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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.

The constant scalar-field shift symmetry has momentum-space actionFor the canonical commutation relation $[\varphi (\mathbf{x}),\Pi (\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$, its Noether charge can be writtenIndeed, $i[Q,\varphi (\mathbf{x})]=c$.

If a scalar vacuum is invariant under an exact scalar-field shift symmetry, the Ward identity makes the leading soft bispectrum vanish: