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Density of States at the Fermi Energyby nboogerz
Tags: fermi energy 
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#1
Mar213, 03:53 PM

P: 4

The density of states at the fermi energy is given by
D(E_F)=(3/2)n/E_F I understand the density of states is the number of states per energy per unity volume, accounting for n/E_F. I don't understand how the 3/2 multiplying factor accounts for the volume? 


#2
Mar213, 09:45 PM

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 [tex] \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} [/tex] 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 [tex] N(k)=\frac{k^{3}}{3\pi^{2}} [/tex] The density of states (for the isotropic case) is given by [tex] \begin{eqnarray} D(E) &=& \frac{dN(E)}{dE}\\ &=& \frac{dN(k)}{dk}\left(\frac{dE}{dk}\right)^{1} \end{eqnarray} [/tex] For a parabolic dispersion we have [tex] E=\frac{\hbar^{2}k^{2}}{2m^{*}} [/tex] Therefore, at ##k=k_F## we have [tex] \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} [/tex] From the above expressions you can make the appropriate substitutions [tex] D(E_{F}) = \frac{3}{2}nE_{F}^{1} [/tex] 


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