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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:

Quantum Field Theory without LSZ, how is it possible?

LSZ reduction vs adiabatic hypothesis in perburbative calculation of interacting fields However, these questions have not been solved and I explain this question in detail.

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    I am just as confused as you here, but, I think turning off interactions can change states too much to match them up to the interacting states. In QCD, your in and out states are hadrons; if you turn off interactions, they fall apart. In the electroweak theory, if you turn off interactions, you lose the longitudal modes of the W and Z bosons because they no longer have mass, and so on. I think in simpler theories like $\phi^3$ or QED it should work fine, though. – knzhou Mar 24 '16 at 05:09

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