Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 55 / 2 / b / i / Solution

For a centred Gaussian random field, Wick theorem expresses each even correlation function as a sum over all pairings of two-point functions; odd moments vanish. For example,For six fields there are $5\cdot 3\cdot 1=15$ pairings. In a tree-level connected three-point calculation with one cubic vertex, each of the three external fields must contract with a different vertex field. This gives $3!=6$ connected Wick contractions. Pairings connecting two external fields instead are disconnected tadpole contributions, omitted from the primordial bispectrum.

Put $\mathcal{F}=(1-c_s^2)\mathcal{A}/c_s^2$. Using $dt=ad\tau$ and $\dot{\zeta }={\zeta }^{\prime }/a$, the integrated interaction Hamiltonian becomesThe unequal-time Wick contraction required by the displayed operator order isThe conjugate is essential: the equal-time power spectrum alone does not specify the erroneous unstarred replacement printed in the target expression. The in-in bispectrum conjugation rule fixes the vacuum convergence and final sign.

Fourier transforming the vertex gives $(2\pi {)}^3{\delta }^{(3)}({\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3)$ with three measures $d^3{p}_j/(2\pi {)}^3$. The six pairings supply the product of three external-vertex delta functions, each carrying $(2\pi {)}^3$. Integrating all three internal momenta leaves one overall $(2\pi {)}^3{\delta }^{(3)}({\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3)$. Thus the normalized connected contraction formula isThis spells out what the printed permutations and delta functions must represent: all six permutations, full Fourier normalization, and conjugated vertex modes. Three cyclic permutations alone would miss a factor of two.