what growth conditions should the potential inside the Hamiltonian $ H=p^{2}+ V(x) $ has in order to get ALWAYS a discrete spectrum ??

for example how can we know for teh cases $ |x|^{a} $ , $ exp(b|x|) $ and so on with real parameters a,b and c

how can we prove that for theses potentials we will have only a discrete set of eigenvalues ??

another question is it possible to find a potential which have the eigenvalues $ E_{n}=n^{c} $ for some positive real number 'c' , i have heard that no potential in QM could have eigenvalues bigger than $ E_{n}=n^{2} $

  • $\begingroup$ I guess the only thing you need is that $V(x) \to \infty$ for $|x|\to\infty$. $\endgroup$ – Fabian Oct 16 '12 at 21:44

For the first question, I leave it up to the experts in point spectra and stuff like that (even though I think that you only need $V(x)\to\infty$ for $|x|\to\infty$).

Regarding your second question: spectra with $E_n = n^c$. I present below the reason why $c>2$ should be impossible. In fact you can easily find a potential where $E_n \sim n^c$ for $n\to\infty$ by using the correspondence principle:

The differences of energies scale like $\Delta E_n \sim n^{c-1}$ which should correspond to $h/T$ where $T$ is the period of the classical motion in the potential. Given that $$ T(E) = \int_{V\leq E} \frac{dx}{\sqrt{2(E- V)/m}} $$ we find that with $V(x) \sim x^{\alpha}$, we have $T(E) \sim R^{(\alpha-2)/\alpha} \sim E^{(2-\alpha)/2\alpha}$ with $R$ the turing point such that $V(R) = E$.

So in order to have $E_n = n^c$, we need to have $\Delta E_n \sim n^{c-1} \sim E^{(c-1)/c}$ to match $T^{-1} \sim E^{(\alpha-2)/2\alpha}$ which leads to $$\frac{c-1}{c} = \frac{\alpha-2}{2\alpha}$$ or $$\alpha= \frac{2c}{2-c}.$$

For $c\to 2^-$, we have $\alpha$ approaching $\infty$. So for the steepest possible potential, we only get $c=2$.

Note that:

  • for $c=1$, we get $\alpha=2$ (harmonic oscillator)
  • for $c=-2$, we get $\alpha=-1$ (H-Atom)

By the way, looking at the inverse relation: $$c = \frac{2\alpha}{2+\alpha}$$ we see that there is also something happening at $\alpha=-2$: this is the point at which Heisenbergs uncertainty principle is not enough to hinder the particle from falling into the centre.


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