Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 1

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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