I find there are two methods to calculate the amplitude in QFT.  First method:  Use LSZ reduction formula  $$\langle p_1\cdots p_m;out|k_1\cdots k_n;in\rangle=\big(\frac{i}{\sqrt{Z}}\big)^{n+m}\int d^4x_1\cdots d^4y_m e^{-ik_1\cdot x_1}\cdots e^{ip_m\cdot y_m}(\Box_{x_1}+m^2)\cdots (\Box_{y_m}+m^2)\langle0|T(\phi(x_1)\cdots\phi(y_m))|0\rangle$$  Then we only need to solve $n+m$-point correlation function. In general, $n$-point function is  $$G(x_1,\cdots x_n)=\frac{\langle 0|T(\phi_I(x_1)\cdots\phi_I(x_n)\exp(-i\int H_I(t^\prime)dt^\prime))|0\rangle}{\langle 0|T(\exp(-i\int H_I(t^\prime)dt^\prime))|0\rangle}$$  And we can solve the correlation function perturbatively.  Second method:  Using adiabatic switch and interacting picture, one can derive  $$\langle p_1\cdots p_m|S|k_1\cdots k_n\rangle=\langle0|a_{p_1}\cdots a_{p_m}\exp(-i\int H_I(t^\prime)dt^\prime)a^\dagger_{k_1}\cdots a^\dagger_{k_n}|0\rangle$$  And we also solve it perturbatively with Wick theorem.  Then my question is:  (1)Are these two methods equivalent up to any order? How to prove.  (2)What's the difference between $|k_1\cdots k_n;in\rangle$ in the first method and $|k_1\cdots k_n\rangle$ in the second method?   (3)If these two methods are equivalent, obviously the second method is easier. What's the motivation to create the first method. Or are there some examples in which the 2nd method fails and we can only use 1st method?   This question may be similar to following question:     However, these questions have not been solved and I explain this question in detail.
Most modern texts spend some time deriving the  that connects S matrix elements to time ordered field correlation functions. It seems essential, and really helps clear up what you are calculating. Yet some earlier texts and even some modern texts (e.g.  by R. Klauber) seem to skip right past this, working everything out in the "interaction" picture. It seems there must be something going wrong with this latter procedure, but I am not quite able to put it together.
The Cartesian product of a finite number of countable sets is countable The Cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod\limits_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i\}$$ Let $I_n=\{i \in \mathbb N \mid i \le n\}$ for $n\in \Bbb N$ and $A_i$ be countable for all $i\in I_n$. Then $\prod\limits_{i\in I_n}A_i$ is countable.   My attempt: Lemma 1: $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ and $\prod\limits_{i\in I_{n+1}}A_i$ are equinumerous for all $n \in \mathbb N$. (I presented a proof ) Lemma 2: If $A$ and $B$ are countable, then $A \times B$ is countable. (I presented a proof ) I will prove the theorem by induction on $n$. The theorem is trivially true for $n=0$. Assume that the theorem is true for $n=k$, then $\prod\limits_{i\in I_k}A_i$ is countable. Since $\prod\limits_{i\in I_k}A_i$ is countable (by inductive hypothesis) and $A_{k+1}$ is countable, then by Lemma 2 $(\prod\limits_{i\in I_n}A_i) \times A_{n+1}$ is countable. 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