Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 4 / 5D / ii

The sign of the Dirac delta distribution in the equation is positive after the outer minus sign is expanded. Integrating through zero, the nonsingular potential contributes a term tending to zero, whereas the delta term contributes $3k\psi (0)$. The jump condition for an impulse is thereforeContinuity at zero gives $C_{+}=C_{-}=C$. The two one-sided derivatives are $-3kC/2$ and $+3kC/2$, whose difference is exactly $-3kC$. Hence the matching condition introduces no further restriction on $C$, and the globally bounded solution isThis continuous solution has precisely the derivative jump required for its second distributional derivative to cancel the delta term. Its amplitude remains arbitrary because no normalization has been imposed.