to a patch of width $dE=2k\,dk$.
Thus
$$\frac{dA}{dE}=\pi.$$
But $A\propto N$, so the density of states is constant.
A: It is easy to show that the total number of electrons in a 3D fermi sphere is : 
$$N(e)=\frac{V}{3\pi^2}*k_F^3$$
Where $k_F$ is the Fermi wave vector and $V$ is the real space volume of your sphere.
Now if you rearrange for $k_F$ in terms of the total number of electrons you'll get a particular equation.
It is know that 
$$E=\frac{p^2}{2m}=\frac{\hbar^2k^2}{2m}$$
where $k$ is an arbitrary wave vector.
By substituting $k$ in the above for $k_F$ you can then obtain an equation for Energy in terms of $\frac{N(e)}{V}$ i.e. the number density.
Rearrange this Energy equation to make $N(e)$ the subject and differentiate with respect to energy and voila! You have the density of states (as a function of the energy) for a 3D Fermi sphere and a conclusive explanation as to why it has an $E^\frac{1}{2}$ dependency.
EDIT : I recommend doing this on an actual piece of paper so as to better understand my explanation. I've purposely left bits out so you're forced to do so ! Additionally, I suggest trying the same idea for the arguably easier 1D case.
