Photons: energy, momentum, polarisation and helicity

Topic 11.03 wrote the transverse field in a periodic box of volume V as A(r) = Σ Nₖ[a(k, λ)ε(k, λ)e^(ik⋅r) + h.c.], with Nₖ = √(ħ/(2ε₀ωV)), ω = c|k|, k⋅ε = 0, E = −∂A/∂t and B = ∇ × A, and found H = Σ ħω(a†a + ½). That is quoted. Classically the field also carries momentum ε₀∫E × B d³r and angular momentum ε₀∫r × (E × B) d³r, and this lesson feeds the same expansion into both. Units are SI throughout, and the one four-vector statement uses the metric (+, −, −, −). One care is new: E and B at the same point do not commute, so the product is ambiguous as an operator, and we take the Hermitian form ε₀(E × B − B × E)/2, which reduces to ε₀E × B when the fields are numbers. Each integral then collapses through ∫exp[i(k − k′)⋅r] d³r = V δ(k, k′), which pairs a mode with itself, giving a†a terms, or with its mirror −k, giving a a and a†a† terms. Dimensions say what must come out: ε₀EB times a volume is kg m s⁻¹, and the only momentum a mode owns is ħk.

With ε = ε(k, λ) and ε′ = ε(k, λ′), the self-paired terms carry ε × (k × ε′*) = k(ε⋅ε′*) − ε′*(ε⋅k) = k δ(λ, λ′), transversality removing the second term, and the prefactor ε₀ωNₖ²V = ħ/2. They give Σ(ħk/2)(a a† + a†a) = Σ ħk(a†a + ½). The mirror-paired terms carry a(k, λ)a(−k, λ′) times a vector that changes sign under k → −k with λ and λ′ swapped; the operators commute, so over the symmetric set of box modes the sum is its own negative and vanishes at every instant, though each term oscillates as e^(−2iωt). In H the same terms cancel differently, electric against magnetic. The zero-point part, Σ ħk/2, vanishes for the same reason: for every k the box also holds −k. So P = Σ ħk a†a, derived, and the vacuum carries no momentum. Contrast the energy: Σ ħω/2 is positive term by term and grows as the fourth power of a cutoff, while any cutoff symmetric under k → −k, sphere or cube, leaves the momentum sum exactly zero. No regulator is needed for this result to survive.

H and P are sums of number operators, so they commute and share eigenstates. The state a†(k, λ)|0⟩ lies ħω above the vacuum and carries momentum ħk; n quanta in one mode carry nħω and nħk. Then E = ħω = c|ħk|, so E² − c²|p|² = 0: the four-momentum is null and the rest mass is zero, the free dispersion ω = c|k| of 11.02 read one quantum at a time. Because the creation operators commute, a†(k, λ)a†(k′, λ′)|0⟩ is unchanged when the two photons' labels are exchanged, and any number can share a mode: photons are bosons, as Unit 9's spin-statistics connection requires of an integer-spin field quantised with commutators. Two checks. Classical limit: a beam of power P_b delivers P_b/ħω photons per second, each with ħω/c, so an absorber feels P_b/c, 3.34 nN per watt, the classical radiation pressure with ħ gone. Measurement: a ⁸⁷Rb atom absorbing a 780.24 nm photon recoils at h/(mλ) = 5.885 mm s⁻¹, and atom interferometers that count such recoils measure h/m to better than a part in 10⁹ (quoted).

Put B = ∇ × A into ε₀∫r × (E × B) d³r and integrate by parts with ∇⋅E = 0: the angular momentum splits into an orbital part ε₀∫Eⱼ(r × ∇)Aⱼ d³r and a spin part S = ε₀∫E × A d³r. S has no ordering problem, since the equal-time commutator of Aᵢ and Eⱼ is symmetric in i and j and the cross product contracts it antisymmetrically. The mode sum runs as for P, with the same ħ/2: mirror terms cancel between k and −k, and in a real basis e₁, e₂ with e₁ × e₂ = k̂ no zero-point term survives, as e₁ × e₁ = 0. What remains per mode is S⋅k̂ = iħ[a†(k, 2)a(k, 1) − a†(k, 1)a(k, 2)], not diagonal: a linearly polarised photon has no definite spin. The unitary change to a± = [a(k, 1) ∓ i a(k, 2)]/√2 gives S = Σ ħk̂(n₊ − n₋), derived. For a plane wave (r × k)⋅k̂ = 0, so the orbital part has no component along k̂, and the helicity J⋅k̂ = S⋅k̂ is +ħ or −ħ. Per unit energy that is ±1/ω, the classical U/ω of a circular wave; Beth's 1936 wave-plate torque matched it to about 10% (quoted).

A massive spin-1 particle, the Z boson say, has three states in its rest frame, and in motion they include helicity 0, the longitudinal state with ε along k. The photon has no rest frame, and k⋅ε = 0 removes that state: the longitudinal part of E is no oscillator but the instantaneous Coulomb field of the charges, as 11.01 showed. That the two helicities are the same in every frame is Wigner's classification of massless particles, quoted: for m = 0 helicity labels the representation, no proper Lorentz transformation changes it, and only parity exchanges +ħ with −ħ, so a parity-invariant theory needs both. A massive particle lacks that protection: an observer can overtake it, reversing p but not J along the boost. With the measured bound mc² below 1 × 10⁻¹⁸ eV, overtaking a 2.0 eV photon would need a Lorentz factor above 2 × 10¹⁸. The count of two is measured too. Planck's law with two states per mode gives σ = 2π⁵k_B⁴/(15h³c²) = 5.670 × 10⁻⁸ W m⁻² K⁻⁴; a third state in equilibrium would multiply it by 3/2, and radiometric measurements agree with two to better than 0.1% (quoted).

The derivation gives a photon an energy, a momentum and a helicity, and stops short of a position. Newton and Wigner asked for an operator whose eigenstates sit at a point, whose components commute and which rotates as a vector, and found that for a massless particle of helicity ±ħ none exists; that is named here, not derived. A hint of why: a packet localised in space needs every direction of k, and at each k̂ the polarisation must be perpendicular to it, a tangent vector on the sphere of directions. A smooth tangent field that vanishes nowhere on a sphere does not exist, so transversality and sharp localisation fight. Nothing measurable is lost. A detector reports a click rate proportional to ⟨E⁽⁻⁾(r)⋅E⁽⁺⁾(r)⟩ at its own position, a field correlation, not |ψ(r)|² of a photon wavefunction. The model also stops on the other side: all of this is the free field, and how an atom makes or absorbs a photon needs the coupling of 11.08 and the decay of Unit 12.