University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.8
The fermion loop's sign checked by unitarity
This is where the closed fermion loop's −1 stops being a rule out of Wick's theorem and becomes a sign you can test. Cut the bubble open and unitarity names its imaginary part in advance: the φ → ψψ̄ rate you already computed. Then count powers to see which couplings the loop's real part forces on you.
Build the model
Connect the measurement to the mechanism.
Topic 7.04 derived a −1 for every closed fermion loop from the order of anticommuting fields; here it meets a test it cannot fudge. In Yukawa theory the one-loop φ self-energy is a trace over a fermion and an antifermion propagator with that −1 in front. Doing its energy integral by residues, as topic 5.08 did for the φ⁴ bubble, leaves 1/(√s − 2Eₖ + iε), whose −iπδ part puts both lines on shell once p² passes 4m², and only that makes Σ complex.
Unitarity has fixed the answer in advance: by topics 5.07 and 5.09, Im Σ must be minus half the spin-summed |M(φ → ψψ̄)|² over two-body phase space, so Im Σ(mφ²) = −mφΓ. On the cut the antifermion line runs at negative energy, where SF's numerator is minus the spin sum Σvv̄, so the trace is minus a sum of squares; the loop's −1 restores the sign and gives Im Σ = −g²sβ³/(8π) for the scalar coupling and −g²sβ/(8π) for the pseudoscalar, topic 7.07's widths exactly, derived. Without the −1 every width would be negative.
The imaginary part is finite and the real part is not: one power of momentum per fermion line makes Σ diverge quadratically and the four-scalar box logarithmically, so the model needs a φ⁴ coupling, and φ³ too for the scalar coupling. The check is order g², good while g²/8π is small and Γ ≪ mφ, and silent on the real part, which waits for renormalisation conditions.
- Simple definition
- The unitarity check of the fermion loop computes the imaginary part of the one-loop φ self-energy, closed-loop −1 included, and shows that it equals −mφ times the width Γ(φ → ψψ̄), as the optical theorem demands.
- Example
- For g = 1, mφ = 2 GeV and a fermion of 0.5 GeV, β = 0.866 and the scalar loop gives Im Σ(mφ²) = −0.103 GeV² = −mφ × 51.7 MeV, the φ → ψψ̄ width; without the −1 it would be +0.103 GeV², a width of −51.7 MeV.
The (−1) and the trace belong to the closed loop. No ½: the arrow tells the ψ line from the ψ̄ line. The full propagator is i/(p² − mφ² − Σ).
SF(k) = i(k̸ + m)/(k² − m² + iε). Σ in GeV² (mass dimension 2); g dimensionless; k, p, m in GeV. From topic 7.04's rules.
From the −iπδ(√s − 2Eₖ) of the energy integral. Unitarity (topic 5.07) holds only if Ncut = Σₛₚᵢₙₛ |M(φ → ψψ̄)|², which is never negative.
N: −iΣ's integrand times its two denominators, in GeV²; Ncut: N with both lines on shell; kcm = ½√s β in GeV. Derived.
SF's negative-energy residue is minus the spin sum, topic 7.01's ordering sign. So the cut trace is −Σ|ū(p₁)v(p₂)|² = −2sβ², and only the loop's −1 makes Ncut positive.
p₂ = (Eₖ, −k) is the antifermion's on-shell momentum, in GeV. Spin sum quoted from Advanced QM (UAQM-04).
At s = mφ² it equals −mφΓ with Γ = g²mφβ³/(8π), topic 7.07's width, as topic 5.09's Breit-Wigner needs. Without the −1 it is +mφΓ.
ΣS in GeV²; s = p² in GeV²; β = √(1 − 4m²/s), dimensionless; θ the unit step. Derived here.
Equals −mφΓP at s = mφ². Its ratio to the scalar's is 1/β²: the ¹S₀ pair's s wave against the ³P₀ pair's p wave. The two agree as m → 0.
Vertex ±gγ⁵, its sign squared away; the γ⁵'s turn +m² into −m² in the numerator, so Ncut = 2g²s. Derived here.
φφ: D = 2, so Re Σ ∝ Λ². φφφ: 1. φφφφ: 0, the box ∝ ln Λ. Hence a φ⁴ coupling always and φ³ for the scalar coupling; ψψ̄ and ψψ̄φ renormalise m and g.
Eφ, Eψ: external scalar and fermion lines. From D = 4L − 2Iφ − Iψ, a ψ line costing one power of k. Shown.
The bubble carries a trace, a −1 and no ½
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. With Lᵢₙₜ = −gφψ̄ψ, topic 7.04's rules, quoted, give the one-loop φ self-energy as a fermion loop: −ig at each vertex, SF(k) = i(k̸ + m)/(k² − m² + iε) along the arrow, and a −1 and a trace for the closed loop. Call the sum of 1PI insertions −iΣ, as topic 5.09 did, so the full propagator is i/(p² − mφ² − Σ). Then −iΣ(p²) = (−1)(−ig)² ∫ d⁴k/(2π)⁴ Tr[i(k̸ + p̸ + m) i(k̸ + m)]/(D₁D₂), with D₁ = (k + p)² − m² + iε and D₂ = k² − m² + iε. The trace is 4[k⋅(k + p) + m²] and the constants give (−1)(−g²)(−1) = −g², so −iΣ = −4g² ∫ d⁴k/(2π)⁴ [k⋅(k + p) + m²]/(D₁D₂), derived. No ½ appears: topic 5.08's φ⁴ bubble had one because its two lines could be swapped, but here the arrow tells the fermion line from the antifermion line. Dimensions: d⁴k is mass⁴, the numerator mass², the denominators mass⁻⁴ and g a pure number, so Σ is a mass², as a shift of p² − mφ² must be.
The energy integral puts both lines on shell
Work in the φ rest frame, p = (√s, 0), and do the k⁰ integral by residues as topic 5.08 did. Let N stand for everything in −iΣ's integrand but 1/(D₁D₂), here −4g²[k⋅(k + p) + m²]. With Eₖ = √(|k|² + m²), D₂ has poles at k⁰ = ±(Eₖ − iε) and D₁ at k⁰ = −√s ± (Eₖ − iε). Close below. The pole at k⁰ = Eₖ leaves D₁ = s + 2√sEₖ, never zero. The pole at k⁰ = Eₖ − √s puts line k + p on shell and leaves D₂ = √s(√s − 2Eₖ + iε), whose −iπδ(√s − 2Eₖ) part is the only source of an imaginary part. It fires only if √s = 2Eₖ ≥ 2m: below threshold Σ is real and φ cannot decay. On the cut |k| = kcm = ½√s β with β = √(1 − 4m²/s), and the δ function over d³k gives Im Σ(s) = −kcm Ncut/(8π√s), derived, Ncut being N with both lines on shell. As kcm/(8π√s) = ½ × β/(8π), and β/(8π) is two-body phase space, topic 5.07's optical theorem holds exactly when Ncut = Σₛₚᵢₙₛ |M(φ → ψψ̄)|² for a φ of mass √s, a sum of squares.
On the cut the trace is minus a sum of squares
On the cut, line k + p carries p₁ = (Eₖ, k), a fermion of positive energy (k now the three-momentum), while line k carries (−Eₖ, k) = −p₂, with p₂ = (Eₖ, −k) the antifermion's momentum. There SF's numerator is k̸ + m = −(p̸₂ − m) = −Σₛ v(p₂, s)v̄(p₂, s): the negative-energy part of the Feynman propagator is minus the antiparticle spin sum, the minus sign topic 7.01's time ordering attaches when y⁰ comes after x⁰. With p₁⋅p₂ = s/2 − m², the trace on the cut is 4(m² − p₁⋅p₂) = −Tr[(p̸₁ + m)(p̸₂ − m)] = −Σₛₚᵢₙₛ |ū(p₁)v(p₂)|² = −2sβ². Every other factor, two vertices and two i's, is what a boson loop has, and a boson loop needs no sign: in topic 3.06's scalar Yukawa model Ncut = (−ig)²i² = g², already |M|². The fermion trace brings one extra −1, from the backward line, and the closed loop's −1 cancels it: Ncut = −g² × (−2sβ²) = 2g²sβ² = Σ|M(φ → ψψ̄)|², derived. Both minus signs are anticommutation, met once in time-ordering the propagator and once in closing the loop.
Both couplings give Im Σ = −mφΓ, each with its threshold
Put Ncut = 2g²sβ² and kcm/√s = β/2 into the cut formula: Im ΣS(s) = −g²sβ³/(8π) above s = 4m², derived. At s = mφ² that is −mφ times g²mφβ³/(8π), topic 7.07's width, quoted, so Im Σ(mφ²) = −mφΓ as topic 5.09's Breit-Wigner requires. For i g φψ̄γ⁵ψ the vertex is ±gγ⁵, whose square is +g² where (−ig)² was −g², and γ⁵(k̸ + p̸ + m)γ⁵ = −k̸ − p̸ + m flips the momenta: together, N = −4g²[k⋅(k + p) − m²], with only m² changed in sign. On the cut that is −4g²(−s/2) = 2g²s, again Σ|M|² and positive, so Im ΣP = −g²sβ/(8π) = −mφΓP, derived, the same −1 needed. With g = 1, mφ = 1.2 GeV and m = 0.5 GeV, β = 0.553, ΓS = 8.06 MeV and ΓP = 26.4 MeV, a ratio β² = 0.306. Limits check both: as β → 0 the scalar's β³ is the ³P₀ pair's p wave and the pseudoscalar's β the ¹S₀ s wave; as m → 0 they coincide at −g²s/(8π), since a massless fermion cannot tell ψ̄ψ from iψ̄γ⁵ψ.
Drop the −1 and every width turns negative
Without the closed loop's −1, Ncut = −Σ|M|² and Im Σ(mφ²) = +mφΓ for both couplings. Resummed as in topic 5.09, the propagator near its pole becomes i/(p² − mφ² − imφΓ): the pole crosses the real axis, the state's amplitude goes as e(−imφt + Γt/2), and its probability grows by a factor e every ħ/Γ, 8.2 × 10⁻²³ s for the 8.06 MeV width above. The optical theorem fails with it, since the forward amplitude, −Σ, would acquire a negative imaginary part, and a total rate cannot be negative. No redefinition rescues it: g is real for a Hermitian Lagrangian, so g² is positive and the sign is not a phase. Nor does another quantisation: with commutators in UAQM-09.06's reading d|0⟩ = 0, SF is unchanged, the norm −1 of d†|0⟩ now supplying the backward line's minus, but commuting fields close loops with no −1, so the width comes out negative, the cut summing over pair states of negative norm. Nature has the sign: every measured width is positive, the Higgs boson's included, a 0⁺ scalar decaying mostly to b b̄, at 3.2 (+2.4, −1.7) MeV from CMS's 2022 off-shell analysis, quoted.
Counting powers: what the real parts demand
Im Σ is finite, because the δ function fixes |k|; the real part is not. Topic 6.01's count, with a fermion propagator costing one power of momentum against the scalar's two, gives D = 4L − 2Iφ − Iψ, and V = 2Iφ + Eφ with 2V = 2Iψ + Eψ turn it into D = 4 − Eφ − (3/2)Eψ, shown. The self-energy has D = 2: at large k its integrand is −4g²/k², and with ∫ d⁴k/(2π)⁴ k⁻² = −iΛ²/(16π²) for a Euclidean cutoff Λ, Re Σ ≈ −g²Λ²/(4π²), shown, plus p² ln Λ and m² ln Λ terms, taken up by the φ mass and field-strength counterterms the theory already has. The four-scalar box has D = 0: its six orderings, each 4(k²)²/(k²)⁴ at large k, add to M ≈ −(3g⁴/(2π²)) ln(Λ²/s) for either coupling, shown, and Yukawa theory has no φ⁴ vertex to absorb that, so −(λ/4!)φ⁴ must be added. The triangle has D = 1; Tr(k̸k̸k̸) vanishes, but the scalar coupling leaves 12mk², a logarithm in proportion to m, so a φ³ coupling is needed too. For the pseudoscalar, parity sends φ → −φ and forbids φ³, stated, and its triangle's trace, one γ⁵ against at most three slashes, is zero.
Change one variable at a time
Make the relationship visible.
Flip the loop factor to +1: the solid curve leaves the dashed one and mirrors below the axis, a negative width. Then set mφ/2m to 1.1 and switch η: near threshold the scalar's β³ lies far below the pseudoscalar's β.
β AT p² = mφ²0.745
−Im Σ(mφ²), LOOP0.148 g²m²
½Σ∫dΠ₂|M|², DECAY SIDE0.148 g²m²
Γ = −Im Σ / mφ0.0494 g²m
Live interpretationβ AT p² = mφ²: 0.745. −Im Σ(mφ²), LOOP: 0.148 g²m². ½Σ∫dΠ₂|M|², DECAY SIDE: 0.148 g²m². Γ = −Im Σ / mφ: 0.0494 g²m
Catch the common trap
Explain before calculating.
Yukawa theory with coupling g φψ̄ψ, g = 1.0, a fermion of mass 0.48 GeV and mφ = 1.00 GeV. What is the imaginary part of the one-loop fermion-loop self-energy at p² = mφ², closed-loop −1 included?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyYukawa theory with g φψ̄ψ, g = 0.60, a fermion of m = 0.12 GeV and mφ = 0.40 GeV. At p² = mφ², evaluate the loop's trace 4[k⋅(k + p) + m²] on the cut and the spin sum Σ|ū(p₁)v(p₂)|², then use Im Σ = −kcm Ncut/(8π√s) to find Im Σ(mφ²), the width and the lifetime, and say what dropping the −1 would give.
- β = √(1 − 4m²/mφ²) = √(1 − 0.0576/0.16) = √0.64 = 0.800, so kcm = ½ × 0.40 × 0.800 = 0.160 GeV, and Eₖ = √(0.160² + 0.12²) = 0.200 GeV = mφ/2, as the δ function requires.
- p₁⋅p₂ = s/2 − m² = 0.0800 − 0.0144 = 0.0656 GeV². Spin sum: 4(p₁⋅p₂ − m²) = 4 × 0.0512 = +0.2048 GeV², which is 2sβ² = 2 × 0.16 × 0.64. Loop trace with k = −p₂: 4(m² − p₁⋅p₂) = −0.2048 GeV², its exact negative.
- Ncut = (−1)(−ig)²i² × trace = −g² × (−0.2048) = 0.36 × 0.2048 = 0.07373 GeV², which is Σ|M|² = g² × 0.2048, positive as unitarity requires.
- Im Σ(mφ²) = −0.160 × 0.07373/(8π × 0.40) = −0.011796/10.053 = −1.173 × 10⁻³ GeV². Dimensions: GeV × GeV² / GeV = GeV².
- Γ = −Im Σ/mφ = 2.93 × 10⁻³ GeV = 2.93 MeV, matching g²mφβ³/(8π) = 0.36 × 0.40 GeV × 0.512/25.13 = 2.93 × 10⁻³ GeV. With ħ = 6.582 × 10⁻²⁵ GeV s, τ = ħ/Γ = 2.24 × 10⁻²² s.
- Dropping the −1 leaves Ncut = −0.0737 GeV², Im Σ = +1.17 × 10⁻³ GeV² and Γ = −2.93 MeV: a probability that grows by e every 2.24 × 10⁻²² s.
AnswerCut trace −0.205 GeV² against spin sum +0.205 GeV²; Im Σ(mφ²) = −1.17 × 10⁻³ GeV², Γ = 2.93 MeV, τ = 2.24 × 10⁻²² s. Without the −1, Γ = −2.93 MeV.
MediumIn Yukawa theory, with g dimensionless, derive the superficial degree of divergence D of a 1PI function with Eφ external scalars and Eψ external fermions, and list the functions with D ≥ 0. Then estimate the fermion box's large-k part for g = 1.2, a cutoff Λ = 1 TeV and external momenta of order √s = 10 GeV, and say what doubling Λ does to it.
- Each loop brings d⁴k, four powers; each φ propagator removes two and each ψ propagator, i(k̸ + m)/(k² − m²) ~ 1/k, removes one: D = 4L − 2Iφ − Iψ, with L = Iφ + Iψ − V + 1.
- Each vertex has one φ end and two ψ ends, so V = 2Iφ + Eφ and 2V = 2Iψ + Eψ. Substituting, D = 4 − Eφ − (3/2)Eψ, with V gone because g carries no dimension.
- D ≥ 0 for: (Eφ, Eψ) = (1, 0), D = 3; (2, 0), D = 2; (3, 0), D = 1; (4, 0), D = 0; (0, 2), D = 1; (1, 2), D = 0. Everything else is superficially convergent, for example ψφ → ψφ with D = −1 and ψψ → ψψ with D = −2.
- Box: the 3! = 6 orders in which a directed fermion loop can visit the four vertices each give (−1)(−ig)⁴ Tr[(ik̸)⁴]/(k²)⁴ = −4g⁴/(k²)² at large k, so iM ≈ −24g⁴ ∫ d⁴k/(2π)⁴ (k²)⁻² = −24g⁴ × i ln(Λ²/s)/(16π²), and M ≈ −(3g⁴/(2π²)) ln(Λ²/s).
- Numbers: 3g⁴/(2π²) = 3 × 2.0736/19.739 = 0.315 and ln(Λ²/s) = ln(10⁴) = 9.21, so M ≈ −2.90 at leading logarithm. Doubling Λ adds −0.315 × ln 4 = −0.437.
- A shift of 0.44 from a regulator change is as large as a φ⁴ coupling of order 0.5, so the four-scalar amplitude depends on Λ unless a −(λ/4!)φ⁴ term with its counterterm absorbs ln Λ: the box makes λ compulsory.
AnswerD = 4 − Eφ − (3/2)Eψ; superficially divergent: φ (3), φφ (2), φφφ (1), φφφφ (0), ψψ̄ (1), ψψ̄φ (0). The box is M ≈ −2.9 at Λ = 1 TeV, and doubling Λ shifts it by −0.44.
HardTake the pseudoscalar coupling, Lᵢₙₜ = −i g φψ̄γ⁵ψ, with g = 0.50, a fermion of m = 0.35 GeV and mφ = 2.5 GeV. In the φ rest frame find kcm, build Ncut with the loop's −1 and the γ⁵ vertices, evaluate Im Σ(mφ²) from the cut formula, and check it against the decay rate. Then give the scalar coupling's value at the same masses and the sign with no −1.
- β = √(1 − 4 × 0.1225/6.25) = √0.9216 = 0.960 and kcm = ½ × 2.5 × 0.960 = 1.200 GeV; Eₖ = √(1.44 + 0.1225) = 1.250 GeV = mφ/2.
- The vertex is i × (−igγ⁵) = gγ⁵. So N = (−1) × g² × i² × Tr[γ⁵(k̸ + p̸ + m)γ⁵(k̸ + m)] = g² Tr[(−k̸ − p̸ + m)(k̸ + m)] = −4g²[k⋅(k + p) − m²], using γ⁵k̸γ⁵ = −k̸.
- On the cut k + p = p₁ and k = −p₂, so k⋅(k + p) = −p₁⋅p₂ = −(s/2 − m²) = −(3.125 − 0.1225) = −3.0025 GeV². Then Ncut = −4 × 0.25 × (−3.0025 − 0.1225) = +3.125 GeV², equal to Σ|M|² = 2g²s = 2 × 0.25 × 6.25.
- Im Σ(mφ²) = −kcm Ncut/(8π√s) = −(1.200/(8π × 2.5)) × 3.125 = −0.019099 × 3.125 = −0.0597 GeV².
- Decay side: ΓP = g²mφβ/(8π) = 0.25 × 2.5 × 0.960/25.133 = 0.02387 GeV, so mφΓP = 0.0597 GeV², matching. τ = 6.582 × 10⁻²⁵/0.02387 = 2.76 × 10⁻²³ s.
- Scalar coupling: Im ΣS = β² Im ΣP = 0.9216 × (−0.0597) = −0.0550 GeV², only 8% smaller since m ≪ mφ, as the massless limit predicts. Without the −1, Ncut = −3.125 GeV² and Im ΣP = +0.0597 GeV², a width of −23.9 MeV.
Answerkcm = 1.20 GeV, Ncut = +3.125 GeV², Im Σ(mφ²) = −0.0597 GeV² = −mφΓ with Γ = 23.9 MeV (τ = 2.76 × 10⁻²³ s); scalar coupling −0.0550 GeV²; without the −1, Γ = −23.9 MeV.