Why does the attractive Dirac delta distribution (function) potential $V = \alpha\delta(x)$ (for negative $\alpha$) yield both bound AND scattered states? Is this due to the definition of the Dirac delta distribution? Or is it for a different reason?
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6$\begingroup$ What sort of answer beyond "well, because these solutions to the energy eigenvalue problem (aka Shrodinger equation) exist" are you looking for? Perhaps it's physical intuition you're after? $\endgroup$– joshphysicsCommented Aug 6, 2013 at 20:11
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4$\begingroup$ Is there a finite potential which doesn't yield scattered states? $\endgroup$– ZachMcDarghCommented Aug 6, 2013 at 20:49
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1$\begingroup$ In re Zach's comment, Weston showed in 63 a wide range of scattering bodies in which there is zero backscattered field. Then it was a curiosity but now with metamaterials you can do some really crazy stuff. The DOI for Weston's paper is 10.1109/TAP.1963.1138082 $\endgroup$– user27777Commented Aug 6, 2013 at 22:46
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1 Answer
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Since the potential has a lowest value of -Infinity, the particle can be found in negative and positive energy. Now, for a bound state to occur, its corresponding energy of the eigenstate must be lower than both the value of the potential at $x\to +\infty$ and $x\to -\infty$.
So, the negative energy eigenstates have corresponding energies that are lower than the value of the potential at minus and plus infinity, so the negative energy eigenstates are also bound eigenstates.
On the other side, the positive energy eigenstates are scattered states because their corresponding energies are higher than the value of the potential at plus and minus infinity.
So, we get both bound and scattered states.
answered Mar 17, 2016 at 15:55