Problem with infinitesimal factor in the expression for $G^{+}(x,y,E)$ In my book on QFT (Lancaster & Blundell) they give the following expression for the Green's function: $$G^{+}(x,y,E) = \sum \frac{i\phi _{n}(x)\phi ^{*}_{n}(y)}{E-E_{n}}.$$ However, they then state that "to obey causality" $E_{n}$ needs to be replaced with $E_{n} - i\epsilon$ where $\epsilon$ is an infinitesimal factor, to give us $$G^{+}(x,y,E) = \lim_{\epsilon\rightarrow 0}\sum \frac{i\phi _{n}(x)\phi ^{*}_{n}(y)}{E-E_{n}+i\epsilon}.$$ However, I don't understand how the omitting of the infinitesimal factor in the denominator would cause the expression for the green's function to violate causality.  Any explanation of this, or any other explanation of the necessity of including a factor of $i\epsilon$ in the denominator, would be greatly appreciated.
 A: Causality means, that $G(x, y, t - t^{'}) = 0$ for $t > t^{'}$ in the present context, which matches the common definition, that the subsequent action cannot influence the preceeding one. The expression you wrote is a Fourier transform by the variable $t$, returning back to the space-time variables $x, t$ via inverse Fourier transform, one obtains:
$$
G(x, y, t - t^{'}) = \int \frac{d E}{2 \pi} e^{-i E(t - t^{'})} \lim_{\varepsilon \rightarrow 0}\sum \frac{i \phi_n(x) \phi_n^{*}(y)}{E - E_n + i \varepsilon}
$$
Here the expression has poles only in the lower plane $E = E_n - i \varepsilon$. In order to make sense of the integration, one needs to take the semicircular contour in the lower part of complex plane $\mathbb{C}$ for $t < t^{'}$ and in the upper for $t > t^{'}$.
With this choice of contour the integration will pick all the poles for $t < t^{'}$, and nothing for $t > t^{'}$.
This choice corresponds to the retarded Green's function.
If one had chosen $E - E_n - i \varepsilon$ - this would lead to advanced  Green's function, vanishing for $t < t^{'}$.
