Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 304 / 4

Choose Hermitian Lie-algebra generators satisfying $[T^a,T^b]=i{f}^{abc}{T}^c$, with real totally antisymmetric Lie algebra structure constants in an orthonormal compact-group basis. The Adjoint representation generators are $(T_{\mathrm{ad}}^b{)}_{ac}=-i{f}^{bac}$. For a canonically normalized Yang-Mills action, the coupling is in the gauge covariant derivative, rather than an overall $1/g^2$ kinetic prefactor. The conventions $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }^b{T}_{\mathrm{ad}}^b$ giveEquivalently, for the Lie-algebra-valued spinor, $D_{\mu }\psi ={\partial }_{\mu }\psi -ig[A_{\mu },\psi ]$. This is the adjoint covariant derivative with the Hermitian-generator and explicit-coupling convention specified. Changing the sign convention for $D$ changes the accompanying gauge-transformation convention consistently.

Take Euclidean gamma matrices with $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\delta }_{\mu \nu }$, and independent Grassmann fields $\psi ,\bar{\psi }$. The minimally coupled Dirac action isIts mass term is the Dirac mass term, and the interaction is $-ig{\bar{\psi }}^a{\gamma }_{\mu }{A}_{\mu }^b(T_{\mathrm{ad}}^b{)}_{ac}{\psi }^c$. With Fourier convention $e^{ip\cdot x}$, the free kinetic operator is $i/p+m$. The Euclidean adjoint Dirac Feynman rules are thereforeThe vertex sign follows by expanding $e^{-S_{\psi }}$, not $e^{i{S}_{\psi }}$. Momentum conservation supplies the usual Dirac delta function at each vertex. There is only a one-gluon fermion vertex, with no two-gluon seagull because the fermionic covariant derivative is first order. A closed fermion loop has the usual fermionic sign $-1$. The identity $(i/p+m)(-i/p+m)=(p^2+m^2)I$ checks the Dirac propagator.

Write $d=4-ϵ$ and $C_A=C_2(G)$. The pole term is independent of $m$ and external momentum beyond the transverse tensor: ${\int }_0^{1}x(1-x)dx=1/6$, $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $({\mu }^2/\Delta {)}^{ϵ/2}=1+O(ϵ)$. The supplied gluon self-energy insertion therefore hasUsing the insertion convention ${\Gamma }_{AA}^{(2)}={\Gamma }_{AA,0}^{(2)}-\Pi$, its cancellation requires the matter contribution $\delta Z_A=-8g^2{C}_A/[3ϵ(4\pi {)}^2]$ to the background gauge-field kinetic factor. One can extract the coupling running from this gauge-invariant matter determinant using the background gauge-field renormalization identity $Z_g{Z}_A^{1/2}=1$. This is a background-field Ward identity, not a claim that the ordinary quantum-gluon wave-function factor alone always determines the full coupling renormalization. For this one adjoint Dirac spinor,Differentiate at fixed $g_0$. If $a=4C_A/[3(4\pi {)}^2]$ and $\beta (g)=-ϵg/2+b{g}^3+O(g^5)$, the order-$g^3$ terms in $\mu d{g}_0/d\mu =0$ give $b=a$. Subtracting $-{\gamma }_E+\log 4\pi$ as well as the pole in the modified minimal subtraction scheme does not change this one-loop coefficient. Thus the adjoint Dirac contribution to the Yang-Mills beta function isThis is screening. The pure Yang-Mills one-loop beta function coefficient is $-11C_A{g}^3/[3(4\pi {)}^2]$. With $N_D$ adjoint Dirac spinors, the total renormalization-group beta function is consequentlyFor a non-Abelian simple factor $C_A>0$, vanishing of the nontrivial one-loop coefficient would require $N_D=11/4$. It cannot be a nonnegative integer, proving the integer obstruction to one-loop marginality with adjoint Dirac fermions. For a product group the same argument applies to every non-Abelian factor; the statement concerns the beta-function coefficient, not the trivial fixed point $g=0$. The modified minimal subtraction scheme is mass independent, so a massive spinor has this ultraviolet contribution; decoupling below its mass instead requires effective-theory threshold matching.