1
$\begingroup$

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)?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.