Suppose we have a 1D potential $V(x)$ of finite range, i.e.,

$$ V(x) ~=~0 $$

for $|x| > b $. The potential is assume to support at least two bound states, but might have more, say $n\geq 2$.

The shallowest one $E_n<0$ can be infinitesimally small. However, how about the second shallowest one $E_{n-1}<E_n$?

Question: What is the smallest possible value for $|E_{n-1}|$? Is it on the order of $\hbar^2 / 2m b^2$ definitely?

To make the question clear, I emphasize that the shallowest bound state $E_n$ is the highest excited bound state. Do not confuse it with the ground state $E_1$! It is apparent that this state $E_n$ can be arbitrarily close the continuum edge.

  • $\begingroup$ Comment to the question (v4): The bound is likely saturated for systems with precisely two bound states $n=2$, and this means that semi-classical WKB approximations cannot be trusted. $\endgroup$ – Qmechanic Nov 27 '14 at 23:49

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