For a stationary nonrelativistic state, $\boxed{[-\hbar^2\nabla^2/(2m)+V(\mathbf r)]\psi=E\psi}$. 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 $d^2/dx^2$.

Solution

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

 Solution (source code)