Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 (Advanced Quantum Field Theory)

The first two terms are the kinetic term and mass term of a real scalar field; together they determine the free quantum field theory propagator. The term $-\lambda {\varphi }^4/4!$ is its quartic field interaction term. The remaining three terms are counterterms: ${\delta }_Z$ renormalizes the field normalization, ${\delta }_m$ renormalizes the mass, and ${\delta }_{\lambda }$ renormalizes the quartic coupling. Their regulator dependence cancels the ultraviolet divergences of loop diagrams, while their finite parts implement the chosen renormalization conditions.

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Through order ${\lambda }^2$, the quartic one-particle-irreducible Feynman diagrams are the tree-level four-point vertex, the local ${\delta }_{\lambda }$ counterterm vertex, and three one-loop bubble diagrams. The bubbles have two quartic vertices and differ by whether the momentum through the loop is the $s$, $t$, or $u$ Mandelstam variables; each has symmetry factor $1/2$. External-leg self-energies are not part of the 1PI four-point vertex, and ${\delta }_Z$ begins beyond the required order in this theory.

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For a bubble carrying momentum $p$, a Feynman parameter and a shift of the loop momentum give, after Wick rotation and cutoff regularization,where $F(p^2)=m^2-x(1-x)p^2-i0$. The quantum effective action therefore containsThe renormalization condition ${\Gamma }_4(M^2,M^2,M^2)=\lambda$ fixesSubstitution cancels both the cutoff and the scheme-dependent constant and leavesas required.

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Regard $\lambda$ as a running coupling $\lambda (M)$. At fixed $s,t,u$ and to order ${\lambda }^2$, differentiating the one-loop answer with respect to $\log M$ givesbecause $m\ll M$ implies $F(M^2)\simeq -x(1-x)M^2$. Thus the beta function isSolving this ordinary differential equation with $\lambda (M_0)={\lambda }_0$ givesto leading-logarithmic order.

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The leading contributions to ${\delta }_2$ and ${\delta }_m$ come from the one-loop electron self-energy: one internal electron line and one internal photon line join two electron-photon vertices. Its coefficients of ${\gamma }^{\mu }{p}_{\mu }$ and $m$ determine the field-strength and mass counterterms. The leading contribution to ${\delta }_1$ comes from the one-loop electron-photon vertex correction, with two internal electron propagators and one internal photon propagator, together with the corresponding local counterterm vertices. The Ward identity gives ${\delta }_1={\delta }_2$ in a gauge-invariant scheme.

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Use dimensional regularization with $d=4-ϵ$ and choose Feynman gauge. Combining the electron and photon denominators with a Feynman parameter, shifting the loop momentum, and discarding the odd term leaves the numeratorThe pole of the rotationally symmetric integral consequently givesWith the inverse-propagator conventionthe minimal subtraction scheme chooses the counterterm pole to equal ${\Sigma }_{\mathrm{pole}}$. Hence

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The leading two-point diagram has one sextic vertex, two external legs, and its remaining four legs paired into two tadpole loops; it determines the mass counterterm. The leading six-point diagram has two sextic vertices joined by three internal lines, leaving three external legs at each vertex; it is a two-loop diagram and determines the coupling counterterm. The corresponding local ${\delta }_m$ and ${\delta }_{\lambda }$ vertices complete the counterterm calculation.

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For the two-point graph, $L=2$ and $I=2$, so its superficial degree of divergence in three dimensions isIts additive mass-squared counterterm is therefore proportional to $\lambda {\Lambda }^2$; writing that counterterm as $m^2{\delta }_m$ gives the dimensionless scaling ${\delta }_m\sim \lambda {\Lambda }^2/m^2$, or ${\Lambda }^2/m^2$ when coupling factors are suppressed as in the question. For the six-point graph, $L=2$ and $I=3$, so $D=0$ and its ultraviolet divergence is logarithmic:Suppressing powers of $\lambda$ yields the two stated estimates.

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The only two-point diagram through second loop order is the double tadpole from one sextic vertex. It has no route by which the external momentum can pass through an internal propagator, so its value is independent of $p$ and renormalizes only the mass. A field-strength counterterm is determined by the coefficient of $p^2$ in the two-point 1PI function, hence ${\delta }_Z=0$ at one and two loops. The first momentum-dependent two-point topology uses two sextic vertices joined by five internal lines and has four loops.

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Complete the square in the heavy variable:The shifted Gaussian integral is independent of $\varphi$. For $M>0$, choosing $N=M/\sqrt{2\pi }$ removes it and gives

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The induced four-$\varphi$ interaction is the tree-level diagram with two $\lambda {\varphi }^2\chi$ vertices joined by one internal $\chi$ propagator. The zero-dimensional propagator is $M^{-2}$, and the second-order expansion of the exponential supplies the factor $1/2$. The resulting effective-action coefficient is therefore $-{\lambda }^2/(2M^2)$, exactly the value of $\tilde{\lambda }$ found by the Gaussian integral.

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Let $K_M=-{\partial }^2+M^2$ and let ${\Delta }_M=K_M^{-1}$ be its Euclidean quantum field theory propagator. With the source-sign convention suited to the expression in the question, definewhere $N_0$ makes $Z_0[0]=1$. Since inserting $\chi (x)$ is equivalent to acting with $-\delta /\delta J(x)$, the integral over $\chi$ gives

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The interaction is linear in the Gaussian field $\chi$, so completing the functional square makes its order-${\lambda }^2$ contribution exact:In momentum space, ${\Delta }_M(p)=(p^2+M^2{)}^{-1}$. If every external momentum satisfies $p^2\ll M^2$, the derivative expansionstarts with the local effective interactionwhich is the field-theory version of the zero-dimensional result.

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For a gauge orbit near a root ${\alpha }_0$ of $G(A^{\alpha })=0$, functional linearization givesThe multidimensional delta-function change-of-variables formula therefore givesComparison with the defining identity provesup to the usual field-independent normalization and a choice of determinant sign on the gauge patch.

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For axial gauge,so the Faddeev-Popov determinant is $\det (n^{\mu }{D}_{\mu })$. Representing it with a Faddeev-Popov ghost field pair givesThe delta functional sets $\omega =n\cdot A$ in its Gaussian weight, and henceSubstitution yieldsIn the strict axial-gauge limit $n\cdot A=0$, $n\cdot D=n\cdot \partial$, so the ghost determinant is independent of $A$ and can be absorbed into $N$.

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Write the gauge-fixing and ghost action as the BRST transformation of the gauge-fixing fermion:with harmless rescalings of $B$ accommodating the convention $s{\bar{c}}^a=\xi B^a$ used in the question. The gauge-invariant action obeys $sS[A]=0$, while nilpotence givesThus the entire gauge-fixed action is BRST invariant. Eliminating the auxiliary field $B^a$ by its algebraic field equation reproduces the axial gauge-fixing term and the ghost action found in part (b).

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Changing $n^{\mu }$ to $u^{\mu }$ changes the gauge-fixing fermion byThe corresponding change of the action is the BRST-exact term $s({\Psi }_u-{\Psi }_n)$. For a BRST-closed observable $\mathcal{O}$, invariance of the functional measure implies that the variation of its expectation value is the expectation of a total BRST variation and vanishes:Therefore physical states and observables, which are classes in BRST cohomology, do not depend on the fixed axial-gauge vector, provided there is no BRST anomaly or boundary contribution.

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