Quantising the field: A, E and B as operators

Units are SI throughout, and where a four-vector appears the metric is (+, −, −, −). Topic 11.02 wrote the free transverse field in a periodic box of volume V as A = Σₖₛ Nₖ[εₖₛαₖₛe^{(i(k⋅r − ωₖt))} + c.c.], with two polarisations per k obeying k⋅εₖₛ = 0 and a classical energy 2ε₀Vωₖ²Nₖ²|αₖₛ|² in each mode. Quantisation replaces each amplitude αₖₛ by an operator aₖₛ with [aₖₛ, a†ₖ′ₛ′] = δₖₖ′δₛₛ′: quoted from Quantum Mechanics I and imposed, not derived. Nₖ is then no longer free. The classical |α|² becomes the symmetrised (aa† + a†a)/2 = a†a + 1/2, and the mode energy is ħωₖ(a†a + 1/2) only if Nₖ² = ħ/(2ε₀ωₖV), which is the same as demanding [q, p] = iħ for each of 11.02's canonical pairs. For a 1.0 cm cube and its lowest mode, λ = 1.0 cm, Nₖ = 5.6 × 10⁻¹⁵ T m. The time dependence is consistent, not assumed: with H_rad from the next section, [aₖₛ, H_rad] = ħωₖaₖₛ gives aₖₛ(t) = aₖₛe^(−iωₖt) in the Heisenberg picture.

With A in hand the other fields are definitions. E⊥ = −∂A/∂t brings down iωₖ on each a term and −iωₖ on each a† term, and B = ∇ × A brings down ik × and −ik ×, giving amplitude ℰₖ = √(ħωₖ/(2ε₀V)) for E⊥ and ℰₖ/c for B. Substitute both into (ε₀/2)∫(E⊥² + c²B²)d³r. The volume integral pairs each k with k′ = ±k. The aa and a†a† pieces, which would change the photon number by two, arrive from E⊥² and c²B² with equal size and opposite sign, because B pairs k with −k and (k × ε)⋅(−k × ε′) = −k²ε⋅ε′ for transverse vectors, while c²k² = ωₖ²; they cancel. The a†a pieces arrive with the same sign: the electric and magnetic parts each give (ħωₖ/2)(a†a + 1/2), so H_rad = Σħωₖ(a†a + 1/2), derived. In a number state ⟨n|E⊥|n⟩ = 0, yet one running mode gives ⟨n|E⊥²|n⟩ = ℰₖ²(2n + 1): a field with no mean value and a sharp energy. The quanta are measured, not inferred: Gleyzes and colleagues (2007) watched a single 51 GHz cavity mode jump between n = 1 and n = 0, one ħω at a time.

The Lagrangian density (ε₀/2)(E² − c²B²) gives A the conjugate momentum Π = ∂ℒ/∂(∂A/∂t) = −ε₀E⊥. Their equal-time commutator, computed from the expansions, contains the polarisation sum Σₛεᵢε*ⱼ = δᵢⱼ − k̂ᵢk̂ⱼ, true because the two polarisations and k̂ form an orthonormal triad. So [Aᵢ(r), −ε₀E⊥ⱼ(r′)] = iħδ⊥ᵢⱼ(r − r′), derived. The scalar-field answer δᵢⱼδ³ is impossible here: ∂/∂rᵢ of the left side is zero because ∇⋅A = 0, and of δᵢⱼδ³ it is ∂ⱼδ³. The price is nonlocality. With the electrostatic identity ∂ᵢ∂ⱼ(1/r) = (3r̂ᵢr̂ⱼ − δᵢⱼ)/r³ − (4π/3)δᵢⱼδ³(r), quoted, δ⊥ᵢⱼ(r) = (2/3)δᵢⱼδ³(r) − (δᵢⱼ − 3r̂ᵢr̂ⱼ)/(4πr³): a contact term plus a dipole-shaped tail. At 1.0 m along the field component the tail is +0.159 m⁻³, and perpendicular to it −0.080 m⁻³. Its trace is 2δ³(r), one per polarisation: the field has two degrees of freedom at each point, not three.

Write δ⊥ᵢⱼ = δᵢⱼδ³ + ∂ᵢ∂ⱼ(1/(4πr)). B = ∇ × A, and the curl annihilates the second, nonlocal piece because εⱼₗₘ∂ₗ∂ₘ = 0, so what survives is local: [Eᵢ(r, t), Bⱼ(r′, t)] = −(iħ/ε₀)εᵢⱼₗ∂ₗδ³(r − r′), derived, while [Eᵢ, Eⱼ] and [Bᵢ, Bⱼ] vanish at equal times. Parallel components commute; Eₓ and B_y do not, so no state makes both sharp. Values at a point are too singular to be operators, so ask about averages. Average Eₓ over a cube of side a and B_y over the same cube shifted by a/2 along z: only the shifted cube's lower face lies inside the first, and [Ēₓ, B̄_y] = −iħ/(ε₀a⁴). For a = 1.0 mm, ΔĒₓΔB̄_y ≥ ħ/(2ε₀a⁴) = 6.0 × 10⁻¹² V m⁻¹ T; for unshifted cubes the face terms cancel and the averages commute. The a⁻⁴ is the classical limit at work: averages over large regions commute ever more nearly. Bohr and Rosenfeld (1933) showed that ideal measurements with charged test bodies reach exactly these limits, a result quoted here.

At unequal times the same polarisation sum gives [Aᵢ(r, t), Aⱼ(0, 0)] = −(iħ/ε₀)(δᵢⱼ − ∂ᵢ∂ⱼ∇⁻²)D(r, t), where (1/V)Σₖ sin(ωₖt)e^(ik⋅r)/ωₖ is D = [δ(r − ct) − δ(r + ct)]/(4πcr), which is −1/c times Unit 9's real Pauli-Jordan function Δ at m = 0 and x⁰ = ct, quoted, and supported on the cone c²t² − r² = 0. In [Eᵢ, Eⱼ] two time derivatives meet ∇⁻², and since ∂²D/∂t² = c²∇²D the inverse Laplacian cancels, leaving (iħ/ε₀)(δᵢⱼ∂²/∂t² − c²∂ᵢ∂ⱼ)D: zero wherever r exceeds c|t|, derived, and the same holds for [Eᵢ, Bⱼ] and [Bᵢ, Bⱼ]. In A's commutator nothing cancels. ∇⁻²D is the Coulomb potential of a shell of radius c|t| carrying total charge ∫D d³r = t, so outside it equals −t/(4πr), and wherever r exceeds c|t| the commutator is (iħt/(4πε₀r³))(δᵢⱼ − 3r̂ᵢr̂ⱼ). At r = 1.5 m and t = 3.0 ns, perpendicular to the components, that is i × 8.4 × 10⁻³⁴ T² m². A in Coulomb gauge is fixed by an instantaneous, nonlocal condition and is not an observable; covariant quantisation keeps A local at the price of unphysical states, named only.

Dimensions first. A is in T m and ε₀E in C m⁻², so [A, ε₀E] carries T C m⁻¹ = J s m⁻³, ħ times the m⁻³ of δ⊥; [E, B] carries V m⁻¹ T, ħ/ε₀ times m⁻⁴. Limits next. As t → 0, D → 0 and ∂D/∂t → δ³(r), so the unequal-time [Eᵢ(r, t), Bⱼ(0, 0)] = −(iħ/ε₀)εᵢⱼₗ∂ₗ∂D/∂t becomes the equal-time card. One mode gives a commutator spread smoothly over the cavity (the Medium example); only the sum over every mode concentrates it at a point. In a standing mode holding n photons the bound is a fraction 1/(2n + 1) of ΔEₓΔB_y, the classical limit. Then measurement. Two polarisations per k, the trace of δ⊥, enter Planck's law, quoted from statistical mechanics, and give σ = π²k_B⁴/(60ħ³c²) = 5.670 × 10⁻⁸ W m⁻² K⁻⁴, the measured value; a longitudinal third state would make it 8.51 × 10⁻⁸. And Brune and colleagues (1996) saw atoms in a 51 GHz cavity Rabi-oscillate at frequencies in the ratios 1 : √2 : √3 : 2 for n = 0 to 3, the √(n + 1) that a† puts into ⟨n + 1|E|n⟩.