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I've not come across any expression involving $\langle\Omega|f|\Omega\rangle$ in Srednicki's QFT book (please correct me if these exist there). On the other hand, they are abound in Chapter 7 of Peskin&Schroeder, in relation to the LSZ reduction formula. Here $|\Omega\rangle$ is the ground state of the interacting theory and $f$ is anything you can imagine that makes sense.

Srednicki only uses $\langle0|f|0\rangle$ where $|0\rangle$ is the ground state of the free theory.

How come this is so? Can one always go from one expression to the other by rewriting in terms of creation and annihilation operators?

Srednicki also derives the LSZ but he only uses the above-mentioned $|0\rangle$.

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    $\begingroup$ I am not familiar with Srednicki. By $$ |\Omega \rangle = \lim_{T \rightarrow \infty(1-i \varepsilon ) } ( e^{-iE_0T} \langle \Omega | 0 \rangle )^{-1} e^{-iHT} | 0 \rangle $$ as Eq. (4.27) in P&S, as well as for the bra, couldn't we just plug in $f$, got $\langle \Omega | f | \Omega\rangle = \langle 0 | \cdots f \cdots | 0 \rangle$? I don't think in general $\langle \Omega | f | \Omega\rangle = \langle 0 | f | 0 \rangle$ since the vacua for the free and interacting theories are different. $\endgroup$ – user26143 Feb 21 '14 at 18:02
  • $\begingroup$ Not exactly sure, but 1) one of the assumption of Peskin is "$\Omega$ has some overlap with 0, if this were not the case, $H_I$ would in no sense be a small perturbation" and 2) in general, QFT book use the assumption that all reactions are asymptotically free. It might not be the case for exact solutions. I'm not sure about that, though it might be interesting to look at some exactly solvable QFT models like the Thirring model. $\endgroup$ – Slereah Feb 21 '14 at 19:15
  • $\begingroup$ The answer to this question is: "sort of" and can be found in equation 4.31 PS. $\endgroup$ – Your Majesty Feb 21 '14 at 22:17

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