Physics Asked on November 25, 2020
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?
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 $$ V(x)=left{begin{array}{cc} 0&hbox{if } xle 0, , V_0&hbox{if } x> 0, . end{array}right. $$ 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, $$ V(r)=left{begin{array}{cc} -V_0&hbox{if } r<r_0, , 0&hbox{if } rge 0, ,end{array}right. $$ with $V_0>0$ and $rge 0$, 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.
Answered by ZeroTheHero on November 25, 2020
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