Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 40 / 1 / vi

There are two closely related objects to distinguish. Inverting the quadratic Proca action gives the usual covariant Proca propagator. After integration by parts, its kernel isThe matrix inverse with the Feynman i-epsilon prescription givesMultiplication by $K^{ab}$ gives $i{\delta }_b^a$ in the distributional limit. The numerator agrees with the polarization sum for a massive vector boson at the poles, but is not a transverse linear projection at arbitrary four-momentum.

For the literal canonical time-ordered product of $A_a$ and $A_b$, the nondynamical component $A_0$ produces the Proca time-ordering contact term. In the chosen time coordinate the full answer isTo see the local term directly, the three physical polarization vectors give ${\sum }_{\lambda }{ϵ}_{\lambda 0}^2=|\mathbf{q}{|}^2/m^2$. The canonical $00$ two-point correlation function therefore has Fourier transformBy contrast, ${\eta }_{00}+q_0^2/m^2=|\mathbf{q}{|}^2/m^2-(q^2+m^2)/m^2$, so the covariant expression contains an additional $i/m^2$. The mixed and spatial components have no additional contact term. ThusIf $T$ is used to mean covariant time ordering, commonly denoted $T^{\ast }$, the conventional answer is instead just ${\Delta }^{\mathrm{cov}}$. Both conventions have the same propagating poles and agree away from coincidence; explicitly separating them respects the printed definition as an ordinary time-ordered product.