Past exam of the theoretical physics II course of the University of Cambridge / 2017 / 3 / b / Solution

Away from the delta potential, the decaying bound state obeys $E_0=-{\hslash }^2{\kappa }^2/(2m)$. Integrating the Schrödinger equation across the attractive interaction givesFor an even exponential, the left side is $-2\kappa \phi (0)$. ThereforeThe normalized bound state in a delta potential is ${\phi }_0(x)=\sqrt{\kappa }{e}^{-\kappa |x|}$, 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.