Vacuum fluctuations: ⟨E²⟩ and its spectrum

SI units throughout; where a four-vector appears the metric is (+, −, −, −). Quoted from topic 11.03: E(x, t) = i Σₖₛ √(ħωₖ/(2ε₀V)) [aₖₛêₖₛ e^{(ik⋅x − iωₖt)} − aₖₛ†êₖₛ* e^{(−ik⋅x + iωₖt)}], with unit polarisation vectors êₖₛ ⊥ k in a periodic box of volume V. Quoted from Unit 9: each mode is an oscillator whose ground state has energy ½ħω. Since aₖₛ|0⟩ = 0 and ⟨0|aₖₛ† = 0, ⟨0|E|0⟩ = 0: derived. In ⟨0|E⋅E|0⟩ every product with an a on the right or an a† on the left vanishes, leaving ⟨0|aₖₛaₖ'ₛ'†|0⟩ = δₖₖ'δₛₛ', and êₖₛ⋅êₖₛ* = 1 gives Σₖₛ ħωₖ/(2ε₀V), the same at every x and t: derived. The same steps give c²⟨B²⟩ = ⟨E²⟩. Check: one mode's electric energy (ε₀V/2)⟨E²⟩ₖₛ is ħω/4, its magnetic energy another ħω/4, total ½ħω, Unit 9's zero point. In the one-mode limit the variance is ħω/(2ε₀V), falling as 1/V: a bigger box gives each mode a smaller share and holds more modes to share it.

A periodic box holds V d³k/(2π)³ wavevectors in d³k; with two polarisations and ω = ck that is Vω² dω/(π²c³) modes between ω and ω + dω. Each brings ħω/(2ε₀V), so ⟨E²⟩ = ∫ dω ħω³/(2π²ε₀c³): derived. Dimensions: ħω³ dω/(ε₀c³) is (J s⁻³)/(F m² s⁻³) = J/(F m²) = V² m⁻², and V has cancelled, as it must for a local quantity. The ω² mode density is the one in Planck's law, which the COBE FIRAS spectrum of the microwave background confirms to 50 parts per million of its peak; times ε₀, the spectrum is exactly the ½ in Planck's (n̄ + ½)ħω³/(π²c³). This is the zero-point spectrum Quantum Mechanics II assumed for Welton's estimate, now a consequence of quantising the field; Unit 13 puts it to use. Its shape matters more than its normalisation. The share of the variance below ω grows as ω⁴, so half of everything up to a cutoff ω_c comes from above 0.841ω_c: whatever cutoff is chosen controls the answer.

Keep the modes up to ω_c = cK and the integral gives ⟨E²⟩ = ħcK⁴/(8π²ε₀), derived. The number of modes grows as ω² and each one's share as ω, so nothing stops it. Dimensions say it had to. ħ, c and ε₀ combine into ħc/ε₀, in V² m², with no length left over, so the variance at a point can only be ħc/ε₀ times the fourth power of a wavenumber supplied from outside. A crystal has such a number, its lattice spacing, which ends the phonon spectrum at the Debye frequency; the free Maxwell field has none. Each halving of the shortest wavelength kept multiplies the r.m.s. field by four, from 2.65 × 10⁵ V m⁻¹ with every mode down to 1 µm to 4.5 × 10¹⁶ V m⁻¹ down to the electron's Compton wavelength, √(α/2π) times the Schwinger field m²c³/(eħ). A cutoff is not a property of the vacuum. It stands for whatever limits the question, the size of a probe, the bandwidth of a detector or the scale where the free-field model stops, and an answer that depends on it asked about the field at a point.

No instrument reads E at a point. Let one register Ē = ∫f(x)E(x) d³x with a Gaussian weight f = (πL²)^(−3/2) e^{(−|x|²/L²)}, normalised to 1. Each mode is then multiplied by f's Fourier transform, e^{(−k²L²/4)}, so ⟨Ē²⟩ is ħc/(2π²ε₀) times the integral of k³ e^{(−k²L²/2)} over k from 0 to ∞, which is 2/L⁴: ⟨Ē²⟩ = ħc/(π²ε₀L⁴), derived, finite, and free of any cutoff. Modes with kL ≫ 1 oscillate many times across the weight and cancel; the weighted spectrum peaks at k = √3/L. With a sharp cutoff K as well, the result carries the factor 1 − (1 + K²L²/2)e^{(−K²L²/2)}, which passes 99% at KL = 3.6 and then stops changing. Limits: L → 0 returns the K⁴ divergence, and L → ∞ gives zero, as it must for a field with no mean averaged over all space. For L = 1.00 µm the r.m.s. field is 1.90 × 10⁴ V m⁻¹. Electro-optic sampling with femtosecond pulses (Riek et al., 2015) measured this variance directly and found it scaling inversely with the space-time volume sampled, as ħc/(ε₀L⁴) does when all four extents grow together.

The divergent piece is a number, not an operator. Normal ordering, every a† moved left of every a, writes E⋅E = :E⋅E: + ⟨0|E⋅E|0⟩ with ⟨0|:E⋅E:|0⟩ = 0; applied to the Hamiltonian it drops E₀ = Σ½ħω, as Unit 9 did. Nothing measured is lost. Every photon energy, every atomic transition frequency and every rate in this course is a difference between two states of the same field, and the constant cancels from it. It survives only where the set of mode frequencies changes with a parameter: move a conducting plate and every ωₖ shifts, so Σ½ħωₖ changes by an amount that, regulated, is finite and independent of the cutoff. That difference is the Casimir pressure, topic 11.06's calculation, measured to a few per cent since 1997. Gravity responds to absolute energy and so would see E₀ itself. Cut off at the Compton wavelength, ħcK⁴/(8π²) is already 1.8 × 10²² J m⁻³, against a measured dark-energy density near 5 × 10⁻¹⁰ J m⁻³; this course leaves that mismatch where Unit 9 left it.

Split E = E⁽⁺⁾ + E⁽⁻⁾, with E⁽⁺⁾ holding the a's. In the vacuum ⟨0|E⋅E|0⟩ = ⟨0|E⁽⁺⁾⋅E⁽⁻⁾|0⟩: the whole variance is anti-normally ordered. A detector that works by absorption, an atom or a photodiode, gains energy, so by energy conservation in the golden rule, quoted, the field must lose it, which only E⁽⁺⁾ can arrange. Its rate is therefore proportional to ⟨E⁽⁻⁾⋅E⁽⁺⁾⟩ (Glauber, quoted), and E⁽⁺⁾|0⟩ = 0 makes it zero: a ground-state atom in the vacuum is never excited, and no photodiode counts zero-point photons. The variance shows itself wherever no energy is taken from the field: as the noise floor of a balanced homodyne detector with its signal port blocked, against which squeezed light has been measured 15 dB below (Vahlbruch et al., 2016); in level shifts such as the 1057.8 MHz Lamb shift, Unit 13; in forces between moving boundaries, topic 11.06; and in spontaneous emission, which Unit 12 derives.