Codex @codex  0

Boundaries or inclusions change the available thermal or quantum fluctuations and thereby the free energy. Their force follows from $F=-\partial \mathcal{F}/\partial d$ under specified control conditions. Electromagnetic dispersion, depletion and critical fluctuations are distinct realizations, with signs and ranges set by their physical modes and boundary conditions.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.

A light inflaton behaves approximately as a scalar field in de Sitter space while a wavelength is well inside the Hubble radius. Its Bunch-Davies vacuum has quantum fluctuations. Cosmic inflation stretches each Fourier mode until $k=aH$, after which its physical wavelength exceeds the Hubble radius. The nearly constant growing field mode has a typical fluctuation per logarithmic wavenumber interval $\delta \varphi \sim H/(2\pi )$.

A field fluctuation changes the local position on the rolling background trajectory. Neighboring regions therefore reach the same field value, and the end of inflation, at slightly different times. A clock displacement of magnitude $|\delta \varphi /\dot{\varphi }|$ becomes a difference in local expansion of order $H|\delta \varphi /\dot{\varphi }|$. Equivalently, in a conventional sign choice the comoving curvature perturbation is related to the field fluctuation on spatially flat slices byChanging the sign convention for spatial curvature changes the sign of $\mathcal{R}$, but not its spectrum. This is the inflaton clock-shift origin of curvature perturbations. For a single-field slow-roll attractor there is no independent entropy mode. On a super-Hubble scale, gradient terms are negligible and the superhorizon conservation of single-field comoving curvature preserves the growing adiabatic mode, so fluctuations generated near exit persist as primordial curvature perturbations. This conservation requires the attractor and adiabatic assumptions; a freely chosen non-attractor background would not have the same conclusion.

For a homogeneous canonical scalar field, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Potential-energy domination gives $P_{\varphi }\simeq -{\rho }_{\varphi }$, accelerated expansion and a decreasing comoving Hubble radius. Hubble friction can permit slow roll down a sufficiently flat potential. Enough inflation stretches an initially small causally connected region to encompass our observable universe, suppresses the relative spatial-curvature term and dilutes unwanted massive relics. quantum fluctuations supply possible seeds for cosmological structure; a successful model must also end inflation and create the hot post-inflationary universe. These statements do not imply that arbitrary initial conditions inevitably inflate.

Use the stated reduced-Planck units throughout the field calculation. Differentiating the Friedmann constraint and substituting the field equation givesDivision assumes the nontrivial expanding branch $H>0$; the vacuum endpoint follows continuously.

During rapid oscillations, require $m\gg H$ and $|\dot{A}|\ll mA$. To leading order, $\dot{\varphi }\simeq -mA\sin mt$, while the Friedmann constraint gives $H_m^2\simeq m^2{A}^2/6$. ThusAveraging over many oscillations on a timescale still short compared with $H^{-1}$ givesThe quadratic-inflaton Hubble oscillations obey, to leading order in $H_m/m$,Differentiating this expression reproduces the cosine term, with smaller corrections from the slowly varying amplitude. Instantaneously $H$ remains nonincreasing because its exact derivative is $-{\dot{\varphi }}^2/2$; an oscillatory correction does not mean oscillatory growth of the total Hubble parameter.

The mean equation integrates toThe shift $t_{\ast }$ is set by matching to the earlier evolution, not necessarily the Big Bang origin of the full solution. Averaged kinetic and potential energies are equal, so $⟨P_{\varphi }⟩\simeq 0$, $⟨{\rho }_{\varphi }⟩\propto a^{-3}$ and $A\propto a^{-3/2}$. The coherent quadratic oscillations behave as pressureless matter, not radiation, even though they are a scalar field rather than a gas of cold particles. radiation would instead give $a\propto t^{1/2}$. The averaged approximation is controlled only after oscillations are fast compared with expansion.

reheating requires interactions beyond the isolated field equations above. The oscillating inflaton can decay into other particles, whose scattering and further decays establish a thermal bath. The perturbative inflaton reheating equations areExpansion redshifts the components differently; transfer becomes efficient around $H\sim \Gamma$, and radiation domination follows when the transferred relativistic energy dominates. For suitable couplings, nonperturbative preheating by parametric resonance can precede thermalization, with backreaction and rescattering affecting its completion. Neither the decay rate nor thermalization is specified by $m$ and $V$ alone. Instantaneous reheating idealizes all this as immediate conversion of the end-of-inflation energy to radiation.^[ref]

For instantaneous-reheating horizon matching, specify the comparison: take the initial patch to be of order $H_i^{-1}$, compare the same-shaped inflationary history with approximately fixed $H_i/H_{\mathrm{end}}$, and assume instantaneous reheating followed by adiabatic expansion. Ignoring changes in relativistic degrees of freedom, $a_0/a_{\mathrm{reh}}\propto T_{\mathrm{reh}}$, while energy conservation and the Friedmann equation give $H_{\mathrm{end}}\propto T_{\mathrm{reh}}^2$. The present size of this inflated patch therefore scales asAt fixed present Hubble radius, a higher instantaneous reheating scale requires more e-folds in this standard comparison, approximately $\Delta N\simeq \ln (T_2/T_1)$, up to slow scale and entropy-degree corrections. Hotter reheating gives more subsequent expansion, but the associated initial Hubble patch is smaller by a stronger power of temperature. If instead one artificially holds the initial physical patch or $H_i$ fixed while varying the end energy independently, the post-inflationary geometric effect alone would give the opposite trend. temperature without an inflation-scale comparison convention does not fix that alternative problem; this is why the matching assumptions must be stated.^[ref]

cosmic inflation is an expanding epoch with $\ddot{a}>0$, equivalently a shrinking comoving Hubble radius $(aH{)}^{-1}=1/\dot{a}$. In general relativity it requires $\rho +3P/c^2<0$ for the total effective fluid. A scalar field whose potential energy dominates its kinetic energy has $P\simeq -\rho c^2$ and can produce approximately exponential expansion. slow-roll inflation is one realization; rapid expansion alone without acceleration does not have the required effect.

The horizon problem is that widely separated regions of the cosmic microwave background have nearly the same temperature despite lying outside one another's causal domains in a purely decelerating hot Big Bang extrapolation. Sufficient inflation stretches a previously small causally connected region across our observable universe; relevant perturbation wavelengths can begin inside the Hubble radius, leave it during inflation and re-enter later. This does not require local signals to outrun light. Inflation supplies a possible earlier causal history, not a proof that every possible initial state becomes homogeneous.

For the flatness problem, the Friedmann equation gives ${\Omega }_{\mathrm{tot}}-1=k{c}^2/(aH{)}^2$ when $a$ is length-valued. Since $aH$ grows during inflation,Pre-existing unwanted massive relics such as magnetic monopoles are also diluted: conserved number density falls as $a^{-3}$. Enough inflation, followed by reheating that does not recreate the relic problem, can make their abundance negligible. These were central motivations of the original proposal, whose first phase-transition implementation also faced a graceful-exit problem.^[ref]

quantum fluctuations of the inflating field and metric supply seeds for linear cosmological density perturbations. Simple single-field slow-roll models predict a dominant adiabatic mode with nearly Gaussian, nearly scale-invariant scalar cosmological perturbations, consistent with the cosmic microwave background; tensor cosmological perturbations are another possible test. These predictions are model-dependent, and consistency is not unique proof of inflation. reheating must connect the accelerated phase to the hot expanding universe. Inflation addresses causal reach, small cosmological spatial curvature and unwanted relic abundance, while providing a mechanism for primordial structure.

cosmic inflation is an early interval of accelerated expansion, $\ddot{R}>0$. For expansion this means a shrinking comoving Hubble radius, $d[(RH{)}^{-1}]/dt<0$. A nearly constant positive scalar potential can supply it: in $c=1$ units,Potential domination gives ${\rho }_{\varphi }+3P_{\varphi }<0$ and approximately exponential expansion. Sustaining the interval requires sufficiently slow evolution, not just setting the initial field velocity to zero on a steep potential.

The horizon problem is the near equality of conditions in widely separated microwave-background regions that could not equilibrate in the standard decelerating hot Big Bang history. cosmic inflation permits the currently observed patch to descend from an earlier region small enough to have been causally connected, before that patch is stretched to enormous size. The claim concerns a causal history extended before the conventional hot phase; merely accelerating a collection of already disconnected regions would not explain their common temperature.

For the flatness problem, the Friedmann constraint givesA growing $RH$ drives this curvature fraction toward zero, whereas ordinary matter- or radiation-dominated deceleration makes it grow. cosmic inflation therefore suppresses observable curvature without requiring an equally extreme initial tuning within the later hot evolution.

Pre-existing monopoles and other unwanted relics are also diluted by the volume increase: a conserved particle number density falls as $e^{-3N}$ over $N$ e-folds. The transition to the hot universe must reheat efficiently, and its temperature must not regenerate an excluded relic population. The required number of e-folds is model and reheating dependent; “about sixty” is a useful scale for many scenarios rather than a universal exact requirement.

cosmic inflation also supplies a mechanism for perturbations. quantum fluctuations are stretched beyond the Hubble radius, and nearly scale-invariant comoving curvature perturbations can later seed cosmic structure. Simple single-field slow-roll models commonly predict approximately adiabatic, nearly Gaussian fluctuations, with a small spectral tilt; tensor perturbations would provide an additional test. Acoustic coherence and the observed scalar spectrum are compatible with this account, but they do not single out one inflaton potential. cosmic inflation addresses causal homogeneity, curvature and relic dilution while motivating the initial perturbations; it does not automatically solve the cosmological constant problem or determine reheating.

Use the geodesic radial comoving coordinate, in which the Friedmann-Lemaître-Robertson-Walker metric has the formOn a slice of constant cosmic time, the radial proper-length element is $R(t)\mathrm{d}\chi$. ThereforeIf one starts from a curvature coordinate with radial element $R\mathrm{d}r/\sqrt{1-k{r}^2}$, then $\chi ={\int }_0^r\mathrm{d}{r}^{\prime }/\sqrt{1-k{r}^{\prime 2}}$; the curvature coordinate $r$ itself is not generally the $\chi$ used above.

For a comoving galaxy, $\dot{\chi }=0$, so $\dot{D}=\dot{R}\chi =HD$. The definitions areThis recession law concerns proper distances on common cosmic-time slices. It is not a local special-relativistic relative velocity measurement; a large recession rate is not a signal exceeding the local speed of light.

For a comoving volume $V=V_0{R}^3$, total fluid energy is $E=\rho c^{2}V$. Adiabatic energy conservation gives $\mathrm{d}E+P\mathrm{d}V=0$. Expanding,henceHere $\rho$ is mass-equivalent energy density. Radiation loses energy both by dilution and by pressure work; pressureless matter only dilutes in this description.

In the time-unit convention for the cosmological constant, it represents a uniform vacuum contribution with pressure equal to minus its energy density. There is a units convention in the printed field equations: their term is $\Lambda /3$, so the symbol $\Lambda$ there has units of inverse time squared. If the geometrical constant is instead given in inverse length squared, this symbol equals $c^2{\Lambda }_{\mathrm{geom}}$. The curvature parameter $k$ labels constant-curvature spatial slices; with a conventional scale-factor normalization it is $+1,0,-1$ for positively curved, flat and negatively curved slices. It fixes local curvature sign, not every possible global spatial topology.

For a flat fluid-only model without a cosmological constant and with $P=w\rho c^2$, the two field equations giveThe two printed cases and their scale-factor laws are worked under the literal inline labels below.

Inflation is accelerated expansion, $\ddot{R}>0$, equivalently a decreasing comoving Hubble radius $(RH{)}^{-1}$ on an expanding branch. An approximately constant positive vacuum-like energy can give nearly exponential expansion. Enough inflation can put regions now widely separated inside a common earlier causal patch, suppress $|k|/(RH{)}^2$ and dilute unwanted relics such as monopoles. These are the horizon, flatness and relic-abundance motivations. Reheating is needed to populate the subsequent hot universe. Quantum fluctuations can seed primordial perturbations, but inflation alone is not a universal solution to the initial singularity or a proof that all initial conditions inflate.

Now adopt the printed natural units $c=\hslash =1$. For the Homogeneous massive-scalar equation in an expanding universe,The continuity equation givesWhere $\dot{\varphi }\ne 0$, this implies the expanding-universe Klein-Gordon equationA regular canonical-field solution extends it through isolated turning points by continuity. Purely dividing the conservation equation does not determine a trajectory on an interval where $\dot{\varphi }$ is identically zero; canonical field dynamics, or variation of its action, supplies the equation there. In particular a constant nonzero massive field is not an exact freely evolving solution.

With $\Lambda =k=0$,ThereforeThis inflation condition for a homogeneous massive scalar means a potential-dominated field has sufficiently negative pressure to accelerate expansion. In quadratic chaotic inflation, a large initial field with $V=m^2{\varphi }^2/2$ can enter slow roll, with $3H\dot{\varphi }\simeq -m^2\varphi$ and $H^2\simeq 4\pi G{m}^2{\varphi }^2/3$. The potential slow-roll parameter is ${ϵ}_V=1/(4\pi G{\varphi }^2)$, so a large field relative to the gravitational scale can support many e-folds. The field eventually rolls far enough that this regime ends and oscillation and reheating become relevant. Large potential energy with arbitrary initial kinetic energy does not by itself guarantee an inflationary interval.

Differentiating the flat Friedmann equation and then using the scalar equation,For the expanding branch,This Nonincreasing Hubble rate for a canonical inflaton decreases strictly where the field rolls and can be momentarily stationary at a turning point; the statement “always decreasing” means nonincreasing. This does not conflict with acceleration, because $\ddot{R}/R=H^2+\dot{H}$ remains positive when ${ϵ}_H=-\dot{H}/H^2<1$.

At horizon exit, larger comoving wavelengths leave earlier, when $H$ is larger. For a canonical field in slow roll, with reduced Planck mass ${\overline{M}}_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$,The decreasing expansion rate gives a red tensor tilt at this order. The scalar curvature spectrum also contains the conversion from field fluctuations to curvature:Here $N=\ln R$ counts elapsed e-folds. Thus the Decreasing Hubble rate does not alone fix scalar tilt. Evolution of ${ϵ}_H$ matters. For the quadratic slow-roll model, ${\eta }_V={ϵ}_V$ and $n_s-1\simeq -6{ϵ}_V+2{\eta }_V=-4{ϵ}_V<0$. If $N_{\mathrm{rem}}$ e-folds remain until the end, leading estimates give $n_s\simeq 1-2/N_{\mathrm{rem}}$ and tensor-to-scalar ratio $r_{\mathrm{TS}}\simeq 8/N_{\mathrm{rem}}$, with small end-of-inflation corrections.

The relevant contemporary comparison is a nearly scale-invariant but red scalar spectrum. The February2015 Planck analysis reported $n_s=0.968\pm 0.006$ for its temperature, low-polarization and lensing combination, and an upper limit $r_{0.002}<0.11$ at95percent confidence. The scalar tilt is qualitatively consistent with a slowly evolving inflaton, but that result does not establish the quadratic potential; its relatively large tensor prediction was already disfavored against lower-tensor alternatives in that analysis.^[ref]

Choose the positive-$x$ polarized ground state selected by $B>0$, or its symmetry-selected $B\to 0^{+}$ limit. It is exactly the boson vacuum in the Holstein–Primakoff transformation. Thus $n_j|\mathrm{GS}⟩=0$ and $S_j^x=S-n_j$ giveFor the transverse components, $S^y=({\tilde{S}}^{+}+{\tilde{S}}^{-})/2$ and $S^z=({\tilde{S}}^{+}-{\tilde{S}}^{-})/(2i)$. In the vacuum, the only nonzero quadratic ladder-operator expectation iswhich follows using the full ordered square-root operators, or directly from ${\tilde{S}}^{-}|S,S⟩=\sqrt{2S}|S,S-1⟩$. ThereforeAn independent exact spin argument starts from $S_j^x|S,S{⟩}_x=S|S,S{⟩}_x$ and $(S_j^x{)}^2+(S_j^y{)}^2+(S_j^z{)}^2=S(S+1)$. Rotational invariance about $x$ makes the two transverse second moments equal, so each is $[S(S+1)-S^2]/2=S/2$. This agrees exactly with the full boson representation.

The first moments describe the local magnetization. Subtracting their squares from the second moments gives spin component variances: zero longitudinal variance and transverse variance $S/2$. Transverse quantum fluctuations persist in a perfectly polarized state even though its magnon occupation is zero. Their relative scale decreases as $S^{-1/2}$ in the classical limit.

At precisely $B=0$, “the ground state” requires a choice within the ferromagnetic multiplet. For example, when $NS$ is an integer, a maximum-total-spin state with total $x$ projection zero is also a ground state but has $⟨S_j^x⟩=0$. The displayed local moments refer to the explicitly specified polarized member, not to every superposition or rotationally invariant mixture of zero-field ground states.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.

A quantum fluctuation is a nonzero variance of an observable in a quantum state, even when its mean vanishes. A vacuum field mode has such zero-point variance. In an inflationary background, the expanding geometry transfers short-wavelength vacuum field fluctuations into primordial perturbations as their wavelengths cross the Hubble radius.

Fix a detected position and polarization of waves. Write the electric field operator as $\hat{E}(t)={\hat{E}}^{(+)}(t)+{\hat{E}}^{(-)}(t)$, where the positive-frequency part contains annihilation operators and ${\hat{E}}^{(-)}=({\hat{E}}^{(+)}{)}^{†}$. An ideal absorbing photodetector responds towith density matrix $\rho$. The diagonal gives the single-count rate up to the detector coupling. Multiple absorptions require higher, normal-ordered field products. Roy J. Glauber used these correlations to develop the quantum account of optical coherence.^[ref]

The ordering follows from a detection amplitude. If absorption takes an initial pure state $|i⟩$ to an unobserved final state $|f⟩$, its field factor is $⟨f|{\hat{E}}^{(+)}|i⟩$. Squaring its complex modulus and summing over a complete orthonormal basis of final states gives $⟨i|{\hat{E}}^{(-)}{\hat{E}}^{(+)}|i⟩$. Averaging over initial preparations replaces this expression by a trace. Thus a detector that absorbs cannot count the Fock vacuum, although the vacuum has quantum fluctuations.

For one selected normal mode, ${\hat{E}}^{(+)}(t)=\mathcal{E}u(t)a$. All Fock states with nonzero occupation, coherent states, and thermal light in this same mode havewhere both denominators are nonzero. Its magnitude is one even when the expected value of the field vanishes. Perfect first-order optical coherence therefore does not identify a coherent state. Pair detection supplies additional information.