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

1. Apr 30, 2015

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:
$$G_{0}(\omega)=\int_{-\infty}^{\infty}d\epsilon\nu(\epsilon)\frac{1}{\omega - \epsilon + i\delta\text{sign(\omega)}}$$
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 $$\nu(\epsilon)$$ 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 $$\delta\to 0$$ becomes a delta function inside. Therefore:
$$G_{0}(\omega)=-i \nu\int_{-\infty}^{\infty}d\epsilon\delta(\omega - \epsilon)=-i\nu\text{sign(Re(\omega))}$$

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

Now, taking $$\Lambda\to\infty$$ 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. May 1, 2015