Scattering and bound states

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SUMMARY

In quantum mechanics, bound states (E < 0) consistently yield discrete energy spectra, while scattering states (E > 0) result in continuous energy spectra. This conclusion is supported by examples from Griffiths' "Introduction to Quantum Mechanics," including the harmonic oscillator, free particle, infinite square well, and finite square well. Bound states are characterized by normalizable wave functions, typically sinusoidal, whereas scattering states exhibit non-normalizable wave functions, often resembling Gaussian distributions. The relationship between energy and wave function behavior is crucial for understanding these concepts.

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  • Knowledge of discrete vs. continuous spectra
  • Basic concepts of wave functions and normalization
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Niles
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In all the possible potentials I have encountered so far, it seems that the bound states (i.e. E < [V(-infinity) and V(infinity)]) always results in a discrete spectrum of energies, whereas the scattering states (E > [V(-infinity) and V(infinity)]) always results in a continuous spectrum of energies.

I can't seem to find a logical explanation for this. If we use the anove definition of bound and scattering states: The potential at plus/minus infinity of the harmonic oscillator is infinite, but so is the energy (for infinite n). But the harmonic oscillator has a bound spectrum.

I can't quite see this. I've taken the above from Griffith's, and sadly he never mentions whether "bound states = discrete spectrum" or not.
 
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Ok, I can see that my question is poorly formulated. I'll refraise it:

First of all, a bound states is defined as the energy E < 0 and a scattering state is defined as E > 0. My questions on this topic are the following:

1) Will a bound state (i.e. E < 0) always result in a solution of the Schrödinger equation with a discrete energy spectrum?

2) Will a scattering state (i.e. E > 0) always result in a solution of the Schrödinger equation with a continuous energy spectrum?

In all the potentials I've encountered so far (harmonic oscillator, free particle, infinite square well and finite square well) it is so. But no where in the book (Griffiths) is an explanation of why. Can you guys enlighten me?
 
1) Will a bound state (i.e. E < 0) always result in a solution of the Schrödinger equation with a discrete energy spectrum?

Yes. I think the easy way to see this is to note that every bound wave, classical or quantum, is limited to a discrete set of possible frequencies/ wave lengths. So if you span a cord, fix it on both ends, then pluck it, you will see you can not make it wave in every possible frequency.

The discrete wave spectrum translates in a discrete energy spectrum.

2) Will a scattering state (i.e. E > 0) always result in a solution of the Schrödinger equation with a continuous energy spectrum?

If the scattered particle is allowed to go to infinity, is not bound, then yes.
 
Will a scattered particle always have an oscillating wave function? And similarly, will a bound particle always have an exponential wave function?

Again, these things I am taking from the examples in Griffith's QM book.

EDIT: And are bound states always normalizable, whereas scattering states are not?
 
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bound states will always have sinusoidal time independent wave functions i think. and likewise I'm going to take a guess and say that an unbound scattering state will always have a guassian like time independent wave function.
 
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