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

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: $$$$\lim\limits_{t\rightarrow \pm\infty}\int d\alpha\, e^{-iE_{\alpha}t}g(\alpha)|\psi_{\alpha}^{\pm}\rangle=\int d\alpha\, e^{-iE_{\alpha}t}g(\alpha)|\phi_{\alpha}\rangle,$$$$ where the Hamiltonian is split into a free and an interacting part $$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.

• Following this post physics.stackexchange.com/q/41439 I found that Theorem 3.4.4 of Thirrings textbook: "Quantum mechanics of atoms and molecules" answers both my questions. If the interaction term is compact relative to the free Hamiltonian then the late time particle states approach eigenstates of the free Hamiltonian. The proof requires some topology which I am not familiar with.
– Luke
Dec 28, 2017 at 10:01

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.