- #1
gonadas91
- 75
- 5
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 dependence 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?
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 dependence 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?