= Solution
Let $g=aU_0\gg1$ and $z=ka$. A denominator zero away from the removable zero at $z=0$ obeys
$$
2z+ig-ig e^{2iz}=0,\qquad e^{2iz}=1-\frac{2iz}{g}.
$$
Near $z=\pi$, write $z=\pi+\delta$ and choose the <logarithm> branch with $\delta\to0$. Expanding gives
$$
\delta=-\frac z g-\frac{iz^2}{g^2}+O(g^{-3}),\qquad
z=\pi-\frac\pi g+\frac\pi{g^2}-\frac{i\pi^2}{g^2}+O(g^{-3}).
$$
Consequently
$$
\boxed{\alpha=-\frac\pi{aU_0}+O((aU_0)^{-2}),\qquad
\gamma=\left(\frac\pi{aU_0}\right)^2+O((aU_0)^{-3}).}
$$
The lower-half-plane pole is a <scattering resonance>: an odd <standing wave> is nearly confined between the two high repulsive barriers but leaks out slowly. If $E=E_R-i\Gamma/2=\hbar^2k^2/(2m)$, then $\Gamma\simeq2\hbar^2\pi\gamma/(ma^2)$ and its probability lifetime is $\hbar/\Gamma$. It is not a genuine <bound state> of the repulsive potential.
Back to article page
