One Particle State in Interacting QFT (4.88 4.89 in Peskin)

Question

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

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

How to derive equation 4.89?

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

Comments

The first eqn is an assumption (which can only be proven for specific given Hamiltonians), that when two particles are far apart their mutual interaction should be negligible, hence an eigenstate of the free Hamiltonian; the second is the Dyson series, you should be able to find its derivation in QM textbooks, say Sakurai's modern QM. 

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@Jia Yiyang Thanks a lot. Does it mean that the interaction term should be $e^{-\epsilon |t|}H_{int}$?

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no, $H_{int}+i\epsilon$ for positive times and $H_{int}-i\epsilon$ for negative times.

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@Arnold Neumaier Thanks a lot.

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@Arnold Neumaier Could you elaborate? 

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It is only ''roughly'' equivalent. $t\to\pm\infty(1-i\epsilon)$ says $t=s(1-i\epsilon)$ with $s\to\pm\infty$. Substitute this into $tH$ and note that in the limit where $\epsilon$ vanishes, $H\epsilon$ and $\epsilon$ behave similarly (if $H$ is nonnegative).