Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2019 / 2 / a

For a real local parameter $\alpha (x)$, the Dirac adjoint varies as $\delta \bar{\psi }=ie\alpha \bar{\psi }$. The matter kinetic term varies byThe QED interaction varies by $-e({\partial }_{\mu }\alpha )\bar{\psi }{\gamma }^{\mu }\psi$, since the phases in $\bar{\psi }{\gamma }^{\mu }\psi$ cancel. These two variations cancel pointwise. The mass term is invariant for the same reason, and $\delta F_{\mu \nu }={\partial }_{\mu }{\partial }_{\nu }\alpha -{\partial }_{\nu }{\partial }_{\mu }\alpha =0$.

Only the gauge fixing term varies. Put $B={\partial }_{\mu }{A}^{\mu }$; then $\delta B={\partial }^2\alpha$, givingFor arbitrary local $\alpha$, the gauge-fixed action is therefore not invariant under this transformation of its dynamical fields. The restricted parameters satisfying ${\partial }^2\alpha =0$ give a residual invariance of the displayed density.

Let $ϵ$ here be a constant infinitesimal transformation parameter, unrelated to the dimensional regularization parameter in question 1. The matter and Maxwell terms undergo an ordinary infinitesimal gauge transformation with $\alpha =ϵ\varphi$, and their variations cancel as before. The gauge fixing contribution isThe extra real scalar field kinetic term varies byAdding the two gives a total derivative:The last equality uses vanishing boundary contributions, for example sufficiently fast decay of the fields. No field equations were used. This gauge-scalar global symmetry of covariantly fixed QED leaves the action invariant even though the density need not be invariant pointwise; the scalar's variation supplies exactly the missing boundary-term cancellation.