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Dielectric Function

  1. Dec 26, 2011 #1
    i was reading Random Phase Approximation and encouter with dielectric function

    ε(q,ω)= 1- v(q) ∏(q,ω)
    where v(q) is coulumb potential and ∏(q,ω) is density-density correlation function without coulumb interaction
    now i do not understand this dielctric function can any one explain it physically for me?????
    thanks in advance :)
  2. jcsd
  3. Dec 27, 2011 #2


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    the general definition of the polarization is [itex]\int_{-\infty}^t j dt [/itex] (up to maybe some minus) whose longitudinal part (the expression you gave is only for the longitudinal dielectric function) can be expressed in terms of rho (and q and omega). In linear response rho will be induced by the coupling of the charges to the electromagnetic field V in the hamiltonian proportional to rho V.
    So the polarization will become proportional to a charge-charge correlation function and all the q´s and omegas flying around can be shown to give the Coulomb factor v.
  4. Feb 3, 2012 #3
    Hi there,

    I am also curious about this. In particular the electron part of the dielectric function, which can be described by the Linhard dielectric function.

    This function assumes a NFE electron model, which uses the FD distributions to calculate the electron component of the dielectric function (attached equations).

    I am just a little unclear about how to use this;

    is q the initial wave-vector?
    is p the final wave-vector?
    omega is the photon frequency I guess.

    I also am not sure how to represent the FD distributions in wave vector form, I know the k is proportional to the square root of E for electrons. But why do we have to use the FD distributions in k vector form, as then we deal a k^2 in the integration making it tricky.

    Overall, I am not clear on the relationship between k and q, as we need to have this to get a solution. Right?

    Sorry for the confused questions!


    Attached Files:

  5. Feb 3, 2012 #4


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    q is the wavevector of the electromagnetic field.
  6. Feb 3, 2012 #5
    so the energy of p + q ~ p^2 + h*c*q

    so hw = h*c*q, where h is the Planck constant, c speed of light, and q the photon wave-vector?

    and I should use this in FD distribution in place of energy for FD(p+q) ?
  7. Feb 3, 2012 #6


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    in the FD distribution p+q (or p-q) has to be interpreted as the momentum of an electron so that c should not enter.
    also hw = h*c*q would only hold for a free solution of maxwell equations in the medium.
    As the Lindhard function only describes the longitudinal response of the medium, these correspond to plasma oscillations.
    However, in general the frequency and the wavevector of the electromagnetic field can be varied independently of each other. E. g. you can modulate a coulombic field at any rate and the spatial fourier transform of a coulomb field will contain all wavevectors.
  8. Feb 3, 2012 #7
    so how should this be done for the FD distribution. Should it be E(p + q) ~ p^2 + q^2 ?

    [/QUOTE] also hw = h*c*q would only hold for a free solution of maxwell equations in the medium. [/QUOTE]

    but how do we know the relationship between q and w if we don´t yet know the refractive index?
  9. Feb 3, 2012 #8


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    also hw = h*c*q would only hold for a free solution of maxwell equations in the medium. [/QUOTE]

    but how do we know the relationship between q and w if we don´t yet know the refractive index?[/QUOTE]

    first note that p and q are vectors. Also E(p+q)=(p+q)^2/2m.
    In general there is no relation between w and q! Such a relation only exists for free solutions of the Maxwell equations. These can be derived from the Maxwell equations [itex]\epsilon (q,\omega)\omega^2 E(q,\omega)=q^2 E(q,\omega)[/itex].
    Even for light in a non-metallic medium the refractive index is not only a function of w but also a function of k, i.e. there may be light waves with the same frequency but different values of q. E.g. a left and a right circularly polarized wave will have a slightly different refractive index in a chiral medium and will propagate with different wavevectors.
    In most isolators this effect (spatial dispersion) is very weak, but in a metal, where electrons are free to travel over larger distances, it can be quite pronounced.
  10. Feb 3, 2012 #9
    so, how does one go about actually using this formulation for obtaining the refractive index of the electron system?

    Many people use this I think, but, for example, if I want to plot the εr, the real part of the dielectric function, as a function of the photon frequency, how do I go about this. I have seen some books that take k as the Fermi wavevector. But I am not sure what to do with q...
  11. Feb 3, 2012 #10


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    To start with, you should look up the Lindhard expression for the transversal dielectric function.
    The one you were writing down is the longitudinal one and does not describe photons.
    The refractive index n(q,w)=qc/w=sqrt(epsilon(q,omega))
  12. Feb 3, 2012 #11
    actually, the form is quite similar from what I can see Kliewer et al., Phys. Rev: 181, 2, 1969

    It basically involves integrating the FD functions in terms of k, and k+q, so can anyone tell me what forms these FD functions take? and in particular what is the relationship between the q and k vectors in this case?
  13. Feb 10, 2012 #12
    hi again,

    Looking at PRB, 40, 6, I see a treatment of the Lindhard function.

    Here they describe the response function of the free electrons (equation 1). But I am still confused about the symbols and what to put in. They have the FD functions in terms of k and k + q. k is the electron energy, and q is the "wave vector associated with the electron density".

    With the analytical solutions (eqns. 7 and 9), you need to define q. I have no idea what this wave vector is or what value it should take. Can someone help me out here?

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