Normalizable States in QM: Bound vs. Scattering

[asdf60]

I'm kind of self studying from Griffiths's QM book, and I'd like to clarify a few things I find confusing.
As I understand it, for any potential V, there can exist bound states or scattering states. In the case of the bound states, the solutions to the time-independent schrodinger equation are normalizable, whereas with scattering states they are not. Is this true for all potentials? Why is that true?
Anyway, in that case, you can immediately reject the solutions which are not normalizable for the bounded states, since they do not represent physical states. But for the scattering states, there is no reason to throw out the non normalizable solutions.

 

[Gfunction]

Normalizing is equivalent to being "bound". All being bound means is that you can find the particle within a definite region in space (even if that region is infinite like in the case of the single bound state for the Dirac potential). But remembering that the squared wavefunction gives the probability density for the particle reveals that this is just the same as having a function whose squared integral returns a finite value.

Also not every potential has scattering states. For instance, the infinite square well and the simple harmonic oscillator potentials only yield bound states (note that the potential goes to infinity at $x=\pm \infty$ in both these examples). Not every potential has bound states either. A trivial example is $V(x) = a$ where "a" is a constant i.e. the free particle solution. I'm actually not sure though if there is a nontrivial potential that only yields scattering states (or some weird mathematical function equivalent to the trivial case such as the indicator function on the set of rationals).

 

[PeroK]

Normalizing is equivalent to being "bound". All being bound means is that you can find the particle within a definite region in space (even if that region is infinite like in the case of the single bound state for the Dirac potential). But remembering that the squared wavefunction gives the probability density for the particle reveals that this is just the same as having a function whose squared integral returns a finite value.

Also not every potential has scattering states. For instance, the infinite square well and the simple harmonic oscillator potentials only yield bound states (note that the potential goes to infinity at $x=\pm \infty$ in both these examples). Not every potential has bound states either. A trivial example is $V(x) = a$ where "a" is a constant i.e. the free particle solution. I'm actually not sure though if there is a nontrivial potential that only yields scattering states (or some weird mathematical function equivalent to the trivial case such as the indicator function on the set of rationals).

Maybe you're 11 years too late with that answer!

 

[Gfunction]

Maybe you're 11 years too late with that answer!

Yeah, I'm new here and saw a question that I actually knew something about. I didn't realize how old it was. Who knows, maybe someone will have a similar question 11 years from now and my response will help them out.

 

[DrClaude]

Thanks for the answer. We will now close the thread.