Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 14E / c / Solution

Under the gauge transformation,because the curl of a gradient vanishes, andThe transformed Lagrangian isA total time derivative changes the action only by endpoint terms, whose variation vanishes for fixed endpoints, so the Euler-Lagrange equations are unchanged.

The new momentum isUse the type-two generating function for a canonical transformationIt gives $\mathbf{p}=\partial F_2/\partial \mathbf{r}=\tilde{\mathbf{p}}-q\nabla f$, $\tilde{\mathbf{r}}=\partial F_2/\partial \tilde{\mathbf{p}}=\mathbf{r}$, andThis equals $|\tilde{\mathbf{p}}-q\tilde{\mathbf{A}}{|}^2/(2m)+q\tilde{\varphi }$, so the gauge change is a time-dependent canonical transformation.