= Solution
A gauge symmetry is a local redundancy in the fields used to describe one physical state. With metric signature $(+---)$, define
$$
D_\mu=\partial_\mu+ieA_\mu,
\qquad
F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.
$$
The <quantum electrodynamics> Lagrangian is
$$
\mathcal L_{\mathrm{QED}}
=-\frac14F_{\mu\nu}F^{\mu\nu}
+\overline\psi(i\gamma^\mu D_\mu-m)\psi
=-\frac14F^2+\overline\psi(i\not\partial-m)\psi
-e\overline\psi\gamma^\mu\psi A_\mu.
$$
Under the local <U(1) gauge symmetry>
$$
\psi\mapsto e^{-ie\chi(x)}\psi,
\quad
\overline\psi\mapsto\overline\psi e^{ie\chi(x)},
\quad
A_\mu\mapsto A_\mu+\partial_\mu\chi,
$$
one has $D_\mu\psi\mapsto e^{-ie\chi}D_\mu\psi$ and $F_{\mu\nu}\mapsto F_{\mu\nu}$. Both terms in the Lagrangian are therefore invariant.
In <Feynman gauge>, the momentum-space <QED Feynman rules> are:
* an internal electron line of momentum $r$: $i(\not r+m)/(r^2-m^2+i\epsilon)$;
* an internal photon line of momentum $k$: $-i\eta_{\mu\nu}/(k^2+i\epsilon)$;
* an electron-photon vertex: $-ie\gamma^\mu$, with four-momentum conserved;
* incoming and outgoing electrons: $u_s(p)$ and $\overline u_s(p)$;
* incoming and outgoing positrons: $\overline v_s(p)$ and $v_s(p)$ along the oriented fermion chain;
* incoming and outgoing photons: $\epsilon_\mu^{(\lambda)}(k)$ and $\epsilon_\mu^{(\lambda)*}(k)$.
One integrates each undetermined loop momentum, includes a factor $-1$ for each closed fermion loop, and imposes overall momentum conservation. None of the following tree diagrams contains a loop.
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