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?
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2What is a "bound physical system"? – DanielC Oct 19 '17 at 19:17
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A bound system like an atom? Or what type of bound system? – Nemo Oct 19 '17 at 19:28
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1Related: http://physics.stackexchange.com/q/65636/2451 and links therein. – Qmechanic Oct 19 '17 at 19:30
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By bound, I meant confined to a compact region. – Patrick Oct 19 '17 at 19:40
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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.
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