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

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.

For the proper four-point interaction vertex, retain one-particle-irreducible Feynman diagrams with two labelled external photons and the two ends of a directed scalar line. Put $C$ for a cubic scalar–photon vertex and $S$ for a seagull vertex. The one-loop counting identities require $V_C+2V_S=4$. There are three possibilities, giving the nine proper scalar-QED seagull-vertex diagrams:

Each graph has Feynman-diagram symmetry factor one with these external labels: the internal photon and scalar are different species, and the charged scalar endpoints cannot be exchanged. The list is complete because the internal photon must join the ends of the open scalar chain for the graph to remain connected after any one internal line is cut. If it joined two interior points, an outside scalar segment would be a bridge and the graph would be one-particle reducible.

External-leg self-energy insertions contribute to an unamputated four-point correlation function, not this proper vertex. The local four-point counterterm from i is included separately when constructing the renormalized answer; it is not another one-loop topology. Abelian Faddeev-Popov ghosts do not interact in this linear gauge and add no diagram.

Set the external four-momenta to zero after identifying the proper diagrams. In each box or triangle, at least one endpoint of the internal photon is a cubic vertex adjacent to an external scalar of zero momentum. That vertex is proportional to $l_{\rho }$, whereas the supplied numerator obeys $l^{\rho }{P}_{\rho \sigma }(l)=0$. All seven such graphs vanish algebraically. This use of the transverse numerator is stronger than merely saying that one has chosen a generic Lorenz gauge.

The two surviving seagull vertex bubbles have vertex factors $2i{e}_D^2{g}^{\mu \rho }$ and $2i{e}_D^2{g}^{\sigma \nu }$. Combining their two propagators givesThis scalar-QED four-point vertex pole in transverse gauge calculation determines the proportionality constant as well as the required tensor structure. Keep $m>0$ while extracting the ultraviolet pole. Lorentz covariance inside the regulated integral replaces $l^{\mu }{l}^{\nu }$ by $g^{\mu \nu }{l}^2/D$, giving the factor $(D-1)g^{\mu \nu }/D$.

For its scalar integral, Feynman parameter givesThere is no infrared divergence at $l=0$ with nonzero $m$. At pole order the two photon assignments and the factor $3/4$ therefore yieldThe common engineering scale factor is included in $e_D^2$; finite logarithms are immaterial to the pole. Adding $2i{e}_D^2{\delta }_4{g}^{\mu \nu }$ cancels it whenThis is minimal subtraction scheme, up to optional finite terms. It agrees with the transverse-gauge scalar QED wave-function pole and $Z_4=Z_2$. Setting $m=0$ at the outset would instead make the zero-momentum integral scaleless and obscure its separate ultraviolet divergence.

The tree-level Feynman diagrams are the two ordered quark-exchange graphs and the $s$-channel gluon-exchange graph with a three-gluon vertex. There is no two-gluon-two-quark contact term in the ordinary QCD Lagrangian density. Let $q=k_1+k_2=p_1+p_2$, $r_t=p_1-k_1$, $r_u=p_1-k_2$, and $s=q^2$, $t=r_t^2$, $u=r_u^2$. Define $i\mathcal{M}$ as the product of the stated Feynman rules.

Suppressing external colour indices, the quark graphs areThe massless Dirac equations give $\bar{u}/p_1=0$ and $/p_{2}v=0$. ConsequentlyThe last equality uses $\bar{u}/qv=0$. The quark-exchange sum is therefore $-g^2\bar{u}/k_1[T^a,T^b]v$.

For the gluon graph, use all incoming momenta $(k_1,k_2,-q)$ in the three-gluon tensor:Its double contraction is $k_{1\mu }{k}_{2\nu }{V}^{\mu \nu \rho }=s(k_2-k_1{)}^{\rho }/2$. The vertex and gluon propagator giveCombining the graphs yieldsThus the requested spinor proportionality factor is $-g^2\bar{u}(p_1)/k_{1}v(p_2)$ when the colour matrix is displayed separately; retaining it inside the spinor bilinear makes its external colour indices explicit. For a convention calling the Feynman-rule product itself $\mathcal{M}$, rather than $i\mathcal{M}$, the entire result gains one factor of $i$.

The double-longitudinal Ward identity for gluon fusion uses the defining Lie algebra identity $[T^a,T^b]=i{f}^{abc}{T}^c$ makes the boxed amplitude vanish:In QED, both external gauge bosons are photons, the charge matrices commute, and there is no three-photon vertex. The two fermion-exchange graphs then cancel without a third graph. In QCD, omitting the three-gluon interaction would leave a generally nonzero colour commutator: gauge-boson self-interaction is essential to the cancellation.

This is a tree-level manifestation of a Ward identity and, more generally in a gauge-fixed non-Abelian theory, the Slavnov-Taylor identities. They connect the matter representation, the structure constants and the gauge interactions, ensuring that auxiliary longitudinal polarizations do not enter physical probabilities. With all other external legs physical and on shell, replacing one external massless gauge-boson polarization by its momentum leaves zero. The double-contraction result proved here is one consequence, not by itself a proof that every open-index amplitude tensor is transverse off shell. At loop order the corresponding non-Abelian identities also involve Faddeev-Popov ghosts and a consistent symmetry-preserving renormalization.