Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2018 / 1 / a / i / Solution

Let ${\varphi }_0=\sqrt{Z_2}\varphi$, $A_{0\mu }=\sqrt{Z_3}{A}_{\mu }$, and define the renormalized coupling by $e_0=e_D{Z}_1/(Z_2\sqrt{Z_3})$. Write $Z_2{m}_0^2=m^2+\delta m^2$ and $Z_2^2{\lambda }_0={\mu }^{ϵ}(\lambda +\delta \lambda )$. The complete renormalized Lagrangian density, with its counterterms already included, isSeparating off the classical renormalized terms leaves the counterterm LagrangianThe seagull vertex factor is thus $2i{e}_D^2{Z}_4{g}^{\mu \nu }$. The bare covariant derivative fixes $Z_4=Z_1^2/Z_2$, rather than supplying an unrelated fourth charge parameter. The scalar QED Ward identity, for a symmetry-preserving regularization, gives $Z_1=Z_2$ and hence $Z_4=Z_2$.

For an invertible quadratic gauge field operator, also include ${\mathcal{L}}_{\mathrm{gf}}=-(\partial \cdot A{)}^2/(2\xi )$. One may choose ${\xi }_0=Z_3\xi$, so this term needs no independent counterterm. The strict transverse limit is $\xi \to 0$. Even if the renormalized quartic coupling is set to zero at one scale, its counterterm need not vanish: photon graphs generate a four-scalar interaction. The renormalized Lagrangian density must retain this allowed operator.