# 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 are approach superpositions of direct products of eigenstates of the $\textit{free}$ 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 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 seperated 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 '17 at 10:01

## 1 Answer

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.