Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 304 / 2

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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