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Why are these states called as such, and how do they differ? I vaguely understand that when $E > 0$ you obtain a scattering state, but when $E < 0$ you have a bound state.

Qmechanic
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wrongusername
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    Hey, it is not necessary to have a bound state for $E<0$. The condition for bound state is that $E<V(-\infty)$ and $E<V(+\infty)$ while for scattering state, the condition is $E>V(-\infty)$ or $E>V(+\infty)$. Just take the example of Harmonic oscillator to understand this. And the physical meaning is correctly given in the answer below that bound states die out far away while scattering states do not. – cleanplay Jul 14 '13 at 09:22

2 Answers2

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These terms apply when you're solving the Schrodinger equation with a potential that goes to zero at large distances. In this situation, the solutions with $E<0$ have the property that $\psi$ dies away to zero for large distance. So the particle is, with high probability, guaranteed to be in a confined region (not at large distance). So those are bound states.

The solutions with $E>0$, on the other hand, do not die away to zero at large distances -- instead, they go like $e^{ikr}$ where $k=\sqrt{2mE}/\hbar$. So these solutions represent particles that have high probability to be arbitrarily far away. Physically, they are useful when describing particles that start far away, approach the scattering center, and end up far away again. Hence the name "scattering states."

Ted Bunn
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    $E<0$ is not precisely the condition for bound states. Rather $E<V(\pm\infty)$ is the correct condition since without it, you cannot even discuss the bound states of the Harmonic oscillator. It is misleading to call $E<0$ as the bound states. – cleanplay Jul 14 '13 at 09:24
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    @cleanplay The answer clearly states in its first sentence "These terms apply when you're solving the Schrodinger equation with a potential that goes to zero at large distances." This is quite a reasonable way of defining things. – not all wrong May 09 '16 at 10:39
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Let me explain with a simple example. Consider a particle in a finite potential well. There will be two cases:

i) $E<V$;

ii) $ E>V $.

If the energy of the particle its smaller than the magnitude of the potential, the particle will be confined in the box forever, that is, the particle its bounded to the thing that is generating the potential. In this case where the particle its confined to a finite space, in a bound state, the energy will be quantized, that is, only multiples of a certain quantity of energy will be allowed.

But if the energy of the particle it's greater than the intensity of the potential the still will "feel" the "hole" below it, a portion of the "particle" will be reflected, and will go back, and the other portion will cross the well.

The energy did not need to be necessarily smaller than zero to a bound state occur, as a matter of fact it only need to be smaller than the intensity of the potential.

All of the information described above can be obtained by solving the Schrödinger Equation to the potential in question, as was done in the link in the top of the answer.

A very good introductory book in this subject its "Introduction to Quantum Mechanics" by David J. Griffiths. Read it, it is very nice!

Rodrigo Thomas
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  • Thanks for the book recommendation! I am reading it and it is indeed very nice. – wrongusername May 13 '11 at 03:55
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    $E<V(\pm\infty)$ is the condition for bound states and not $E<V$. If you refer Griffiths, it says for bound state $E<V(\pm\infty)$ and for normalizability, $E>Vmin$ where $Vmin$ is the minimum value of the potential. (-Vo in case of finite well)(this is problem 2.2 in Griffiths ) – cleanplay Jul 14 '13 at 09:28
  • @cleanplay Here, the point is that (in the Wiki conventions) $V = V(\pm \infty)$. :) – not all wrong May 09 '16 at 10:40