Kinetic momentum tensor and causal propagator I can't find anywhere how the kinetic operator defined as $K_{\mu \nu}$ in the term $A^{\mu}K_{\mu \nu}A^{\nu}$ of the lagrangian density is related to causal propagator (for a vector field). I mean, I know I can obtain causal propagator by calculating kinetic operator, but i cannot find anywhere in my books why, the demonstration and what's the relationship between them. Can someone explain this to me?
Thank you
 A: In the path integral formalism, the propagator is defined to be the inverse of $K^{\mu\nu}$, so there is nothing to prove. By definition,
$$
K_{\mu\nu}\Delta^{\nu\sigma}_\mathrm{PI}\equiv\delta^\mu_\sigma\tag{1}
$$
In the operator formalism, the propagator is defined to be the contraction of two free fields,
$$
\Delta_\mathrm{OF}^{\mu\nu}(x-y)\equiv\overline{A^\mu(x)A}{}^\nu(y)\tag{2}
$$
To calculate $\Delta_\mathrm{OF}^{\mu\nu}$ you have to expand $A^\mu$ in terms of ladder operators, and use the canonical commutators $[a,a^\dagger]\sim 1$. When you do this, you end up with
$$
\Delta^{\mu\nu}_\mathrm{OF}(p)\sim\frac{-\eta^{\mu\nu}}{p^2+i\epsilon}+``p^\mu p^\nu\ \text{terms}"\tag{3}
$$
The function above has the expected property that
$$
K_{\mu\nu}\Delta^{\nu\sigma}_\mathrm{OF}=\delta^\mu_\sigma\tag{4}
$$
as can be checked by direct computation (left to the reader).
In the operator formalism, the equation $K_{\mu\nu}\Delta^{\nu\sigma}_\mathrm{OF}=\delta^\mu_\sigma$ is a corollary, so to speak. You  can only conclude that this relation holds a posteriori, once you've explicitly calculated the propagator.
After all this, we end up with
$$
\Delta_\mathrm{OF}^{\mu\nu}=\Delta_\mathrm{PI}^{\mu\nu}\tag{5}
$$
The important point is that in the operator formalism, there is no general proof that
$$
K_{\mu\nu}\Delta^{\nu\sigma}_\mathrm{OF}\equiv\delta^\mu_\sigma\tag{6}
$$
but, rather, you have to check this case by case. It can be proven that this relation always holds as long as Wick's theorem is satisfied, but this isn't necessarily true for any theory. For more details, see Are the path integral formalism and the operator formalism inequivalent?.
