= Solution
{c}
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.
\Image[/past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2024-photon-loop-diagrams.png]
{title=Original <Feynman diagrams> for the two orientations of the three-photon <fermion loop> and a representative four-photon <box diagram>}
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>.
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