Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S18 / 2 / e / Solution

Multiply the expression for $\psi$ by $F^{\ast }(x)$ and integrate. Cancelling the nonzero $C$ givesThe kernel is real and symmetric, so relabelling the integration variables puts the expression in the printed order:This is the scalar consistency condition for the rank-one potential, rather than an independent normalization equation. The double integral is real and nonnegative because the Fourier transform of $e^{-\kappa |x|}$ is $2\kappa /(q^2+{\kappa }^2)>0$. Thus attractive $\alpha >0$ is needed for a nontrivial solution. Under the usual integrability assumptions the corresponding free resolvent expectation decreases with $\kappa$, giving at most one bound state for this rank-one attraction.