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?


2 Answers 2


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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.