Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 301 / 2 / a

Integrating by parts givesbecause the $b$ and $c$ terms differ only by a total derivative. Hence

The differential operators are self-adjoint under integration by parts, so the Euler-Lagrange field equations are

Using the original first-derivative Lagrangian, its momentum current isTranslation invariance and Noether theorem give the canonical stress-energy tensoron shell. The Hamiltonian is

When $\tilde{b}=0$, integration by parts makes the kinetic term $-\tilde{a}{\partial }_{\mu }{V}_{\nu }{\partial }^{\mu }{V}^{\nu }$. The Lorentzian contraction over the field index gives the time component and the three spatial components opposite kinetic-energy signs. Reversing the sign of $\tilde{a}$ merely exchanges which component is a ghost field; no nonzero choice makes all four energies positive. At $\tilde{a}=0$ there is no healthy kinetic term. Thus the Hamiltonian is never positive definite.

Positivity selectsfor which the kinetic terms combine, up to normalization and a total derivative, into the Proca field form $-F_{\mu \nu }{F}^{\mu \nu }/4$. Taking the divergence of the equation of motion givesFor the selected relation and $m\ne 0$, this implies the transverse vector field condition ${\partial }_{\mu }{V}^{\mu }=0$. The remaining three massive polarizations have positive on-shell energy when $\tilde{a}>0$.