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research28 min studyLesson 864 in Physics

Normalize a primordial mode before choosing its vacuum

Separate the canonical oscillator, expanding pump field, initial state, and late-time limit.

Before you begin
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The question for today

What you’ll understand

  • Write the canonical quadratic scalar action under explicit single-field assumptions.
  • Normalize a positive-frequency mode with its Wronskian.
  • Explain why horizon crossing is not an instantaneous exact freeze-out rule.

Inflationary fluctuation calculations quantize a constrained perturbation of matter and geometry, rather than treating a density contrast as an arbitrary oscillator. After solving lapse and shift constraints, a canonical mode has a time-dependent pump term. Its differential equation, commutator normalization, and initial quantum state are separate ingredients. A familiar analytic mode is useful in a slowly varying background, but it does not establish that every inflationary history has the same vacuum or that a growing canonical amplitude directly represents growing physical curvature.

Choose a regular single-field scalar variable

Assume general relativity, one minimally coupled canonical scalar field, a flat expanding background, positive small ε=−Ḣ/H², and a nonzero background field velocity. Define the comoving curvature variable ℛ as minus the spatial ψ on slices where the field perturbation vanishes. In a general scalar gauge it is −ψ−ℋδφ/φ′. Its spatial metric has the positive conformal factor exp(2ℛ); some sources define R with the opposite sign. Uniform-density ζ agrees at leading large-scale order for the regular adiabatic solution under the appropriate constraints.

The reduced quadratic action becomes canonical after setting v=zℛ, where z=a√(2ε)Mpl and Mpl is the reduced Planck mass. The overall sign choice for ℛ does not change this quadratic action or its power. Integrating by parts generates z″/z in the canonical equation. This term is background pumping, not an adjustable positive spring constant. Exact ε=0 makes this scalar clock definition degenerate; the familiar de Sitter-like equation is used as a controlled approximation to a regular rolling background.

S2=12∫dη d3x ⁣[(v′)2−(∇v)2+z′′zv2],v=zR,z=a2ϵ MPl,MPl−2=8πGS_2=\frac12\int d\eta\,d^3x\!\left[(v')^2-(\nabla v)^2+\frac{z''}{z}v^2\right],\qquad v=z\mathcal R,\quad z=a\sqrt{2\epsilon}\,M_{\rm Pl},\quad M_{\rm Pl}^{-2}=8\pi G
Natural units ℏ=c=1 are used. ℛ is dimensionless, z has mass units, and v is a canonical field. The action assumes the stated matter content and solved gravitational constraints.↔ Scroll wide equations horizontally.

Fix the commutator and then the initial state

Each Fourier mode obeys v″k+[k²−z″/z]vk=0. Promoting the field to an operator introduces creation and annihilation operators with the same Fourier measure used above. Canonical commutation requires a nonzero Wronskian normalization. That requirement alone allows Bogoliubov mixtures of positive and negative frequencies; it does not uniquely select a vacuum. An additional physical state prescription is therefore indispensable.

The Bunch–Davies prescription selects the positive-frequency Minkowski-like branch when a mode has an early adiabatic regime k² much larger than the pump term. It is a specified initial-state assumption, not a measurement showing that the entire universe began in one unique state. Finite-duration inflation or rapid background changes require care in applying an asymptotic boundary condition. Excited states must satisfy normalization and consistency conditions, including backreaction and the validity of the effective description.

vk′′+(k2−z′′z)vk=0,vkvk∗′−vk∗vk′=i,vk⟶e−ikη2k in an early adiabatic regimev_k''+\left(k^2-\frac{z''}{z}\right)v_k=0,\qquad v_kv_k^{*\prime}-v_k^*v_k'=i,\qquad v_k\longrightarrow\frac{e^{-ik\eta}}{\sqrt{2k}}\ \text{in an early adiabatic regime}
The Wronskian uses ℏ=1 and [ak,ak′†]=(2π)³δ³(k−k′). The positive-frequency boundary condition is additional to the commutator normalization.↔ Scroll wide equations horizontally.

Take the late-time limit in the correct variable

When z″/z is approximated by 2/η² with η negative, the normalized positive-frequency solution has a constant-frequency factor times one minus i/(kη). The canonical amplitude grows at small |kη|, while division by the expanding z can leave ℛ approximately constant in an attractor background. Evaluating the leading late-time spectrum with background quantities near k=aH is a slow-variation approximation; it does not say the exact amplitude at crossing already equals the final amplitude.

The oscillator lab compares phase evolution in a different canonical system. Reset and set damping ratio 0 and frequency 1, then Step twice to time 1; predict x=cos1 and velocity −sin1. Reset and set damping 0,frequency 2, then Step twice; predict x=cos2 and velocity −2sin2. This classical oscillator starts at x=1,v=0 and has positive constant stiffness. It contains no quantum vacuum, Wronskian readout, expanding pump, or primordial mode solver.

z′′z≃2η2,vk=e−ikη2k(1−ikη),∣vk∣2a2≃H22k3(1+k2η2)(a≃−1Hη)\frac{z''}{z}\simeq\frac2{\eta^2},\qquad v_k=\frac{e^{-ik\eta}}{\sqrt{2k}}\left(1-\frac{i}{k\eta}\right),\qquad \frac{|v_k|^2}{a^2}\simeq\frac{H^2}{2k^3}(1+k^2\eta^2)\quad\left(a\simeq-\frac1{H\eta}\right)
These formulas use the leading quasi-de Sitter pump approximation, not a regular scalar clock at exactly ε=0. The last expression approaches a constant only as |kη| becomes small.↔ Scroll wide equations horizontally.

Work it through

Compare crossing with a later amplitude

Use the displayed approximate mode with k=2,H=1,a=−1/η in common natural units. Define χ=v/a. Find |χ|² at η=−1/2 and −1/4, and its asymptotic late-time value.

  1. At η=−1/2, kη=−1 and |v|²=[1+1/(kη)²]/(2k)=2/4=1/2. Since a=2, |χ|²=(1/2)/4=1/8=0.125.

  2. At η=−1/4, kη=−1/2 and |v|²=5/4. Now a=4, so |χ|²=(5/4)/16=5/64=0.078125, despite the larger canonical amplitude.

  3. The late-time value is H²/(2k³)=1/16=0.0625. The crossing value was twice this limit. The example uses the approximate pump solution and does not set a rolling scalar clock’s ε exactly to zero.

The result

The rescaled mode powers are 0.125 and 0.078125, approaching 0.0625; canonical growth does not imply growing late-time curvature.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Compare two fixed-frequency phases

Reset and set ζ=0,ω₀=1, then Step twice to t=1; predict x=cos1,v=−sin1. Reset and set ζ=0,ω₀=2, then Step twice; predict x=cos2,v=−2sin2.

Compare two fixed-frequency phasesUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis classical positive-stiffness oscillator has no expanding pump term, quantum vacuum, or inflationary mode calculation.

Pause & reason

Check your understanding

What does the canonical Wronskian condition fix by itself?

Your turn

Try it yourself

Relative to a normalized positive-frequency basis, a candidate mode has real Bogoliubov coefficients α=5/4 and β=3/4. Does it satisfy |α|²−|β|²=1? Does that make it the same vacuum choice as α=1,β=0?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Check the difference of the squared coefficients; normalization and vacuum selection are distinct.

Compare with the solution

The difference is 25/16−9/16=1, so the canonical normalization is satisfied. The nonzero negative-frequency coefficient still describes a different state relative to the chosen basis. This algebra alone does not establish acceptable backreaction or a physically justified early-time preparation, and it does not equate that state with the positive-frequency vacuum.

Keep the idea

Take it with you

  • Canonical normalization and initial-state selection are separate steps.
  • The pump approximation requires a specified rolling-background regime.
  • A rescaled physical mode can freeze while its canonical amplitude grows.

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