Scattering states and their physical significance

Question

I am following the QM book by Sakurai and currently dealing with the chapter involving scattering theory. I am wondering about what the physical significance of the scattering states $|\psi^{(\pm)}\rangle$ is.

My confusion stems from the following two facts:

1. Under the assumption that for $t \to -\infty$, we start off with the initial state $|i\rangle$, the transition probability from $|i\rangle$ to a final state $|n\rangle$ is proportional to the matrix element of the $\hat{T}$ operator, i.e. $$P(i\to n) \propto |\langle n|\hat{T}|i\rangle|^2 = |T_{ni}|^2.$$ This suggests that $\hat{T} |i\rangle$ is the state of the particle after the scattering process. The definition of the "out" scattering state is $\hat{T}|i\rangle = \hat{V}|\psi^+\rangle$, where $\hat{V}$ is the potential of the scatterer. Does this mean that $\hat{V}|\psi^+\rangle$ is the state of the particle after the scattering process?

2. Delving more into the state $|\psi^{+}\rangle$ for elastic scattering with the initial and final states being momentum eigenkets, Sakurai shows that its position representation for large distances from the target ($r \to \infty$) is $$\langle \bf{x}|\psi^+\rangle\propto e^{i\bf{k}\cdot\bf{x}}+\frac{e^{ikr}}{r} f(\bf{k^\prime}, \bf{k}),$$ where $f$ is the scattering amplitude. This would make sense if this was the final state after scattering, since we have one contribution from the incoming harmonic wave and an outgoing spherical wave. However, as said in 1., wouldn't we still have to apply $\hat{V}$ before taking the position representation to get the actual final state after scattering?

The same problem arises later when the scattering state is expanded in terms of partial waves. Everything said there would make sense to me if $|\psi^+\rangle$ actually was the final state after scattering. I would be thankful if someone could clear my confusion.

Answer 1

The $|\psi^+\rangle$ state is an eigenstate of the Hamiltonian. It does not change with time, and is therefore not going to describe the scattering as a time dependent phenomenon.

If you want to look at the time dependence like in an experiment, the incoming particle is localized away from the target. To do this you could make a wave packet and expand it in plane waves. Most scattering experiments have a collimated beam with a reasonably specified energy. That says that most of the packet plane waves have ${\bf k}$ values in the beam direction and peaked around the beam energy. Taking the $z$ direction as the direction of the incoming beam at large negative times, before the scattering, the packet is peaked at negative z and has no substantial components at the target at the origin. If you now use the same plane wave coefficients as the coefficients of the $|\psi^+\rangle$ state with the same $e^{i{\bf k}\cdot {\bf x}}$ term. This term will give the same beam component. The outgoing $e^{ikr}/r$ term will give no contibution, i.e. it would be peaked mathematically approximately at the $r$ value which would be equal to the $z$ value, but since $r$ is positive, it gives no contribution. At large positive times, after the scattering, the plane wave plus the forward scattering part of the outgoing spherical wave combine to give the original beam minus the total scattering (i.e. the optical theorem) and an outgoing spherical packet, described by the $e^{ikr}/r$ term.

The $|\psi^+\rangle$ is the eigenstate of the Hamiltonian which has incoming components equal to those of the plane wave. They are exactly the states which make the wave packet calculation easy. The $|\psi^-\rangle$ similarly have the outgoing components equal to those of a plane wave and they also form a complete set of scattering states. The unitary transformation between them is the S-matrix.