# "In"- and "out"-states in scattering theory

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?

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.

• 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 $\psi^{in}_{\alpha}$ May 11, 2021 at 15:35
• @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 May 11, 2021 at 15:41

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.