Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 44 / 4

Use Hermitian generators $T^a$ of the special unitary group, with $[T^a,T^b]=i{f}^{ab}{}_c{T}^c$. An adjoint field is the Lie-algebra-valued matrix $\psi ={\psi }^a{T}^a$, transforming as $\psi \mapsto U\psi U^{-1}$. The adjoint covariant derivative isIn components, $(D_{\mu }\psi {)}^a={\partial }_{\mu }{\psi }^a+g{f}^{abc}{A}_{\mu }^b{\psi }^c$. With $A_{\mu }\mapsto U{A}_{\mu }{U}^{-1}-(i/g)({\partial }_{\mu }U)U^{-1}$, direct substitution gives $D_{\mu }\psi \mapsto U(D_{\mu }\psi )U^{-1}$. This is covariance in the Adjoint representation of a Lie group. In the following BRST symmetry formulas, absorb the coupling into the connection, so $D_{\mu }={\partial }_{\mu }-i[A_{\mu },\cdot ]$.

Factor the odd parameter on the left, $\delta =ϵs$. The resulting left-acting BRST differential obeys the graded Leibniz ruleFor the odd Grassmann field $c$, the bracket in the transformation is a graded commutator: $[c,c{]}_{\mathrm{gr}}=2c^2$, not the identically zero ordinary commutator of a matrix with itself. Thus $sc=i{c}^2$, while $s{A}_{\mu }=D_{\mu }c$, $s\bar{c}=h$ and $sh=0$.

On the ghost,On the gauge field, variation of the connection and the adjoint covariant derivative givesHere $D_{\mu }$ is even and therefore obeys the ordinary product rule. Also $s^2\bar{c}=sh=0$ and $s^{2}h=0$, without using any field equation; this is off-shell nilpotence supplied by the Nakanishi-Lautrup field.

Applying the graded Leibniz rule twice cancels the two cross terms:The square is consequently an even graded derivation. Since it vanishes on every generator, it vanishes inductively on every polynomial in the fields. Hence $s^2\mathcal{O}=0$ for every such operator. This genuine result is stronger than the automatic vanishing obtained by merely setting ${ϵ}^2=0$; two independent transformation parameters also give a vanishing commutator.

Use the Minkowski metric, path-integral weight $e^{iS}$ and the Abelian gauge theory transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\omega$. The gauge functional $F[A]=\partial \cdot A+A^2$ varies asThus the Faddeev-Popov operator is ${\mathcal{M}}_A=\square +2A\cdot \partial$. It depends on the gauge field despite the gauge group being Abelian: the ghosts interact because this gauge condition is nonlinear.

Choose the gauge-fixing fermion $\Psi =\int d^{4}x\bar{c}(F[A]+\xi h/2)$. The gauge-fixed action $S=S_{\mathrm{Maxwell}}+s\Psi$ isHere the tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is distinct from the scalar gauge functional $F[A]$. With $\xi =0$, integrating over $h$ imposes the exact printed constraint. For nonzero $\xi$, eliminating $h=-F[A]/\xi$ instead givesThis version displays the additional gauge-dependent cubic and quartic gauge-field vertices as well as the ghost interaction; the strict condition is its $\xi \to 0$ limit.

For the Fourier transform convention $c(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}c(p)$, the quadratic ghost kernel is $p^2$. With the ordering $⟨c(p)\bar{c}(q)⟩$,The term $-2\bar{c}{A}^{\mu }{\partial }_{\mu }c$ has Fourier coefficient $2i{p}_{\mu }$, where $p$ is the incoming ghost momentum. Multiplication by $i$ in the Feynman rule givesThese signs refer to the displayed action, Fourier convention and ghost ordering. Reversing the ghost/antighost convention changes corresponding signs consistently. A closed ghost loop has the additional minus sign from Grassmann variables.

A gauge-invariant observable is BRST-closed: replacing its infinitesimal gauge parameter by the ghost gives $s\mathcal{O}=0$. Change the gauge functional continuously, or interpolate between two admissible choices, by changing the gauge-fixing fermion to ${\Psi }_t$. The action changes by the BRST-exact operator ${\partial }_{t}S=s({\partial }_t{\Psi }_t)$.

For a normalized correlator of $\mathcal{O}={\prod }_i{\mathcal{O}}_i$ with each insertion BRST-closed, differentiation of the functional integral givesThe BRST Ward identity says $⟨sX⟩=0$ for an invariant measure and action, with appropriate boundary conditions. Since $s\mathcal{O}=0$, the graded Leibniz rule makes the first insertion an exact variation of $\mathcal{O}{\partial }_t{\Psi }_t$ up to its harmless parity sign, and both terms vanish. ThusPhysical gauge-invariant correlation functions are independent of this gauge condition, even though individual gauge-field and ghost Feynman diagrams change. This argument assumes an admissible perturbative gauge fixing, a BRST symmetry-preserving regulator/measure and no uncanceled boundary contribution. A global failure of those assumptions is not settled by the formal local calculation.