University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.9
Para-positronium's lifetime: Γ = α⁵m/2
This is where Unit 10's tree-level QED meets a clock. Positronium's singlet ground state lives 125 ps, and the job here is to earn that number: the atom's density at contact times a threshold annihilation rate, evaluated on the right spin state, which turns out to be worth a factor of four.
Build the model
Connect the measurement to the mechanism.
Para-positronium decays by annihilation, and its lifetime can be computed from two things already in hand: a hydrogen-like wavefunction and topic 10.08's threshold amplitude for e⁺e⁻ → γγ. What licenses combining them is a separation of scales. The internal electron line of either annihilation diagram is off shell by 2m², so annihilation acts within about 1/m = 386 fm, while the atom's Bohr radius 2/(mα) is 274 times larger. Over that range the wavefunction is flat, and the pair's momenta, about mα/2, are negligible on the scale m over which the amplitude varies.
So the bound pair decays at its contact density |ψ(0)|² times the free pair's σv at threshold, provided σv is taken on the right spin state. That is the second ingredient. The threshold amplitude is proportional to the pair's overlap with the spin singlet, so the singlet's σv is four times the unpolarised average and the three triplet states make no photon pairs at all. The product is Γ = α⁵m/2, three powers of α from the contact density and two from the vertices: 8.03 × 10⁹ s⁻¹, or 124.5 ps.
The model stops at relative order α. Radiative corrections at the annihilation point, which non-relativistic QED organises, move it by 0.6 per cent, to within 0.07 per cent of the measured 125.14 ps; and it needs ψ(0) ≠ 0, which excludes P states, and a coupling small enough that 2/α is a large number.
- Simple definition
- Para-positronium is the spin-singlet 1S state of an electron and a positron; it annihilates into two photons at Γ = |ψ(0)|² σv(singlet) = α⁵m/2, its density at contact times the free singlet pair's annihilation rate per unit density at threshold.
- Example
- With |ψ(0)|² = 1/(πa³) = 2.685 × 10²⁹ m⁻³ for a = 105.84 pm, and σv(singlet) = 4πrₑ²c = 2.992 × 10⁻²⁰ m³ s⁻¹, Γ = 8.03 × 10⁹ s⁻¹: a lifetime of 124.5 ps, against the measured 125.14 ps.
Holds while annihilation's range 1/m is far below the Bohr radius 2/(mα); non-relativistic QED makes it systematic, with corrections from relative order α.
|ψ(0)|² the density at zero separation (m⁻³); σv the free pair's annihilation rate per unit density (m³ s⁻¹); Γ in s⁻¹
A positron of spin state ζ has η = −iσ²ζ*, so η†ξ = ζ₁ξ₂ − ζ₂ξ₁: −√2 times the overlap with the singlet, and zero on all three triplets.
n̂ the first photon's direction; ε₁, ε₂ transverse polarisations; ξ, η the two-spinors of e⁻ and e⁺ at rest; M dimensionless
Four times topic 10.08's spin average πα²/m², because the triplets contribute nothing and the whole spin sum, Σ|M|² = 16e⁴, is the singlet's.
v the relative speed; with ħ and c, 4πrₑ²c = 2.992 × 10⁻²⁰ m³ s⁻¹ for rₑ = 2.818 fm; the 1/2 for identical photons is included
Three powers of α: the density at contact scales as a⁻³. Excited S states carry an extra 1/n³; P states have none at all.
hydrogen with reduced mass m/2: Bohr radius a = 2/(mα) = 105.84 pm, so |ψ(0)|² = 1/(πa³) = 2.685 × 10²⁹ m⁻³
A mass, as a width must be when ħ = 1. Its ratio to the binding energy mα²/4 is 2α³ = 7.8 × 10⁻⁷, so the level is sharp.
with units Γ = α⁵mc²/(2ħ) = 8.03 × 10⁹ s⁻¹, a width ħΓ = 5.29 μeV and a lifetime τ = 124.5 ps
One-loop QED at the annihilation point, its Coulomb part already inside |ψ(0)|²: tree level's rate is 0.52 per cent too fast, this 0.07 per cent too slow.
(5 − π²/4)α/π = 0.00588; Γ = 7985.2 μs⁻¹, τ = 125.23 ps, against the measured 7990.9 ± 1.7 μs⁻¹ (125.14 ps)
Two lengths a factor of 274 apart make the rate factorise
Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Para-positronium is the 1¹S₀ state of e⁻e⁺, hydrogen with reduced mass m/2: Bohr radius a = 2/(mα) = 105.8 pm, binding energy mα²/4 = 6.80 eV. Annihilation is far more compact. At threshold the internal electron of either e⁺e⁻ → γγ diagram carries (p₁ − k₁)² − m² = −2m², off shell by 2m², so the two vertices lie within about 1/m = 386 fm of each other, 2/α = 274 times less than a. Over that distance |ψ|² is flat, and the pair's momenta, q ≈ mα/2 = 1.9 keV, are negligible beside m, the scale on which the amplitude varies. Write the atom, of mass 2m, as free pairs weighted by ψ̃(q) with relativistic normalisation √(2 × 2m)/(2m) = 1/√m; its amplitude is then ψ(0)M(0)/√m, and Γ = (1/(4m))(1/2)∫|ψ(0)M(0)|²dΦ₂/m = |ψ(0)|²|M(0)|²/(64πm²), exactly |ψ(0)|² times σv at threshold. That factorisation is argued, not derived: non-relativistic QED (NRQED) makes it systematic, and its corrections start at relative order α.
The threshold amplitude in two-component form
Unit 9's rules, as topic 10.08 used them, give M = −e²v̄(p₂)[ε̸₂*(p̸₁ − k̸₁ + m)ε̸₁*/((p₁ − k₁)² − m²) + (1 ↔ 2)]u(p₁). At threshold p₁ = p₂ = (m, 0) and k₁,₂ = m(1, ±n̂), so both denominators are −2m². Take transverse polarisations with ε⁰ = 0, so that p₁⋅ε = 0; then (p̸₁ + m)ε̸₁*u = 2(p₁⋅ε₁*)u − ε̸₁*(p̸₁ − m)u = 0 by the Dirac equation, only −k̸₁ε̸₁*u survives, and M = −(e²/(2m²))v̄[ε̸₂*k̸₁ε̸₁* + ε̸₁*k̸₂ε̸₂*]u. In the Dirac representation at rest u = √(2m)(ξ, 0) and v = √(2m)(0, η), so v̄Γu = v†γ⁰Γu keeps only the block of γ⁰Γ that carries upper components into lower. The γ⁰ inside each k̸ gives a block-diagonal product and drops out; three spatial γ's give a product of σ's, which σ⋅a σ⋅b σ⋅c = (a⋅b)σ⋅c − (a⋅c)σ⋅b + (b⋅c)σ⋅a + i a⋅(b × c) reduces. The σ⋅n̂ pieces of the two diagrams cancel and their cross products add, leaving M = −2ie² n̂⋅(ε₁* × ε₂*) η†ξ, derived; explicit 4 × 4 matrices reproduce it for every spin and polarisation.
Only the singlet makes two photons
Every spin dependence now sits in η†ξ. A positron of spin state ζ has η = −iσ²ζ*, so η†ξ = ζᵀ(iσ²)ξ = ζ₁ξ₂ − ζ₂ξ₁. That is antisymmetric under exchanging the two spin states, hence −√2 times the overlap ⟨S|ξ, ζ⟩ with the singlet S = (↑↓ − ↓↑)/√2 and zero on all three triplet states, shown. For any spin state χ of the pair, M = 2√2 ie² n̂⋅(ε₁* × ε₂*)⟨S|χ⟩. Its form is the parity-odd scalar that UAQM-08.07's argument demanded of a 0⁻ state, crossed linear polarisations included, now with its coefficient. Summed over polarisations, |n̂⋅(ε₁* × ε₂*)|² gives 2 in every direction, an isotropic decay, as J = 0 requires. And because the triplets give zero, any complete spin sum is all singlet: Σ|M|² = 16e⁴ is the singlet's own |M|², while the average over four states is 4e⁴, topic 10.08's threshold value. Hence σv(singlet) = 16e⁴/(64πm²) = 4πα²/m², four times the average, derived. Tree level shows the triplet's two-photon amplitude vanishing at threshold; why it vanishes to all orders, through C-parity, is topic 11.09's.
Assembling α⁵m/2 and putting the units back
Multiply: Γ = |ψ(0)|² σv(singlet) = [(mα)³/(8π)] × [4πα²/m²] = α⁵m/2, derived. Three powers of α come from the contact density, which scales as a⁻³, and two from the squared vertices. Dimensions: |ψ(0)|² is a mass cubed and σv an inverse mass squared, so Γ is a mass, an energy width, as a rate must be when ħ = 1. Restore units: Γ = α⁵mc²/(2ħ) = 2.0693 × 10⁻¹¹ × 0.255499 MeV ÷ 6.5821 × 10⁻²² MeV s = 8.03 × 10⁹ s⁻¹, so τ = 124.5 ps. In SI the factorisation can be watched: 1/(πa³) = 2.685 × 10²⁹ m⁻³ for a = 105.84 pm, and 4πrₑ²c = 2.992 × 10⁻²⁰ m³ s⁻¹ with rₑ = 2.818 fm, and their product is 8.03 × 10⁹ s⁻¹. Two limits check it. As α → 0 the rate vanishes as α⁵, and at fixed α it grows in proportion to m, as dimensions demand. And the width ħΓ = 5.29 μeV is 2α³ = 7.8 × 10⁻⁷ of the 6.80 eV binding energy, so the level is sharp: treating the atom as a stationary state with a known wavefunction was consistent.
Against the measured lifetime
The measured rate, quoted, is 7990.9 ± 1.7 μs⁻¹, a lifetime of 125.14 ps. Tree level's 8032.5 μs⁻¹ is 0.52 per cent too fast, 24 standard deviations off. The discrepancy has the size the factorisation warned of, order α, and its leading term is known. One-loop QED at the annihilation point, the vertex and self-energy corrections and the photon exchanged between the annihilating pair, with that exchange's Coulomb part removed because |ψ(0)|² already contains it, multiplies the rate by 1 − (5 − π²/4)α/π, quoted. The coefficient, 2.5326 × α/π = 0.00588, moves the prediction to 7985.2 μs⁻¹, 125.23 ps: the 0.52 per cent excess becomes a 0.07 per cent deficit. Terms of order α² ln(1/α) and α², which NRQED computes, close the rest to within the measurement's error. The loops themselves belong to Unit 12. The lesson here is the bookkeeping: tree-level factorisation fixes the lifetime to half a per cent, and every further digit is a radiative correction, not a better wavefunction.
Where the formula stops
Three conditions bound the result. First, ψ(0) ≠ 0. Excited singlets qualify, with |ψₙ₀₀(0)|² carrying 1/n³, so the 2¹S₀ state lives 8 × 124.5 ps = 1.0 ns; P states have ψ(0) = 0, the leading term vanishes, and they decay first by emitting a photon. Second, 1/m ≪ a, which needs α ≪ 1: raise α in the figure and the atom shrinks toward the annihilation zone, while the fall of |ψ|² across that zone, about α for every S state, measures the corrections. Third, vacuum. In matter the positron can annihilate with a foreign electron, the pick-off process that shortens ortho-positronium's lifetime in polymers and porous glasses and is used to size their pores; that rate is set by the material, not by α⁵m/2. Finally, the spin bookkeeping fixes what the triplet cannot do: its three states make no photon pairs, and they decay to three photons with a measured lifetime of 142 ns, some 1100 times longer, a story that waits for topic 11.09.
Change one variable at a time
Make the relationship visible.
Hold n = 1 and raise α: the solid curve slides toward the band and the drop of |ψ|² across it grows from 0.73 per cent to 25 per cent at 40α₀. Then return to α₀ and try n = 2 and 3: the atom swells as n² and grows nodes, yet stays flat over the band, so only |ψ(0)|² ∝ 1/n³ changes the rate.
SCALE RATIO a ÷ (1/m) = 2/α274.1
DROP OF |ψ|² ACROSS 1/m0.73 %
CONTACT DENSITY |ψ(0)|²2.685 × 10²⁹ m⁻³
SINGLET σv AT THRESHOLD2.992 × 10⁻²⁰ m³ s⁻¹
Live interpretationSCALE RATIO a ÷ (1/m) = 2/α: 274.1. DROP OF |ψ|² ACROSS 1/m: 0.73 %. CONTACT DENSITY |ψ(0)|²: 2.685 × 10²⁹ m⁻³. SINGLET σv AT THRESHOLD: 2.992 × 10⁻²⁰ m³ s⁻¹
Catch the common trap
Explain before calculating.
An electron and a positron meet almost at rest with their spins prepared antiparallel along z, electron up and positron down: the product state |↑↓⟩. At tree level, what is σv for their annihilation into two photons, with v their relative speed?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTrue muonium is the μ⁺μ⁻ atom, with mμc² = 105.658 MeV. Using this topic's result, find the two-photon lifetime of its spin-singlet ground state, its Bohr radius and that radius's ratio to the annihilation range ħ/(mμc), and decide whether muon decay, with lifetime 2.197 μs, competes.
- At fixed α, Γ = α⁵mc²/(2ħ) is proportional to the constituent mass, so Γ = (105.658/0.510999) × 8.0325 × 10⁹ s⁻¹ = 206.77 × 8.0325 × 10⁹ s⁻¹ = 1.661 × 10¹² s⁻¹.
- τ = 1/Γ = 6.02 × 10⁻¹³ s = 0.602 ps.
- Bohr radius a = 2ħc/(αmμc²) = 2 × 197.327 MeV fm ÷ (0.0072974 × 105.658 MeV) = 511.9 fm. Annihilation range ħ/(mμc) = 197.327 MeV fm ÷ 105.658 MeV = 1.868 fm. Ratio 511.9/1.868 = 274, which is 2/α.
- The scale separation depends on α alone, so the factorisation is exactly as good as in positronium; the atom is simply 207 times smaller and decays 207 times faster.
- Muon decay first: probability about 2 × 0.602 ps ÷ 2.197 μs = 5.5 × 10⁻⁷, the 2 because either muon may decay. Annihilation wins outright.
Answerτ = 0.602 ps; a = 512 fm, 274 times ħ/(mμc) = 1.87 fm; muon decay pre-empts annihilation in only about 5.5 × 10⁻⁷ of atoms.
MediumThe singlet positronium decay rate has been measured as 7990.9 ± 1.7 μs⁻¹. Compute the tree-level α⁵mc²/(2ħ) in μs⁻¹ and its deviation in standard deviations; apply the quoted one-loop factor 1 − (5 − π²/4)α/π, then add the quoted next term +2α² ln(1/α), giving the deviation each time; and estimate the naive relative-momentum correction ⟨q²⟩/m².
- α⁵ = (7.29735 × 10⁻³)⁵ = 2.06932 × 10⁻¹¹ and mc²/2 = 0.255499 MeV, so ħΓ₀ = 5.28709 × 10⁻¹² MeV. Dividing by ħ = 6.58212 × 10⁻²² MeV s gives Γ₀ = 8.0325 × 10⁹ s⁻¹ = 8032.5 μs⁻¹.
- Γ₀ exceeds the measured rate by 41.6 μs⁻¹: 0.52 per cent, or 41.6/1.7 = 24 standard deviations. Tree level fails at this precision.
- (5 − π²/4)α/π = 2.53260 × 0.00232282 = 0.0058828, so Γ₁ = 8032.5 × 0.994117 = 7985.2 μs⁻¹, now 5.7 μs⁻¹ below the measurement: −3.3 standard deviations, a 0.07 per cent deficit.
- 2α² ln(1/α) = 2 × 5.32514 × 10⁻⁵ × 4.92024 = 5.240 × 10⁻⁴, adding 4.2 μs⁻¹: Γ = 7989.5 μs⁻¹, 1.4 μs⁻¹ or 0.8 standard deviations below, a lifetime of 125.16 ps against the measured 125.14 ps.
- The pair's momentum spread is ⟨q²⟩ = (mα/2)², so ⟨q²⟩/m² = α²/4 = 1.33 × 10⁻⁵, worth 0.11 μs⁻¹: sixteen times below the error bar. The pair's motion is not what tree level misses; the radiative corrections are.
AnswerTree level 8032.5 μs⁻¹ (+24 σ); with the one-loop factor 7985.2 μs⁻¹ (−3.3 σ); adding 2α² ln(1/α), 7989.5 μs⁻¹ (−0.8 σ), or 125.16 ps. The naive momentum correction is about 0.1 μs⁻¹.
HardIn a field B along z, the Zeeman term's only matrix element within ground-state positronium's spin space links the singlet S to the m = 0 triplet T₀, and equals gμB B. Take g = 2.00, μB = 5.788 × 10⁻⁵ eV T⁻¹, T₀ lying ΔE = 8.41 × 10⁻⁴ eV above S, the singlet's α⁵m/2 = 8.03 × 10⁹ s⁻¹ and the triplet's three-photon rate 7.04 × 10⁶ s⁻¹. At B = 1.00 T find the lifetime of the mostly-T₀ state and its share of two-photon decays.
- In the basis (S, T₀) the Hamiltonian is [[0, V], [V, ΔE]] with V = gμB B = 2.00 × 5.788 × 10⁻⁵ eV × 1.00 = 1.158 × 10⁻⁴ eV. The m = ±1 triplets have no singlet partner with the same spin projection, so they are not mixed.
- The upper eigenstate is (|T₀⟩ + y|S⟩)/√(1 + y²) with y = x/(1 + √(1 + x²)) and x = 2V/ΔE = 0.2753, so y = 0.2753/2.0372 = 0.1351 and y² = 0.01826.
- By this topic only the singlet component has a two-photon amplitude, and it decays at the full singlet rate: two-photon rate = [y²/(1 + y²)] × 8.03 × 10⁹ s⁻¹ = 0.01793 × 8.03 × 10⁹ s⁻¹ = 1.440 × 10⁸ s⁻¹.
- The triplet component keeps its three-photon rate with weight 1/(1 + y²): 7.04 × 10⁶ s⁻¹ ÷ 1.01826 = 6.91 × 10⁶ s⁻¹. The two final states differ, so the rates add: 1.509 × 10⁸ s⁻¹, and τ = 6.63 ns.
- Checks: as B → 0, y → 0 and the field-free 142 ns returns. The mostly-singlet partner loses the same 1.8 per cent of singlet weight and lives 126.8 ps instead of 124.5 ps, and the m = ±1 states keep 142 ns.
Answerτ ≈ 6.6 ns for the m = 0 state at 1.00 T, 21 times shorter than 142 ns, with 1.440/1.509 = 95 per cent of its decays giving two photons; the m = ±1 states are unmixed and keep 142 ns.