= Solution
Away from the <delta potential>, the decaying <bound state> obeys $E_0=-\hbar^2\kappa^2/(2m)$. Integrating the <Schrödinger equation> across the attractive interaction gives
$$
\varphi'(0^+)-\varphi'(0^-)=-\frac{2mg}{\hbar^2}\varphi(0).
$$
For an even exponential, the left side is $-2\kappa\varphi(0)$. Therefore
$$
\boxed{\kappa=\frac{mg}{\hbar^2},\qquad E_0=-\frac{mg^2}{2\hbar^2}.}
$$
The normalized <bound state in a delta potential> is $\varphi_0(x)=\sqrt\kappa e^{-\kappa|x|}$, since $2\kappa\int_0^\infty e^{-2\kappa x}dx=1$. The single positive solution for $\kappa$ gives one <bound state>; an odd decaying wave cannot be both continuous and nonzero with this interaction.
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