University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.7

Furry's theorem: odd-photon fermion loops vanish

Furry's theorem is the first time a symmetry deletes Feynman diagrams before you compute them. Charge conjugation pairs every closed fermion loop with its reverse at a cost of −1 per photon, so odd-photon loops cancel, the three-photon function needs no counterterm, and what is left over is light-by-light scattering.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

QED is invariant under charge conjugation: C leaves the vacuum alone, commutes with the Hamiltonian, and reverses the sign of the current j^μ = :ψ̄γ^μψ: and of the photon field. Sandwich a time-ordered product of n currents between vacuum states, slip C⁻¹C between its factors, and it returns multiplied by (−1)ⁿ, so for odd n it is zero, in the free theory and at every order of the interacting one. In diagrams the same fact is a pairing.

Every closed electron loop can be read with its arrow either way round, and transposing its Dirac trace with C₀ = iγ²γ⁰, which turns each γ^μ into −(γ^μ)^T and each propagator S_F(p) into S_F(−p)^T, shows that the reversed loop is (−1)ⁿ times the original for n photon vertices. One triangle is not zero; the pair is. The consequences are concrete: no photon tadpole and no three-photon function, so no counterterm for either, nine two-loop g − 2 diagrams cut to seven, and a nucleus's vacuum polarisation running in odd powers of Zα, while four-photon loops survive and scatter light by light.

The theorem stops where its premises do. It counts only C-odd vertices, so scalar or axial couplings on the loop do not count; it concerns closed loops and vacuum correlators, so it says nothing about photons radiated from an open electron line; and it needs C, so a Z boson, coupled to a mixture of C-odd and C-even currents, escapes it.

Simple definition: Furry's theorem: in QED, which conserves charge conjugation, the vacuum expectation value of an odd number of currents vanishes, so every closed fermion loop with an odd number of photon vertices cancels against the same loop with its arrow reversed.
Example: Three photons on one electron loop give two diagrams, one per arrow direction; the second is (−1)³ = −1 times the first, so they sum to zero. Four photons give (4 − 1)! = 6 diagrams in 3 pairs, and each pair doubles instead.

Odd current correlators vanish⟨Ω|T j^μ₁(x₁) ⋯ j^μₙ(xₙ)|Ω⟩ = 0 for odd n

C|Ω⟩ = |Ω⟩, CHC⁻¹ = H and Cj^μC⁻¹ = −j^μ make the correlator (−1)ⁿ times itself, at every order in e. Photon fields obey the same rule.

j^μ = :ψ̄γ^μψ: has mass dimension 3 (e j^μ is a current density, SI A m⁻²); |Ω⟩ the interacting vacuum; xᵢ in GeV⁻¹ (SI m)

The charge-conjugation matrixC₀γ^μC₀⁻¹ = −(γ^μ)^T, with C₀ = iγ²γ⁰

Transposing a product reverses its order, so C₀ turns a string of γ's into a transposed string read backwards, one minus sign per γ.

C₀ is a dimensionless 4 × 4 matrix with C₀² = −1, so C₀⁻¹ = −C₀; topic 11.02's CψC⁻¹ = iγ²γ⁰ψ̄^T uses the same matrix

The propagator under C₀C₀S_F(p)C₀⁻¹ = S_F(−p)^T, where S_F(p) = i(p̸ + m)/(p² − m² + iε)

From C₀p̸C₀⁻¹ = −p̸^T: conjugating a propagator reverses its momentum, the step that turns a loop into the same loop run backwards.

p the four-momentum and m the electron mass, both in GeV; S_F in GeV⁻¹; the denominator is even in p

Reversing a fermion loopTr[γ^μ₁S_F(p₁)γ^μ₂S_F(p₂) ⋯ γ^μₙS_F(pₙ)] = (−1)ⁿ Tr[γ^μ₁S_F(−pₙ)γ^μₙ ⋯ γ^μ₂S_F(−p₁)]

The right side is the loop with its arrow reversed: the same vertices visited backwards with every momentum flipped. The closed-loop −1, the couplings and the i's are common to both.

pᵢ is the momentum, along the arrow, of the propagator entering vertex i, the one written after γ^μᵢ; n photon vertices; both sides in GeV⁻ⁿ before the loop integral

Other vertices on the loopC₀ΓC₀⁻¹ = η_Γ Γ^T: η = +1 for 1, γ⁵, γ^μγ⁵ and η = −1 for γ^μ, σ^μν

Only C-odd vertices count: two photons and a scalar ψ̄ψ vertex give +1 and survive; one photon and two scalar vertices give −1 and cancel.

Γ a dimensionless vertex matrix; a loop with vertices Γ₁ … Γₙ reverses with the factor η₁η₂ ⋯ ηₙ

Summing the two orientationsMₐ + M_b = [1 + (−1)ⁿ] Mₐ

Odd n: every pair cancels. Even n: every pair doubles, so the four-photon loop's 6 orderings are 3 integrals each counted twice. For n ≤ 2 the reverse is the same diagram.

Mₐ an ordering of n ≥ 3 photon legs, M_b its reverse; mass dimension 4 − n; (n − 1)! orderings make (n − 1)!/2 pairs

C turns an odd correlator into minus itself

Work in natural units ħ = c = 1 with the metric (+, −, −, −). Three facts are quoted from this unit: C|0⟩ = |0⟩ and Cj^μC⁻¹ = −j^μ for j^μ = :ψ̄γ^μψ:, the sign supplied by anticommutation (topic 11.02), and CA^μC⁻¹ = −A^μ (topic 11.04). The interaction e j^μA_μ is then C-even, so C commutes with the full Hamiltonian H. Its ground state |Ω⟩ is unique, so C|Ω⟩ = λ|Ω⟩ with |λ| = 1, and the Heisenberg-picture currents inherit Cj^μ(x)C⁻¹ = −j^μ(x) at every time. C acts on operators and time ordering on c-number times, so the two commute, and currents are bosonic, so reordering them costs no sign. Insert C⁻¹C between every pair of factors: ⟨Ω|T j^μ₁(x₁) ⋯ j^μₙ(xₙ)|Ω⟩ equals |λ|²(−1)ⁿ times itself, derived, and vanishes for odd n. Nothing was assumed small, so this holds at every order in e. The same steps kill ⟨Ω|T A^μ₁ ⋯ A^μₙ|Ω⟩ for odd n: the photon one- and three-point functions vanish, and a mixed correlator needs an even total of currents and photon fields.

Transposing the trace reverses the loop

Order by order the Gell-Mann–Low formula (topic 2.06) turns the correlator into free-field traces, and there the cancellation becomes visible. Take C₀ = iγ²γ⁰, the matrix in topic 11.02's CψC⁻¹. In the Dirac representation γ⁰ and γ² are symmetric while γ¹ and γ³ are antisymmetric, and C₀² = −1; a line per index then gives C₀γ^μC₀⁻¹ = −(γ^μ)^T, so C₀(p̸ + m)C₀⁻¹ = (−p̸ + m)^T and C₀S_F(p)C₀⁻¹ = S_F(−p)^T, derived. Now write C₀⁻¹C₀ between every factor of a loop's trace Tr[γ^μ₁S_F(p₁) ⋯ γ^μₙS_F(pₙ)]. Each vertex becomes −(γ^μ)^T and each propagator S_F(−p)^T; a product of transposes is the transpose of the reversed product, and a trace ignores both transposition and cyclic shifts. What remains is (−1)ⁿ Tr[γ^μ₁S_F(−pₙ)γ^μₙ ⋯ γ^μ₂S_F(−p₁)]: the same vertices visited backwards with every momentum reversed, which is exactly the loop drawn with its arrow the other way, shown. Topic 7.04's closed-loop −1, the vertex factors −ieγ^μ and the propagators' i's are shared by both orientations, so (−1)ⁿ is the whole relative sign.

Counting γ matrices: one orientation is not zero

Each propagator numerator p̸ + m adds either a slashed momentum or a mass to the trace. With n photon vertices and j slashed momenta the trace holds n + j γ's and survives only when n + j is even, so for odd n every surviving term is odd in the momenta. For the triangle, four of the eight terms survive: three ∝ m², each a four-γ trace such as m²Tr[γ^μγ^νγ^ρk̸₃] = 4(g^μν k₃^ρ − g^μρ k₃^ν + g^νρ k₃^μ)m², and one ∝ m⁰ with six γ's. One orientation is a nonzero integrand, superficially linearly divergent by the count of topic 12.01. Reversal flips every momentum, and the reversal rule Tr[γ^a₁ ⋯ γ^a₂ₖ] = Tr[γ^a₂ₖ ⋯ γ^a₁], which C₀ also proves, restores the order, so each term changes sign: (−1)³ again, a second derivation. Route the reversed loop's momentum as the mirror of the original and the two integrands cancel point by point, before any regulator is chosen. With one or two vertices the reverse is the same diagram: the one-photon tadpole equals minus itself and vanishes, while topic 13.02's vacuum-polarisation loop equals plus itself and is counted once, not twice.

Only the C-odd vertices count

The trace argument never used that the vertices were photons, only that C₀ΓC₀⁻¹ = η_Γ Γ^T. For 1, γ⁵ and γ^μγ⁵, η = +1; for γ^μ and σ^μν, η = −1, derived, γ⁵ coming out even because four minus signs and the reversal of four anticommuting γ's both give +1. These are topic 11.04's C signs of the bilinears. A loop reverses with the product of its η's, (−1)^(n_γ) when n_γ photons share it with scalar or pseudoscalar vertices. Two photons and one Yukawa vertex −gφψ̄ψ give +1, so the top-quark loop in H → γγ, a channel in which the Higgs boson was discovered in 2012 (quoted), survives; one photon and two scalar vertices give −1 and cancel, as C demands for a C-even scalar. A classical background potential is a C-odd photon too, since C carries QED in a field Aₑₓₜ into QED in −Aₑₓₜ. So the vacuum polarisation around a nucleus runs in odd powers of Zα, Uehling's α(Zα) and then Wichmann and Kroll's α(Zα)³ with no α(Zα)² between them (named and quoted), and a photon can split in a magnetic field only through an odd number of field insertions.

What it removes and what survives

Power counting in topic 12.01 will list the one-, two-, three- and four-photon functions as superficially divergent. Furry's theorem removes the one- and three-photon functions at every order, so renormalised QED in Unit 12 needs neither a photon tadpole nor a three-photon counterterm; the photon effective action is even in A, as the first section implies. In the electron's g − 2 it prunes diagrams before any integral: of the nine two-loop vertex diagrams, the two with the external photon on a closed loop are one orientation pair and cancel, leaving the seven behind A₂ = −0.328479, quoted. Four-photon loops pair with + signs and survive: at three loops they add 0.371005(α/π)³ = 4.65 × 10⁻⁹ to aₑ (coefficient quoted, product computed), 3.6 × 10⁴ times the 1.3 × 10⁻¹³ uncertainty of the measured aₑ = 0.00115965218059. γγ → γγ itself, with quasi-real photons from passing lead nuclei, was observed by ATLAS in 2019 at 8.2 standard deviations, quoted; at low energy its cross-section (973/10125π)α⁴ω⁶/m⁸, quoted, is 7.3 × 10⁻⁷⁰ m² for ω = 1 eV in the centre of mass.

Checks, and where the theorem stops

Dimensions: the two orientations are one integrand with its momenta reversed, so they share mass dimension, −n from n propagators and 4 − n after ∫d⁴k as an n-photon amplitude needs, and (−1)ⁿ is a pure number, as a relative sign must be. Free-field limit: at e = 0 Wick's theorem writes ⟨0|T j^μ(x)j^ν(y)j^ρ(z)|0⟩ as two contraction cycles, x → y → z → x and x → z → y → x, which are the triangle's two orientations, and the operator argument says they sum to zero: the two derivations agree at lowest order. Massless limit: at m = 0 the m² terms go, but the six-γ term still cancels in pairs. The theorem stops with its premises. It concerns vacuum correlators and closed loops; a matrix element between states that contain charged particles is constrained only when those states are C eigenstates, as positronium will be in topic 11.08, and an electron radiating one photon is not forbidden. And without C it fails: the Z boson couples to a mixture of the C-odd vector and the C-even axial current, so in a loop with one Z and two photons the theorem removes the vector part but not the axial one, named only.

Change one variable at a time

Make the relationship visible.

Interactive model

Photon vertices on the loop3

Scalar ψ̄ψ vertices on the loop0

Step the photons from 3 to 4: b flips from −a to +a and the pair turns from cancelling to doubling. Then take one photon with two scalars, which cancels, and two photons with one scalar, the H → γγ loop, which survives: only C-odd vertices count.

REVERSED ÷ ORIGINAL-1

CYCLIC ORDERINGS (n + s − 1)!2

ORIENTATION PAIRS1

SURVIVING DIAGRAMS0

Live interpretationREVERSED ÷ ORIGINAL: −1. CYCLIC ORDERINGS (n + s − 1)!: 2. ORIENTATION PAIRS: 1. SURVIVING DIAGRAMS: 0

Catch the common trap

Explain before calculating.

Misconception check

The tempting proof is one line: the three-photon loop contains three γ^μ, the trace of an odd number of γ matrices vanishes, so each triangle is zero by itself. It is wrong because the propagators sit inside the trace too. Each numerator p̸ + m contributes a slashed momentum or a mass, and the terms with one or three slashed momenta carry four or six γ's: m²Tr[γ^μγ^νγ^ρk̸₃] = 4(g^μν k₃^ρ − g^μρ k₃^ν + g^νρ k₃^μ)m² is not zero. A single orientation is a nonzero, linearly divergent integrand. What vanishes is its sum with the reversed loop, because every surviving term is odd in the momenta and the reversed loop carries them backwards. The distinction matters once the vertices change: put γ^ργ⁵ at one corner and the trace still holds three γ^μ's, yet η = (−1)(−1)(+1) = +1 and the two orientations add.

QED is extended by a neutral scalar φ coupled through −gφψ̄ψ. Which one-loop contribution is forced to vanish by charge conjugation once both orientations of every fermion loop are summed?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA closed electron loop has photon vertices γ^μ, γ^ν, γ^ρ and propagators carrying k₁, k₂, k₃. Expand the numerator trace Tr[γ^μ(k̸₁ + m)γ^ν(k̸₂ + m)γ^ρ(k̸₃ + m)] by the number of slashed momenta, decide which terms survive and what reversing the loop does to each, then count the diagrams and orientation pairs when five photons share one loop.

Picking k̸ or m from each numerator gives 2³ = 8 terms. A term with j slashed momenta carries m^{(3 − j)} and a trace of 3 + j γ matrices.
A trace of an odd number of γ's vanishes (UAQM-04.03, quoted), so j must be odd. j = 1 gives three terms ∝ m², each a four-γ trace such as m²Tr[γ^μγ^νγ^ρk̸₃] = 4(g^μν k₃^ρ − g^μρ k₃^ν + g^νρ k₃^μ)m²; j = 3 gives one six-γ term ∝ m⁰. Four of the eight survive, so one triangle is not zero.
Every surviving term has one or three momenta, an odd degree. The reversed loop carries −k₃, −k₂, −k₁ through the vertices in the opposite order, and Tr[γ^a₁ ⋯ γ^a₂ₖ] = Tr[γ^a₂ₖ ⋯ γ^a₁] restores the order, so each term changes sign: reversed = (−1)³ × original.
Five photons: fix photon 1 and arrange the other four around the loop in 4! = 24 ways. Reversal maps each ordering to a different one, making 12 pairs, and each pair carries 1 + (−1)⁵ = 0.
Check with four photons: 3! = 6 orderings in 3 pairs, each with 1 + (−1)⁴ = 2, so even loops double rather than cancel, as the light-by-light loop must.

AnswerFour of the eight terms survive, three ∝ m² and one ∝ m⁰, all odd in the momenta, so the reversed triangle is −1 times the original; five photons give 24 diagrams in 12 cancelling pairs.

MediumA photon crosses a uniform magnetic field B far below the critical field B_cr = m²c²/(eħ) and splits into two photons through an electron loop that also absorbs k insertions of the background field. Compute B_cr, use Furry's theorem to find the allowed k, and, given that the k = 1 box vanishes for collinear photons (Adler, quoted), find the factor by which the splitting rate grows from B = 2.2 × 10⁸ T to 6.6 × 10⁸ T.

B_cr = m²c²/(eħ) = (9.109 × 10⁻³¹ kg)²(2.998 × 10⁸ m s⁻¹)² / [(1.602 × 10⁻¹⁹ C)(1.055 × 10⁻³⁴ J s)] = 4.41 × 10⁹ T. Units: kg² m² s⁻² ÷ (C kg m² s⁻¹) = kg C⁻¹ s⁻¹ = T. In natural units B_cr = m²/e.
Under C, QED in a background Aₑₓₜ becomes QED in −Aₑₓₜ, so each field insertion is a C-odd photon vertex. The loop carries 3 + k of them and its two orientations differ by (−1)^{(3 + k)}, so k must be odd: k = 0, 2 and 4 cancel, the k = 2 pentagon included.
The k = 1 box survives Furry's theorem but vanishes for collinear photons (quoted), so the hexagon, k = 3, leads: the amplitude goes as B³ and the rate, ∝ |M|², as B⁶, while B ≪ B_cr.
The two fields are 0.050 B_cr and 0.150 B_cr, a ratio of 3.0, so the rate grows by 3⁶ = 729. A pentagon, had Furry's theorem allowed it, would have led with B⁴ and given only 3⁴ = 81, a factor 9 apart.

AnswerB_cr = 4.41 × 10⁹ T; only odd k survive, the B³ hexagon leads, and the splitting rate grows by 3⁶ ≈ 7.3 × 10², against the 81 a forbidden B² pentagon would give.

HardThe electron anomaly is aₑ = α/2π + A₂(α/π)² + A₃(α/π)³ + ⋯. Build the two-loop vertex diagrams by attaching the external photon to every charged line of the three two-loop self-energies and count those that contribute. Then decide the fate of three-loop diagrams whose electron loop holds the external photon and three internal ones, evaluate them with the quoted coefficient 0.371005, and compare with the measured aₑ = 0.00115965218059(13).

The two-loop self-energies are the rainbow and the crossed diagram, each with three internal electron propagators, and the one-loop self-energy with a vacuum-polarisation bubble: one electron propagator outside and two inside the closed loop. Attaching the photon everywhere gives 3 + 3 + 1 + 2 = 9 diagrams.
The last two put the external photon on the closed loop, which then has three photon vertices. They are its two orientations, related by (−1)³, so they cancel and 7 remain, the set behind A₂ = −0.328479 (quoted): −0.328479 × (α/π)² = −1.772 × 10⁻⁶.
The three-loop light-by-light loop carries four photon vertices, so reversal gives (−1)⁴ = +1 and the orientations add. With α/π = 2.32282 × 10⁻³, (α/π)³ = 1.25327 × 10⁻⁸ and the term is 0.371005 × 1.25327 × 10⁻⁸ = 4.650 × 10⁻⁹.
That is 4.0 ppm of aₑ, and 4.650 × 10⁻⁹ ÷ 1.3 × 10⁻¹³ = 3.6 × 10⁴ times the experimental uncertainty, so the surviving four-photon loop is tested tens of thousands of times over.
Check the bookkeeping against the measurement: α/2π = 1.161410 × 10⁻³ overshoots aₑ by 1.758 × 10⁻⁶; the seven two-loop diagrams remove 1.772 × 10⁻⁶, leaving 1.48 × 10⁻⁸ for three loops, which A₃ = 1.181241 (quoted) supplies as 1.480 × 10⁻⁸, light by light being 31% of it. Every term is dimensionless, as aₑ must be.

AnswerSeven of the nine two-loop diagrams contribute, the triangle pair cancelling; the four-photon loops survive and add 4.65 × 10⁻⁹ to aₑ, 4.0 ppm of it and about 3.6 × 10⁴ times the 1.3 × 10⁻¹³ uncertainty.