Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 46 / 2 / i

The cubic scalar field theory has a one-particle-irreducible Feynman diagram with two cubic vertices, one external leg at each vertex, and two internal lines joining them:

The Euclidean Feynman rule for each cubic interaction vertex is $-g{\mu }^{(6-d)/2}$: the minus sign comes from expanding $e^{-S_{\mathrm{int}}}$, and $1/3!$ in the action cancels the permutations of its three fields. The mass dimension of the cubic coupling is $(6-d)/2$, so ${\mu }^{(6-d)/2}$ permits $g$ to remain dimensionless. Two vertices supply $(-g{)}^2{\mu }^{6-d}$.

Each internal scalar propagator contributes $1/(p^2+m^2)$ with its own momentum. Conservation leaves one independent loop momentum; choose the two propagator momenta to be $p$ and $p+k$. The remaining Fourier integration measure is $d^{d}p/(2\pi {)}^d$. Finally, the factor $1/2$ is the Feynman-diagram symmetry factor for exchanging the two identical internal lines. Direct Wick contraction counting gives the same factor: two choices for which vertex receives a labeled external leg, $3^2$ choices of the incident fields and $2!$ pairings of the remaining fields, divided by $2!(3!{)}^2$, yield $1/2$.

The displayed loop integral is the amputated two-point insertion. For the correction to the full propagator, multiply it by the external scalar propagators, giving $D_0(k{)}^{2}I(k)$ with $D_0(k)=1/(k^2+m^2)$.