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Density of state

  1. Dec 18, 2008 #1


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    Consider electrons in metal as a quantum ideal gas, the quantized energy would be

    E(n_x, n_y, n_z) = \frac{h^2}{8m}\left(\frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} + \frac{n_z^2}{L_z^2}\right)

    Now, letting

    \alpha = \frac{n_x}{L_x}, \qquad \beta = \frac{n_y}{L_y}, \qquad \gamma = \frac{n_z}{L_z}


    \rho^2 = \alpha^2 + \beta^2 + \gamma^2

    and setting up an coordinate with [tex]\alpha, \beta, \gamma[/tex]. One could see that the volume b/w E-E+dE is proportional to the volume b/w [tex]\rho - \rho + d\rho[/tex]

    Note that the volume within [tex]\rho - \rho + d\rho[/tex] is just one-eighth of the volume of the spherical shell with thickness [tex]d\rho[/tex], namely,

    [tex]V_\rho = \frac{4\pi \rho^2 d\rho}{8}[/tex]

    Hence, the density of states within [tex]\rho - \rho + d\rho[/tex] becomes

    [tex]N(\rho)d\rho = \frac{V_\rho}{\textnormal{volume per state}} = \frac{4\pi\rho^2d\rho /8}{1/V} = \frac{\pi V \rho^2 d\rho}{2} [/tex]

    Well, we know that

    [tex]E = \frac{h^2}{8m}\rho^2[/tex]

    in addition

    N(\rho)d\rho = N(E)dE

    which gives

    N(E) = \frac{1}{2}\frac{\pi V}{2}\left(\frac{8m}{h^2}\right)^{3/2}\sqrt{E}

    But the correct answer for density of state in energy should be (two times of the above reuslt), i.e.

    N(E) = \frac{\pi V}{2}\left(\frac{8m}{h^2}\right)^{3/2}\sqrt{E}

    Where is the '2' come from?

    Someone suggests

    [tex]N(\rho)d\rho = 2\times\frac{\pi V \rho^2 d\rho}{2} [/tex]

    This will give the correct answer, but where is "2" come form?
  2. jcsd
  3. Dec 18, 2008 #2


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    Science Advisor

    Each electron has two possible spin states.
  4. Dec 18, 2008 #3


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    Oh, how can I forget that! Thanks a lot :)
  5. Dec 9, 2009 #4
    Can someone help me calculating the density of states from phonon dispersion relations.
    I have computed the phonon dispersion relations from considering the vibrations of crystals. Now as a next step, i want to compute the density of states. not sure how.
    many thanks in advance for your help
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