Let $g_E$ solve the trapped stationary equation in the question and useWriting $y=x-X$, the harmonic potential splits as $(y+X{)}^2/2=y^2/2+Xy+X^2/2$. Matching the coefficient of $g_E^{\prime }$ and then the linear and constant coefficients of $g_E$ givesThus the soliton center obeys $\ddot{X}=-X$. These are exactly Hamilton's equations for the classical Hamiltonianwith $p=u$, while the extra phase evolves by $-E$ plus the classical Lagrangian $(u^2-X^2)/2$.

Let $y=x-X(t)$ and substituteSeparating the coefficient of $f_E^{\prime }$ and the constant and linear coefficients multiplying $f_E$ givesThis is the Galilean boost of the stationary soliton.

Write $b=\sqrt{2|E|}$. Its conserved squared norm isUsinggivesThereforeThe moving soliton behaves like a classical particle of inertial mass $N(E)$, with negative internal binding energy $-N(E{)}^3/24$.

Put $\mu =|E|=-E>0$ and seek $\varphi =e^{-iEt}f(x)$ with real $f$. The equation becomesFor $b=\sqrt{2\mu }$, the identityshows thatsatisfies the stationary equation. Hence $e^{-iEt}{f}_E(x)$ is the bright soliton of the focusing nonlinear Schrödinger equation.

Write ${\mathbf{x}}_j={\mathbf{x}}_0+({\eta }_0-{\eta }_j)\hat{\mathbf{n}}$. The delta functions in $g$ evaluate the local last-scattering sources, while the piecewise $e^{-\tau }$ weights the integrated gravitational source. ThusThe first two lines are weighted Sachs-Wolfe combination and Doppler sources at the two last-scattering surfaces; the final lines are the corresponding Integrated Sachs-Wolfe effect.

To predict these quantities from primordial perturbations, one solves the coupled linear Einstein field equations and Boltzmann equation hierarchy for photons, baryons, cold dark matter, and neutrinos, together with the ionization history that determines $\Gamma$ and hence $g$. The solutions provide ${\Theta }_0$, ${\mathbf{v}}_e$, $\Phi$, and $\Psi$ at both surfaces and along the line of sight.

The cosmological visibility function is the probability density for last scattering, so the stipulated two instantaneous populations giveSince $g={\partial }_{\eta }{e}^{-\tau }$ and $e^{-\tau }$ vanishes before the first transparent epoch,Its plot is a nondecreasing step function: it jumps from zero to $p$ at recombination and from $p$ to one at ${\eta }_1$. The cosmological optical depth interpretation is that $e^{-\tau (\eta )}$ is the probability that a photon present at time $\eta$ reaches the observer without any later scattering.

Let $\hat{\mathbf{n}}$ point from the observer toward the emission point, so the photon propagation direction is $\hat{\mathbf{p}}=-\hat{\mathbf{n}}$ andAlong this path $D=d/d\eta ={\partial }_{\eta }-\hat{\mathbf{n}}\cdot \nabla$. Substituting $\hat{\mathbf{p}}=-\hat{\mathbf{n}}$ into the Photon Boltzmann equation with Thomson scattering givesBecause ${\partial }_{\eta }{e}^{-\tau }=\Gamma e^{-\tau }$, multiplication by the integrating factor giveswhereIntegrating along the ray and using $e^{-\tau }\to 0$ at sufficiently early time yields the Cosmic microwave background line-of-sight solution

For Gaussian linear perturbations, a connected tree on five external density insertions has four Wick-contraction edges and therefore containsThus $n=4$. The three unlabeled tree-level connected matter five-point function topologies correspond to the perturbative-order partitionsThey are respectively an $F_4$ star with four $F_1$ leaves, an $F_3$ vertex joined to an $F_2$ vertex with three $F_1$ leaves, and a chain with three $F_2$ vertices and two endpoint $F_1$ vertices.

For one labeling of the last topology, let ${\mathbf{k}}_{12}={\mathbf{k}}_1+{\mathbf{k}}_2$ and ${\mathbf{k}}_{123}={\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3$. Its algebraic contribution isThe factor $2^3=8$ counts the interchange of the two linear legs at each symmetric $F_2$ vertex. The full correlator adds all distinct permutations of external labels and the overall momentum-conserving delta function.

The ultraviolet softness of the second-order density kernel requiresfor fixed $\mathbf{k}$ as $q\to \infty$. In this limit the two hard vectors are opposite,The constant part of the stated kernel is thereforeIt must vanish, givingThis recovers the standard standard perturbation theory density kernel $F_2$.

Let $\theta =\nabla \cdot \mathbf{v}$. Vanishing vorticity implies in Fourier spaceFourier transforming $\nabla \cdot (\delta \mathbf{v})$ in the continuity equation giveswith the alpha mode-coupling kernelTaking the divergence of $(\mathbf{v}\cdot \nabla )\mathbf{v}$ and symmetrizing its two velocity arguments giveswhere the beta mode-coupling kernel is

Expand the quantized field using the normalized modeThe canonical commutation relation fixes the Wronskian $M_{\mathrm{Pl}}^2{z}^2(u_k{u}_k^{\ast \prime }-u_k^{\ast }{u}_k^{\prime })=i$. Writing $x=k\eta$, the primed correlators areEach unprimed expression includes $(2\pi {)}^3{\delta }^{(3)}(\mathbf{k}+{\mathbf{k}}^{\prime })$. The imaginary part of the mixed correlator givesConsequently the classicality parameter of a cosmological perturbation isIt tends to zero on superhorizon scales, so the perturbation becomes effectively classical even while its amplitude grows as $a^3$. In ordinary slow-roll inflation, $\mathcal{R}$ instead freezes to a constant; its analogous normalized commutator also vanishes, but behaves as $|k\eta |$ for the leading de Sitter mode rather than $|k\eta {|}^3$.

Varying the quadratic action givesFor a Fourier mode,At late times $ϵ\propto a^{-6}$, soThe equation becomes ${\mathcal{R}}_k^{\prime \prime }+4{\mathcal{R}}_k^{\prime }/\eta +k^2{\mathcal{R}}_k=0$. At $k=0$, its independent solutions are a constant and ${\eta }^{-3}$. The leading late-time solution in the form given in the question is thereforeThis is the growing curvature perturbation in ultra-slow-roll inflation.

The second Friedmann equation gives the first Hubble slow-roll parameterBecause $\dot{\bar{\varphi }}\propto a^{-3}$,ThusAs $a\to \infty$, the kinetic energy decays as $a^{-6}$ and the constant $\Lambda$ dominates. Hence $H$ approaches a constant,This non-attractor background is ultra-slow-roll inflation.

The massless field equation isFor a homogeneous solution this becomesThereforeSince $\dot{\bar{\varphi }}={\bar{\varphi }}^{\prime }/a$, the cosmic-time velocity decays as $a^{-3}$.

The exchange graph has one quartic vertex attached to ${\mathbf{k}}_1,{\mathbf{k}}_2,{\mathbf{k}}_3$, a second attached to ${\mathbf{k}}_4,{\mathbf{k}}_5,{\mathbf{k}}_6$, and an internal line of momentumSchematically,The Schwinger-Keldysh conjugation relation gives

Set ${\eta }_0=0$ only in integration limits and phases, retaining its leading explicit power from the six external propagators. The scale factors, six external propagators, and one internal propagator leave the common coefficientFor two right-branch vertices, split the time-ordered integral into ${\eta }_1>{\eta }_2$ and ${\eta }_2>{\eta }_1$:The two right-vertex factors contribute the compensating minus sign, soFor one vertex on each branch, the two integrals factorize:Thus the quantities requested in the question areAt late time these expressions are real. Summing all four in-in formalism assignments gives

Forthe specified cosmological bulk-to-boundary propagator isThe right-right cosmological bulk-to-bulk propagator is time ordered:The mixed Wightman function in the order stated in the question isThe suppressed factor in each expression is $(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}+{\mathbf{p}}^{\prime })$.

Varying the scalar action and Fourier transforming the spatial coordinates gives the Klein-Gordon equationIn de Sitter spacetime, $a=-1/(H\eta )$ and $m^2=2H^2$, soDirect substitution shows that $\eta e^{-ik\eta }$ and $\eta e^{ik\eta }$ are independent solutions. HenceThe mass value is the one that gives a conformally coupled scalar field in four-dimensional de Sitter spacetime.

For a canonical scalar field with positive kinetic energy, the Friedmann equation givesso the first Hubble slow-roll parameter satisfies $ϵ\ge 0$. The tensor tilt is consequentlyA blue primordial spectrum of tensor modes cannot arise in standard canonical single-field slow-roll inflation; it requires changing assumptions, such as the matter content, kinetic structure, initial state, or gravitational dynamics.

The scalar tilt instead obeysIt can be positive in a standard single-field model if $\eta <-2ϵ$, meaning that $ϵ$ decreases sufficiently rapidly with the number of e-folds. Equivalently, in potential slow-roll parameters the first-order condition is $2{\eta }_V>6{ϵ}_V$. Thus a blue scalar spectrum is allowed by the framework, although it is not produced by every slowly rolling potential.

At horizon exit, the primordial tensor power spectrum is proportional to $H^2$. ThereforeThusThe ratio of the stated tensor and scalar spectra isEliminating $ϵ$ gives the single-field inflation consistency relation

The right-hand side of the slow-roll curvature power spectrum must be evaluated separately for each mode at cosmological horizon exit, $k=aH$. Taking logarithms givesFor $N=\log a$,HenceThis is the scalar spectral index in Hubble slow-roll parameters.

The creation and annihilation operators obey the canonical bosonic commutation relationsSince $\delta \varphi =\hat{f}/a$, the vacuum two-point correlation function isForone has $|f_k{|}^2=(1+1/(k^2{\tau }^2))/(2k)$. Comparison with the definition of the dimensionless power spectrum giveswhere $a=-1/(H\tau )$. On superhorizon scales, $k\ll aH$ or $|k\tau |\ll 1$, and therefore