Consider the reduced radial Schrodinger equation:
$$-\frac{1}{2}\frac{\text{d}^2}{\text{d}r^2}\phi(r)+V(r)\phi(r)=E\phi(r).$$
We try to find a bound state (i.e. $\phi(0)=\phi(+\infty)=0$).
Here $V(r)=1/r$ or $1/r^2,1/r^3$ etc.
A numerical approach showed that the energy $E$ tends to zero when $a$ (i.e. the upper bound for $r$, use it to represent $+\infty$ in the numerical method) increases.
(In this table $V(r)=1/r^{\alpha}$. $E_1$ is the calculated energy.)
My question is, how to explain this result physically?
Is that the real $E$ should be $0$, or allowed $E$s actually are continuous?
I think it is because of the repulsive potential $V(r)$, that $E$ should be $0$.
(P.S. I'm interested in the "ground state", if it exists. My numerical procedure tries to find the minimum energy. )