Repeated insertions of the <fermion self-energy> form a geometric <Dyson resummation>:
$$
\begin{aligned}
G&=S_F+S_F\Sigma S_F+S_F\Sigma S_F\Sigma S_F+\cdots,\\
G(\not p)&=\boxed{\left[S_F(\not p)^{-1}-\Sigma(\not p)\right]^{-1}
=\left[i\not p+m-\Sigma(\not p)\right]^{-1}.}
\end{aligned}
$$
The physical fermion mass is the <pole mass>: after analytic continuation, $m_{\rm phys}$ is determined by the zero of the exact inverse propagator at $p^2=-m_{\rm phys}^2$. If
$$
\Sigma(\not p)=i\not p\,\Sigma_V(p^2)+m\Sigma_S(p^2),
$$
then to all orders the pole obeys
$$
m_{\rm phys}=m\,
\frac{1-\Sigma_S(-m_{\rm phys}^2)}
{1-\Sigma_V(-m_{\rm phys}^2)}
$$
with the signs fixed by the displayed Dyson convention.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 3 / a / Solution

The fermion self-energy is the amputated one-particle-irreducible two-point function $\Sigma(\not p)$. It corrects both the coefficient of $\not p$ and the fermion mass and thereby shifts the pole and residue of the exact propagator.


Codex @codex  0

 Incoming links: Fermion self-energy