**How are one-point functions evaluated in causal perturbation theory?**

I'm not sure where my mistake is in following the standard procedure.

Take the first-order coupling $T_1=\lambda \phi^3$. Within the framework of causal perturbation theory, the corresponding tadpole diagram would be $D^+_m(x_1-x_2):\phi:$ where $D^+_m$ is the positive frequency part of the massive Jordan-Pauli distribution. The singular order is simply that of the given distribution $\omega=-2$, so one can proceed with the splitting procedure trivially:

$$t(x)=R_1-R'_1=D^{+(ret)}_m-D^+_m=D^{+(adv)}_m.$$

For two-point functions, the splitting procedure reproduces the usual Feynman rules if $\omega<0$, and cases where $\omega\geq 0$ are split differently, which avoids UV divergences.

The issue is that the above both does not give a Feynman propagator while having $\omega<0$, yet in the "standard" theory with Feynman rules the tadpole loop would be divergent, suggesting $\omega\geq 0$. Is that discrepancy a feature of the causal theory or have I misconstrued the procedure? In the latter case, what is the procedure to evaluate such diagrams?

This post imported from StackExchange Physics at 2020-03-18 13:31 (UTC), posted by SE-user Quantumness