Difference between stationary states, collision states, scattering states, and bound states

Question

A few weeks ago, I was presented one-dimensional systems in my QM class, and of course one-dimensional potentials too. Nonetheless, I'm still a bit unclear about the terminology my professor uses. Also, before closing this question as a duplicate of this one, please consider I've already had a look at it and the comments seem to disagree with almost every answer, let alone they are overly concise.

Let us consider a general potential, like the one in this drawing:

To my understanding, there are 3 regions in this potential we can separate:

1. $V_{min}<E<V_+$: Bound states
2. $V_+<E<V_-$: Stationary states? Collision states?
3. $E>V_+$: Collision states?

I don't quite understand the difference between "bound" and "stationary" states, and my professor pretty much uses them interchangeably, although he admits they're not. I think the difference between them is bound states have a point spectrum whilst stationary states have a continuous spectrum, but they are not degenerate (that is to say, bound states are in region 1, stationary states in region 2).

Also, I have no idea what a "collision state" is, and I can't seem to find it online.

Please help me distinguish these concepts, and of course bonus points for pointing out where each of them is in my graph.

Answer 1

I always refer to them as "scattering states". It is difficult for me to tell what your potential is doing at infinity, so assuming it is finite as it looks in the example, then this example of potential is one for which the WKB method was invented, because even for case I, the wave function will exhibit tunneling into the non-classical regions (places where $E\lt V$). Thus in all of the examples, the solutions are scattering states.

Bound states are characterized by eigenvalues, which are generated by subjecting the Schrodinger equation to boundary conditions. In this case the energies take on discrete values with some spacing between them and the spectra may or may not be degenerate. Scattering states, on the other hand, are evocative of the free-particle, the energy make take on any allowable value.

So how do you tell which are scattering states? Griffiths' Introduction to Quantum Mechanics $2^{nd}$ed., elucidates the problem nicely; the conditions you want are: $$E\lt [V(-\infty)\;\text{and}\; V(+\infty)]\implies\; \text{bound state},$$ $$E\gt [V(-\infty)\;\text{or}\; V(+\infty)]\implies\;\text{scattering state}$$ In cases where the potential goes to zero at infinity, which is the majority case in everyday life, then we have that: $$E\lt 0\;\implies\;\text{bound state},$$ $$E\gt 0\implies\;\text{scattering state}.$$ Even if a particle looks like it is trapped in a well as in example, I, if the potential doesn't grow to infinity, then the particle will leak by tunneling and the problem becomes a scattering problem in the technical sense. If you can get a hold of Griffith's book then take a look at section 2.5, it should clear up all of your doubts.