= Solution
{c}
The first panel is the <one-particle-irreducible Feynman diagram> for <photon vacuum polarization>. The two internal <Dirac propagators> carry $k$ and $k+q$, and the external <photons> carry $q$. The other panels fix the routings used in parts <2/c> and <2/d>.
\Image[/past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2025-qed-one-loop-diagrams.png]
{title=Original labelled one-loop <QED> <Feynman diagrams>: <photon vacuum polarization>, <fermion self-energy>, and the electromagnetic <interaction vertex>}
The <closed fermion loop sign> contributes a minus sign; there is no extra <symmetry factor>. Applying the given <Feynman rules>, with the external <propagators> amputated,
$$
i\Pi_{\mu\nu}(q)=(-1)\int_k\operatorname{tr}\left[
(-ie\gamma_\mu)\frac{i(\not k+m)}{k^2-m^2+i0}
(-ie\gamma_\nu)\frac{i(\not k+\not q+m)}{(k+q)^2-m^2+i0}\right].
$$
Equivalently, the required <amplitude> is
$$
\boxed{\Pi_{\mu\nu}(q)=ie^2\int_k
\frac{\operatorname{tr}[\gamma_\mu(\not k+m)\gamma_\nu(\not k+\not q+m)]}
{(k^2-m^2+i0)((k+q)^2-m^2+i0)}.}
$$
The <Feynman slash notation> is $\not k=\gamma^\rho k_\rho$ and $\operatorname{tr}\mathbf1=4$ in the chosen <dimensional regularization> convention.
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