Energy spectrum for a step potential Most of the books tend to give this explanation that for a bound physical system, the energy and momentum eigen values have discrete spectrum and otherwise, they have a continuous spectrum, which I seem to understand after seeing the solution to the infinite potential well problem where energy takes discrete values. But, what about the potential step problem? What determines whether a system is bound or not? And is the energy spectrum continuous or discrete for the case of step potential? 
 A: One ought to clarify the kind of step you are referring to, in particular in connection with boundary conditions.
On the real line (extending from $-\infty$ to $+\infty$), the potential step is defined typically as
$$
V(x)=\left\{\begin{array}{cc}
0&\hbox{if } x\le 0\, ,\\
V_0&\hbox{if } x> 0\, . \end{array}\right.
$$
In this case this potential does not accommodate bound states and discrete values of energy for any $V_0$ (positive or negative).  The solutions are plane waves with different wave vectors $k_+$ and $k_-$ (depending on $V_0$) on the positive or negative portion of the axis.  This setup is typically studied for scattering by the step.
If, on the other hand,
$$
V(r)=\left\{\begin{array}{cc}
-V_0&\hbox{if } r<r_0\, ,\\
0&\hbox{if } r\ge r_0\, ,\end{array}\right.
$$
with $V_0>0$ and $r\ge 0$, the restriction on $r$ functions as an infinite barrier at $r=0$.  This is then a finite potential well, which can accommodate one bound state and possible more depending on the depth $V_0$ and the range $r_0$.  This kind of potential is a crude approximation to the Wood-Saxon potential of nuclear physics.
