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, $H$.


1 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 $\epsilon > 0$

  • $\begingroup$ 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 $i\sin(-Et)=-i\sin(Et)$? $\endgroup$
    – Charlie
    Oct 4, 2020 at 19:05
  • $\begingroup$ @Charlie there's no real purpose in expanding the exponential out into sine and cosine, other than to make explicit that the leading factor in the size of the term was the part containing epsilon. And yes, that is a sign error, I'll fix it in the answer. $\endgroup$ Oct 4, 2020 at 20:35
  • 1
    $\begingroup$ Great! Thanks for your time, all solved. $\endgroup$
    – Charlie
    Oct 4, 2020 at 21:12

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