 Solution

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.