Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 304 / 4

Write $U(x)=e^{i{\alpha }^a(x)T^a}$ and $A_{\mu }=A_{\mu }^a{T}^a$. With the conventionthe gauge field must transform asThen $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. The conjugate field has $D_{\mu }\bar{\psi }={\partial }_{\mu }\bar{\psi }+ig\bar{\psi }{A}_{\mu }$. For the adjoint scalar,or $(D_{\mu }\varphi {)}^a={\partial }_{\mu }{\varphi }^a+g{ϵ}^{abc}{A}_{\mu }^b{\varphi }^c$. It transforms as $D_{\mu }^{\prime }{\varphi }^{\prime }=U(D_{\mu }\varphi )U^{-1}$. The gauge field strengthsimilarly transforms by conjugation.

Up to Euclidean sign conventions, the parity-even renormalizable Lagrangian isThe Yang-Mills term is invariant because $F_{\mu \nu }$ transforms by conjugation and the matrix trace is cyclic. Covariance of $D_{\mu }\psi$ makes $\bar{\psi }{\gamma }_{\mu }{D}_{\mu }\psi$ invariant, and the fermion mass is invariant because the factors $U^{-1}U$ cancel. Likewise $D_{\mu }\varphi$ transforms by conjugation, so its trace norm, ${\varphi }^a{\varphi }^a=2\mathrm{tr}{\varphi }^2$, and every power of that norm are invariant. Finally,which proves gauge invariance of the Yukawa interaction.

Gauge invariance and power counting also permit the parity-odd Yukawa interaction $i{y}_5\bar{\psi }{\gamma }_5\varphi \psi$, a pseudoscalar fermion mass $i{m}_5\bar{\psi }{\gamma }_5\psi$, and the Yang-Mills theta term. They are absent if parity and CP are imposed. There is no nonzero cubic scalar invariant: $\mathrm{tr}{\varphi }^3$ vanishes for $SU(2)$, equivalently ${ϵ}^{abc}{\varphi }^a{\varphi }^b{\varphi }^c=0$ for commuting scalar components.

The quadratic Yang-Mills operator has zero directions along gauge orbits, so it has no propagator until one chooses a gauge fixing. The Faddeev-Popov determinant generated by this choice is represented by anticommuting Faddeev-Popov ghost fields, which cancel unphysical gauge-field contributions in loop calculations. A Nakanishi-Lautrup field imposes the gauge condition algebraically and lets the gauge-fixing plus ghost action be written as a BRST-exact term with off-shell nilpotency.

Remove the common Grassmann parameter and write the BRST transformation as the odd derivation $s$:The last two equations immediately give $s^2{\bar{c}}^a=s^2{B}^a=0$. Applying $s$ to the ghost and using the graded product rule givesThe cancellation is the Jacobi identity for the $SU(2)$ structure constants together with anticommutation of the ghost fields. For the gauge field, the component calculation isThe graded product rule givesso the two terms cancel. Thus $s^2$ vanishes on every elementary field.

For two independent Grassmann parameters, ${\delta }_1={\eta }_{1}s$ and ${\delta }_2={\eta }_{2}s$ satisfy ${\delta }_1{\delta }_{2}O={\eta }_1{\eta }_2{s}^{2}O=0$. Since $s$ obeys the graded Leibniz rule, induction extends $s^{2}O=0$ from the generators $A,c,\bar{c},B$ to every polynomial $O(A,c,\bar{c},B)$. Hence the BRST transformations are nilpotent.