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Use Green's Function calculate photonic density of state

  1. Jul 2, 2014 #1
    Hi Everyone:

    I think some of you who familiar with quantum-optics know that the local photonic density of state can be calculated by the imaginary part of electromagnetic Green's function.

    The Green's function can be further presented by the dipole's mode pattern as

    G = E(r)*p0*ε(r)*c^2/ω^2

    , where E(r) is the electric field profile, p0 is the dipole moment, ε is the dielectric function, ω is the frequency

    You can find these formulas in Lukas' book "Principle of nano-optics"

    However, I'm confused by the calculation's process. E(r) contain both the real and imaginary part, and so dose ε. Therefore, the final imaginary part of G will contain the cross-product item.

    The dielectric function will have a negative real part if there has metal material. But this will lead a negative imaginary part of Green's function in metal area, as also a "NEGATIVE DENSITY OF STATE"!

    Dose this reasonable?
  2. jcsd
  3. Jul 2, 2014 #2


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    Looking at equation 8.111, 8.113, 8.114, I don't see how it could be negative.
    However, the track from 8.111 to 8.114 seems a little bit obscure.
    IMHO, I think the author tried to avoid handling the singularity from the beginning (near 8.109).
    IMHO, solving for the Green function must be done by using the Bromwich coutour properly in order to account for the imaginary part.
    (this is similar to the way Landau damping in plasma physics is derived)

    My best advice is to read about Landau damping and about the Bromwhich contour.
    In this way, you will probably be able to clarify the origin of the imaginary part and see why it is always positive.
    (in a stable media)
  4. Jul 2, 2014 #3
    Thanks for your comments

    My problem came from equation 16.29 which use the volume-integral method to calculate Green's Function. In this method, the system's Green's Function can be linked to the electric mode excited by a electric dipole. However, the dielectric function can be negative, if there is a metal material, and finally you will get a negative imaginary part of Green's function.

    Of course, these negative imaginary part of Green's function will only exist in the region of metal.
    Photonic density of state is negative in metal??
    Last edited: Jul 2, 2014
  5. Jul 2, 2014 #4


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  6. Jul 2, 2014 #5
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