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For a centered Gaussian random field, Wick's theorem says that every odd moment vanishes and every even moment is the sum over all pairings of products of two-point functions. For six fields there are $15$ pairings. In the connected three-point function at nonzero external momenta, each of the three external fields must pair with a distinct field at the cubic interaction. There are $3!=6$ such Wick contractions. Choosing the undifferentiated field gives three possibilities; interchanging the two differentiated fields gives a further factor of two. The remaining nine pairings involve an external-external pair and an internal pair and belong to tadpole/disconnected contributions. Define the background so the one-point function vanishes, or equivalently subtract these contributions.

It is useful to keep a coefficient $\mathcal{C}$ multiplying the cubic Hamiltonian: its literal printed value is $1$, whereas the standard dimensionally normalized curvature interaction has $\mathcal{C}=M_{\mathrm{Pl}}^2$. This distinction will matter for the final amplitude. Withand $dt=ad\tau$, $\dot{\zeta }={\zeta }^{\prime }/a$, the interaction entering the time integral isIn the interaction picture, the unequal-time vacuum contraction required by the in-in formalism isFor a differentiated internal field replace $u_k^{\ast }$ by $u_k^{\ast \prime }$. The equal-time power spectrum fixes its magnitude; the displayed free De Sitter curvature mode functions and vacuum choice fix its unequal-time phase.

Performing the three momentum integrations imposes ${\mathbf{p}}_a=-{\mathbf{k}}_a$ for each assignment and leaves one overall momentum delta function. Put $U=u_{k_1}(0)u_{k_2}(0)u_{k_3}(0)$, which is real here, and defineThe upper early-time contour ${\mathcal{C}}_{+}$ runs from $-\infty (1-i\delta )$ to $0$, with $\delta >0$. The Hamiltonian's minus sign and the six connected Wick contractions then giveThis is equivalently $\mathrm{Re}[-4i\mathcal{C}{ϵ}^{2}U\sum I_i]$ with unconjugated De Sitter curvature mode functions and the conjugate lower contour ${\mathcal{C}}_{-}$. Before momentum integration, the same result consists of the three cyclic delta assignments, each with the extra factor of two for the identical differentiated legs.

The PDF's intermediate formula writes only three cyclic assignments without that factor of two. Taken literally it undercounts the connected contractions. Its unconjugated modes must also use the conjugate contour, rather than the upper contour of the original in-in expression. Both points are required for a consistent in-in bispectrum conjugation rule.

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.