Density of States at the Fermi Energy

  1. 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. jcsd
  3. 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]
     
    Last edited: Mar 2, 2013
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