Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 (Advanced Quantum Field Theory)

The Quantum effective action contains the following diagrams. For ${\Gamma }_2$, there are the tree inverse propagator, the one-loop tadpole at order $\lambda$, and at order ${\lambda }^2$ the two-loop sunset and the two-loop tadpole with a tadpole insertion. For ${\Gamma }_4$, there are the tree vertex $-i\lambda$ and, at order ${\lambda }^2$, the three one-loop bubble diagrams in the $s$, $t$, and $u$ channels. Disconnected and one-particle-reducible graphs do not occur in the quantum effective action.

After Wick rotation, the zero-momentum bubble integral with cutoff regularization isThe three channels contribute $3{\lambda }^{2}I(0)/2$. For $\Lambda \gg m$, this is $\frac{3{\lambda }^2}{32{\pi }^2}[\log ({\Lambda }^2/m^2)-1]$, so the zero-momentum renormalization condition is enforced byand hence $F=e{m}^2$. Keeping the exact cutoff expression merely replaces $F$ by the finite cutoff-dependent quantity defined by $\log ({\Lambda }^2/F)=16{\pi }^{2}I(0)$.

Introducing a Feynman parameter and subtracting at zero external momentum giveThe bare coupling ${\lambda }_0=\lambda +{\delta }_{\lambda }$ is independent of the renormalization scale. Differentiating its logarithmic counterterm at fixed ${\lambda }_0$ gives the leading beta function

Completing the square in the Gaussian functional integral and choosingnormalizes $Z_0[0]=1$. The result iswhere the Feynman propagator in the convention of the question is

The analogous Grassmann Gaussian integral, normalized by $Z_0[0,0]=1$, giveswith the free Dirac propagatorIndeed $(i/{\partial }_x-m)S_F(x-y)=i{\delta }^{(4)}(x-y)$.

With left functional derivatives for the Grassmann sources, replace fields in the interaction by $\varphi \mapsto i^{-1}\delta /\delta J$, $\psi \mapsto i^{-1}\delta /\delta \bar{\eta }$, and $\bar{\psi }\mapsto -i^{-1}\delta /\delta \eta$. ThusThe ordering displayed fixes the Grassmann signs and makes this an explicit source functional with no dynamical fields.

For momenta much smaller than $M$,Expanding the source functional to order $g^2$, equivalently integrating out the heavy scalar by its field equation, producesThus the four-spinor coefficient is $g^2/(2M^2)$ in this normalization. Since $[\psi ]=3/2$ and $[g]=0$ in four dimensions, $(\bar{\psi }\psi {)}^2$ has dimension six and its coefficient has the required dimension $-2$.

The first three terms are the scalar kinetic, mass, and $n$-point interaction terms. The remaining terms are the field-strength, mass, and coupling counterterms. The momentum-space rules arefor a propagator, an $n$-leg interaction vertex, a two-leg counterterm insertion, and an $n$-leg counterterm vertex, respectively, together with momentum conservation at every vertex.

Put $d=6-ϵ$. The one-loop two-point bubble has symmetry factor $1/2$ and, after a Feynman parameter and momentum shift, its pole is proportional toCancelling the pole in the minimal subtraction scheme gives, with $ϵ=6-d$,If one defines dimensional regularization by $d=6-2ϵ$, these same poles are written with $2ϵ$ in place of $ϵ$.

The scalar has canonical dimension $[\varphi ]=(d-2)/2$, soThe ${\varphi }^6$ coupling is therefore marginal in $d=3$. Its leading two-point correction is the order-$\lambda$ diagram with one six-leg vertex, two external legs, and the remaining four legs paired into two tadpole loops. This diagram is independent of external momentum, so it renormalizes the mass but has no $p^2$ pole and does not contribute to ${\delta }_Z$.

Substitution of the gauge transformations and $U{U}^{†}=I$ givesA direct commutator calculation givesCovariance of the left side then implies $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{†}$. The cyclic property of the trace makes the traces of $F_{\mu \nu }{F}^{\mu \nu }$, $D_{\mu }\varphi D^{\mu }\varphi$, and ${\varphi }^3$ invariant, so the Lagrangian is gauge invariant.

The product of generators satisfiesand thereforeBecause ${\varphi }^a{\varphi }^b{\varphi }^c$ is symmetric, its contraction with the antisymmetric structure constant of a Lie algebra $f_{abc}$ vanishes. Hence

For the Grassmann-odd ghost $c$, the BRST transformations areup to a simultaneous convention-dependent sign for $Q$ and $c$. The ghost transformation makes the BRST charge nilpotent: $Q^2{A}_{\mu }=Q^2\varphi =0$.

Gauge invariance gives $Q\int \mathcal{L}=0$. Nilpotence gives $Q^2\Psi =0$, and thereforeThus writing the gauge-fixing contribution as a BRST-exact term makes the complete action BRST invariant.

An infinitesimal change of gauge-fixing fermion, $\Psi \mapsto \Psi +\delta \Psi$, changes the action by the BRST-exact term $Q\delta \Psi$. If $\mathcal{O}$ is gauge invariant, then $Q\mathcal{O}=0$. BRST invariance of the measure gives the BRST Ward identity $⟨QX⟩=0$, so the normalized variation isTherefore correlation functions of gauge-invariant operators do not depend on the choice of gauge, provided there is no BRST anomaly or boundary contribution.