To compute the vacuum expectation value(adsbygoogle = window.adsbygoogle || []).push({});

##\langle \Omega | T\{q(t_{1})\cdots q(t_{n})\}|\Omega\rangle##

in the path integral formalism, we start with the time-ordered product in the path integral representation

##\langle q_{f},t_{f}|\ T\{q(t_{1})\cdots q(t_{n})\}\ |q_{i},t_{i}\rangle=\int\ \mathcal{D}\phi\ e^{iS}\ q(t_{1})\cdots q(t_{n}),##

use the fact that

##|\psi\rangle = \int\ dq_{i}\ |q_{i},t_{i}\rangle\ \langle q_{i},t_{i}|\psi\rangle##

and the projection trick

##\lim\limits_{T \to\infty}e^{-iHT(1-i\epsilon)}|\Omega\rangle = \sum\limits_{n}e^{-iE_{n}T(1-i\epsilon)}|n\rangle\langle n|\Omega\rangle##

to project out all the states with ##n \neq 0## and obtain

##\langle \Omega | T\{q(t_{1})\cdots q(t_{n})\}|\Omega\rangle = \frac{\int\mathcal{D}q(t)\ e^{iS[q]}q(t_{1})\cdots q(t_{n})}{\int \mathcal{D}q(t)\ e^{iS[q]}}.##

How can we be sure that the projection trick is a legitimate step in the calculation and not some sleight of hand? Is the projection trick performed in the limit that ##\epsilon \rightarrow 0##?

I would also like some help in deriving

##\langle \Omega | T\{q(t_{1})\cdots q(t_{n})\}|\Omega\rangle## using the above hints. Do I substitute for ##|\Omega\rangle## using the expansion of ##|\psi\rangle## I have from above?

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# A Projection trick to obtain time-ordered correlator

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