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Use renormalized propagators of mass $m$ and tree vertex $-ig{\mu }^{ϵ}$. The interaction counterterms have Feynman factors $-i{\delta }_{m^2}$ for a two-leg mass insertion and $-i{\mu }^{ϵ}g{\delta }_g$ for a four-leg insertion. At one loop the two-point one-particle-irreducible vertex isThe directed tadpole has $S=1$: interchanging its two ends would reverse the field labels, so the real-scalar tadpole's factor $1/2$ does not apply. It is independent of external momentum, so no wave-function counterterm is required at this order.

For the four-point vertex, label the two unstarred incoming fields by $1,2$ and the starred ones by $3,4$, with all momenta incoming. The tree contact, the three oriented bubbles and the contact counterterm giveThe first bubble has two equivalent arrows along the same direction between its vertices, giving $S=2$. In the other two channels the arrows oppose one another, and interchanging the lines does not preserve their oriented attachments, giving $S=1$. This is the oriented complex-scalar quartic graph factors rule.

The mass tadpole is order $g{m}^2$, while the four-point loop is order $g^2$. In the multiplicative coupling convention of the source,The distinction between ${\delta }_g$ and the whole additive four-leg coefficient $g{\delta }_g$ matters for the requested order counting.

Work in the symmetric vacuum and continue with the action-consistent kernel $C(u,v)=2D_F(u-v)$. Under global U(1) symmetry, a correlator with $r$ fields $\varphi$ and $4-r$ fields ${\varphi }^{\ast }$ picks up a phase $e^{i\alpha (2r-4)}$. Invariance for every $\alpha$ requires $r=2$. Consequently ten of the sixteen ordered field-type choices vanish, and the six placements with two fields of each type can be nonzero. This is the charge selection rule for complex scalar correlators, not an assertion about a symmetry-breaking vacuum with a nonzero one-point function.

Choose external fields $\varphi (x_1)\varphi (x_2){\varphi }^{\ast }(y_1){\varphi }^{\ast }(y_2)$. Its zeroth-order full correlator is $C(x_1,y_1)C(x_2,y_2)+C(x_1,y_2)C(x_2,y_1)$ and is disconnected; its connected part starts at order $\lambda$. DefineThe tree term has Feynman-diagram symmetry factor 1 and is $-i\lambda \int d^{D}zP(z)$.

At second order, a connected graph has two quartic vertices, two internal lines and one loop. The three bubble integrals areThe oriented complex-scalar quartic graph factors distinguish the channels. In $B_s$ the two internal lines have the same orientation and can be interchanged, giving symmetry factor $S_s=2$. In each mixed channel the two orientations are opposite; interchanging them is not an oriented graph automorphism, so $S_t=S_u=1$. Equality of the numerical scalar kernels $C(z,w)=C(w,z)$ does not erase the field-slot distinction used in counting contractions.

There are also four connected external-leg tadpole diagrams. They must be included because a connected correlation function is neither amputated nor restricted to one-particle-irreducible Feynman diagrams. Representative integrals, with the other two obtained by the indicated external permutations, areEach oriented external-leg graph has symmetry factor 1: a directed self-loop has no independent reversal symmetry. Altogether,These expressions retain every external propagator. The physical normalization is explicit: each $C$ is $2D_F$, not $D_F$. Equivalently set $\chi =\varphi /\sqrt{2}$; its quadratic action has unit mixed propagator and its interaction is $-\lambda |\chi {|}^4$, giving vertex $-4i\lambda$. Multiplying a four-$\chi$ correlator by four converts it to the four-$\varphi$ correlator, confirming the powers of two.

Completeness can also be seen without a picture. With two vertices and four external legs, $I=(8-4)/2=2$. Connectivity allows either two intervertex lines with a two-and-two external split, or one bridge and one self-loop with a one-and-three split. The charge assignments give exactly the three bubble attachments and four leg choices above. Direct labelled Wick contraction counting gives respectively $16,32,32,32,32,32,32$ contractions. Dividing by the expansion denominator $2!4^2=32$ reproduces $1/2,1,1,1,1,1,1$. Vacuum factors cancel, and there are no additional connected topologies at this order.