Potential energy function for high energy continuum?

Question

For the hydrogen atom the quantised energy levels are:

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

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)?

Qmechanic · Answer

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 $rto infty$ (but still keeps, say, a hydrogen-like $1/r$ dependence of $V(r)$ for $rto 0^+$), then there will be only a finite number of bound states.

Potential energy function for high energy continuum?

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