Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 46 / 4

Choose Lie-algebra generators with $[T_b,T_c]=f^a{}_{bc}{T}_a$ and an invariant orthonormal color metric. Since the coupling is placed outside the gauge kinetic term, it is absorbed into $A_{\mu }$ in the connection convention. The gauge field strength and adjoint covariant derivative areThe structure constants are antisymmetric in $b,c$ and obey the Jacobi identity. These definitions give $[D_{\mu },D_{\nu }]v=[F_{\mu \nu },v]$ on an adjoint-valued field. The field $c$ is a Grassmann-odd Faddeev-Popov ghost field; the Lie bracket in $D_{\mu }c$ contracts its color components in the same way as for any adjoint field.

Put the anticommuting parameter on the left and write $\delta \Phi =ϵs\Phi$. Then $s$ is a left-acting BRST differential: it is odd and satisfies the graded product rule $s(UV)=(sU)V+(-1{)}^{|U|}U(sV)$. The Faddeev-Popov ghost field $c$ and antighost field $b$ are odd, whereas $A_{\mu }$ and the Nakanishi-Lautrup field $h$ are even. Use $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $sb=h$, $sh=0$.

A total derivative in the Lagrangian variation must be included in the Noether current. For constant $ϵ$, the ghost covariant derivative has $s(D_{\mu }c)=0$, by the Jacobi identity and the odd statistics of $c$. The Yang-Mills term is invariant, and the remaining variation isThe term proportional to $\xi$ does not vary because $sh=0$.

To obtain the signs without an ambiguity about fermionic canonical momenta, now allow $ϵ=ϵ(x)$. In particular,The coefficients of ${\partial }_{\mu }ϵ$ in the Yang-Mills, ghost, and multiplier terms are respectivelyThe positive sign of the last ghost expression results from moving ${\partial }_{\mu }ϵ$ past the odd ${\partial }^{\mu }{b}^a$. Subtracting the total-derivative term $K^{\mu }$ therefore gives the Yang-Mills BRST Noether currentIndeed, the full localized variation is $\delta \mathcal{L}=({\partial }_{\mu }ϵ)j^{\mu }+{\partial }_{\mu }(ϵK^{\mu })$. The Noether theorem then gives ${\partial }_{\mu }{j}^{\mu }=0$ on the field equations. With spatial boundary terms vanishing, the corresponding BRST charge in four spacetime dimensions isIt is odd and has ghost number one. Overall generator phases depend on the convention relating this Noether charge to quantum commutators; one may use $s\mathcal{O}=i[Q_{\mathrm{BRST}},\mathcal{O}\}$. The displayed current fixes the classical Noether normalization for the left-parameter convention.

Let $G_1,\dots ,G_r$ be the separated gauge-invariant insertions. Their BRST symmetry variations vanish: the gauge field strength and its gauge covariant derivatives transform by adjoint commutators, and invariant color contractions remove these commutators. Thus $s{G}_i=0$.

Make the infinitesimal change of integration variables $\Phi \mapsto \Phi +ϵs\Phi$ in the normalized expectation of $\mathcal{O}(x){\prod }_i{G}_i$. The action is invariant up to its boundary term and, by assumption, the measure has no Jacobian anomaly. ThereforeSince $s\mathcal{O}$ is generated by the BRST charge, this is the graded BRST Ward identityHere $[Q,\mathcal{O}\}=Q\mathcal{O}-(-1{)}^{|\mathcal{O}|}\mathcal{O}Q$ is the graded commutator. For even $\mathcal{O}$ it is the ordinary commutator printed in the question; for odd $\mathcal{O}$ it is an anticommutator. Separation of the other insertions from $x$ avoids the additional coincident-point contact terms. An operator need not itself be BRST-closed for this identity to hold.

A BRST-exact insertion has zero correlation with physical, gauge-invariant insertions. This is the decoupling of BRST-exact insertions in physical correlation functions. In the associated BRST cohomology, physical information is represented by closed states or operators modulo exact ones, schematically $\ker Q/\mathrm{im}Q$, using the nilpotence of the BRST charge in the anomaly-free theory. Gauge-fixing fields thereby do not supply additional physical observables.

For example, the gauge parameter dependence is exact:Differentiating a normalized physical correlation function with respect to $\xi$ inserts this expression; the BRST Ward identity makes that derivative zero under the same invariant-measure and boundary assumptions. This illustrates gauge-fixing parameter independence from BRST symmetry.