University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.7
Scalar and pseudoscalar decays: β³ against β
This is the first rate in the course where the spin of the products decides the answer. Two bosons that differ only by a γ⁵ in their coupling decay to the same fermion pair at rates that part company near threshold. You will get that from a trace, see it in the rest-frame spinors, and predict it from parity alone.
Build the model
Connect the measurement to the mechanism.
Couple a spin-zero boson φ to a Dirac fermion in the two ways a renormalisable Yukawa vertex allows, g φ ψ̄ψ or i g φ ψ̄γ⁵ψ. At tree level the decay φ → f f̄ is one vertex with two external spinors, and Casimir's trick turns the spin sum into a trace in which the only difference is the sign of m in the antifermion's projector: Σ|M|² = 4g²(p₁⋅p₂ ∓ m²). Fed into topic 5.03's rate with topic 5.02's phase space β/(8π), that sign becomes a power of β, the speed of either fermion in the boson's rest frame: Γ = g²mφβ³/(8π) for the scalar and g²mφβ/(8π) for the pseudoscalar.
The extra β² is not an accident of algebra. A fermion and its antifermion have opposite intrinsic parity, so a pair with orbital L has parity (−1)(L+1), and with J = 0 the only states are ¹S₀, odd, and ³P₀, even. A 0⁺ boson must therefore make a p wave, whose amplitude grows as |p|, and a 0⁻ boson an s wave.
The result is a lowest-order rate in a model whose couplings are free: it assumes Γ ≪ mφ, ignores loops of relative size g²/(16π²) and the pair's mutual attraction near threshold, and far above threshold, where β → 1, it cannot tell the two couplings apart at all, so there parity has to be read from spin correlations instead.
- Simple definition
- A spin-zero boson coupled to a fermion pair through g φ ψ̄ψ decays at the rate g²mφβ³/(8π), and through i g φ ψ̄γ⁵ψ at g²mφβ/(8π), where β = √(1 − 4m²/mφ²) is each fermion's speed in the boson's rest frame.
- Example
- For g = 1, mφ = 1 GeV and m = 0.3 GeV, β = 0.8: the scalar width is 39.8 MeV × 0.512 = 20.4 MeV and the pseudoscalar's 39.8 MeV × 0.8 = 31.8 MeV, lifetimes of 3.2 × 10⁻²³ s and 2.1 × 10⁻²³ s.
Derived by Casimir's trick with Σuū = p̸₁ + m and Σvv̄ = p̸₂ − m. It vanishes at threshold, mφ = 2m: the first sign of a p wave.
p₁, p₂ the fermion and antifermion four-momenta, m their mass, mφ the boson's, all in GeV; g dimensionless; |M|² in GeV²
Derived. Independent of the fermion mass and nonzero at threshold, the signature of an s-wave amplitude.
The vertex γ⁵, moved through p̸₂ − m, flips the sign of m in the antifermion projector; symbols and units as above
Topic 5.03's 1/(2mφ) times topic 5.02's phase space β/(8π). Both couplings share it, so any difference in the power of β lives in |M|².
Γ in GeV; β the speed of either fermion in the rest frame, with |p| = mφβ/2; no ½, since f and f̄ are distinct
Computed. g = 1 and mφ = 1 GeV with β → 1 give 39.8 MeV, τ = 1.65 × 10⁻²³ s. Near threshold it creeps in as |p|³.
For g φ ψ̄ψ; ΓS and mφ in GeV; lifetime τ = ħ/ΓS with ħ = 6.582 × 10⁻²⁵ GeV s
Computed. It switches on linearly in |p|, so ΓS/ΓP = β² tells the boson's parity whenever β is well below 1.
For i g φ ψ̄γ⁵ψ; ΓP and mφ in GeV; equal to ΓS as m → 0, where a chiral rotation turns one coupling into the other
Derived: a decay into a wave of orbital L has an amplitude carrying |p|L, and two-body phase space adds one more |p|. L = 1 gives β³, L = 0 gives β.
|p| the fermion momentum in the rest frame, GeV; L = 1 (³P₀) for 0⁺ and L = 0 (¹S₀) for 0⁻, by UAQM-08.07's parity (−1)(L+1)
One vertex, two spinors, one trace
In natural units, ħ = c = 1, with the metric (+, −, −, −), topic 7.04's rules give the decay φ(k) → f(p₁) f̄(p₂) a single vertex. Reading against the fermion arrow, the outgoing fermion supplies ū(p₁) and the outgoing antifermion v(p₂), so iMS = −ig ū(p₁)v(p₂) and iMP = ±g ū(p₁)γ⁵v(p₂), the sign being topic 7.04's convention and invisible in |M|². φ has no spin to average; sum the final spins by topic 7.06's Casimir trick with Σvv̄ = p̸₂ − m. The scalar gives g² Tr[(p̸₁ + m)(p̸₂ − m)] = 4g²(p₁⋅p₂ − m²). For γ⁵ the conjugate vertex is γ⁰γ⁵†γ⁰ = −γ⁵, and γ⁵(p̸₂ − m)γ⁵ = −(p̸₂ + m), so the γ⁵ only flips the mass in the antifermion projector: 4g²(p₁⋅p₂ + m²). With (p₁ + p₂)² = mφ², p₁⋅p₂ = mφ²/2 − m², and the sums are 2g²(mφ² − 4m²) and 2g²mφ², derived. For mφ = 1 GeV and m = 0.3 GeV they are 1.28g² and 2.00g² GeV².
From |M|² to a width in seconds
Topic 5.03 gives Γ = (1/(2mφ))∫Σ|M|² dΠ₂ for a parent at rest, and topic 5.02's two-body phase space integrates to |p|/(4π mφ). Each fermion carries |p| = mφβ/2, so ∫dΠ₂ = β/(8π), and |M|² is isotropic because a spin-zero parent at rest offers no direction. Hence Γ = βΣ|M|²/(16π mφ), with no ½ since f and f̄ are distinct: ΓS = g²mφβ³/(8π) and ΓP = g²mφβ/(8π), computed. Dimensions check: [φ] = 1 and [ψ] = 3/2 make g dimensionless, so far above threshold mφ is the only scale a width can have; in topic 5.03 g carried a mass and Γ went as g²/m instead. For g = 1 and mφ = 1 GeV with β → 1, Γ = 1/(8π) GeV = 39.8 MeV, and ħ = 6.582 × 10⁻²⁵ GeV s turns that into τ = 1.65 × 10⁻²³ s. At m = 0.3 GeV, β = 0.8 and the widths are 20.4 MeV and 31.8 MeV.
The rest-frame spinors show which wave each vertex makes
The traces give the size of Σ|M|²; writing the spinors out shows its structure. In the Dirac representation, quoted from UAQM-04, u(p) = (√(E+m) ξ, σ⋅p ξ/√(E+m)) and v(p) = (σ⋅p η/√(E+m), √(E+m) η), normalised to ūu = 2m and v̄v = −2m. In the boson's rest frame p₁ = p and p₂ = −p, and multiplying out gives ū(p)v(−p) = −2ξ†(σ⋅p)η and ū(p)γ⁵v(−p) = 2E ξ†η, derived. The scalar bilinear is linear in p and dots the spin operator σ into the direction of p, spin 1 with L = 1 coupled to J = 0: the ³P₀ wave, vanishing as |p| at threshold. The pseudoscalar bilinear survives at p = 0 and holds only ξ†η, a spin singlet with no orbital factor: ¹S₀. Summing over a basis of ξ and η, Σ|ξ†Aη|² = Tr(AA†) gives 8|p|² and 8E², which equal 2mφ²β² and 2mφ² because mφ = 2E and β = |p|/E: the traces confirmed.
Parity picks the wave, and the wave sets the power of β
Two spin-½ particles make J = 0 in only two ways: L = S = 0, the ¹S₀ state, or L = S = 1 coupled to zero, the ³P₀ state. UAQM-08.07, quoted, gives a fermion-antifermion pair the parity (−1)(L+1), the extra −1 from their opposite intrinsic parities, so ¹S₀ is 0⁻ and ³P₀ is 0⁺; both have C = (−1)(L+S) = +1, so charge conjugation cannot separate them. ψ̄ψ is a parity scalar and ψ̄iγ⁵ψ a pseudoscalar, so each coupling conserves parity once φ is 0⁺ or 0⁻ to match, and the decay reaches only the pair state of the same parity. Now the threshold law, derived: the decay acts within about 1/mφ, where a wave of orbital L is suppressed as (|p|r)L, so the amplitude carries |p|L and the rate |p|(2L) times the phase space's |p|. L = 1 gives |p|³, the β³; L = 0 gives |p|, the β. Measured pairs follow the pattern: charmonium's ηc at 2984 MeV is the 0⁻ ¹S₀ c c̄ state and χc0 at 3415 MeV the 0⁺ ³P₀ one, quoted.
Far above threshold the two couplings become one
As m/mφ → 0, β → 1 and ΓS/ΓP = β² → 1, a limit with a reason. At m = 0 the chiral rotation ψ → e(iθγ⁵/2)ψ, which sends ψ̄ to ψ̄e(iθγ⁵/2), leaves ψ̄i∂̸ψ unchanged and turns ψ̄ψ into cos θ ψ̄ψ + i sin θ ψ̄γ⁵ψ, so at θ = 90° the scalar theory is the pseudoscalar theory. Only the mass term breaks the rotation, and 1 − β² = 4m²/mφ² measures by how much. A boson coupled as g φ ψ̄(cos α + i sin α γ⁵)ψ has no definite parity and decays at g²mφβ(β² cos²α + sin²α)/(8π): the cross term is a trace of γ⁵ with two gammas and vanishes, as ³P₀ and ¹S₀ are orthogonal. For the Higgs boson's τ coupling, g = mτ/v with v = 246.2 GeV, quoted from a later course and confirmed at the LHC to about 10%, the width is 0.259 MeV and the two pure cases differ by only 4mτ²/mₕ² = 8 × 10⁻⁴. So CMS read α from the τ decay planes instead, finding it near 0° and disfavouring a pure pseudoscalar at about three standard deviations, quoted.
Where the tree-level width stops
These are order-g² widths in a model where g, m and mφ are free parameters. They assume a narrow particle: Γ/mφ = g²/(8π) far above threshold is 4.0% at g = 1, reaches 10% at g = 1.59, and is 36% at g = 3, where topic 5.09's Breit-Wigner peak and the very idea of a decaying particle blur. Loops correct the rate by relative amounts of order g²/(16π²), 0.6% at g = 1, and topic 7.08 checks this same width against the imaginary part of the fermion loop, where the loop's −1 is what keeps it positive. Near threshold the slow pair also feels the attraction that topic 7.05 showed scalar exchange produces, a final-state interaction absent at tree level that rescales both rates there. Because that force has the finite range 1/mφ, it changes their coefficients but not the powers β³ and β, by Wigner's threshold law, quoted; only a long-range force, such as the Coulomb attraction of a charged pair, would alter the powers themselves. Finally the clean powers assume the coupling conserves parity; a boson with both couplings follows the incoherent mixture above, not either law.
Change one variable at a time
Make the relationship visible.
Push 2m/mφ towards 1: the upper s-wave curve plunges into threshold while the lower p-wave curve arrives flat, a factor β² below. Sweep α from 0° to 90° and the solid curve slides between them with no interference. Raise mφ and every width grows in proportion, since g is dimensionless.
PAIR SPEED β0.800
WIDTH Γ AT α, g = 120.4 MeV
LIFETIME τ = ħ/Γ3.23 10⁻²³ s
SCALAR ÷ PSEUDOSCALAR = β²0.640
Live interpretationPAIR SPEED β: 0.800. WIDTH Γ AT α, g = 1: 20.4 MeV. LIFETIME τ = ħ/Γ: 3.23 10⁻²³ s. SCALAR ÷ PSEUDOSCALAR = β²: 0.640
Catch the common trap
Explain before calculating.
A spin-zero boson of mass 2.5 GeV couples to a Dirac fermion of mass 1.0 GeV as g φ ψ̄ψ; a second boson of the same mass couples to it as i g φ ψ̄γ⁵ψ with the same g. At tree level, what is Γ(scalar)/Γ(pseudoscalar)?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA scalar boson of mass 2.0 GeV couples as g φ ψ̄ψ, with g = 0.40, to a fermion of mass 0.50 GeV. Find its width into the pair and its lifetime in seconds, then repeat for a pseudoscalar of the same mass and coupling.
- Speed: β = √(1 − 4m²/mφ²) = √(1 − 4 × 0.25/4.0) = √0.75 = 0.866, so β³ = 0.650.
- Common prefactor: g²mφ/(8π) = 0.16 × 2.0 GeV ÷ 25.13 = 0.01273 GeV = 12.7 MeV, the width either boson would have far above threshold.
- Scalar: ΓS = 12.73 MeV × 0.650 = 8.27 MeV, and τ = ħ/Γ = 6.582 × 10⁻²⁵ GeV s ÷ 8.27 × 10⁻³ GeV = 7.96 × 10⁻²³ s.
- Pseudoscalar: ΓP = 12.73 MeV × 0.866 = 11.0 MeV, and τ = 6.582 × 10⁻²⁵ GeV s ÷ 1.103 × 10⁻² GeV = 5.97 × 10⁻²³ s.
- Checks: ΓS/ΓP = 8.27/11.03 = 0.750 = β², as the p wave requires; each width is under 1% of mφ, so treating φ as a narrow particle is safe.
AnswerScalar: Γ = 8.3 MeV, τ = 8.0 × 10⁻²³ s. Pseudoscalar: Γ = 11 MeV, τ = 6.0 × 10⁻²³ s.
MediumA spin-zero boson of mass 2.00 GeV couples with the same g = 0.30 to a light fermion of mass 0.10 GeV and a heavy one of mass 0.90 GeV, and decays to nothing else. An experiment finds Γ(heavy pair)/Γ(light pair) = 0.085 ± 0.010. Is the boson a scalar or a pseudoscalar? Find its total width and lifetime.
- Speeds: β₁ = √(1 − 4 × 0.01/4.00) = √0.990 = 0.9950 for the light pair and β₂ = √(1 − 4 × 0.81/4.00) = √0.190 = 0.4359 for the heavy one.
- With equal couplings the prefactor g²mφ/(8π) cancels in the ratio. A scalar predicts (β₂/β₁)³ = 0.4381³ = 0.0841; a pseudoscalar predicts β₂/β₁ = 0.438.
- The measured 0.085 ± 0.010 agrees with 0.084 and lies 35 standard deviations from 0.438: the heavy pair pays the extra β² of a p wave, so the boson is a 0⁺ scalar decaying into ³P₀ pairs.
- Total width: g²mφ/(8π) = 0.09 × 2.00 GeV ÷ 25.13 = 7.162 MeV, times β₁³ + β₂³ = 0.9850 + 0.0828 = 1.0678, gives Γ = 7.65 MeV.
- Lifetime: τ = ħ/Γ = 6.582 × 10⁻²⁵ GeV s ÷ 7.648 × 10⁻³ GeV = 8.61 × 10⁻²³ s. The heavy branching fraction is 0.0828/1.0678 = 7.8%, where a pseudoscalar would have given 30%.
AnswerA scalar: its predicted ratio 0.084 matches, and the pseudoscalar's 0.44 is excluded. Γ = 7.65 MeV, τ = 8.6 × 10⁻²³ s, with a heavy-pair branching fraction of 7.8%.
HardA boson of mass 1.00 GeV couples to a fermion of mass 0.40 GeV through g φ ψ̄(cos α + i sin α γ⁵)ψ, with g = 0.50 and α = 30°. Show that the spin-summed rate has no interference term, find the width, the lifetime and each partial wave's share, and explain why the same formula cannot fix α from the Higgs boson's τ⁺τ⁻ rate.
- Write the vertex matrix O = cos α + i sin α γ⁵. Its conjugate γ⁰O†γ⁰ = cos α − i sin α γ⁰γ⁵γ⁰ = O, since γ⁰γ⁵γ⁰ = −γ⁵, so Casimir's trick gives Σ|M|² = g² Tr[O(p̸₂ − m)O(p̸₁ + m)].
- The cos²α term is the scalar trace 4(p₁⋅p₂ − m²) and the sin²α term the pseudoscalar's 4(p₁⋅p₂ + m²). The cross terms are i sin α cos α times traces of γ⁵ with at most two other gammas, which topic 7.06 showed vanish: ³P₀ and ¹S₀ cannot interfere.
- So Σ|M|² = 2g²mφ²(β² cos²α + sin²α) and Γ = g²mφβ(β² cos²α + sin²α)/(8π), which falls back to the β³ law at α = 0° and the β law at α = 90°.
- Numbers: β = √(1 − 0.64) = 0.600, β² cos²α + sin²α = 0.36 × 0.75 + 0.25 = 0.520 and g²mφ/(8π) = 0.25 GeV ÷ 25.13 = 9.95 MeV, so Γ = 9.95 × 0.600 × 0.520 MeV = 3.10 MeV and τ = 6.582 × 10⁻²⁵ GeV s ÷ 3.104 × 10⁻³ GeV = 2.12 × 10⁻²² s.
- Shares: ¹S₀ carries 0.25/0.52 = 48% of the rate and ³P₀ 0.27/0.52 = 52%, although sin²α is only 0.25 against cos²α = 0.75: the p wave's β² = 0.36 nearly evens them up.
- For h → τ⁺τ⁻, with g = mτ/v quoted, mτ = 1.777 GeV and mₕ = 125.2 GeV, 1 − β² = 4mτ²/mₕ² = 8.1 × 10⁻⁴, so the rate changes by at most 0.08% as α runs from 0° to 90°, far below any measurement; α has to come from spin correlations in the τ decays.
AnswerNo interference term. Γ = 3.1 MeV and τ = 2.1 × 10⁻²² s, shared 48% ¹S₀ and 52% ³P₀. For h → τ⁺τ⁻ the rate moves by at most 0.08% as α varies, so it cannot measure α.