Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 304 (Advanced Quantum Field Theory)

After integration by parts, the quadratic action isWith the pole prescription appropriate to the conventions in the question, its inverse kernel isFor $t>0$ close the contour in the lower half-plane and for $t<0$ close it in the upper half-plane. The enclosed pole in each case giveswhich equivalently satisfies $({\partial }_t^2+{\omega }^2)D(t)=i\delta (t)$.

The source-dependent Gaussian functional integral is evaluated by translating the integration variable by the classical sourced solution. Completing the square givesChanges in the sign of the source term or of the path-integral phase move factors of $i$ between $D$ and the exponent but leave the contraction rules equivalent.

Replace each occurrence of $x(t)$ in the interaction by the functional derivative that inserts it. In standard Minkowski source conventions,and henceThe factors of $i$ are adjusted together if one uses the source convention of part a directly.

This is a perturbation series, generally an asymptotic rather than convergent series because the number of Wick contractions grows factorially. For real cubic coupling the potential is also unbounded on one side, so the real-axis theory does not possess a stable nonperturbative ground state without a contour prescription or further stabilizing interactions.

The time-domain Feynman rules are:

There is no order-$\lambda$ connected two-point correction. Through order ${\lambda }^2$, the two connected topologies are the two-vertex fish graph and the one-particle-reducible tadpole graph. With the displayed vertex convention,up to the common factors of $i$ associated with the propagator convention. Vacuum normalization removes disconnected vacuum bubbles.

These integrals are ultraviolet finite in one time dimension: a harmonic-oscillator propagator behaves as $E^{-2}$ at large frequency, and the loop-frequency integrals have negative superficial degree of divergence. They are also infrared finite because $\omega >0$ supplies a gap. The cubic instability affects nonperturbative convergence but does not create a divergence in these fixed-order integrals.

For massless phi-fourth theory in four dimensions, the momentum-space rules areWritingthe two-point counterterm insertion is $i({\delta }_Z{p}^2-\delta m^2)$ and the four-point counterterm is $-i\delta \lambda$.

The renormalized one-particle-irreducible correlation function ${\Gamma }_4$ through order ${\lambda }^2$ contains the tree quartic vertex, the quartic counterterm, and three one-loop bubble diagrams. The bubbles are the $s$, $t$, and $u$ channels and each has symmetry factor $1/2$.

For one channel with momentum $p$, introduce a Feynman parameter and shift the loop momentum:In $d=4-ϵ$, with dimensional-regularization scale $\bar{\mu }$ before the usual modified-minimal-subtraction redefinition, the bubble at $p^2=-M^2$ isSince ${\int }_0^1\log \{x(1-x)\}dx=-2$, summing the three equal channels gives the form displayed in the question, beginning with $6/ϵ-3\gamma$.

ConsequentlyThe momentum-subtraction scheme condition ${\lambda }_{\mathrm{eff}}(M)=\lambda$ is enforced byChoosing $\bar{\mu }=M$ removes the logarithm. A minimal-subtraction scheme keeps only the pole and therefore defines a different finite renormalized coupling.

For a renormalized $n$-point one-particle-irreducible function, the Callan-Symanzik equation iswith a convention-dependent sign on ${\gamma }_{\varphi }$. At this order ${\gamma }_{\varphi }=0$.

At a general Euclidean momentum scale $Q$, the one-loop four-point function containsRequiring a physical amplitude to be independent of the arbitrary subtraction scale givesand therefore

For the convention in the question, a finite Yang-Mills gauge transformation acts covariantly on the field strength:or with $U$ and $U^{†}$ exchanged if the opposite convention is used for $D_{\mu }$. Cyclicity of the matrix trace givesso the Yang-Mills theory Lagrangian is gauge invariant.

The quadratic gauge-field operator has zero directions $A_{\mu }\sim A_{\mu }+D_{\mu }\alpha$ along each gauge orbit. It therefore has no inverse on the full field space. Gauge fixing removes this degeneracy and produces a propagator, while the Faddeev-Popov determinant accounts for the corresponding Jacobian.

Requiring $D_{\mu }\varphi$ to transform as $U^{†}(D_{\mu }\varphi )U$ and using the gauge-field transformation law givesIndeed, differentiating $U^{†}\varphi U$ produces two inhomogeneous derivative terms, and those cancel against the inhomogeneous part of the transformed connection.

For $U=1+i{\alpha }^a{T}_a+O({\alpha }^2)$,If $[T_a,T_b]=i{f}_{ab}{}^c{T}_c$, thenThis is the infinitesimal Adjoint representation of a Lie algebra.

At lowest derivative order, a general invariant effective Lagrangian through fourth order in the fields has the schematic formHere ${\lambda }_{abcd}$ is any invariant symmetric rank-four tensor, the ${\mathcal{I}}_A$ run over invariant contractions such as $(\mathrm{Tr}{\varphi }^2)\bar{\psi }\psi$ and $\bar{\psi }{\varphi }^2\psi$, and the ${\mathcal{J}}_B$ run over gauge- and Lorentz-invariant four-fermion contractions. The two sign symmetries forbid scalar cubic terms and Yukawa terms $\bar{\psi }\varphi \psi$. The covariant kinetic terms automatically contain the allowed cubic and quartic interactions involving $A_{\mu }$.

The canonical dimensions areand henceA coupling is relevant, marginal, or irrelevant when its dimension is positive, zero, or negative at the Gaussian fixed point. Thus gauge and scalar-quartic interactions are marginal in $d=4$, the mixed two-scalar fermion bilinear is marginal in $d=3$, and four-fermion interactions are marginal in $d=2$. Quantum corrections replace this engineering classification near an interacting fixed point by the eigenvalues of its RG stability matrix.

The one-loop four-scalar one-particle-irreducible diagrams, together with permutations of the external scalar labels, are:

The first four are forced by the scalar covariant derivative and scalar potential. There is no ordinary Yukawa fermion box because the imposed $\varphi \mapsto -\varphi$ symmetry forbids a one-scalar fermion vertex.

Treat the BRST transformation $Q$ as an odd graded derivation. In matrix notation the first three transformations areThenafter substituting $Qc$ and using the Jacobi identity. Similarly, the two terms incancel because the ghost components anticommute and only the Lie-algebra commutator survives. Applying $Q$ once more to $Qc$ gives a sum proportional to $f^d{}_{e[a}{f}^e{}_{bc]}{c}^a{c}^b{c}^c$, which vanishes by Jacobi. Finally,Thus $Q^2=0$ on every field.

The gauge-invariant Yang-Mills and matter part ${\mathcal{L}}_0$ is BRST invariant because a BRST variation is a gauge transformation with ghost-valued parameter. For the gauge-fixing fermionthe graded Leibniz rule givesup to the common sign convention used to define the ghost term. This is precisely the gauge-fixing, auxiliary-field, and ghost sector of the displayed Lagrangian. Therefore

Under $\Psi \mapsto \Psi +\delta \Psi$, the transition amplitude changes to first order by an insertionIfwith the corresponding condition on the bra, the two terms vanish after moving $\hat{Q}$ to the external states. The amplitude is then independent of the gauge-fixing choice.

Conversely, invariance for arbitrary changes of the gauge-fixing fermion requires physical external states to be BRST closed. States differing by a BRST-exact state have identical matrix elements against closed states, so the physical state space is the cohomology