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For a real phi cubed theory with labelled external points, the connected three-point function at third order in the coupling constant has seven labelled one-loop Feynman diagrams: one triangle, three external-branch bubble insertions, and three tadpole-ended chains. The symmetry factors are respectively $1,2,2$, with graph weight $1/S$. To prove completeness, use $3V=2I+3$ and $L=I-V+1$: $V=3$ gives $I=3$ and $L=1$. The unique internal cycle is then a triangle, a pair of parallel edges, or a self-loop; the degree-three condition fixes the external attachments. Only the triangle remains one-particle irreducible after amputation of external propagators. A tadpole-free prescription removes the three chains but must be stated separately.

The elastic scattering of two photons into two photons. In vacuum QED there is no photon self-coupling at tree level; its leading scattering amplitude is the sum of six oriented charged-fermion box diagrams. The superficial degree of divergence is zero, but the photon Ward identities forbid the required local dimension-four $A^4$ counterterm. The complete one-loop Feynman diagram sum is consequently ultraviolet finite. This does not remove its finite four-momentum dependence or its physical thresholds.

There are two one-loop Feynman diagrams: the two orders of the labelled photons around a top quark fermion loop, with one Yukawa interaction insertion for the outgoing Higgs boson. They are added, rather than identified by a Feynman-diagram symmetry factor. Each has the same color charge trace $N_c=3$.

In the first routing, traverse the fermion loop from the $p_2$ insertion to the $p_1$ insertion with momentum $l$, then to the Higgs boson insertion with momentum $l+p_1$, then back to the $p_2$ insertion with momentum $l-p_2$. Conservation at the Higgs boson insertion gives $p_h=p_1+p_2$. The second routing exchanges $(p_1,\mu )$ and $(p_2,\nu )$: its three internal momenta are $l$, $l+p_2$ and $l-p_1$. Reversing the orientation is another way to represent this second ordering. Two vector insertions give no Furry theorem cancellation.

The two one-loop Feynman diagrams are fermion loops with three photon interaction vertices. Fix vertex 1; the two cyclic orders are $(1,2,3)$ and $(1,3,2)$. A cyclic rotation does not produce another Feynman diagram. Keeping all three external labels in the same positions, these two orders can be represented by the same triangle with opposite fermion flow, as in the first two panels.

The photons are labelled $(k_1,\mu )$, $(k_2,\nu )$, $(k_3,\rho )$, with all four-momenta incoming and $k_1+k_2+k_3=0$. The amputation of external propagators removes the three external photon propagators, leaving their Lorentz indices open. There are no other connected one-loop Feynman diagrams with only three external photons in QED.

For an $n$-vertex fermion loop, reversing the fermion flow reverses the ordered product inside its trace. The same charge-conjugation matrix calculation transforms every Dirac propagator by $p\mapsto -p$ and contributes $-1$ for each vector photon interaction vertex. Relabel the loop momentum so that the external labels are fixed. The reversed Feynman diagram therefore obeysFor odd $n\ge 3$, the $(n-1)!$ cyclic orders pair into opposite orientations and cancel. For $n=1$, the single photon tadpole maps into itself with a minus sign and is zero. The common closed fermion loop sign is present on both sides and does not supply the factor $(-1{)}^n$. This proves Furry's theorem for all the requested one-loop Feynman diagrams. The charge conjugation argument also gives vanishing odd-photon correlation functions to all orders in vacuum QED, provided no charge conjugation symmetry breaking background or charged external state is introduced.

For a connected cubic Feynman diagram with $V$ internal vertices, $I$ internal lines and three external lines, $3V=2I+3$. Hence $V$ is odd and the loop order is $L=I-V+1=(V-1)/2$. Through order ${\lambda }^3$, there is one tree-level Feynman diagram at $V=1$ and seven labelled one-loop Feynman diagrams at $V=3$:

For an explicit check, use the propagator notation of solution and define, with $\{i,j,k\}=\{1,2,3\}$,There is no ordering of $j,k$ in this notation. The complete connected correlation function through this order isFor completeness of the classification, a one-loop connected internal graph either has a three-edge cycle, a pair of parallel edges and a bridge, or a self-loop and two bridges. The cubic degree condition forces exactly the external attachments just listed. Direct Wick contraction counting gives 1296 pairings for the triangle and 648 for each of the six other labelled graphs, divided by $3!(3!{)}^3=1296$.

Connected does not mean one-particle irreducible. After amputation of external propagators, only the triangle is one-particle irreducible at this order; the six other loop graphs belong to the requested connected answer. The tadpole diagrams cannot be discarded without an additional subtraction prescription.