Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 1 / 32D / Solution

Insert the position and momentum resolutions of the identity to obtain the unitary Fourier transformThe kinetic term is multiplication by $p^2/(2m)$, and the potential has momentum kernel $⟨p|V|p^{\prime }⟩=(2\pi \hslash {)}^{-1}\int V(x)e^{-i(p-p^{\prime })x/\hslash }dx$. Thus the Schrödinger equation becomesFor the attractive delta potential, $V(x)=-({\hslash }^2\lambda /m)\delta (x)$, the kernel is the constant $-\hslash \lambda /(2\pi m)$, giving the stated positive right-hand side. For $\tilde{\psi }=N/(p^2+{\alpha }^2)$ with $\alpha >0$, constancy of the left side requires $E=-{\alpha }^2/(2m)$. Since $\int dp/(p^2+{\alpha }^2)=\pi /\alpha$, its coefficient equation givesFinally, transforming $\sqrt{\lambda }{e}^{-\lambda |x|}$ gives $\tilde{\psi }=2{\lambda }^{3/2}{\hslash }^2/[\sqrt{2\pi \hslash }(p^2+{\hslash }^2{\lambda }^2)]$, verifying the same momentum-space form and energy.