= Solution
The one-loop graph is a fermion line that emits and reabsorbs one internal photon:
\Image[/media/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-304-qed-self-energy.png]
{title=One-loop QED fermion self-energy}
{description=An external fermion of momentum p emits an internal photon of momentum p minus k, propagates with loop momentum k, and reabsorbs the photon.}
{height=380}
The <QED Feynman rules> assign $(-ie\gamma^\mu)(-ie\gamma^\nu)$ to the two vertices, the <Feynman-gauge photon propagator> contracts $\mu$ and $\nu$ and contributes $\delta_{\mu\nu}/(p-k)^2$, and the internal <Dirac propagator> is $(-i\not k+m)/(k^2+m^2)$. Hence
$$
\boxed{\Sigma(\not p)=(-ie)^2\int\frac{d^4k}{(2\pi)^4}
\gamma^\mu\frac{-i\not k+m}{k^2+m^2}\gamma_\mu
\frac1{(p-k)^2}.}
$$
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