 Solution

At the delta potential, the bound equation becomes $\psi (x)=\beta \psi (0)e^{-\kappa |x|}/\kappa$. A nonzero solution therefore requires $\kappa =\beta$, givingThere is one bound level. The transmission amplitude $t(k)$ has its physical-sheet pole at $k=i\beta$, which produces exactly this decaying state. The real-axis probability's rational continuation $T(k)=k^2/(k^2+{\beta }^2)$ has both $+i\beta$ and $-i\beta$ poles; the lower pole is not an additional bound state. It is essential to distinguish the analytic amplitude from an even continuation of its modulus squared.