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In scattering theory the Hamiltonian $H$ can be written as the sum

$H = H_0 + V$

where $H_0$ is the Hamiltonian of free particles and $V$ shall contain the interaction between particles. We can find $\psi^{in}_{\alpha}$, which represent eigenstates of the "full" Hamiltonian, and $\phi_{\alpha}$ as eigenstates of the free-particle Hamiltonian $H_0$. If we consider very early times ($t \rightarrow -\infty$) it becomes obvious, that

$\psi^{in}_{\alpha} \rightarrow \phi_{\alpha}$

due to the fact, that $ V \rightarrow 0$.

My problem is that I don´t understand why the $\psi^{out}_{\alpha}$-states are needed at all. If $\psi^{in}_{\alpha}$ is an eigenstate of $H$ why should it evolve into a different state?

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You are right, in the sense that, if you consider the Hamiltonian eigenstates, the time evolution of the state just yields a meaningless phase factor. The true definition of the asymptotic states starts from a generic superposition of energy eigenstates

$$ \int d\alpha\, g(\alpha) \psi^{\pm}_\alpha $$

where $\alpha$ is the collection of all the momenta, spins and other quantum numbers of the state $\psi$ and $+$ = in, $-$ = out.

The definition of asymptotic state is then that, when evolving this state to the infinite past or infinite future, your state looks like the same superposition of free states.

$$ e^{-i H t}\int d\alpha\, g(\alpha) \psi^{\pm}_\alpha \to e^{-i H_0 t}\int d\alpha \,g(\alpha) \phi_\alpha$$

for $t\to \mp \infty$ respectively.

With this definition the time evolution of the states you are considering is in general non-trivial.

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  • $\begingroup$ Honestly I think, that I still don´t get it: If $\psi^{in}_{\alpha}$ is an eigenstate of $H$ the time evolution of the state - as you said - yields just a phase factor. A wave package, which is obviously not an eigenstate of $H$, has a more complex time evolution. But if $\psi^{in}_{\alpha}$ and $\psi^{out}_{\alpha}$ are both eigenstates of the Hamiltonian and if both are a basis to the same Hilbert space why do we even need the $\psi^{out}_{\alpha}$-states? From my point of view we could describe the wave package before and after the scattering with a superposition of $\psi^{in}_{\alpha}$ $\endgroup$
    – maxxam
    May 11, 2021 at 15:35
  • $\begingroup$ @maxxam. Sure. Any out-state is a linear combination of in-states. The coefficients of this expansion is exactly the definition of the S-matrix $\endgroup$
    – FrodCube
    May 11, 2021 at 15:41
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The in-state is an asymptotic state. Let $\psi$ be the energy eigenstate of the full Hamiltonian $H \psi = E \psi$ around $t=0$. Its energy $E$ as is conserved during the scattering process.

The in-state (in the book by Weinberg, and by Peskin and Schroeder) is the $\psi$ state in the limit infinite past limit $$ \psi_{\rm in} \equiv \lim_{t \to -\infty} e^{-iHt} \psi.$$

Although it goes out of the influence of the potential $V$, if $V$ becomes weak fast enough, it cannot be completely free state. Conversely, a free state is by definition the eigenstate of the free Hamiltonian, e.g. a plane wave, therefore it never interacts.

However it is useful if we introduce really free state having the same energy eigenvalue $H_0 \phi = E \phi$, because we do the perturbation using this. That is, we find the solution of the interacting Schr"odinger equation expanded in terms of such $\phi$'s. The scattering, in-field $\psi_{\rm in}$ becomes arbitrarily close to this free field $\phi$, like the converging function in the analysis. (In other textbooks like the renowned J. R. Taylor, the free field $\phi$ itself is called the in-state.)

Likewise, the out-field is the scattering field $\psi$ in the limit $t \to +\infty$. If we know the time evolution of the field connecting $\psi_{\rm in}$ to $\psi_{\rm out}$, we understand the scattering problem. It is encoded in the $S$-matrix, the inner product of the in- and the out-states.

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