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A damped mode $\delta =e^{-k^2/k_D^2}(A\cos q+B\sin q)$ does not alone imply oscillations in its ensemble power spectrum. Averaging its modulus squared gives the displayed formula. Equal independent initial variances, $P_A=P_B$ and $P_{AB}=0$, erase the oscillatory variance through ${\cos }^{2}q+{\sin }^{2}q=1$. A coherent initial phase, as supplied by a regular growing adiabatic mode, selects one common oscillator phase and retains acoustic peaks. Its leading density and velocity oscillations are a quarter-period apart, and angular projection also preserves their correlated contribution.

The equipartition theorem says that each independent quadratic term in a Hamiltonian described by classical mechanics has mean energy $k_{B}T/2$ in a canonical ensemble. Here $k_B$ is the Boltzmann constant and $T>0$ the absolute temperature. Count terms in the energy, rather than vaguely counting everything that can move: one harmonic oscillator has one position degree of freedom but TWO quadratic terms, one in momentum and one in displacement.

A precise sufficient setting is $H(y,\xi )=a{y}^2/2+H_0(\xi )$ with $a>0$, where $y$ is one of the canonical variables, $\xi$ denotes all remaining variables, and the phase space integration measure factors as $dyd\xi$. The range of $y$ is the whole real line, independently of $\xi$; there is no constraint coupling its allowed values to other variables. The partition function must be finite, and the relevant expected values must exist. The Boltzmann factor then givesThe probability density function ${\rho }_y$ is a normal distribution with variance $k_{B}T/a$. A larger coefficient narrows the fluctuations enough to cancel its greater energy cost. This is why the mean contribution is independent of mass or spring constant.^[ref]

For a proof outline, the one-variable Gaussian integral is $Z_y={\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2/2}dy=\sqrt{2\pi /(\beta a)}$, so $-{\partial }_{\beta }\ln Z_y=1/(2\beta )$. More generally, integration by parts in a canonical variable gives $⟨y{\partial }_{y}H⟩=k_{B}T$ when its boundary term $[y{e}^{-\beta H}]$ vanishes and the integrals converge. For $a{y}^2/2$ this is twice its mean energy. At a finite hard boundary the boundary term need not vanish; hard-wall correction to harmonic equipartition illustrates the resulting departure. A constraint, a nonquadratic potential energy, or a noncanonical integration measure therefore requires its own calculation. The theorem is an ensemble average, not equality of instantaneous energies, and applying it to a recorded trajectory requires equilibration and sufficient sampling.

For a classical one-dimensional harmonic oscillator, with positive mass $m$ and spring constant $k$,The origin is the equilibrium position. Its heat capacity is $k_B$ at fixed $m,k$, not $k_B/2$. Although the two mean energies agree, their measured amplitudes do not: $⟨p^2⟩=m{k}_{B}T$ and $⟨x^2⟩=k_{B}T/k$. In a harmonic optical trap, the position measurement observes the potential energy contribution; it does not measure both contributions by observing $x$ alone.^[ref]

For $N$ particles in a three-dimensional classical monatomic gas, neglect interactions and thermally accessible internal excitations. Its three translational momentum terms give internal energy $U=3N{k}_{B}T/2$, heat capacity at constant volume $C_V=3N{k}_B/2$, and heat capacity at constant pressure $C_p=C_V+N{k}_B=5N{k}_B/2$. Per mole, replace $N{k}_B$ by the molar gas constant $R$. For example, using stipulated $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the molar heat capacity at fixed volume is $12.471\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These results assume a dilute, nonrelativistic ideal gas with classical translational statistics, rather than a quantum degeneracy regime.

For an ideal gas of diatomic molecules, modeled as two chemical atoms with a bond, there are three translational quadratic terms, two accessible rotational kinetic energy terms for changes of bond orientation, and one bond-stretching harmonic oscillator with two quadratic terms. Rotation about the bond axis is not an additional active mode in this linear-molecule model. The three familiar limits of heat capacity at constant volume areCorresponding molar heat capacities are $3R/2$, $5R/2$ and $7R/2$. Counting the stretching harmonic oscillator only once as $k_{B}T/2$ would give the wrong last value.^[ref]

The frozen regimes require quantum mechanics. In the simple molecular rotation model, $E_j=k_B{\theta }_{r}j(j+1)$ with state degeneracy $2j+1$, ${\theta }_r={\hslash }^2/(2I{k}_B)$, and moment of inertia $I$. For the bond-stretching quantum harmonic oscillator, ${\theta }_v=\hslash \omega /k_B$ with angular frequency $\omega$ and $\hslash =h/(2\pi )$ from the Planck constant. When ${\theta }_r\ll {\theta }_v$, the three limits above correspond to $T\ll {\theta }_r$, ${\theta }_r\ll T\ll {\theta }_v$, and $T\gg {\theta }_v$, provided translational motion remains classical. They are smooth crossovers, not abrupt switches. Real molecules can have different allowed rotational populations; at high temperature, thermal dissociation, internal excitations and nonharmonic bond motion can invalidate the fixed-molecule model before its last limit is reached.^[ref]

For a quantum harmonic oscillator with fixed $\omega$, put $z={\theta }_v/T$. Its exact mean energy and heat capacity areAt $T\gg {\theta }_v$, the mean energy approaches $k_{B}T$ and $C_{\mathrm{osc}}$ approaches $k_B$. At $T\ll {\theta }_v$, thermal excitation is exponentially suppressed and $C_{\mathrm{osc}}$ tends to zero, although the zero-point energy $\hslash \omega /2$ survives. Frozen thus means little thermal response, not zero motion or zero energy.^[ref]

A motivating measurement is thermal calibration of an optical trap. A focused laser through a microscope objective confines a bead in water; imaging or a quadrant photodetector records a position time series. Gieseler and colleagues describe a polystyrene bead of radius $1.03\mu \mathrm{m}$ at $293.15\mathrm{K}$, a $1064\mathrm{nm}$ trapping laser, and photodetector acquisition at $10^5\mathrm{Hz}$. Their recorded bead continues undergoing Brownian motion; higher trapping power produces a narrower equilibrium position distribution. Infer the centered variance, not a squared distance from an arbitrary camera origin:The normal distribution position histogram and agreement with dynamical calibration methods support the local harmonic optical trap interpretation. For a stipulated example, $T=300\mathrm{K}$ and ${\sigma }_x=20\mathrm{nm}$ imply $k\simeq 1.04\times 10^{-5}N/m=0.0104pN/nm$. This calculated scale is not an experimental reading.^[ref]

Position calibration is essential: if detector output $v$ has conversion $x=s(v-v_0)$, its physical variance is $s^2\mathrm{Var}(v)$. An unknown scale $s$ cannot be determined together with $k$ from that one variance. A complementary calibration demonstrated by Tolić-Nørrelykke and colleagues combines the power spectrum of thermal motion with the bead response to known sinusoidal motion of a translation stage. The driven response supplies an additional length scale, and their method does not require bead size, liquid dynamic viscosity or wall distance as inputs. It still uses a thermal dynamical model; agreement between methods is a consistency check, not a theory-free proof of equipartition theorem.^[ref]

The readout itself can hide fluctuations. Wong and Halvorsen used a $1064\mathrm{nm}$ optical trap and adjustable-exposure CCD imaging of a bead in water. They varied exposure and trapping power, held the bead away from the nearest wall, adjusted illumination to maintain comparable images, and measured tracking noise using a stuck bead. The measured variance fell as exposure increased; corrected results agreed with their time-averaging model. For a stationary overdamped harmonic optical trap, with optical trap relaxation time $\tau$, an exposure of length $W$ measures ${\bar{x}}_W=W^{-1}{\int }_0^{W}x(t)dt$. Its ideal noise-free variance isFor $\alpha >0$, $S(\alpha )<1$: using the blurred variance in the uncorrected formula overestimates $k$ by $1/S(\alpha )$. At $W=\tau$, the calculated factor is $e/2\simeq 1.36$. Additive tracking noise instead raises the measured variance and biases the inferred $k$ downward if left uncorrected.^[ref]

Useful controls are to record over many optical trap relaxation times, compare blocks for drift, calibrate the position scale, measure fixed-bead noise, vary exposure, and check the position histogram for departures from a normal distribution. A sparse sequence of accurate instantaneous positions can estimate an equilibrium variance; a long exposure averages away motion even if many frames are collected. Flow, appreciable laser heating, escape from the harmonic region or nonconservative forces can break the assumed thermal equilibrium. Inferring $k$ by imposing equipartition theorem is a calibration, whereas testing the relation requires an independently established stiffness and position scale. The beautiful practical connection is that microscopic randomness can set a force scale: once calibrated, a small constant applied force displaces the mean by $f/k$. The same quadratic-term counting predicts macroscopic heat capacity and makes microscopic fluctuations useful measurements.

For a planar elastic filament under positive constant extension force $F$, the quadratic force-controlled energy isThe higher-order Euler-Lagrange equation is $B{\zeta }^{\prime \prime \prime \prime }-F{\zeta }^{\prime \prime }=B{\kappa }_0^{\prime \prime }$. On a periodic interval or in a bulk Fourier transform treatment, every nonzero mode obeys the displayed transfer relation. The length deficit is $\frac{1}{2}\int ({\zeta }^{\prime }{)}^{2}dx$, so its ensemble value is determined by the power spectrum of the quenched intrinsic curvature. Boundary conditions must be specified for a finite filament.

The second term in the equal-time two-point function of a de Sitter scalar is independent of time for constant $H$. It dominates on superhorizon scales and has constant dimensionless power spectrum, giving a scale-invariant inflationary power spectrum. The infrared qualification of a scale-invariant covariance still applies to the formally infinite-volume coincident variance.

Assume all relevant modes begin outside the horizon at common conformal time ${\tau }_i$, grow as ${\tau }^2$ until radiation-era horizon entry, then freeze until matter-radiation equality, and again grow as ${\tau }^2$ in matter domination. With horizon entry ${\tau }_h=2\pi /k$, $k_{\mathrm{eq}}=2\pi /{\tau }_{\mathrm{eq}}$ and final time ${\tau }_0$, direct multiplication of the growth factors gives the displayed transfer with $G=({\tau }_0/{\tau }_i{)}^2$. A primordial density power spectrum $P_i=Ak$ becomes $P_0=A{G}^{2}k$ below equality wavenumber and $A{G}^2{k}_{\mathrm{eq}}^4{k}^{-3}$ above it. This approximation neglects the true logarithmic radiation-era matter growth, baryonic acoustic structure and late acceleration; a conventionally normalized transfer divides out $G$.

Sampling with a homogeneous Poisson point process of mean comoving number density $\bar{n}$ adds a constant $1/\bar{n}$ to the dimensional galaxy power spectrum. Subtracting its expectation leaves sampling uncertainty. Correlated tracer populations and survey weights modify this simple formula; survey depth alone does not specify $\bar{n}$.

For the Fourier transform convention $C(\mathbf{s})=\int \Phi (\mathbf{\kappa })e^{i\mathbf{\kappa }\cdot \mathbf{s}}{d}^2\kappa /(2\pi {)}^2$, angular integration of an isotropic power spectrum gives the displayed Hankel transform. A purely radial Hankel transform without the prefactor corresponds to defining its spectrum as $\Phi /(2\pi )$.

Write the growing density contrast as ${\delta }_{m,\mathbf{k}}=A_{\mathbf{k}}{\tau }^2$, with $A_{\mathbf{k}}$ having the indicated $k^{n/2}$ spectral amplitude. In the dust-comoving synchronous gauge, $h^{\prime }=-2{\delta }_m^{\prime }$, and the supplied $h_s^{\prime }=h^{\prime }$ therefore implies $h_s^{\prime \prime }=-4A_{\mathbf{k}}$. On the same comoving density slicing, the cosmological Poisson equation isbecause $\mathcal{H}=2/\tau$. This is the comoving-density form of the relativistic scalar constraint, so it can be used on these large scales; replacing its density by the Newtonian-gauge density would not be justified there. Combining the two results givesThus the large-angle Sachs-Wolfe effect has source $\Phi /3$. The potential is constant for this matter-era growing mode, and its power spectrum obeys $P_{\Phi }(k)\propto k^{n-4}$.

For subhorizon pressureless matter during radiation domination, expansion is too rapid for significant self-gravitating growth. Neglecting the small matter-gravity term gives ${\ddot{\delta }}_m+t^{-1}{\dot{\delta }}_m\simeq 0$, hence a constant plus a logarithm of time; the exact coupled radiation problem supplies the detailed amplitudes. This is the Mészáros effect. Modes entering the horizon during radiation domination are suppressed relative to modes entering later, while after equality ordinary matter growth resumes.

The matter power spectrum consequently has a turnover near the equality horizon scale $k_{\mathrm{eq}}\sim a_{\mathrm{eq}}{H}_{\mathrm{eq}}$ in units with $c=1$. Shorter-scale modes retain the radiation-era suppression. This statement concerns cold pressureless matter; baryons coupled to photons also show acoustic effects before recombination.

For a general colored medium and unit plane-wave input $E(0,\mathbf{s})=1$, let $U_W(x,0)$ be the propagator of ${\partial }_x=i{\Delta }_{\perp }/(2k)+ik\mu W(x,\mathbf{s})$. Then the solution isA time-ordered operator exponential defines $U_W$. Its expectation depends on the full longitudinal covariance; the initial Gaussian and transverse-stationarity assumptions do not specify a universal closed expression.

With the stated longitudinal Gaussian white noise and transverse statistical homogeneity, the moment depends only on $\mathbf{\rho }={\mathbf{s}}_1-{\mathbf{s}}_2$. Hence $({\Delta }_1-{\Delta }_2)\Gamma =0$. The closed equation in the previous part and initial value $\Gamma (0)=1$ giveIn particular $\Gamma (x,0)=1$: the ensemble mean intensity remains constant for plane-wave input, even though coherent attenuation in a white-noise random medium reduces the mean complex field.

Use a consistent transverse power spectrum convention,For an isotropic spectrum, angular integration gives the isotropic transverse covariance spectrum, $C(\rho )=(2\pi {)}^{-1}{\int }_0^{\infty }\Phi (\kappa )J_0(\kappa \rho )\kappa d\kappa$. ThereforeThe supplied two-dimensional Fourier transform definition has angular factor $2\pi$, whereas the displayed radial transform pair omits it. The latter is a normalized Hankel transform pair. If its radial spectrum is denoted ${\Phi }_H=\Phi /(2\pi )$, use ${\Phi }_H$ in the integral and omit the prefactor $1/(2\pi )$. These are equivalent conventions; mixing them changes the attenuation rate.

Introduce a constant filament bending modulus $B$, which sets the mechanical force scale. The filament has intrinsic curvature of an elastic filament ${\kappa }_0(s)$, and its actual signed curvature in a small-slope Monge representation is $\kappa \simeq {\zeta }_{xx}$. To quadratic order one may identify material arc length $s$ with $x$ in ${\kappa }_0$, whileAt fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal{L}-(\mathcal{L}-L)$ leaves, up to a force-dependent constant,Its first variation isThe higher-order Euler-Lagrange equation is thereforeFor a Fourier transform relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary boundary conditions because the question specifies no filament end conditions. Using $f(x)=\int \hat{f}(q)e^{iqx}dq/(2\pi )$, every nonzero wavenumber obeysA zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The Parseval identity givesFor clarity about finite normalization, on a periodic interval of length $L$ define ${\kappa }_n=L^{-1}{\int }_0^L{\kappa }_0(x)e^{-i{q}_{n}x}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula isThis is force-extension of an intrinsically curved filament, determined by its particular fixed shape rather than by a thermal orientation entropy.

For an ensemble with quenched intrinsic curvature, first minimize each realization and then average the last expression, replacing $|{\kappa }_n{|}^2$ by its ensemble mean. For stationary bulk disorder, let $S_{\kappa }(q)$ be the power spectrum, normalized so that its correlation is $⟨{\kappa }_0(x){\kappa }_0(x+r)⟩=\int S_{\kappa }(q)e^{iqr}dq/(2\pi )$. The measurable mean deficit per unit projected length isIf the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the smooth-curvature high-force extension lawFor a stationary ensemble the integral is $L⟨({\kappa }_{0,x}{)}^2⟩$. Equivalently, the bulk slope is ${\zeta }_x\simeq -(B/F){\kappa }_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament. For example, ideal white intrinsic-curvature disorder with $S_{\kappa }(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $⟨\mathcal{L}-L⟩/L=S_0\sqrt{B}/(8\sqrt{F})$ in that ideal continuum model.

Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation areat each end. These can produce a free-end curvature boundary layer of width $\ell =\sqrt{B/F}$. For a constant intrinsic filament curvature $\kappa$, integration of the higher-order Euler-Lagrange equation with zero transverse force gives $B({\zeta }_x{)}_{xx}-F{\zeta }_x=0$ and henceThe last formula follows by integrating ${\sinh }^2[(x-L/2)/\ell ]$, and tends to ${\kappa }^2{\ell }^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.

For the freely jointed chain, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute temperature, and $\beta =1/(k_{B}T)$ with Boltzmann constant $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos \theta$. Its angular canonical partition function and the full chain partition function areDifferentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacementHere $\mathcal{L}$ is the Langevin function; this is force-extension of a three-dimensional freely jointed chain, in the fixed-force canonical ensemble. At high extension, $\coth \xi =1+2e^{-2\xi }+\cdots$, soThus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional freely jointed chain deficit falls as $F^{-1}$ and depends explicitly on temperature. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational entropy; thermal bending fluctuations of a stiff filament have not been included in the first calculation.

If the comparison chain is also restricted to two dimensions, it is a planar freely jointed chain rather than the three-dimensional model. In this case,The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are Modified Bessel functions of the first kind. At large $\xi$ the angular weight is proportional to $e^{-\xi {\theta }^2/2}$, giving $⟨\cos \theta ⟩\simeq 1-⟨{\theta }^2⟩/2=1-1/(2\xi )$. Consequently $Nb-⟨L⟩\sim N{k}_{B}T/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.

For a statistically homogeneous isotropic dimensionless field, write its two-point autocorrelation function of a random field as $C(r)$ andThe formal real-space scale invariance condition is $C(\lambda \mathbf{r})=C(\mathbf{r})$ for all $\lambda >0$. Changing variable $\mathbf{q}=\lambda \mathbf{k}$ in the first expression and equating the nonzero-mode Fourier transforms gives ${\lambda }^{-3}P(q/\lambda )=P(q)$, equivalently $P(\lambda k)={\lambda }^{-3}P(k)$. Taking $\lambda =k/k_0$ yieldsThus a scale-invariant inflationary power spectrum has equal variance per logarithmic wavenumber interval, rather than equal power per Fourier volume.

There is an important mathematical qualification to the literal covariance condition. A nonzero $k^{-3}$ power spectrum over all scales has $C(0)=A\int dk/k/(2{\pi }^2)$, divergent at both endpoints; it is not the covariance of a finite-variance ordinary field. Moreover, exact dilation invariance of a continuous isotropic $C(r)$ makes it constant for $r>0$ and, by continuity, at zero: its spectrum can then consist only of a zero-mode delta measure. The nontrivial cosmological statement consequently concerns nonzero modes, with cutoffs or subtraction of an unobservable constant. For example the finite subtracted covarianceis invariant under simultaneous rescaling of $r,r_0$. This is the precise infrared qualification of a scale-invariant covariance; a regulated unsubtracted $C(r)$ generally shifts by an additive constant under dilation. The formal derivation above gives the intended nonzero-mode scaling, with this qualification rather than an impossible finite-variance premise.

For a centered Gaussian random field, Wick's theorem says that every odd moment vanishes and every even moment is the sum over all pairings of products of two-point functions. For six fields there are $15$ pairings. In the connected three-point function at nonzero external momenta, each of the three external fields must pair with a distinct field at the cubic interaction. There are $3!=6$ such Wick contractions. Choosing the undifferentiated field gives three possibilities; interchanging the two differentiated fields gives a further factor of two. The remaining nine pairings involve an external-external pair and an internal pair and belong to tadpole/disconnected contributions. Define the background so the one-point function vanishes, or equivalently subtract these contributions.

It is useful to keep a coefficient $\mathcal{C}$ multiplying the cubic Hamiltonian: its literal printed value is $1$, whereas the standard dimensionally normalized curvature interaction has $\mathcal{C}=M_{\mathrm{Pl}}^2$. This distinction will matter for the final amplitude. Withand $dt=ad\tau$, $\dot{\zeta }={\zeta }^{\prime }/a$, the interaction entering the time integral isIn the interaction picture, the unequal-time vacuum contraction required by the in-in formalism isFor a differentiated internal field replace $u_k^{\ast }$ by $u_k^{\ast \prime }$. The equal-time power spectrum fixes its magnitude; the displayed free De Sitter curvature mode functions and vacuum choice fix its unequal-time phase.

Performing the three momentum integrations imposes ${\mathbf{p}}_a=-{\mathbf{k}}_a$ for each assignment and leaves one overall momentum delta function. Put $U=u_{k_1}(0)u_{k_2}(0)u_{k_3}(0)$, which is real here, and defineThe upper early-time contour ${\mathcal{C}}_{+}$ runs from $-\infty (1-i\delta )$ to $0$, with $\delta >0$. The Hamiltonian's minus sign and the six connected Wick contractions then giveThis is equivalently $\mathrm{Re}[-4i\mathcal{C}{ϵ}^{2}U\sum I_i]$ with unconjugated De Sitter curvature mode functions and the conjugate lower contour ${\mathcal{C}}_{-}$. Before momentum integration, the same result consists of the three cyclic delta assignments, each with the extra factor of two for the identical differentiated legs.

The PDF's intermediate formula writes only three cyclic assignments without that factor of two. Taken literally it undercounts the connected contractions. Its unconjugated modes must also use the conjugate contour, rather than the upper contour of the original in-in expression. Both points are required for a consistent in-in bispectrum conjugation rule.

For a statistically homogeneous comoving curvature perturbation, define the connected primordial bispectrum byStatistical isotropy makes $B$ depend only on the three magnitudes, which must form a triangle. A Gaussian random field has zero connected bispectrum. Canonical attractor single-field slow-roll inflation with the usual vacuum produces only slow-roll-sized primordial non-Gaussianity. A measurable signal can arise from additional light fields and nonlinear conversion of isocurvature perturbations, or from enhanced interactions such as a small sound speed, departures from an attractor, or suitable features/excited initial states. The resulting model must still reproduce the observed nearly scale-invariant power spectrum and remain under perturbative control, with acceptable backreaction and late-time conversion to the observed perturbations. Large non-Gaussianity is therefore possible but is not automatic in a viable model.

Put $c=3f_{\mathrm{NL}}/5$ and use the consistent Fourier transform convention with ${\zeta }_G(\mathbf{x})$, rather than the repeated momentum argument printed in the integrand. The quadratic part has transformThe subtraction sets the mean to zero and removes the internal contraction of a single quadratic factor. At first order in $f_{\mathrm{NL}}$, choose one of the three external factors to be quadratic. For example, the two connected Wick contractions at the third leg pair its two fields with the first two legs, givingSumming the three placements provesThis is the leading local primordial bispectrum. The factor $6/5$ is the quadratic coefficient $3/5$ multiplied by the two cross-pairings. Terms of order $f_{\mathrm{NL}}^2$ have an odd Gaussian moment and vanish.

The literal quadratic local model also has a connected cubic-in-$f_{\mathrm{NL}}$ contribution, from one quadratic factor at every leg:This is the loop correction to the local primordial bispectrum; regulators may be needed for idealized spectra. Thus the PDF's displayed formula is the tree/leading-order result, not an exact identity for arbitrary $f_{\mathrm{NL}}$. The weak-non-Gaussian expansion assumes these loop terms are small. The local shape is enhanced in the squeezed bispectrum configuration when a long-wavelength perturbation modulates small-scale power.

Let the medium's autocorrelation function of a random field and power spectrum use the explicit conventionSince $n^2-1=2\mu W+{\mu }^2{W}^2$, only the linear term in $\mu$ is needed to obtain the cross-section through second order:Squaring and taking the expectation givesThis is the finite-volume formula in the Born approximation for scalar wave scattering; the zero mean of $W$ does not make its squared scattered amplitude vanish. For a jointly centered Gaussian random field, the third-order contribution is zero, so the next perturbative contribution is fourth order when that expansion is valid.

Introduce the overlap volume $H_V(\mathbf{s})=|V\cap (V-\mathbf{s})|$. Changing variables rewrites the double integral asWhen the linear dimensions of a regular scattering volume are large compared with the correlation length, and $C$ is integrable, $H_V(\mathbf{s})\simeq |V|$ over the significant covariance range. This gives the finite-volume random scattering spectrum resultThere are two distinct approximations here: the weak-scattering expansion in $\mu$ and the negligible-boundary-overlap approximation in the volume-to-correlation-scale ratio. If $S$ instead denotes the power spectrum of the index fluctuation $n-1=\mu W$, then $S={\mu }^2{S}_W$ and the external factor ${\mu }^2$ is absorbed into $S$.

Statistical homogeneity alone allows an anisotropic spectrum, so its argument is generally the vector ${\mathbf{k}}_s$. If statistical isotropy is additionally assumed, $S_W$ depends only onThe scalar notation $S(k_s)$ is valid in that isotropic case. For the alternative convention $S_{\mathrm{norm}}=(2\pi {)}^{-3}{S}_W$, the bulk formula is $2\pi {\mu }^2{k}^4|V|S_{\mathrm{norm}}({\mathbf{k}}_s)$. Changing the sign of the scattering vector does not change the spectrum of a real stationary scalar field; changing the Fourier transform normalization does change the displayed $2\pi$ factors. Both conventions have therefore been stated explicitly.

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.

Write $K=k_1+k_2+k_3>0$ and $C=ϵ+1-c_s^2$. The inflationary scalar Fourier mode has real late-time value $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{c}_s{k}^3}$ and $u_k^{\prime \ast }(\tau )=A_k{c}_s^2{k}^2\tau e^{ik{c}_s\tau }$. Thus the time integrand, before the factor $-2i$, isThe Bunch-Davies vacuum prescription supplies damping in the past. Equivalently, use a positive regulator $d$ and then take the limit:The real part of $-2i$ times this integral is $4/(c_s^3{K}^3)$. Including the six Wick contractions yields the cubic time-derivative curvature bispectrumThe resulting primordial bispectrum is a homogeneous function of degree $-6$ in the wavenumbers, as expected for scale-independent background parameters.

The printed limit $c_s\ll 0$ is not a physical sound-speed limit for these modes. The normalization and positive-frequency choice assume $c_s>0$; a negative value is outside the positive sound-speed convention used by these modes. The likely intended limit is $0<c_s\ll 1$, which is analyzed explicitly here rather than silently substituted for the PDF.

To compare primordial non-Gaussianity, normalize by the power spectrum, not only by the raw three-point amplitude. HereAt an equilateral bispectrum configuration, the exact normalized amplitude isThus $c_s\to 1$ gives an amplitude of order $ϵ$, consistent with weak non-Gaussianity in single-field slow-roll inflation. For $0<c_s\ll 1$, this vertex gives an amplitude of order one, enhanced relative to the slow-roll limit but not diverging as $c_s^{-2}$ after power-spectrum normalization. At fixed $H,ϵ,M_{\mathrm{Pl}}$ the raw bispectrum does scale as $c_s^{-2}$, but $P_{\zeta }^2$ has the same scaling. The often-quoted larger small-sound-speed amplitudes of more general inflationary actions require their actual interaction coefficients and other vertices; they do not follow from the lone supplied vertex. These conclusions concern its tree contribution within the stated weak-perturbation approximation.

The bispectrum shape function isAt fixed perimeter, the arithmetic-geometric mean inequality bounds this by $1/27$, attained only at an equilateral bispectrum configuration. In a squeezed bispectrum configuration with $k_1=q\ll k_2\simeq k_3=k$, it behaves as $q/(8k)\to 0$. At a flattened bispectrum configuration $(k,k,2k)$ it is finite, $1/32$, with no boundary singularity. Thus the shape prefers equilateral triangles and suppresses squeezed triangles, with a broad finite flattened contribution. This is a shape preference, not a complete signal-to-noise calculation: actual observational weighting also depends on cosmological transfer functions, covariance and accessible triangles. If $C=0$, this vertex produces no bispectrum at any configuration; if $C<0$, the signed shape reverses while its absolute triangle preference is unchanged.

Evaluate all derivatives at zero long perturbation, where $\nu ={\delta }_c/{\sigma }_{G,s}$, and write $\bar{n}(\nu )$ for the unperturbed abundance. The peak-background split changes the halo peak height toThe chain rule in the specified bias expansion therefore givesNo explicit abundance law is specified, so these derivatives are the general answer. For example, if $\bar{n}$ has the Press-Schechter halo mass function dependence $\nu e^{-{\nu }^2/2}$, then $b_{10}=({\nu }^2-1)/{\delta }_c$ and $b_{01}=2f_{\mathrm{NL}}({\nu }^2-1)$.