Why does sending $T\rightarrow \infty(1-i\epsilon)$ in the slightly imaginary direction cause the $n=0$ term to decay slower?

Question

This is in reference to equation 4.27 in Peskin and Schroeder. To derive a formula for the interacting vacuum in terms of the free vacuum we evolve the free vacuum in time with the full Hamiltonian and then take the limit as $T\rightarrow \infty(1-i\epsilon)$. We are taking the limit in a "slightly imaginary direction" so that the exponential factor $e^{-iE_nT}$ factor dies slowest for $n=0$. My question is why this is?

The equation for reference: $$e^{-iHT}|0\rangle=e^{-iE_0T}|\Omega\rangle\langle\Omega|0\rangle+\sum_{n\neq 0}e^{-iE_nT}|n\rangle\langle n|0\rangle. \tag{p.86}$$ In which $|0\rangle$ is the free vacuum and $|\Omega\rangle$ is the interacting vacuum and $|n\rangle$ are eigenstates of the full Hamiltonian, $H$.

Answer 1

From the standard mathematics of complex exponentials:

$$\begin{align} e^{-iE\,t(1-i\epsilon)} &= e^{-iE\,t}e^{-E\,\epsilon\, t} \\ &= e^{-E\,\epsilon\,t}\left(\cos(Et) - i \sin(Et)\right) \end{align}$$

Since, definitionally, $n=0$ is the lowest possible value for $E$, and it appears in a negative exponential in front of a term of magnitude 1, the $n=0$ state falls off most slowly for all $\epsilon > 0$