# Gauge invariance of loop Diagrams

Say we have a gauge-fixed QED Lagrangian: $$\mathcal{L} = - \frac{1}{4}{F}_{\mu\nu}F^{\mu\nu}+ \frac{1}{2a}\left(\partial_\mu A^\mu\right)^2+\bar\psi_1(i\gamma^\mu D_\mu - m_1)\psi_1.$$

My question is how can we check the gauge invariance of loop diagrams. I've seen in https://arxiv.org/abs/hep-ph/0508242 (page 46) that they use an argument

"The photon self-energy at two loops is gauge-invariant, because there are no off-shell charged external particles. Therefore, we may use any gauge; the calculation in the Feynman gauge $$a = 1$$ is easiest."

But I am not sure I understood the argument: What do they mean by off-shell charged external particles?

• You might want to have a look at Weinberg's QFT Volume 1, there's a section on gauge invariance which might give you some insights. – Davide Morgante Jan 23 at 13:31

For any given diagram with on-shell fermion legs (we do not need external photons to be on-shell!), i.e. the fermion legs are treated according to the LSZ formula, sum over all possible insertions of an external photon into this diagram, we denote this by $$\sum_{\text{insertions}}\epsilon^{\mu}(k) \mathcal{D}_{\mu},$$ where $$\mathcal{D}_{\mu}$$ is the resulting diagram of a given insertion with the photon leg amputated. Then we have $$\sum_{\text{insertions}} k^{\mu} \mathcal{D}_{\mu} = 0.$$ Now how does this imply gauge invariance? Consider some arbitrary $$S$$-matrix element at some given order, denote $$\mathcal{M}$$. All the diagrams contributing to this will have the same number of internal photons, say $$m$$. Then we can write $$\mathcal{M}= \int d^4k_1...\int d^4k_m~\Pi^{\mu_1 \nu_1}(k_1)...\Pi^{\mu_m \nu_m}(k_m) \mathcal{M}_{\mu_1 \nu_1... \mu_m \nu_m}.$$ Now changing gauge corresponds to the replacement $$\Pi^{\mu_j \nu_j}(k_j) \rightarrow \Pi^{\mu_j \nu_j}(k_j) + \xi k_j^{\mu_j} k_j^{\nu_j}$$ But recall that $$\mathcal{M}$$ contains exactly all the diagrams contributing at this order, hence it contains all the diagrams corresponding to all possible insertions of $$k_j^{\mu_j}$$ (and $$k_j^{\nu_j}$$) and therefore, by the Ward identity, all terms $$\propto \xi$$ vanish. Note that this requires $$\mathcal{M}$$ to have only on-shell external fermion legs, which is true since $$\mathcal{M}$$ is an $$S$$-matrix element. In particular the external photons need not even be on-shell.