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Dielectric function and JDOS

  1. Feb 5, 2009 #1
    Hi. I have been looking at some lecture notes. What is not so clear for me is, how the imaginary part of the dielectric function is related to the joint density of states. Is the "amplitude" of the epsilon2 directly proportional to JDOS? or is JDOS some kind of derivative of epsilon2?

  2. jcsd
  3. Feb 6, 2009 #2
    Epsilon_2 is *almost* directly proportional to the JDoS. It is exactly proportional if the matrix element for the transition is independent of the position in k-space on the surface that defines the energetically allowed transition. For most purposes in crystals, the matrix element is only weakly dependent, and people like to just move it outside of the integral and replace it with an averaged matrix element.
  4. Feb 6, 2009 #3
    Thanks for your clear answer. Just one more question. I see that the formula for Epsilon in books are normally given for isotropic material. What changes in the integral of the formula if we want to know Epsilon in a certain direction for anisotropic material? There is a polarisation vector e in the matrix element for the transition <c|e.p|v>. I guess that for anisotropic material, the matrix element will depend on which e or which direction I take, whereas for isotropic material, it doesn't matter. Is this right?
  5. Feb 6, 2009 #4

    Dr Transport

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    Yes, the coupling between the conduction and valence band will be anisotropic. You also have to include the coupling between the valence band states.
  6. Feb 9, 2009 #5
    Ok. Thanks very much for the help!
  7. Feb 24, 2009 #6
    I noticed that there is a factor of 1/E^2 in the [tex]\varepsilon_2[/tex] equation. Since [tex]\varepsilon_2[/tex] is dependent on E, isn't the JDOS rather *almost* proportional to [tex]E^2\varepsilon_2[/tex]?
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