One-particle-reducible Feynman diagram

A connected Feynman diagram is one-particle reducible if cutting one internal propagator disconnects it. These diagrams occur in connected correlation functions but do not contribute to a quantum effective action defined by a Legendre transform. For a strict Wilsonian effective action, a bridge also cannot be a high-momentum propagator when the sum of external momenta on one side lies below the integrated shell.

In the six-dimensional cubic scalar field theory, use full external-propagator amputation and tadpole subtraction. The exchange part of the zero-momentum four-point amplitude isThe triangle corrects either end of each of three exchanges, and the symmetry factor of an internal bubble is $1/2$. Equivalently insert ${\Gamma }^{(3)}=g+g^3{I}_3$ and ${\Gamma }^{(2)}=m^2-g^2{I}_2/2$ into $3[{\Gamma }^{(3)}{]}^2/{\Gamma }^{(2)}$. Without tadpole subtraction, the shift $v=-g{I}_1/(2m^2)$ adds $3g^4{I}_1/(2m^6)$. None of these is an additional local 1PI quartic coupling; zero-momentum bridges are also excluded from a strict high-mode Wilsonian shell. If bare rather than full external propagators are amputated, external-line decorations additionally contribute $6g^4{I}_2/m^4$, and $6g^4{I}_1/m^6$ without tadpole subtraction. Every displayed term has mass dimension minus two. In a renormalized expansion, the exchange diagrams also receive the counterterm insertions $6g\delta g/m^2-3g^2\delta m^2/m^4$. Without enforcing a zero one-point function, a linear counterterm $\delta h\varphi$ contributes $3g^3\delta h/m^6$ through the background shift; choosing $\delta h=-g{I}_1/2$ cancels the attached tadpole. A kinetic counterterm is proportional to the exchanged momentum squared, so its insertion vanishes at this zero-momentum point.