 Solution

For a negative energy write $E=-{\hslash }^2{\lambda }^2/(2M)$, $\lambda >0$. Away from the delta interaction the decaying solutions are $C{e}^{-\lambda x}$ for $x>0$ and $C{e}^{\lambda x}$ for $x<0$, with the same $C$ by continuity. The derivative jump is $-2\lambda C=-C/\Delta$, giving $\lambda =1/(2\Delta )$. Thus the bound state in a delta potential hasThe normalization follows from ${\int }_{\mathbb{R}}{e}^{-2\lambda |x|}dx=1/\lambda$. This is a genuine square-integrable negative-energy state.