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Usual central potentials produce quantum spectra with energy levels going as $n$, $n^2$, $n^3$ and so on, being $n$ the quantum number of the orbit. In the other extreme we have "dirac-delta" potentials which have only a single discrete eigenvalue. I was wondering, what kind of potential do we need for producing an exponential $e^n$ set of discrete eigenvalues?

arivero
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1 Answers1

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For 1D potentials, the sequence of bound state energy eigenvalues $E_n$ cannot grow faster than what happens in the case of an infinite well, i.e. $E_n$ cannot grow faster than $n^2$.

Qmechanic
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    Note added: This bound also holds in higher dimensions. – Qmechanic Jul 02 '14 at 11:47
  • I am still worried because I can manufacture a potential with any arbitrary finite set of bound states. There is some fine convergence magic here. – arivero Jul 02 '14 at 12:09
  • Guessing... a finite set of bound states growing exponentially could approximate some potential of "tidal" kind, $V(x) \sim 1/x^2$, couldn't it? – arivero Jul 02 '14 at 12:27
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    A negative attractive 1D potential of the asymptotic form $$-V(x) \sim 1/x^2\quad\text{for}\quad |x|\to \infty$$ (where we implicitly assume that a singularity at $x=0$ has been regularized) would lead to exponentially decaying energy levels $$-E_n \sim e^{-\mu n}\quad\text{for}\quad n\to \infty.$$ This potential is also discussed in this Phys.SE answer. – Qmechanic Jul 02 '14 at 12:42