Positivity selects
$$
\widetilde b=-\widetilde a,
$$
for 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 gives
$$
(\widetilde a+\widetilde b)\Box(\partial\cdot V)
+m^2\partial\cdot V=0.
$$
For the selected relation and $m\ne0$, this implies the <transverse vector field> condition $\partial_\mu V^\mu=0$. The remaining three massive polarizations have positive on-shell energy when $\widetilde a>0$.

Solution

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

A vector field is transverse when $\partial_\mu V^\mu=0$ in spacetime, or $\mathbf k\cdot\mathbf V=0$ for a spatial Fourier mode.


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 Incoming links: Transverse vector field