Why would the "in/out'' states asymptotically approach the free Hamiltonian eigenstates?

Question

The "in/out" states of the S-matrix in QFT are defined such that at late times they approach superpositions of direct products of eigenstates of the free Hamiltonian: \begin{equation} \lim\limits_{t\rightarrow \pm\infty}\int d\alpha\, e^{-iE_{\alpha}t}g(\alpha)|\psi_{\alpha}^{\pm}\rangle=\int d\alpha\, e^{-iE_{\alpha}t}g(\alpha)|\phi_{\alpha}\rangle, \end{equation} where the Hamiltonian is split into a free and an interacting part $H=H_0+V,$ and the $|\psi_{\alpha}^{\pm}\rangle$ are eigenstates of the "full" Hamiltonian, and $|\phi_{\alpha}\rangle$ is an eigenstate of the "solvable" Hamiltonian.

1. Consider an electron far from any other particles. Is it an eigenstate of the full or the solvable Hamiltonian?

2. Why would far separated particles not be direct products of eigenstates of the full Hamiltonian? The interaction term $V$ not only describes how two nearby particles time-evolve, but also influences the one-particle states. For instance, regardless of whether other particles are around, the $\bar{\psi}\gamma^{\mu}\psi A_{\mu}$ term affects the charge of the electron.

Answer 1

This looks like Weinberg's definition for scattering states. I posted a similar question, which doesn't have an answer right now, so I can't fully answer you. See: Weinberg's S-matrix and split into free and interacting Hamiltonian

But I can make some clarifications:

Re: (1), I'd clarify that Weinberg would demand you think of a single particle wavepacket (a superposition of eigenstates), not just an eigenstate, because he requires $g(\alpha)$ smooth in his definition.

Re: (2), I'd clarify that you probably don't quite mean that the two-particle state of two far-separated particles is a direct product of the single-particle states. Rather, the two-particle state is obtained by operating on the vacuum with an operator which is the product of the two operators that respectively create the two single particle states when operating on the vacuum. These are the Haag-Ruelle creation operators.

Answer 2

In fact, at least for QED, it is not true because there is still interaction with "vacuum" or said differently, the "in" and "out" particles are "dressed". It is easily seen that we cannot scatter from an electron elastically because it emits photons with the probability 1, i.e. the scattering is always inelastic. Thus the elastic cross section is zero; it means tha target (real electron) is compound. Only inclusive cross sections are meaningful. The corresponding calculations are usually given in chapters devoted to the infrared problem.

I wrote several papers on this subject (available on arXiv).