For the hydrogen atom the quantised energy levels are:

$$E_n = \frac{- 13.6 eV}{n^2}\quad\text{with}\quad n = 1,2,3...$$

One peculiar property of this quantisation is that for large $n$ the energy levels are ever closer together and for $E \geq 0$ (that is $n = \infty$) the energy spectrum becomes a continuum. The electron is then free, of course.

For the hydrogen atom the Potential Energy function is:

$$V(r) = \frac{ - e^2}{4 \pi \epsilon_0 r}$$

Obviously for $r = 0, V = - \infty$, for $r = \infty, V = 0$

Suppose that for a quantum system we construct a Potential Energy of the general form:

$$V(r) = - \frac{V_0}{f(r)},$$

with $f(r)$ a symmetric function of $r$ with a root at $r = 0$ (so that $V(0) = - \infty$).

Lets also assume that $\frac{1}{f(r)}$ tends to $0$ for $r = \infty$ (so that $V(\infty) = 0$).

Intuitively I feel that with such a Potential Energy function, the quantised energy would also smoothly convert to a continuum for high quantum numbers, going fully continuous at $E \geq 0$.

My question is, can this be demonstrated or even proved (or disproved, of course)?


This would not work for an arbitrary smooth central potential $V: \mathbb{R}_+ \to \mathbb{R}$ with

$$V~<~0, \qquad V^{\prime}~>~0,\qquad V(0)~=~-\infty,\qquad V(\infty)~=~0,$$

because there might only be a finite number of bound states, and hence no continuous limit of bound states.

For instance, one can use WKB methods to argue that if $V(r)$ goes faster to zero than $1/r$ for $r\to \infty$ (but still keeps, say, a hydrogen-like $1/r$ dependence of $V(r)$ for $r\to 0^+$), then there will be only a finite number of bound states.

  • $\begingroup$ Qmechanic: If we take a hydrogenic-like $V(r) \propto 1/r^m$ with $1 > m > 0$, for which $V(r)$ tends to $0$ slower than $1/r$, then using the Bohr-Sommerfeld quantization condition I get $E \propto n^\frac{2m}{m-2}$ with $\frac{2m}{m-2} < 0$. $E$ goes to $0$ in the limit and I think there are an infinite number of bound states then. $\endgroup$ – Gert Aug 9 '15 at 2:01
  • $\begingroup$ But the B-S QZ is only approximate for large $n$, assuming I understand well (semi-classical approximation of T = KE + V for high $E$). $\endgroup$ – Gert Aug 9 '15 at 2:09

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