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Why does Mother Nature allow bound states in arbitrarily weak attractive potential in 2D but not in 3D?

See, for example, this article, arXiv:math-ph/0208011.

Qmechanic
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Taiben
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    Possible duplicate: http://physics.stackexchange.com/q/143630/2451 – Qmechanic Sep 01 '14 at 18:45
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    Could you perhaps elaborate on what you mean when you ask Why? here? Are you asking what step of the proof breaks down in higher dimensions? Or are you asking if there are no-go theorems forbidding such bound states in higher dimensions? Or something else? – ACuriousMind Sep 01 '14 at 18:49
  • @ACuriousMind Sure. I'll address your questions one by one. First, I haven't seen a mathematical proof that the statement does not hold in 3D, so it would be great to see such a proof and particularly how it breaks down in higher dimensions. Second, I'm not aware of any such theorems, so it would also be very exciting to know. Third, is there an intuitive explanation, for example, why would higher dimensionality impairs that ability of space to support the bound states? – Taiben Sep 01 '14 at 18:55
  • @ACuriousMind When I asked this question, I was hoping to see an answer to the third aspect -- intuitive reasoning of the underlying physics. – Taiben Sep 01 '14 at 18:56
  • One should probably point out, "mother nature" does not "allow" for arbitrary potentials. This is just mathematicians playing with equations of motion that happen to describe real physical systems for CERTAIN choices of potentials. While the results are interesting, even to physicists, they have very, very little in common with actual physics. – CuriousOne Sep 01 '14 at 22:21
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    It is explained here (pages $159,160$), at least for a simple potential $U(r)= - u \delta(r)$ – Trimok Sep 02 '14 at 09:35

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