Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2017 / S3 / 2 / 1 / Solution

For a stationary nonrelativistic state, $\boxed{[-{\hslash }^2{\nabla }^2/(2m)+V(\mathbf{r})]\psi =E\psi }$. The wavefunction $\psi$ is a probability amplitude, $m$ the particle mass and $E$ its energy eigenvalue; the potential energy $V$ is measured in the same energy units. In one dimension the Laplacian becomes $d^2/d{x}^2$.