I just wrote down the solution for the bound state of the Dirac delta potential well, for $E<0$, and apparently there is only one specific energy for the bound state, and it is negative. I solved for this just assuming $E<0$. What happens if a particle enters with energy $E<0$ but isn't equal to the bound state energy? I don't understand what the solution is telling me, that there is only one bound state, an exact figure. Does that mean we cannot say what the energy of the impinging particle is until we see what it does? I don't understand why you can't have an $E<0$ that doesn't equal the bound state energy, and if you can, what does that do? In the literature they solve for the bound state energy, then solve for the scattering coefficients for the scattering states. I guess they imply $E$ of the particle can be anything, as long as it's greater than zero. Am I wrong to think they are glossing over this too quickly? It doesn't make sense.

  • 3
    $\begingroup$ For unbound states, energy spectrum is continuous, so $E$ can be anything as long as it's $\ge0$. And for bound states it is discrete. A particle with $E<0$ must be already bound, otherwise it's outside of the potential well and is essentially free, thus having $E\ge0$. $\endgroup$ – Ruslan Oct 4 '13 at 7:52
  • 3
    $\begingroup$ The energy of the impinging particle has nothing to do with the Dirac delta potential, or any other one – it's just whatever kinetic energy that particle happens to have. $\endgroup$ – leftaroundabout Oct 4 '13 at 7:53
  • $\begingroup$ So the kinetic energy has nothing to do with whether or not the particle is bound to the potential or it scatters? Physically what does this mean then? If the energy isn't kinetic energy, what is it? $\endgroup$ – walczyk Oct 4 '13 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.