University Quantum Field Theory I · Discrete Symmetries of Quantised Fields: C, P, T and CPT · 11.9

Photon-number selection rules for positronium

Here the section's sign-keeping pays out in nanoseconds. You will count photons by their charge-conjugation parity, see why the spin-triplet atom waits about a thousand times longer than the singlet to annihilate, and learn which forbidden decays experiments still hunt for and what their parts-per-million limits actually test.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The rule has three links, and each is a different kind of argument. The first belongs to the photon field alone: C A^μ C⁻¹ = −A^μ, forced by the C-invariance of e ψ̄γ^μ ψ A_μ (topic 11.04), passes to every creation operator, so n photons have C = (−1)ⁿ whatever their momenta, polarisations or total angular momentum. The second belongs to the atom: positronium is a C eigenstate with C = (−1)^{(L + S)} (topic 11.08), and because C commutes with QED's S-matrix a level can annihilate only into a photon number whose parity matches its C.

Para-positronium, 1¹S₀ with J^PC = 0⁻⁺, goes to 2, 4, … photons; ortho-positronium, 1³S₁ with 1⁻⁻, to an odd number, and since one real photon cannot carry the atom's invariant mass, to at least three. Landau and Yang's theorem, quoted, separately bars two real photons from every J = 1 level. The third link is power counting: each photon is one more vertex and one more e in the amplitude, so the triplet pays an extra α, Γ₃/Γ₂ = 0.123α, and lives 139 ns at lowest order against 124.5 ps.

The rule is exactly as good as C conservation, which QED has and the weak interaction breaks far below any measured limit. And it constrains the atom, not every electron-positron pair: in matter the positron of an ortho atom can annihilate with a different electron in a spin singlet and give two photons.

Simple definition: A photon-number selection rule says that a state of definite charge-conjugation parity C can annihilate, in a C-conserving theory, only into n photons with (−1)ⁿ = C, because every photon carries C = −1.
Example: Ortho-positronium has C = −1: two photons are forbidden by C and one by energy-momentum conservation, so it decays to three and lives 142 ns, about 1100 times longer than para-positronium's 125 ps two-photon decay.

Every photon carries C = −1C |k₁λ₁, …, kₙλₙ⟩ = (−1)ⁿ |k₁λ₁, …, kₙλₙ⟩

Derived from topic 11.04's C A^μ C⁻¹ = −A^μ. The sign depends on n alone, not on momenta, polarisations or J, unlike a fermion pair's (−1)^{(L + S)}.

k in GeV, λ a polarisation label, n the photon number; from C a†(k, λ) C⁻¹ = −a†(k, λ) and C|0⟩ = |0⟩

The selection rule⟨nγ|Ŝ|Ps⟩ = 0 unless (−1)ⁿ = C(Ps) = (−1)^{(L + S)}

Derived: C Ŝ C⁻¹ = Ŝ in QED makes the amplitude equal (−1)ⁿ C(Ps) times itself. Para goes to 2, 4, … photons; ortho to 3, 5, …, since one real photon cannot carry P² = M².

Ŝ the S-matrix; L and S the pair's orbital angular momentum and total spin (topic 11.08); para C = +1, ortho C = −1

Landau and Yang's theoremΓ(J = 1 → γγ) = 0 for two real photons

Quoted (Landau 1948, Yang 1950). It bars two photons from the C-even 2³P₁ level, 1⁺⁺, which C alone would allow; a virtual photon, k² ≠ 0, escapes it.

J the decaying state's total angular momentum; real means k² = 0 for each photon; holds whatever the state's C and P

Ortho-positronium's lowest-order rateΓ₃ = (2(π² − 9)/(9π)) α⁶ m c²/ħ = 7.21 × 10⁶ s⁻¹

Quoted from Ore and Powell (1949). The measured 7.040 μs⁻¹, 142.0 ns, is 2.4% lower; the order-α factor 1 − 10.29α/π, quoted, brings it to 7.039 μs⁻¹.

m c² = 0.511 MeV, ħ = 6.582 × 10⁻²² MeV s, α = 1/137.036; lifetime 1/Γ₃ = 139 ns

One more photon, one more αΓ₃/Γ₂ = (4(π² − 9)/(9π)) α = 0.123α = 8.98 × 10⁻⁴

Computed from the two rates. Each photon adds a vertex e, so Γₙ ∝ α^{(n + 3)} m; three-body phase space and π² − 9 = 0.870 shrink the coefficient. Lifetimes differ by 1114.

Γ₂ = α⁵ m c²/(2ħ) = 8.03 × 10⁹ s⁻¹ from topic 10.09; the ratio is dimensionless

The three-photon energy spectrumdΓ₃/dx ∝ F(x) = 2[x(1 − x)/(2 − x)² − 2(1 − x)² ln(1 − x)/(2 − x)³ + (2 − x)/x + 2(1 − x) ln(1 − x)/x²]

Quoted (Ore and Powell); the integral and mean checked numerically here. Three photons share 1022 keV, so ortho gives a continuum ending in an edge where para gives a line.

x = E/(m c²) from 0 to 1 for a photon of energy E, m c² = 511 keV; ∫₀¹ F dx = π² − 9 and the mean x is 2/3

Every photon carries C = −1, whatever its state

Natural units, ħ = c = 1, and the metric (+, −, −, −) hold throughout. Topic 11.04 found that e ψ̄γ^μ ψ A_μ is C-invariant only if C A^μ C⁻¹ = −A^μ, because the current is C-odd. Insert the mode expansion A^μ(x) = Σ_λ ∫ d³k/((2π)³ 2|k|) [ε^μ(k, λ) a(k, λ) e^(−ik⋅x) + ε^μ*(k, λ) a†(k, λ) e^(ik⋅x)]. C is unitary, hence linear, and leaves the c-number polarisation vectors and exponentials alone, so matching the coefficient of each exponential gives C a(k, λ) C⁻¹ = −a(k, λ) and C a†(k, λ) C⁻¹ = −a†(k, λ). With C|0⟩ = |0⟩, each of the n creation operators that builds an n-photon state contributes one −1: C = (−1)ⁿ, derived. Set this beside topic 11.08's pair, where C exchanged b† with d†, cost −1 to reorder them, then (−1)^L and (−1)^{(S + 1)} to relabel. Photons are their own antiparticles and their operators commute, so no reordering sign appears, and nothing depends on momenta, polarisations or total J. The photon count alone fixes C.

The selection rule, and why ortho needs three

QED's Lagrangian is C-invariant (topic 11.04), so C commutes with the Hamiltonian and with the S-matrix: C Ŝ C⁻¹ = Ŝ. For a level with C|Ps⟩ = C(Ps)|Ps⟩, insert C⁻¹C on each side of Ŝ: ⟨nγ|Ŝ|Ps⟩ = ⟨nγ|C⁻¹(C Ŝ C⁻¹)C|Ps⟩ = (−1)ⁿ C(Ps) ⟨nγ|Ŝ|Ps⟩. Unless (−1)ⁿ = C(Ps) the amplitude equals minus itself and vanishes, to every order in α, derived. With topic 11.08's C = (−1)^{(L + S)}, para-positronium (1¹S₀, 0⁻⁺) has C = +1 and decays to 2, 4, … photons; ortho-positronium (1³S₁, 1⁻⁻) has C = −1 and needs an odd number. One photon is excluded by kinematics, not by C: at rest the atom has P² = M² with M = 2m − 6.8 eV, while a real photon has k² = 0. So three is the minimum, derived. A virtual photon, with k² ≠ 0, is allowed: ortho's virtual annihilation into one photon adds mα⁴/4, 87.6 GHz, to the 1S hyperfine splitting of 203.4 GHz and leaves para untouched, quoted, the selection rule read off a level shift.

Landau and Yang: no J = 1 state reaches two photons

C counts photons modulo two and is silent about angular momentum, so C-even states with J = 1 need a second rule. Landau (1948) and Yang (1950) showed, and this course quotes, that two real photons cannot form a J = 1 state. The reason is Bose symmetry: in the rest frame the photons fly back to back along ±k̂, and every rotationally covariant amplitude linear in the spin-one polarisation ξ and in ε₁ and ε₂, such as (ε₁ × ε₂)⋅ξ or (ε₁⋅ε₂)(k̂⋅ξ), changes sign when the identical photons are exchanged, which an amplitude for identical bosons may not. Apply both rules to the n = 2 levels. The 2³P₀ (0⁺⁺) and 2³P₂ (2⁺⁺) levels may annihilate into two photons; 2³P₁ (1⁺⁺) is C-even but barred from two, so four is its least; 2¹P₁ (1⁺⁻) is C-odd and needs three. In practice every 2P level first falls to 1S by an electric-dipole photon in about 3.2 ns, twice hydrogen's 1.6 ns because the reduced mass is halved, and C holds there too: E1 changes L by one at fixed S, flipping (−1)^{(L + S)}, as one photon's −1 requires.

The extra photon costs a power of α

Annihilation acts within about 1/m, deep inside an atom of radius 2/(mα), so topic 10.09's factorisation serves any photon number: Γₙ = |ψ(0)|² (σv)ₙ with |ψ(0)|² = (mα)³/(8π). At lowest order the n-photon amplitude has n vertices, each carrying e, so (σv)ₙ ∝ αⁿ/m², the m⁻² fixed by dimensions, and Γₙ ∝ α^{(n + 3)} m, of mass dimension one as a rate must be. Topic 10.09 derived Γ₂ = α⁵ m/2 = 8.03 × 10⁹ s⁻¹. For three photons Ore and Powell's 1949 calculation, quoted, gives Γ₃ = (2(π² − 9)/(9π)) α⁶ m. With m c²/ħ = 7.763 × 10²⁰ s⁻¹ and α⁶ = 1.510 × 10⁻¹³ this is 0.0615 × 1.172 × 10⁸ s⁻¹ = 7.21 × 10⁶ s⁻¹, a lifetime of 139 ns. The measured 7.0401 μs⁻¹, 142.04 ns, sits 2.4% lower, and the order-α factor 1 − 10.29α/π, quoted, gives 7.039 μs⁻¹. The ratio Γ₃/Γ₂ = 0.123α is not α itself: three-body phase space and the near-cancellation in π² − 9 cost a further factor of eight. As α → 0 both atoms become stable, the ortho one faster since the ratio itself goes to zero; the lifetimes differ by 1114 at lowest order and 1135 as measured.

Line against continuum: the rule in a detector

The rule leaves fingerprints a detector can read. Para-positronium's two photons fly back to back with exactly m c² = 511 keV each: a line. Ortho's three share 1022 keV, lie in one plane because their momenta sum to zero, and spread over Ore and Powell's continuum, which rises from zero to an edge at 511 keV. The edge is kinematics: if one photon carries E, the other two have invariant mass squared (2m − E)² − E² = 4m(m − E), which cannot be negative, so E ≤ m and each partner carries at least m − E. Experiments therefore tag three-photon decays by counts in a window below the photopeak or by coplanar triple coincidences, and the long-lived component of a lifetime spectrum carries them. Searches for the C-forbidden decays, quoted, set branching fractions below 2.8 × 10⁻⁶ for para → 3γ and below 3.7 × 10⁻⁶ for ortho → 4γ. A limit means something only against the rate C would otherwise allow, about 0.123α of the two-photon rate for an extra photon; the worked examples turn the para limit into a bound on a C-violating amplitude.

One photon makes a C eigenstate: why μ pairs are symmetric

The same C-oddness explains a symmetry topic 9.06 found by computation. At tree level the muon pair is made by the current μ̄γ^μ μ acting on the vacuum, and that current is C-odd. Write |μ⁻(p, s) μ⁺(p′, s′)⟩ = b†(p, s) d†(p′, s′)|0⟩. C swaps b† and d†, and restoring the order of the anticommuting operators costs −1, so C|μ⁻(p, s) μ⁺(p′, s′)⟩ = −|μ⁻(p′, s′) μ⁺(p, s)⟩. Inserting C⁻¹C into ⟨μ⁻μ⁺|J^μ|0⟩, the −1 from J^μ and the −1 from reordering cancel: ū(p, s)γ^μ v(p′, s′) = ū(p′, s′)γ^μ v(p, s), shown, in topic 11.02's phases. With the electron current fixed, |M|² summed over spins is symmetric under p ↔ p′, which in the centre-of-mass frame is θ → π − θ; indeed (α²β/(4s))(2 − β² sin²θ) is even in cos θ, at threshold and at β → 1 alike. An asymmetry needs a C-even amplitude interfering with the C-odd one: two-photon exchange at order α³ in QED, and the Z boson's axial coupling, named, which dominates the asymmetry PETRA measured.

Change one variable at a time

Make the relationship visible.

Interactive model

Three-photon share of annihilations f0.75

Photon energy E340 keV

Start at f = 0, a metal where positrons annihilate without forming positronium: essentially only the line. Raise f to 0.75, positronium formed with random spins in vacuum: 2.75 photons per annihilation, the line keeping 18% of them. Then drag E up and watch the partners' lower bound fall.

PHOTONS PER ANNIHILATION2.75

SHARE IN THE 511 keV LINE18 %

dN/dE AT E5.86 MeV⁻¹

EACH PARTNER AT LEAST171 keV

Live interpretationPHOTONS PER ANNIHILATION: 2.75. SHARE IN THE 511 keV LINE: 18 %. dN/dE AT E: 5.86 MeV⁻¹. EACH PARTNER AT LEAST: 171 keV

Catch the common trap

Explain before calculating.

Ignore their much faster electric-dipole decay to 1S. What is the smallest number of photons into which positronium's 2¹P₁ level, J^PC = 1⁺⁻, and its 2³P₁ level, 1⁺⁺, could each annihilate?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyPositrons stopped in a silica powder annihilate so that 30% of all annihilations proceed through ortho-positronium's three-photon decay and the rest give two photons. Find the mean number of photons per annihilation, the share of photons in the 511 keV line, and the mean and the largest energy of a photon from the three-photon decays.

Each two-photon annihilation gives 2 photons and each three-photon one gives 3, so per annihilation N = 0.70 × 2 + 0.30 × 3 = 1.40 + 0.90 = 2.30 photons.
Two photons from a pair at rest fly back to back and share 2m c² equally, so all 1.40 of them sit in the line: 1.40/2.30 = 0.609, or 61% of all photons.
A three-photon decay shares 2m c² = 1022 keV among three photons (the 6.8 eV binding is negligible), so their mean energy is 1022/3 = 341 keV whatever the spectrum's shape. Ore and Powell's F(x) has mean x = 2/3, which agrees.
For the largest energy let one photon carry E. The other two have invariant mass squared (2m c² − E)² − E² = 4m c²(m c² − E), which cannot be negative, so E ≤ m c² = 511 keV, reached when both partners fly together the opposite way.
Check: every annihilation deposits 1022 keV in either channel, and the continuum ends exactly at the line, so nothing in the spectrum lies above 511 keV.

Answer2.30 photons per annihilation; 61% of photons in the 511 keV line; three-photon photons average 341 keV and never exceed 511 keV.

MediumA search finds that fewer than 2.8 × 10⁻⁶ of para-positronium decays give three photons. Model a C-violating interaction as giving the para state a three-photon amplitude ε times the C-allowed three-photon amplitude of an ortho state, so that Γ(para → 3γ) ≈ |ε|² Γ₃. Using lowest-order rates, bound |ε| and compare it with a weak-interaction amplitude ratio of order G_F m².

Lowest-order rates: Γ₂ = α⁵ m c²/(2ħ) = 8.03 × 10⁹ s⁻¹, derived in topic 10.09, and Γ₃ = (2(π² − 9)/(9π)) α⁶ m c²/ħ = 7.21 × 10⁶ s⁻¹, Ore and Powell's rate, quoted.
Para decays almost entirely to two photons, so Br(para → 3γ) ≈ |ε|² Γ₃/Γ₂ = |ε|² × 8.98 × 10⁻⁴. The size a C-allowed three-photon branch would have is 0.123α.
Br below 2.8 × 10⁻⁶ requires |ε|² below (2.8 × 10⁻⁶)/(8.98 × 10⁻⁴) = 3.12 × 10⁻³, so |ε| below 0.056: a C-violating amplitude no larger than about 6% of a C-conserving one.
The weak interaction breaks C, with an amplitude ratio of order G_F m² = (1.166 × 10⁻⁵ GeV⁻²)(5.11 × 10⁻⁴ GeV)² = 3.0 × 10⁻¹², dimensionless as natural units require.
That predicts Br of order (3.0 × 10⁻¹²)² × 9.0 × 10⁻⁴ = 8 × 10⁻²⁷, about 3.5 × 10²⁰ times below the limit. The search tests C in the electron-photon coupling itself: a para → 3γ signal would mean new physics, not the weak force.

Answer|ε| ≤ 0.056, so |ε|² ≤ 3.1 × 10⁻³; a weak-interaction amplitude of order 3 × 10⁻¹² gives a branching fraction near 10⁻²⁶, some twenty orders of magnitude below the limit.

HardTrue muonium is the hydrogen-like bound state of μ⁺ and μ⁻, with m_μ c² = 105.66 MeV. Find the lifetime of its 1¹S₀ level from Γ = α⁵ m_μ/2. Then show that its 1³S₁ level may decay through one virtual photon to e⁺e⁻, derive that rate from topic 9.06's cross-section near threshold, and find the lifetime and the three-photon branching fraction.

Para: Γ = α⁵ m_μ c²/(2ħ) = 2.069 × 10⁻¹¹ × 105.66 MeV/(2 × 6.582 × 10⁻²² MeV s) = 1.66 × 10¹² s⁻¹, so τ = 0.602 ps, positronium's 124.5 ps divided by m_μ/mₑ = 206.8.
Ortho has C = −1, the same as one photon, so C allows a single photon. A real one is barred by P² = M², but a virtual one with q² = M² = (211.3 MeV)² can become e⁺e⁻, whose invariant mass can match. The singlet, C = +1, cannot couple to one photon at all.
Topic 9.06 gives σ(e⁺e⁻ → μ⁺μ⁻) = (4πα²/(3s)) β(3 − β²)/2, which at s = 4m_μ² and small muon speed β is πα²β/(2m_μ²). Both sides have four spin states, so time-reversal invariance gives σ(μ⁺μ⁻ → e⁺e⁻) = σ(e⁺e⁻ → μ⁺μ⁻) pₑ²/p_μ², with pₑ = m_μ and p_μ = m_μ β: πα²/(2m_μ² β).
The relative speed is 2β, so the spin-averaged σv = πα²/m_μ², finite at threshold. Only the three triplet states annihilate, so the triplet's σv is 4/3 of that. With |ψ(0)|² = (m_μ α)³/(8π), Γ = |ψ(0)|² σv = α⁵ m_μ/6, of mass dimension one: 5.54 × 10¹¹ s⁻¹.
Three photons: positronium's Ore-Powell rate scaled by 206.8, 7.21 × 10⁶ s⁻¹ × 206.8 = 1.49 × 10⁹ s⁻¹, or 6 × 0.0615α = 0.369α = 2.69 × 10⁻³ of the e⁺e⁻ rate. The total is 5.55 × 10¹¹ s⁻¹, τ = 1.80 ps, with a three-photon branching fraction of 0.27%.
Limit check: one virtual photon costs α² in rate, as two real ones do, so ortho and para differ by a factor of 3, not 1100. Switch the e⁺e⁻ channel off, as positronium must since no lighter charged pair exists, and ortho falls back on three photons with τ = 0.671 ns, 1114 times para's lifetime: the positronium pattern.

Answerτ(1¹S₀) = 0.602 ps; τ(1³S₁) = 1.80 ps, decaying through μ⁺μ⁻ → γ* → e⁺e⁻ at Γ = α⁵ m_μ/6, with a three-photon branching fraction of 0.27%.