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

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: $$e^{-iHT}|0\rangle=e^{-iE_0T}|\Omega\rangle\langle\Omega|0\rangle+\sum_{n\neq 0}e^{-iE_nT}|n\rangle\langle n|0\rangle. \tag{4.27}$$ In which $$|0\rangle$$ is the free vacuum and $$|\Omega\rangle$$ is the interacting vacuum and $$|n\rangle$$ are eigenstates of the full Hamiltonian, $$H$$.

\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 $$\epsilon > 0$$
• Ah I see what you mean about the exponential, if I could also just ask, what is the purpose of expanding the exponential into the sine and cosine here? Also should the sine term be $i\sin(-Et)=-i\sin(Et)$? Oct 4, 2020 at 19:05