Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2024 / 2 / f

At large loop momentum, each Dirac propagator scales as ${\ell }^{-1}$. One four-dimensional loop and four internal Dirac propagators therefore give the superficial degree of divergenceThus a possible ultraviolet divergence in the sum is logarithmic. By locality its divergent coefficient is a degree-zero polynomial in the external four-momenta, namely a constant Lorentz tensor,The photon Ward identities apply to the complete sum for arbitrary external four-momenta, including off shell ones: $k_{1\mu }{\mathcal{M}}^{\mu \nu \rho \sigma }=0$, and likewise for the other legs. A constant tensor with $k_{1\mu }{\mathcal{M}}_{\mathrm{div}}^{\mu \nu \rho \sigma }=0$ for every $k_1$ must be zero. Hence $a=b=c=0$ andEquivalently, a logarithmic divergence would require a local dimension-four $A^4$ counterterm, forbidden by Abelian gauge invariance. A gauge-invariant four-photon local operator starts with four gauge field strengths, for example $(F_{\mu \nu }{F}^{\mu \nu }{)}^2$, of mass dimension eight; it cannot be this power counting divergence. Individual boxes need not be finite; their forbidden divergent parts cancel in the sum. There are no divergent proper loop subgraphs at this loop order. With the massive Dirac field in the source, there is also no soft or collinear infrared divergence at generic external kinematics. The argument establishes ultraviolet finiteness, and does not claim the absence of physical threshold singularities.