# Density of States at the Fermi Energy

by nboogerz
Tags: fermi energy
 P: 24 Dimensionally you are correct. But in this case, unfortunately, you have to perform the detailed calculus steps in order to get that factor. First let us determine the expression for ##n##. In ##\bf k##-space you need to count the total number of occupied states. This can be computed as seen in the steps below $$\begin{eqnarray} n&=&2\int_{{\rm FS}}\frac{d^{3}\mathbf{k}}{(2\pi)^{3}} \\ &=& \frac{2}{(2\pi)^{3}}\int_{0}^{k_{F}}dk\int_{0}^{2 \pi}d\phi\int_{0}^{\pi}d\theta\left(k^{2} \sin(\theta)\right) \\ &=& \frac{2}{(2\pi)^{3}}\left(\int_{0}^{k_{F}}dk\, k^{2}\right) \left(\int_{0}^{2\pi}d\phi\right) \left(\int_{0}^{\pi}d\theta\,\sin(\theta)\right) \\ &=& \frac{2}{2\pi^{2}}\int_{0}^{k_{F}}dk\, k^{2} \\ &=& \frac{k_{F}^{3}}{3\pi^{2}} \end{eqnarray}$$ where ##\int_{{\rm FS}}## is an integral from the origin till the (spherical) Fermi Surface (FS). The ##k^{2} \sin(\theta)## in the second step is simply the Jacobian in spherical coordinates. Now, ##n## is the total number of available (and filled) states for ##k\le k_{F}##. The total number of states available up to some arbitrary ##k## is simply $$N(k)=\frac{k^{3}}{3\pi^{2}}$$ The density of states (for the isotropic case) is given by $$\begin{eqnarray} D(E) &=& \frac{dN(E)}{dE}\\ &=& \frac{dN(k)}{dk}\left(\frac{dE}{dk}\right)^{-1} \end{eqnarray}$$ For a parabolic dispersion we have $$E=\frac{\hbar^{2}k^{2}}{2m^{*}}$$ Therefore, at ##k=k_F## we have $$\begin{eqnarray} D(E_{F}) &=& D(E(k_{F}))\\ &=& \frac{m^{*}k_{F}}{\hbar^{2}\pi^{2}}\\ &=& \frac{k_{F}^{3}}{\pi^{2}}\left(\frac{\hbar^{2}k_{F}^{2}}{m^{*}}\right)^ {-1}\\ &=& \frac{3}{2}\left(\frac{k_{F}^{3}}{3\pi^{2}}\right) \left(\frac{\hbar^{2}k_{F}^{2}}{2m^{*}}\right)^{-1} \end{eqnarray}$$ From the above expressions you can make the appropriate substitutions $$D(E_{F}) = \frac{3}{2}nE_{F}^{-1}$$