Discrete or continious spectrum in QM

[meanyack]

Homework Statement

Here is the question: how can we know that if we have discrete or continuous spectrum just by looking at the potential graph?

Specifically, let`s consider the potential V(x)=-F*x (F:const) . After we solve, we can conclude wavefunctin is airy function, and so both continious and discrete spectrum. But, without sloving, how can we decide?

 

[physicsworks]

There are special cases when we can guess what the spectrum would be. For example, when the particle travels in the central field potential it can be shown that: when the energy E of the particle is positive the spectrum is continuous and when E<0 the spectrum is discrete.
For example, if we have repulsive potential energy of the form U \sim 1/r, the total energy E of the particle is positive. (Really,
E=\frac{1}{2m}\int {\psi^* \hat{\mathbf{P}}^2}\psi dV + \int {\psi^*U \psi}dV
which is positive since U is positive and the eigenvalues of \hat{\mathbf{P}}^2} are positive numbers too).
So, in this potential field we have E>0 and continuous spectrum.
If U \sim -1/r we have Сoulomb attraction and to possibilities: E>0 (continuous spectrum, ionized electron) and E<0 (discrete spectrum).
Finally, in case of 6-12 potential we have continuous spectrum for E>0 (dissociated molecule) and discrete spectrum for E<0.
Such examples show that we can guess what the spectrum is without doing anything with Schrödinger equation.

 

[jdwood983]

The way that was explained to me is if there are 2 turning points, then it is discrete. If there is only 1 turning point, it is continuous. I, unfortunately, was never able to ask my professor whether something that had three (or more) turning points would be continuous or not.