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How to derive equation 4.88 in section 4.6, page 108, of Peskin & Schroeder?

$$\left|k_{1}k_{2}\right\rangle\propto\lim_{T\rightarrow+\infty(1-i\epsilon)}e^{-iHT}\left|k_{1}k_{2}\right\rangle_{0}.\tag{4.88}$$

How to derive equation 4.89?

$$\lim_{T\rightarrow+\infty(1-i\epsilon)} {}_{0}\!\left\langle k_{1}k_{2}\right|e^{-iH(2T)}\left|p_{1}p_{2}\right\rangle_{0}$$ $$\propto\lim_{T\rightarrow+\infty(1-i\epsilon)}{}_{0}\!\left\langle k_{1}k_{2}\right|T\left(\exp\left[-i\int_{-T}^{+T}dt\, H_{I}(t)\right]\right)\left|p_{1}p_{2}\right\rangle_{0}.\tag{4.89}$$

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  • $\begingroup$ I think this is a weak point in book's logic. I suggest anticipation LSZ formula from later chapters. $\endgroup$ – Matteo Beccaria May 13 '17 at 21:17
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I would like to hazard a guess here. From equations (4.23) and (4.25) in the book, $$U(T,-T)=T\{\exp[-i\int_{-T}^{T}dtH_{I}(t)]\}=e^{iH_{0}(T-t_{0})}e^{-iH(2T)}e^{iH_{0}(T+t_{0})},$$ where $t_0$ is some reference time at which the Schroedinger and interaction pictures coincide. We then get $${}_{0}\langle k_1k_2\rvert e^{-iH(2T)}\lvert p_1p_2\rangle_0={}_{0}\langle k_1k_2\rvert e^{-iH_{0}(T-t_{0})}T\{\exp[-i\int_{-T}^{T}dtH_{I}(t)]\}e^{-iH_{0}(T+t_{0})}\lvert p_1p_2\rangle_0 \\ =e^{-iE_f(T-t_{0})}e^{-iE_i(T+t_{0})}{}_{0}\langle k_1k_2\rvert T\{\exp[-i\int_{-T}^{T}dtH_{I}(t)]\}\lvert p_1p_2\rangle_0,$$ where $E_i$ and $E_f$ are the energies of the initial and final states.

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