Question regarding Fourier transform of Green functions

Question

The quasiparticle Green's function is defined,

\begin{equation} G^+_{\lambda \lambda'} (\tau) = -i (1-n_\lambda) \delta_{\lambda \lambda'} \begin{cases} -i \ \text{exp} (-i \epsilon_\lambda \tau) , & \tau>0 \\ 0, & \tau<0 \end{cases} \end{equation}

Similarily, the quasihole Green function is,

\begin{equation} G^-_{\lambda \lambda'} (\tau) = -i n_\lambda \delta_{\lambda \lambda'} \begin{cases} -i \ \text{exp} (-i \epsilon^-_\lambda \tau) , & \tau>0 \\ 0, & \tau<0 \end{cases} \end{equation}

Now, the Fourier transforms of these two quantities are taken. We have,

\begin{equation} G_\lambda (\epsilon) = \int_{-\infty}^\infty G_\lambda (\tau) e^{i \epsilon \tau} d\tau , \end{equation}

or explicitly,

$$G^+_\lambda (\epsilon) = -i (1-n_\lambda) \int_0^\infty \text{exp} ( -i \epsilon_\lambda \tau + i \epsilon \tau ) d\tau ,$$

$$G^-_\lambda (\epsilon) = -i n_\lambda \int_0^\infty \text{exp} ( -i \epsilon^-_\lambda \tau + i \epsilon \tau ) d\tau .$$

The result obtained for this is,

$$ G^+_\lambda (\epsilon) = \frac{1 - n_\lambda}{\epsilon - \epsilon_\lambda + i \gamma} , $$

$$ G^-_\lambda (\epsilon) = \frac{n_\lambda}{\epsilon - \epsilon^-_\lambda + i \gamma} , $$

where $\gamma$ is an infinitesimal factor.

Now, my question is both about how this Fourier transform was evaluated, as well as about the appearance of $\gamma$.

I tried inserting the integral,

$$\int_0^\infty \text{exp} ( -i \epsilon_\lambda \tau + i \epsilon \tau ) d\tau ,$$

into mathematica and the result was

$$\frac{1}{\epsilon - \epsilon_\lambda},$$

without the infinitesimal factor $\gamma$. So why does $\gamma$ show up? And how is the integral calculated?

Answer 1

the integral $$ \int_0^{\infty} e^{i\epsilon t}dt$$ is not-defined for $\epsilon \in R$. For your integral, Mathematica probably also gave you the condition that ${\rm Im}\{\epsilon_\lambda - \epsilon\} < 0$. For real frequencies, we have to add a small infinitesimal in order to assure convergence. Thus $$ \int_0^{\infty} dt e^{i\epsilon t} \to \int_0^{\infty} dt e^{i(\epsilon+i\gamma) t} = -\frac{i}{\epsilon+i\gamma}$$ the physical meaning of this $\gamma$ is to give a finite life-time to correlations in the system. The limit $\gamma \to 0^{+}$ assures us that particles and holes (in non-interacting systems) have the limit of infinite life-times as they do not decay. Sometimes you will see that this $\gamma$ is replaced with some finite inverse life-time in a phenomenological style $\gamma \to \tau^{-1}$.

Also: the imaginary part of the Green function is associated with the density of states. Adding this infinitesimal and taking the limit of $\gamma \to 0^{+}$ give us a delta function, as desired. Upon replacing $\gamma \to \tau^{-1}$, in this context, the density of states attains a finite width.