University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.7

Polarisation sums and the −g_μν replacement

Every cross-section with a photon in it sums over that photon's two polarisations, and doing the sum by hand breaks the covariance you worked to keep. This topic licenses the covariant shortcut, −g_μν in place of the sum, shows what it secretly counts, and marks where it fails: any amplitude whose current is not conserved.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A real photon of momentum k has two polarisations, yet the sum Σ_λ ε_μ ε*_ν over them cannot be built from g_μν and k alone: with k along z it is diag(0, 1, 1, 0), and no combination a g_μν + b k_μk_ν reproduces it. Saying which directions are transverse takes a second vector, which can be chosen null, and the exact result is −g_μν + (k_μn_ν + n_μk_ν)/(k⋅n), derived here; tie n to one frame and it is Coulomb gauge's δᵢⱼ − k̂ᵢk̂ⱼ. A squared amplitude meets this tensor as M^μM*^ν, and every extra term carries a factor k⋅M, which topic 8.06's Ward identity sets to zero.

So −g_μνM^μM*^ν is the physical sum, free of n and Lorentz invariant, and that is what lets Units 9 and 10 turn each photon's polarisation sum into a trace. Read in components, −g_μν counts all four Feynman-gauge photons with their norms, and k⋅M = 0 forces |M⁰| = |M³|, so the negative-norm timelike photon cancels the longitudinal one, as the Gupta-Bleuler condition promised. In QED the Ward identity holds for each photon with the other photons' indices left free, so every sum is replaced at once.

The shortcut stops wherever k⋅M ≠ 0: a single diagram, a truncated vertex, a current that forgets one of its charges. There the transverse sum depends on n, and −g can return a negative number of photons.

Simple definition: The polarisation-sum replacement says that, for an amplitude whose photon current is conserved, k_μM^μ = 0, the sum of |ε*_μM^μ|² over the photon's two physical polarisations equals −g_μνM^μM*^ν.
Example: For k = ω(1, 0, 0, 1), the amplitude M^μ = (0.5, 0.3, 0.4, 0.5) has k⋅M = 0, and −g_μνM^μM*^ν = 0.09 + 0.16 + 0.25 − 0.25 = 0.25, exactly |M¹|² + |M²|², the two transverse photons.

Transverse sum with a reference vectorΣ_λ ε_μ(k, λ)ε*_ν(k, λ) = −g_μν + (k_μn_ν + n_μk_ν)/(k⋅n)

An identity, exact for any null n. Its trace is −2, not −4. With n ∝ (1, −k̂) in the lab its spatial block is Coulomb gauge's δᵢⱼ − k̂ᵢk̂ⱼ.

λ = 1, 2 only. k² = n² = 0, k⋅n ≠ 0, ε⋅k = ε⋅n = 0, ε*⋅ε = −1. ε is dimensionless; k, n in GeV, and n's scale cancels. Derived.

The replacementΣ_λ |ε*_μ(k, λ)M^μ|² = −g_μνM^μM*^ν, if k_μM^μ = 0

Both extra terms of the identity carry k⋅M or its conjugate, so they vanish: the sum no longer knows n, and it is a Lorentz scalar that trace technology can evaluate.

M^μ: the amplitude with this photon's ε* removed; same mass dimension as M. k⋅M = 0 is topic 8.06's Ward identity, quoted.

Four polarisations weighted by their norms−g_μνM^μM*^ν = |M¹|² + |M²|² + |M³|² − |M⁰|²

k⋅M = ω(M⁰ − M³) = 0 gives |M⁰| = |M³|, so the timelike photon's negative weight cancels the longitudinal photon's: Gupta-Bleuler, in a cross-section.

M^μ components in the amplitude's units, with k = ω(1, 0, 0, 1). Signs −g_λλ: norms of topic 8.02's Feynman-gauge photons.

Every photon at onceΣ_λ₁λ₂ |ε*₁μ ε*₂ν M^μν|² = (−g_μα)(−g_νβ)M^μνM*^αβ

Each photon's n terms die without help from the other photon's polarisation, so all the sums go at once, for any number of photons. A non-abelian theory fails here.

M^μν: amplitude with both ε* removed, same dimension as M. Needs k₁μM^μν = 0 for all ν and k₂νM^μν = 0 for all μ; QED gives this.

What a non-conserved piece keepsΣ_λ |ε*⋅M_A|² = −g_μνM_A^μM_A*^ν + 2 Re[(k⋅M_A)(n⋅M_A*)]/(k⋅n)

The last term depends on n, so the subset's transverse sum depends on the gauge, and −g alone can come out negative. Only the complete amplitude may be contracted with −g.

M_A: a subset of diagrams or a truncated vertex, with k⋅M_A ≠ 0; dimension of |M|². The identity contracted with M_A.

Soft photons from a suddenly created chargedN/(d ln ω dΩ) = (α/4π²) β² sin²θ/(1 − β cos θ)²

From −g on J = P/(P⋅k) − p/(p⋅k), conserved because the charges sum to zero. The electron term alone would give −(α/4π²)(1 − β²)/(1 − β cos θ)², negative.

Partner at rest, electron of speed β at angle θ to the photon; N counts photons, so it is dimensionless. Needs ω ≪ |p|. Derived.

Two polarisations need a second vector

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. A photon with k² = 0 has two physical polarisations, ε⋅k = 0 and ε*⋅ε = −1 (UAQM-11.03). For k = ω(1, 0, 0, 1), ε(1) = (0, 1, 0, 0) and ε(2) = (0, 0, 1, 0) give Σ_λ ε_με*_ν = diag(0, 1, 1, 0). No a g_μν + b k_μk_ν does that: the 11 entry forces a = −1, and then the 00 and 33 entries need b = +1/ω² and b = −1/ω². Transverse always meant transverse in some frame, and covariantly that is a second null vector n with k⋅n ≠ 0. Take ε(1), ε(2) orthogonal to k and n; then T_μν = Σ_λ ε_μ(λ)ε*_ν(λ) + g_μν − (k_μn_ν + n_μk_ν)/(k⋅n) annihilates k^ν, n^ν and both ε^ν(λ′), using k² = n² = 0 and ε*(λ)⋅ε(λ′) = −δ_λλ′. Those four vectors span spacetime, so T = 0, derived. For n = (1, 0, 0, −1) the last term is diag(1, 0, 0, −1), and diag(0, 1, 1, 0) comes back. For a frame of four-velocity U, n = U − k/(2k⋅U) is null and gives ε⁰ = 0 there; in the lab it yields Coulomb gauge's δᵢⱼ − k̂ᵢk̂ⱼ. Trace check: −2 on the left, −4 from −g_μν alone.

A conserved amplitude cannot see n

An outgoing photon enters as M = ε*_μ(k, λ)M^μ and an incoming one as ε_μM^μ; either way Σ_λ|M|² = M^μM*^ν Σ_λ ε*_μ(λ)ε_ν(λ), and the identity holds for that order too, since its right side is real and symmetric. Substituting it gives Σ_λ|M|² = −g_μνM^μM*^ν + [(k⋅M)(n⋅M*) + (n⋅M)(k⋅M*)]/(k⋅n), derived. Topic 8.06 showed from current conservation, with every charged line on shell, that the complete amplitude obeys k_μM^μ = 0, quoted. Both extra terms die and Σ_λ|M|² = −g_μνM^μM*^ν. The right side no longer knows n, so no choice of transverse directions can matter, and it is a Lorentz scalar built from momenta, the form Casimir's trick turns into a trace from Unit 9 on. Two checks come free. Dimensions: ε is dimensionless, so M^μ carries the amplitude's mass dimension and both sides its square. Positivity: the left side is a sum of squares, so every conserved M^μ for a null k has −M⋅M* ≥ 0. An unpolarised incoming photon is averaged with ½, for its two states, not ¼.

Where the timelike photon goes

Write −g_μν as a sum over topic 8.02's four Feynman-gauge polarisations ε^μ(λ) = δ^μ_λ, each weighted by its norm −g_λλ: −1 for the timelike photon, +1 for the three spacelike ones. This is the one-photon unit operator of the indefinite-metric Fock space, Σ|λ⟩⟨λ|/⟨λ|λ⟩. With k = ω(1, 0, 0, 1), −g_μνM^μM*^ν = |M¹|² + |M²|² + |M³|² − |M⁰|². So −g does not pretend the unphysical photons are absent; it counts them, the timelike one with a negative weight. Now k⋅M = ω(M⁰ − M³) = 0 makes M⁰ = M³, and the last two terms cancel, shown: the longitudinal photon's +|M³|² is eaten by the timelike photon's −|M⁰|². That is topic 8.03's Gupta-Bleuler condition at the level of probabilities: unphysical photons enter only in the zero-norm combination a†(k, 0) − a†(k, 3), which carries nothing. Take M = (0.70, 0.20, 0.10, 0.70): the four terms are 0.04, 0.01, 0.49 and −0.49, and the sum is the transverse 0.05. Drop M⁰ instead, as if Coulomb gauge's ε⁰ = 0 allowed it, and you get 0.54, more than ten times too large, all of it a longitudinal photon that does not exist.

One replacement for each photon

With two external photons, Σ|M|² = M^μνM*^αβ Σ_λ₁ ε*₁μ ε₁α Σ_λ₂ ε*₂ν ε₂β. Replacing the first sum leaves n₁ terms containing k₁μM^μν. If that vanished only against transverse ε₂, it would not be enough: replacing the second sum by −g feeds timelike and longitudinal ε₂ into the first contraction. QED gives the stronger statement k₁μM^μν = 0 for every ν, with photon 2's index free, because topic 8.06's conservation argument never used photon 2's polarisation, only on-shell charged lines; the same holds for k₂. So Σ|M|² = (−g_μα)(−g_νβ)M^μνM*^αβ, shown, and likewise for any number of photons. Vertices built from F_μν obey it automatically: the π⁰γγ amplitude gε^μνρσk₁ρk₂σ vanishes against k₁μ by antisymmetry, as the Medium example uses. Two contrasts mark the edges. In a non-abelian theory k₁μM^μν is only proportional to k₂^ν, and ghosts must cancel what −g lets in, named only. For a massive vector, topic 8.08's −g_μν + k_μk_ν/M² is itself an identity over three states, not a replacement.

A current that forgets one of its charges

A neutral particle at rest decays to a heavy charged partner, which stays at rest with momentum P, and an electron of momentum p and speed β, as in neutron decay. Photons too soft to change either momentum see the classical current j̃^μ(k) = ieJ^μ with J^μ = P^μ/(P⋅k) − p^μ/(p⋅k), from topic 1.09's classical source, quoted. Then k⋅J = 1 − 1 = 0: charge conservation is the Ward identity. With −g, ω²(−J²) = −1 + 2/(1 − β cos θ) − (1 − β²)/(1 − β cos θ)² = β² sin²θ/(1 − β cos θ)², derived, positive and free of n. Now drop the partner, since a charge at rest does not radiate. −g on the electron term gives −(1 − β²)/(1 − β cos θ)²: at β = 0.80 and θ = 60°, −1.00 against the true +1.33. The electron term's transverse sum is positive but depends on n. In the radiation gauge of the decay frame it is the full 1.33, because there ε⋅P = 0. In the electron's rest frame it is zero and the partner carries all of it. Which charge radiates is gauge-dependent; the total is not.

The spectrum, and two limits that check it

The coherent source gives a mean photon number dN = e²Σ_λ|ε*⋅J|² d³k/((2π)³2ω), quoted from topic 1.09. With d³k = ω²dω dΩ and e² = 4πα, dN/(d ln ω dΩ) = (α/4π²)ω²Σ_λ|ε*⋅J|², a pure number per unit ln ω per steradian. Integrating the pattern, ∫dΩ β² sin²θ/(1 − β cos θ)² = 4π[(1/β) ln((1 + β)/(1 − β)) − 2], so dN/d ln ω = (α/π)[(1/β) ln((1 + β)/(1 − β)) − 2], derived: the inner bremsstrahlung of β decay, flat in ln ω, the soft-photon form that QED's infrared cancellation returns to. Slow electron: the logarithm expands to give (2α/3π)β², the classical dipole spectrum of a sudden velocity change, quoted, with the sin²θ pattern of a non-relativistic kick. Fast electron: the bracket grows as 2 ln 2γ − 2, a logarithm of the electron mass, and the peak at cos θ = β closes into a cone of half-angle near 1/γ. At β = 0.80 the bracket is 0.747, so 1.73 × 10⁻³ photons per unit of ln ω. The electron term with −g alone integrates to −4π at every β, a photon number of −α/π per unit ln ω.

Change one variable at a time

Make the relationship visible.

Interactive model

Electron speed β0.60

Frame of n: v/β0.50

Photon angle θ60 °

Set v/β to 0: the dashed curve lies on the solid one and the faint one vanishes, since ε⋅P = 0 in the decay frame. Drag v/β to 1 and they swap. In between neither term is the physics, the solid curve never moves, and −g on the electron term stays negative at every θ.

FULL CURRENT, ANY n0.551 α/4π²

ELECTRON TERM, FRAME v0.191 α/4π²

PARTNER TERM, FRAME v0.093 α/4π²

−g ON ELECTRON TERM-1.306 α/4π²

Live interpretationFULL CURRENT, ANY n: 0.551 α/4π². ELECTRON TERM, FRAME v: 0.191 α/4π². PARTNER TERM, FRAME v: 0.093 α/4π². −g ON ELECTRON TERM: −1.306 α/4π²

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that Σ_λ ε_μ ε*_ν = −g_μν is an identity, the photon's completeness relation, so that −g may be used against anything. It is not. The left side has trace −2 and annihilates k and n; the right side has trace −4 and annihilates nothing. Their difference, (k_μn_ν + n_μk_ν)/(k⋅n), is dropped only because it meets a factor k⋅M, and k⋅M = 0 belongs to the complete, gauge-invariant amplitude with its charged lines on shell, not to a single diagram, a vertex or a current you have truncated. Use −g on such a piece and the answer is wrong: the piece's own transverse sum depends on an arbitrary n, and −g can even come out negative, as it does for the electron term of a β decay at every angle. Before each replacement, set ε → k and watch the amplitude vanish. Nor is −g the massless limit of the massive-vector sum: topic 8.08's −g_μν + k_μk_ν/M² is an exact identity whose last term grows as M → 0.

A neutral scalar of mass m decays to two photons with M^μν = g[(k₁⋅k₂)g^μν − k₂^μk₁^ν]. In its rest frame, with polarisation vectors that have no time component, k₂⋅ε₁ = k₁⋅ε₂ = 0, so a student drops the second term and contracts g(k₁⋅k₂)g^μν with −g_μα and −g_νβ, getting g²m⁴. Which statement is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA photon has k = (2, 0, 0, 2) GeV. Take the reference vector n = (1, 1, 0, 0), find two real polarisation vectors orthogonal to k and n, and check three components of the identity. Then, for M^μ = (0.50, 0.20, 0.30, 0.50) in units of a fixed amplitude, compare the transverse sum, −g_μνM^μM^ν and the sum taken with n̄ = (1, 0, 0, −1). Repeat with M³ = 0.40.

ε = (a, b, c, d) needs ε⋅k = 2a − 2d = 0 and ε⋅n = a − b = 0, so ε = (a, a, c, a) with ε⋅ε = −a² − c². Normalised and orthogonal: ε(1) = (0, 0, 1, 0) and ε(2) = (1, 1, 0, 1), which is the x polarisation plus k/ω, a shift along k as in topic 8.05.
Lower the indices: ε(1)_μ = (0, 0, −1, 0), ε(2)_μ = (1, −1, 0, −1), k_μ = (2, 0, 0, −2) GeV, n_μ = (1, −1, 0, 0), and k⋅n = 2 GeV. Entry 00: left 1 × 1 = 1, right −1 + (2 × 1 + 1 × 2)/2 = 1. Entry 13: left (−1)(−1) = 1, right 0 + (0 × 0 + (−1)(−2))/2 = 1. Entry 11: left 1, right 1 + 0 = 1.
k⋅M = 2(0.50) − 2(0.50) = 0, so M is conserved. ε(1)⋅M = −0.30 and ε(2)⋅M = 0.50 − 0.20 − 0.50 = −0.20, so the transverse sum is 0.09 + 0.04 = 0.13. −g_μνM^μM^ν = −0.25 + 0.04 + 0.09 + 0.25 = 0.13. With n̄ the transverse photons are x and y: 0.20² + 0.30² = 0.13 again.
With M³ = 0.40, k⋅M = 0.20 GeV ≠ 0. Now ε(2)⋅M = −0.10 and the n sum is 0.09 + 0.01 = 0.10, the n̄ sum stays 0.13, and −g gives −0.25 + 0.04 + 0.09 + 0.16 = 0.04: three answers for one amplitude. Each transverse sum is −g plus 2(k⋅M)(n⋅M)/(k⋅n): 0.04 + 0.06 with n, since n⋅M = 0.30, and 0.04 + 0.09 with n̄, since n̄⋅M = 0.90 and k⋅n̄ = 4 GeV.
Traces: g^μν Σ_λ ε_με_ν = ε(1)⋅ε(1) + ε(2)⋅ε(2) = −2, while g^μν(−g_μν) = −4; the n terms supply the +2, since g^μν(k_μn_ν + n_μk_ν)/(k⋅n) = 2.

AnswerFor the conserved M, Σ_λ|ε⋅M|² = −g_μνM^μM^ν = 0.13 with either reference vector. For M³ = 0.40 the answers are 0.10 (n), 0.13 (n̄) and 0.04 (−g). Traces −2 against −4.

MediumThe π⁰ (m_π = 134.977 MeV) decays to two photons with M = g ε^μνρσ ε*₁μ ε*₂ν k₁ρ k₂σ, where g = α/(πf_π) with f_π = 92.1 MeV is quoted from the axial anomaly. Show that both polarisation sums may be replaced by −g, find Σ|M|² that way and with explicit linear polarisations, and compute Γ(π⁰ → γγ).

k₁μM^μν = gε^μνρσk₁μk₁ρk₂σ = 0, since ε is antisymmetric in μ and ρ while k₁μk₁ρ is symmetric. This holds for every ν, with photon 2's index free, and k₂νM^μν = 0 for every μ the same way. Both sums are replaced at once.
Σ|M|² = g²ε^μνρσε_μναβk₁ρk₂σk₁^αk₂^β = −2g²[k₁²k₂² − (k₁⋅k₂)²] = 2g²(k₁⋅k₂)², using ε^μνρσε_μναβ = −2(δ^ρ_αδ^σ_β − δ^ρ_βδ^σ_α). With m_π² = (k₁ + k₂)² = 2k₁⋅k₂, Σ|M|² = g²m_π⁴/2.
Explicitly, in the rest frame k₁ = (m_π/2)(1, 0, 0, 1) and k₂ = (m_π/2)(1, 0, 0, −1), with linear ε along x or y. Parallel pairs give zero; each crossed pair gives |M| = 2g(m_π/2)² = gm_π²/2. Two crossed pairs: 2 × g²m_π⁴/4 = g²m_π⁴/2, agreeing.
Γ = [1/(2m_π)] × ½ × [1/(8π)] × Σ|M|² = g²m_π³/(64π), with ½ for identical photons and 1/(8π) the massless two-body phase space, quoted. Dimensions: g is in MeV⁻¹, so g²m_π³ is in MeV.
g = (1/137.036)/(π × 92.1 MeV) = 2.522 × 10⁻⁵ MeV⁻¹, so Γ = (2.522 × 10⁻⁵)² × (134.977)³/(64π) MeV = 7.78 × 10⁻⁶ MeV = 7.78 eV. Measured (PrimEx-II): 7.80 ± 0.12 eV.

AnswerΣ|M|² = g²m_π⁴/2 by both routes, and Γ(π⁰ → γγ) = g²m_π³/(64π) = 7.78 eV, against the measured 7.80 ± 0.12 eV.

HardIn neutron decay treat the proton as at rest, and take an electron with β = 0.60 (kinetic energy 128 keV). For a soft photon at θ = 60° to the electron, use J^μ = P^μ/(P⋅k) − p^μ/(p⋅k) to find ω²Σ_λ|ε*⋅J|² with −g. Then find the electron term's transverse sum in the radiation gauges of the decay frame, of the electron's rest frame and of a frame moving at v = 0.30 along the electron, and the photon number between 1 and 10 keV.

k⋅J = (P⋅k)/(P⋅k) − (p⋅k)/(p⋅k) = 0: the neutron is neutral, so the outgoing charges sum to zero, and that is the Ward identity. −g may be used on the full J.
With P⋅k = Mω, p⋅k = Eω(1 − β cos θ) and P⋅p = ME: ω²(−J²) = −1 + 2/(1 − β cos θ) − (1 − β²)/(1 − β cos θ)² = β² sin²θ/(1 − β cos θ)². At 60°, 1 − β cos θ = 0.70, so 0.36 × 0.75/0.49 = 0.551.
−g on the electron term alone: −(1 − β²)/(1 − β cos θ)² = −0.64/0.49 = −1.306, a negative sum of squares, because the electron term Jₑ = −p/(p⋅k) has k⋅Jₑ = −1 ≠ 0.
Radiation gauge of a frame moving at v along the electron: n = U − k/(2k⋅U), and ω²Sₑ = −(1 − β²)/A² + 2(1 − vβ)/(AB) − (1 − v²)/B² with A = 1 − β cos θ, B = 1 − v cos θ. v = 0: 0.551, the full answer, since ε⋅P = 0 there. v = β: 0. v = 0.30: −1.306 + 2.756 − 1.260 = 0.191.
At v = 0.30 the partner term gives v² sin²θ/B² = 0.0675/0.7225 = 0.093 and the interference 0.267, and the three add to 0.551 again. Who radiates depends on n; the total does not.
dN/d ln ω = (α/π)[(1/0.60) ln 4 − 2] = (α/π)(0.3105) = 7.21 × 10⁻⁴. From 1 to 10 keV, far below |p| = 383 keV, N = 7.21 × 10⁻⁴ × ln 10 = 1.66 × 10⁻³ photons per decay at this electron energy. The slow-electron form (2α/3π)β² would give 5.57 × 10⁻⁴ per unit ln ω, 23% low at β = 0.60.

Answerω²Σ_λ|ε*⋅J|² = 0.551 at 60°, against −1.31 for −g on the electron term; the electron term's transverse sum is 0.551, 0 or 0.191 as n changes; about 1.66 × 10⁻³ photons between 1 and 10 keV.