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Let $A_k=H/(2\sqrt{ϵ}{M}_{\mathrm{Pl}}{k}^{3/2})$ and $K=k_1+k_2+k_3>0$. For the unconjugated integrand on the lower contour, the De Sitter curvature mode functions giveHere the two derivative factors supply ${\tau }^2$, cancelling $a^2=1/(H^2{\tau }^2)$. The vacuum prescription for inflationary in-in integrals can be implemented by a factor $e^{\delta \tau }$ on the real negative axis, with $\delta >0$, followed by $\delta \downarrow 0$. In this notationThereforeAn undamped boundary evaluation on the real axis at $-\infty$ is not valid. On the upper contour the complex-conjugate integrand instead yields the conjugate value. Conjugating the modes without conjugating the contour would produce exponential growth.

Using all six connected Wick contractions from the preceding part and $U=A_{k_1}{A}_{k_2}{A}_{k_3}$ gives the literal-Hamiltonian resultThe connected correlator is $(2\pi {)}^3{\delta }^{(3)}(\sum {\mathbf{k}}_i)B_{\mathcal{C}}$. The three assignments constitute the curvature bispectrum from a zeta zeta-prime-squared interaction. The numerator contains six powers of $H$ from the six modes, two of which are cancelled by $a^2$; the six powers of $M_{\mathrm{Pl}}$ remain unless the vertex supplies two.

Consequently the printed final expression, with $M_{\mathrm{Pl}}^{-4}$, is obtained for $\boxed{\mathcal{C}=M_{\mathrm{Pl}}^2}$. With the literal printed Hamiltonian $\mathcal{C}=1$, the answer instead has $\boxed{M_{\mathrm{Pl}}^{-6}}$. The Planck normalization of a cubic curvature interaction requires a factor $M_{\mathrm{Pl}}^2$ in the Hamiltonian if its final bispectrum is intended. In addition, integrating the intermediate expression literally with only three assignments gives half the fully contracted amplitude. These are normalization defects, rather than changes in the momentum shape. For dimensionless comoving curvature perturbations, the usual interaction normalization is also required by the mass dimension of the action.

For the standard normalization, $P_{\zeta }(k)=H^2/(4ϵM_{\mathrm{Pl}}^2{k}^3)$, so $B/P_{\zeta }^2$ is of order $ϵ$. This slow-roll suppression means that the primordial non-Gaussianity from this interaction alone is not expected to be detectably large. For example, in the squeezed bispectrum configuration $k_1\ll k_2\simeq k_3$, matching this contribution to the local convention gives $f_{\mathrm{NL}}=5ϵ/24$. This is a single-vertex contribution, not the complete single-field slow-roll inflation prediction; other vertices and field redefinitions contribute at the same slow-roll order. A claim of detectability for enhanced interactions requires a model beyond this approximation.

Write $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{k}^3}$, so $u_k(0)=A_k$. For the cyclic term with the first field undifferentiated, insertion into the in-in formalism giveswhere $K=k_1+k_2+k_3>0$. The two differentiated modes contribute ${\tau }^2$, cancelling the ${\tau }^{-2}$ in $a^2$.

Along the vacuum-selected contour, the lower boundary vanishes. The elementary integrals areand thereforeMultiplying by $4i{M}_{\mathrm{Pl}}^2{ϵ}^2$, taking the real part and summing the three cyclic choices gives the curvature bispectrum from a zeta zeta-prime-squared interaction:This matches the final normalization. The coefficient requires all six connected Wick contractions.

The calculation assumes the Gaussian Bunch-Davies vacuum, tree order in the specified cubic interaction, effectively constant $H$ and $ϵ$ during the integral, the approximate de Sitter scale factor, and observation after the modes freeze as $\tau \to 0^{-}$. The contour tilt is kept until the early boundary has been eliminated; an undamped real-axis integral cannot simply discard that boundary. The reduced Planck mass convention is the one in the supplied De Sitter curvature mode function. Other cubic interactions and nonlinear field redefinitions are outside the specified model, so this is not by itself the complete bispectrum of a general single-field action.

At late time the Gaussian curvature power spectrum is $P(k)=H^2/(4ϵM_{\mathrm{Pl}}^2{k}^3)$. Consider the squeezed bispectrum configuration $k_1=q\ll k_2\simeq k_3=k$. The local-type primordial non-Gaussianity result is dominated byIn the curvature bispectrum from a zeta zeta-prime-squared interaction, the leading cyclic term has the long mode in the undifferentiated slot. Its momentum numerator tends to $k^4/(2k)=k^3/2$, while the other two cyclic terms contain $q^2$. HenceThis is the slow-roll amplitude of the squeezed zeta zeta-prime-squared bispectrum. It retains the local-like $q^{-3}$ enhancement: assigning the soft momentum only to differentiated legs would incorrectly miss the dominant permutation. Comparing squeezed amplitudes givesFor comparison, in the equilateral configuration $B^{\mathrm{sf}}(k,k,k)=(4ϵ/3)P(k{)}^2$, so the signal is slow-roll suppressed away from the squeezed limit as well.

Use the quoted observational bound as the sensitivity benchmark supplied by this problem, rather than as a new current measurement. A local-template sensitivity at the order-ten level is far above this order-$ϵ$ signal; detection of this interaction with comparable CMB measurements is consequently very unlikely. The entire shape is not identical to the local template, so its bound is not an exact constraint on every single-field shape, but the squeezed amplitude already exposes the large suppression. The obstacle is the slow-roll amplitude, even though this specified interaction has a squeezed enhancement.