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In "Introduction to Quantum Mechanics" by Griffiths we discussed the delta potential well. They speak about bound and scattering states for $E<0$ and $E>0$ resp. But before that (Problem 2.2) we proved that $E$ must exceed the minimum value of $V(x)$. Clearly $V(x)$ has minimum value 0, How then, can there exist bound states?

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The one-dimensional delta function well $V(x)= -\lambda \delta(x)$ can be constructed by taking $\delta(x)= \lim_{\epsilon\to 0} \delta_\epsilon(x)$, where $\delta_\epsilon(x)$ is a square potential of width $\epsilon$ and hight $1/\epsilon$. Thus the minimum value of $V(x)$ is $-\lambda/\epsilon$ which is heading to minus infinity.

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  • $\begingroup$ Oh, I did not notice the minus sign... Thank you. Should I delete this ? how do I ? $\endgroup$ – Yannick Stulens Jun 19 '17 at 19:24
  • $\begingroup$ @YannickStulens You can delete your own question by clicking the little gray "delete" word that appears below your question. But I'd leave it up if I were you, it's a reasonable thing to be confused by even if you notice the minus sign, so it might help others later. $\endgroup$ – tparker Jun 19 '17 at 21:05

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