Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 49 / 2 / Solution

Use the mostly-plus Minkowski metric, so $p^2=-p_0^2+{\mathbf{p}}^2$, and the source term $+\int J\varphi$. Define the vacuum-normalized generating functional, connected generating functional, and quantum effective action byThe Legendre transform is taken after expressing $J$ in terms of $\phi$, locally where this relation is invertible. Varying it gives $\delta \Gamma /\delta \phi =J$ and ${\Gamma }^{(2)}=(W^{(2)}{)}^{-1}$. The logarithm generates connected graphs; the Legendre transform removes graphs joined by a single propagator, leaving one-particle-irreducible vertices. This sign convention is important: its classical limit is $\Gamma =-S$, consistent with the positive vertices in this question.

For the free real scalar field, integration by parts puts the action into the form $S=-\frac{1}{2}\varphi A\varphi$, where $A=-\square +m^2$ has Fourier symbol $p^2+m^2$. Include the Feynman i-epsilon prescription $A\mapsto A-i0$. Completing the square yieldsand consequentlyThe physical scalar propagator is $-i{A}^{-1}$, whereas the inverse quadratic kernel appearing in $W_0$ is $A^{-1}$. Differentiating this quadratic functional and Fourier transforming givesThe infinitesimal prescription specifies propagators and is dropped when quoting the local vertex kernel.

At the order represented by tree-level Feynman diagrams, stationary phase gives $W_{mtree}=S[\phi ]+\int J\phi$, with the classical field equation $\delta S/\delta \phi +J=0$. The Legendre transform cancels the source term, soThusOther connected tree graphs with several quartic vertices are one-particle reducible and belong to $W$, not to the proper vertices of $\Gamma$. This explains why no extra tree correction is hidden in these formulas.

The one-loop proper two-point graph is the tadpole diagram. The one-loop proper four-point graphs are the three bubble diagrams, with pairings $(12|34)$, $(13|24)$ and $(14|23)$ of the external legs. Each has Feynman-diagram symmetry factor $1/2$. The figure shows the topologies and every four-point channel; external momenta are incoming.

The tadpole factor can also be counted directly: there are $4\cdot 3=12$ assignments of two labeled external legs to one quartic vertex, while its $4!$ normalization leaves $12/24=1/2$. For a fixed four-point partition, the interchange of the two internal lines gives the same factor $1/2$; there are three distinct partitions. Graphs with an external self-energy insertion are one-particle reducible and are not additional contributions to ${\tau }_4$.

For the two-point divergence, perform a Wick rotation. The Euclidean tadpole contribution to the proper inverse kernel isAssume $m^2>0$ to separate the ultraviolet calculation from infrared issues. A Schwinger parameterization gives $1/(k_E^2+m^2)={\int }_0^{\infty }ds{e}^{-s(k_E^2+m^2)}$. Using the supplied Gaussian integral, initially in the convergence region, givesThe last expression provides the meromorphic dimensional regularization continuation. Set $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,If the coupling is written as ${\mu }^{ϵ}\lambda$, that factor multiplies the intermediate integral but does not change the displayed pole residue. The negative pole is a property of analytic continuation; it must not be inferred from the positive cutoff integrand without carrying out the continuation.

The added mass counterterm contributes $+B$ to ${\tau }_2$, because ${\Gamma }_{mtree}=-S$ and its Lagrangian term is $-B{\varphi }^2/2$. Hence cancellation requiresup to an arbitrary finite subtraction specifying the renormalization scheme. No external momentum occurs in the tadpole, so no one-loop wave-function renormalization is needed. At $m=0$ the unregulated tadpole is scaleless and vanishes in dimensional regularization; the massive calculation identifies the mass-dependent ultraviolet pole without conflating it with a massless infrared limit.

For completeness, the signs of the four-point graphs in this effective-action convention can be read from the Euclidean one-loop functional $\frac{1}{2}\mathrm{Tr}\log (A_E+\lambda {\phi }^2/2)$. Its expansion has terms $\lambda \mathrm{Tr}(A_E^{-1}{\phi }^2)/4$ and $-{\lambda }^2\mathrm{Tr}(A_E^{-1}{\phi }^2{A}_E^{-1}{\phi }^2)/16$. Differentiating four times givesThis is the Euclidean form, analytically continued to the physical external momenta. It confirms the three channel diagrams and their symmetry factors; evaluating its divergence is not needed for the requested mass counterterm.