Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 312 / 1

The operator ${\dot{\varphi }}^2\varphi$ is not invariant under a scalar-field shift symmetry because one field is undifferentiated. An arbitrary large coefficient would therefore radiatively generate equally unsuppressed nonderivative operators, including a mass and a steep scalar potential, and would spoil the approximate shift symmetry and flat potential needed for single-field slow-roll inflation. In a technically natural slow-roll model its coefficient must consequently be small, so it does not produce parametrically large primordial non-Gaussianity. Treated merely as an effective spectator-field interaction, it does generate the tree-level primordial bispectrum computed below, whose size is controlled by the dimensionless interaction strength at the Hubble scale; taking that strength large abandons the controlled slow-roll premise.

Since $dt=ad\tau$ and $\dot{\varphi }=a^{-1}{\varphi }^{\prime }$, the interaction isto first order in $\lambda$. The tree-level in-in formalism givesThere are two Wick contractions for each choice of the undifferentiated field at the vertex. With $K=k_1+k_2+k_3$, the stated Bunch-Davies vacuum mode obeysThe two powers of $\tau$ from the differentiated modes cancel $a^2=1/(H^2{\tau }^2)$, and the remaining integral isRemoving the momentum-conserving delta function, the bispectrum from a time-derivative cubic scalar interaction is therefore

With the Fourier transform convention $\varphi (\mathbf{k})=\int d^{3}x{e}^{-i\mathbf{k}\cdot \mathbf{x}}\varphi (\mathbf{x})$, scale invariance of the position-space three-point function givesBecause ${\delta }^{(3)}(\mu {\sum }_i{\mathbf{k}}_i)={\mu }^{-3}{\delta }^{(3)}({\sum }_i{\mathbf{k}}_i)$, the reduced primordial bispectrum must satisfyIn the result of part b, the bracket has momentum degree five, while $k_1^3{k}_2^3{k}_3^3{K}^2$ has degree eleven. Its total degree is therefore $-6$, exactly as required.