# Discrete or continious spectrum in QM

1. Nov 17, 2009

### meanyack

1. The problem statement, all variables and given/known data

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?

2. Nov 17, 2009

### 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.

3. Nov 17, 2009

### 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.