Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 301 / 3

Varying the displayed gauge-fixed Lagrangian density gives the kinetic operatorIts quantum field theory propagator should satisfyand inversion into transverse and longitudinal projectors gives the numerator

The paper instead prints $-{\eta }_{\mu \nu }+(\alpha -1)k_{\mu }{k}_{\nu }/k^2$. Except at $\alpha =1$, that is not the inverse of the displayed Lagrangian's kinetic operator. Taken literally, for $\alpha \ne 2$ it is the Green function ofand at $\alpha =2$ its longitudinal part is noninvertible. Thus the longitudinal sign in the printed propagator is a typographical error; the two forms coincide in Feynman gauge.

Solved by gpt-5.6-sol high.

Let $G_{\mu \nu }(x,y)=⟨\Omega |T{A}_{\mu }(x)A_{\nu }(y)|\Omega ⟩$. Invariance of the normalized path integral under $A\mapsto A+\delta A$ gives the Schwinger-Dyson equationtogether with $D^{\mu \rho }⟨A_{\rho }{⟩}_j=e{j}^{\mu }$. This assumes the functional measure is translation invariant, boundary terms in field space vanish, and vacuum bubbles are removed by normalization. The current is a fixed c-number and conserved; conservation removes dependence on the longitudinal, gauge-parameter part of the photon propagator. An $iϵ$ prescription and adiabatic switching select the interacting vacuum.

Solved by gpt-5.6-sol high.

Since $D\Delta =i\delta$, the source-induced one-point function isThe action is quadratic and the source is linear, so completing the square gives the exact full two-point functionDiagrammatically these are a free line joining $x$ to $y$ and a disconnected pair of lines, each joining one external insertion to one current cross. The connected two-point function remains exactly ${\Delta }_{\mu \nu }$. The expansion truncates at $e^2$ because a Gaussian integral has no interaction vertices and its mean is linear in $e$.

Solved by gpt-5.6-sol high.

For quantum electrodynamics, $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$ is an operator built from a dynamical Dirac field. Integrating over $\psi ,\bar{\psi }$ generates arbitrarily high powers of $e$, so the Gaussian-source truncation fails.

For the connected photon two-point function, the required diagrams through $e^4$ are: the bare photon line at $e^0$; one fermion-loop photon vacuum polarization insertion at $e^2$; and at $e^4$, a photon line with two successive one-loop polarization insertions together with the two-loop one-particle-irreducible fermion loop whose loop contains one internal photon chord. The latter includes the cyclic placements conventionally interpreted as self-energy and vertex corrections. Counterterm insertions must be added in a renormalized calculation. Disconnected vacuum bubbles cancel against the vacuum normalization.

Solved by gpt-5.6-sol high.