Suppose a system is under the influence of a potential which vanishes at $\pm \infty$. Now we know that if the energy of the system is negative ($E<0$) then the system is in a bound state and the bound state energy spectrum is discrete. For $E>0$, the system is in scattering states (with continuous energy spectrum). My question is:
What happens at $E=0$? Can we consider it a discrete bound state or should we consider it as a scattering state and why?
Yes they can! Though admittedly, the examples are quite awkward.
An explicit example is given here (T. Aktosun and R. G. Newton, 1985, Inverse Problems 1).
The potential is
$$V(x)=\begin{cases} 4\, \frac{(x\sqrt{2}+1)(\sigma(x)-2x+1)}{(\sigma(x)+c)^2} & x>0\\ \frac {4} {(b-x\sqrt{2})^2} & x<0 \end{cases}$$
and the wavefunction is
$$\psi(x)=\begin{cases} \frac {2(1+x\sqrt{2})}{\sigma(x)+c} & x>0\\ \frac {2(b-1)}{b-x\sqrt{2}} & x<0 \end{cases}$$
with $\frac 1 b + \frac 1 c = 1$ and $c\ge 1$. And $\sigma(x)=\frac {2\sqrt{2}}{3}x^3+2x^2+x\sqrt{2}$. Both the wavefunction and the potential vanish in the limit $x\to\pm\infty$, so this is a proper bound state. (In the paper, these are expressed using $\Theta(x)$, the Heaviside function. Also, note that the potential is discontinuous at $x=0$, so $\psi$ has a discontinuous second derivative there.)
The paper additionally discusses what it calls "half-bound" states - e.g., the double delta potential has one for some parameter values (though Wikipedia calls it just a "bound state", which is not entirely accurate.).
This is a graph of $\psi(x)$ and $V(x)$ for the randomly chosen value $c=1.77$.
