Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 312 / 4

For a centered Gaussian random field, Wick theorem takes the classical form of the Isserlis theorem: every even correlator is a sum over pairings of two-point correlators, and every odd correlator vanishes. For example, with $G_i={\zeta }_G({\mathbf{x}}_i)$,A $2n$-point correlator has $(2n-1)!!$ pairings. Equivalently, all connected correlation functions beyond order two vanish. The covariance completely determines the centered Gaussian statistics; higher ordinary correlators can be nonzero, but contain no independent connected information. For a field with nonzero mean, apply these statements to the centered field and then restore its mean.

Use the Fourier transform convention $\zeta (\mathbf{x})=\int d^{3}k{e}^{i\mathbf{k}\cdot \mathbf{x}}\zeta (\mathbf{k})/(2\pi {)}^3$. Statistical homogeneity and isotropy define the primordial bispectrum byThe Dirac delta function enforces momentum conservation; the three magnitudes describe a closed triangle. Set $\alpha =3f_{\mathrm{NL}}/5$. For the centered quadratic Gaussian transformation, its quadratic Fourier term isAt first order in $\alpha$, choose the quadratic field at one of the three external positions. For the first position, Wick contractions pair the two internal fields with the two remaining external fields in two ways. The self-pairing is precisely removed by the subtracted mean. The surviving contribution is $2\alpha P(k_2)P(k_3)$ times the momentum delta. Adding the other positions gives the tree-level local-type primordial non-Gaussianity resultThis expression is at leading order in $f_{\mathrm{NL}}$. The exact quadratic map also has a loop correction to the local primordial bispectrum, from three quadratic vertices:There is no quadratic-in-$\alpha$ bispectrum term, because it would involve five centered Gaussian fields. A regulator may be needed for this higher-order integral. The displayed leading formula uses the Gaussian power $P$, consistently neglecting its order-$f_{\mathrm{NL}}^2$ correction.

Distinguish the slow-roll approximation parameter $ϵ$ from the infinitesimal positive contour regulator $\epsilon$. In the interaction picture, the ordered unequal-time Wick contraction isA differentiated vertex field instead supplies $u_k^{\ast \prime }(\tau )$. These conjugates are fixed by the annihilation operators and creation operators and cannot be dropped while retaining the same contour.

Convert the interaction time integral to conformal time. Since $dt=ad\tau$ and each cosmic-time derivative is $a^{-1}{\partial }_{\tau }$, the integrated cubic vertex isThe in-in formalism supplies $-2i$ multiplying the expectation of the Hamiltonian, so its negative sign gives $+2i$ multiplying this vertex. At nonzero external momenta, the connected contractions join each of the three external fields to one vertex field. There are six bijections, not three. Choosing which external field meets the undifferentiated vertex gives three cyclic choices; swapping the two differentiated fields gives a further factor of two. This is the Wick-pairing multiplicity for a zeta zeta-prime-squared vertex.

For example, before the internal momenta are integrated, one cyclic contribution includesThe factors of $2\pi$ cancel to leave one overall $(2\pi {)}^3$ and the external momentum delta. Thus the properly normalized reduced expression isThe vacuum prescription for inflationary in-in integrals damps the early-time oscillations and selects the Bunch-Davies vacuum. Internal self-contractions correspond to disconnected zero-momentum tadpole terms and are excluded from this connected three-point function.

The printed intermediate expression has unconjugated modes and only three literal cyclic terms with coefficient $-2i$. The conjugate form of the result above would use $-4i$, together with the conjugated contour. A literal $-2i$ with only three terms gives half the final printed answer. The six-contraction expression above is the consistent reduction of the supplied Hamiltonian and leads to that final answer.

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.