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Change of sign on Green's functions (Maths problem)

  1. Apr 30, 2015 #1
    Hi, I am trying to solve a model where Non-interacting Green functions take part it. It has happened something that is spinning my head and I hope someone could help. The non interacting Green function for a chanel of electrons is:
    [tex] G_{0}(\omega)=\int_{-\infty}^{\infty}d\epsilon\nu(\epsilon)\frac{1}{\omega - \epsilon + i\delta\text{sign($\omega$)}}[/tex]
    where I am integrating over the whole energy spectrum since I consider the chanel as a whole. I have to mention that I am at T=0 and [tex] \nu(\epsilon)[/tex] is the density of states on the chanel. Ok, suppose we have a constant density of states. Then, the real part of the integral cancels (gives a cosntant part which is irrelevant) and the imaginary part, since [tex]\delta\to 0[/tex] becomes a delta function inside. Therefore:
    [tex] G_{0}(\omega)=-i \nu\int_{-\infty}^{\infty}d\epsilon\delta(\omega - \epsilon)=-i\nu\text{$sign(Re(\omega))$}[/tex]

    We notice the dependance with the REAL part of the excitation energies [tex]\omega[/tex]. Now, imagine that instead of a constant density of states, this density of states is a Lorentzian of the type:
    [tex]\frac{\Lambda}{\Lambda^{2} + \epsilon^{2}}[/tex]
    Then the integral above can be completed in the complex plane using contour integration, and the result gives:
    [tex] G_{0}(\omega)=\frac{\nu}{\frac{\omega}{\Lambda} + i\text{$sign(Im(\omega))$}} + \frac{i\nu\Lambda\text{$sign(Re(\omega))$}}{\Lambda^{2}+\omega^{2}}[/tex]

    Now, taking [tex]\Lambda\to\infty[/tex] should gives us the result of the flat density of states but instead it gives the sign with the imaginary part!!! Someone can help on this?
  2. jcsd
  3. May 1, 2015 #2
    Ok now doubt is solved I just did it at the end so this post can be closed ;)

    It is just that I took wrong the limit.
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