Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 304 / 4

Substituting $\delta A_{\mu }=g^{-1}{\partial }_{\mu }\alpha -i[A_{\mu },\alpha ]$ into the curvature and collecting terms gives the covariant transformationEquivalently, $\delta F_{\mu \nu }^a=f^{abc}{F}_{\mu \nu }^b{\alpha }^c$ in the stated conventions. Since $F_{\mu \nu }^a{F}^{\mu \nu ,a}$ is proportional to $\mathrm{tr}(F_{\mu \nu }{F}^{\mu \nu })$, cyclicity of the matrix trace givesThus the Yang-Mills action is gauge-invariant.

In components the BRST transformation isThe last two equations immediately give $s^2\bar{c}=s^{2}B=0$. Applying the graded Leibniz rule to $s^{2}c$ produces a sum of three ghost monomials whose coefficient is the Jacobi identity $f^{e[ab}{f}^{c]de}$, so $s^{2}c=0$. Finally,the derivative terms cancel by anticommutation of the Faddeev-Popov ghost fields, and the remaining terms again cancel by the Jacobi identity. Hence $s^2=0$ on every field: $s$ is a nilpotent Grassmann-odd differential.

The Yang-Mills action is BRST invariant because its field strength transforms covariantly, and the gauge-fixing contribution is BRST exact. Nilpotence therefore givesThe gauge-fixing functional itself is generally not closed: $sL[A]=L[Dc]$. Applying the graded Leibniz rule givesThe Nakanishi-Lautrup field $B^a$ is auxiliary. Its algebraic equation $B^a=L[A^a]/\xi$ turns the middle terms into $L[A{]}^2/(2\xi )$, while the final term is the Faddeev-Popov ghost field action.

For $L[A^a]={\partial }^{\mu }{A}_{\mu }^a$, eliminating the Nakanishi-Lautrup field gives the covariant gauge LagrangianThe first term supplies the gluon kinetic term, three-gluon vertex, and four-gluon vertex. The second makes the quadratic gauge-field operator invertible. The last supplies the ghost propagator and ghost-antighost-gluon vertex.

The nonzero one-loop contributions to the gluon propagator are a gluon bubble with two three-gluon vertices and a closed ghost bubble with two ghost-antighost-gluon vertices. A four-gluon tadpole is also present with a cutoff regulator; for massless fields it is a scaleless integral and vanishes in dimensional regularization.

For $L[A^a]=n^{\mu }{A}_{\mu }^a$, eliminating $B^a$ gives the axial gauge term $(n\cdot A^a{)}^2/(2\xi )$ and ghost operatorIn the strict $\xi \to 0$ gauge, $n\cdot A=0$. The gauge-field-dependent part of $n\cdot D$ then vanishes, the Faddeev-Popov determinant becomes field independent, and the ghosts decouple. The one-loop gluon propagator therefore receives the gluon bubble but no ghost bubble. As in the covariant gauge, the four-gluon tadpole can occur with a hard cutoff and vanishes as a scaleless integral in dimensional regularization. Gauge-invariant observables agree between the two gauges even though their individual propagators and diagrammatic decompositions differ.