= Solution
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 is
$$
K^{ab}=\eta^{ab}(\Box-m^2)-\partial^a\partial^b,
\qquad K^{ab}(q)=-(q^2+m^2)\eta^{ab}+q^aq^b.
$$
The <matrix inverse> with the <Feynman i-epsilon prescription> gives
$$
\widetilde\Delta^{\rm cov}_{ab}(q)
=\frac{-i}{q^2+m^2-i0}\left(\eta_{ab}+\frac{q_aq_b}{m^2}\right).
$$
Multiplication by $K^{ab}$ gives $i\delta^a_b$ 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 is
$$
\boxed{\Delta_{ab}(x,y)=\int\frac{d^4q}{(2\pi)^4}\,e^{iq\cdot(x-y)}
\left[\frac{-i(\eta_{ab}+q_aq_b/m^2)}{q^2+m^2-i0}
-\frac{i}{m^2}\delta_a^0\delta_b^0\right]}.
$$
To see the local term directly, the three physical <polarization vectors> give $\sum_\lambda\epsilon_{\lambda0}^2=|\boldsymbol q|^2/m^2$. The canonical $00$ <two-point correlation function> therefore has <Fourier transform>
$$
\widetilde\Delta_{00}(q)=\frac{-i|\boldsymbol q|^2}{m^2(q^2+m^2-i0)}.
$$
By contrast, $\eta_{00}+q_0^2/m^2=|\boldsymbol q|^2/m^2-(q^2+m^2)/m^2$, so the covariant expression contains an additional $i/m^2$. The mixed and spatial components have no additional contact term. Thus
$$
\Delta^{\rm cov}_{ab}(x,y)=\Delta_{ab}(x,y)
+\frac{i}{m^2}\delta_a^0\delta_b^0\delta^{(4)}(x-y).
$$
If $T$ is used to mean covariant time ordering, commonly denoted $T^*$, the conventional answer is instead just $\Delta^{\rm cov}$. 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>.
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