= Solution
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:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-46-cubic-self-energy.png]
{title=One-loop cubic scalar two-point insertion with momenta p and p plus k and symmetry factor one half}
{height=350}
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 $d^dp/(2\pi)^d$. 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 $1/2$.
\b[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 $D_0(k)=1/(k^2+m^2)$.
Back to article page
