University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.4

Yukawa rules and the closed fermion loop

This is where Feynman rules start carrying spin. Take the signed Wick theorem and the LSZ spinors and turn them into a recipe you can run on any Yukawa diagram: which spinor sits at which end of a line, why a propagator cares which way its momentum points, and why every closed fermion loop costs a −1 that no convention removes.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Yukawa theory couples a Dirac field to a scalar through Hᵢₙₜ = gφψ̄ψ, and its Feynman rules are Unit 3's scalar Yukawa rules with spin put on the lines: the arrows are unchanged, but every fermion line now carries a Dirac index. The vertex becomes −ig times the unit matrix in spinor space, the propagator i(p̸ + m)/(p² − m² + iε) is a 4 × 4 matrix whose numerator changes when its momentum is reversed, and the factors along a line multiply as matrices, so their order is part of the rule. An open line is capped by topic 7.03's external spinors, written from the arrow's head back to its tail, and becomes a number.

A closed line has no spinors to end on: its last index wraps round into a trace, and because carrying its last ψ to the front passes an odd number of anticommuting fields, it brings exactly one factor −1 whatever the number of vertices. Both facts are derived from topic 7.02's signed Wick theorem; the pseudoscalar coupling igφψ̄γ⁵ψ changes only the vertex. The rules stop short in three places.

They fix signs within a diagram, not between diagrams that exchange identical fermions, which needs the external states ordered (topic 7.05). Their loops demand a φ⁴ coupling, and for the scalar vertex a φ³ one too, before the theory is renormalisable (topic 7.08). And as a model of pions and nucleons the gγ⁵ vertex overshoots the measured threshold scattering length over a hundredfold, blind to chiral symmetry.

Simple definition: Yukawa theory's Feynman rules turn each fermion line into a chain of Dirac matrices, vertices −ig and propagators i(p̸ + m)/(p² − m² + iε) along the arrow, capped by external spinors on an open line and closed by a trace times −1 on a loop.
Example: The φ self-energy loop has n = 2 vertices, so the last ψ passes 2n − 1 = 3 fields to reach the front and the loop enters as −Tr[(−ig)S_F(q + p)(−ig)S_F(q)]. Its constant (−1)(−ig)²(i)² is −g², or −0.25 for g = 0.5, where a complex-scalar loop gives +g².

Scalar Yukawa vertexHᵢₙₜ = g φ ψ̄_α ψ_α ⇒ vertex −ig δ_αβ

Derived: ψ̄, ψ and φ are three different fields, so no 1/3! survives, and in spinor space the vertex is the unit matrix joining two arrows.

g dimensionless, as [φ] = 1 and [ψ] = 3/2 in GeV; α, β the Dirac indices of the arrow leaving and the arrow arriving

Fermion propagator, momentum along the arrowS_F(p) = i(p̸ + m)/(p² − m² + iε)

Quoted from topic 7.01. The numerator is not even in p: a momentum routed against the arrow gives i(−p̸ + m)/(p² − m² + iε), a different matrix.

p in GeV, along the arrow; m the fermion mass in GeV; S_F a 4 × 4 matrix in GeV⁻¹

An open line, read against its arrowū(p′, s′) (−ig) S_F(p + k) (−ig) u(p, s)

Derived: start where the arrow ends and walk back, so every row index meets a column index and the chain is a number. Reversed, u … ū is a 4 × 4 matrix, not an amplitude.

the s-channel of ψφ → ψφ; u and ū in GeV¹⁄², from topic 7.03; a row, square matrices and a column

A closed fermion loop−∫ d⁴q/(2π)⁴ Tr[V S_F(q₁) V S_F(q₂) ⋯ V S_F(qₙ)]

Derived: the last ψ passes 2n − 1 fields to reach the front, odd for every n, so one −1 per loop; its free index closes into a trace. No ½: the arrow tells the two ways round apart.

n vertices V = −ig or gγ⁵; each qᵢ along the arrow, factors read against it; d⁴q in GeV⁴

Pseudoscalar vertexHᵢₙₜ = i g φ ψ̄γ⁵ψ ⇒ vertex −i(i g γ⁵) = +g γ⁵

Derived: only the vertex changes. The sign convention flips every diagram of a process together, so no |M|² sees it unless a scalar coupling interferes.

γ⁵ = iγ⁰γ¹γ²γ³, Hermitian, (γ⁵)² = 1, (γ⁵, γ^μ) = 0; the opposite sign in Hᵢₙₜ gives −gγ⁵

Forward ψφ → ψφ, either vertexM = −g²[(4κm² + 2mω)/(m_φ² + 2mω) + (4κm² − 2mω)/(m_φ² − 2mω)]

Derived from the s- and u-channel chains with ūu = 2m and ūγ^μu = 2p^μ. At threshold κ = 1 gives 0 and κ = 0 gives −8g²m²/(4m² − m_φ²); both tend to −2g² at high ω.

ω the φ energy in the ψ rest frame, masses in GeV; κ = 1 for −ig, κ = 0 for gγ⁵; M dimensionless

The vertex gains a Dirac index

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. From UQFTI-03.06, quoted: a vertex of −gψ†ψφ joins one arrow in, one arrow out and an undirected φ, so arrows never end inside a diagram and a fermion line either crosses the diagram or closes. Yukawa theory keeps that and adds an index. In Hᵢₙₜ = gφψ̄_αψ_α the ψ̄ carries a row index belonging to the arrow leaving the vertex and ψ a column index belonging to the arrow arriving, so the vertex is −ig δ_αβ, the unit matrix in spinor space. The three fields differ, so no 1/3! appears, and the 1/n! of the Dyson series cancels against relabelling vertices as before. Mass dimensions fix the coupling: [ψ] = 3/2 and [φ] = 1 make [φψ̄ψ] = 4, so g is a pure number, where the scalar model's coupling carried one power of mass. For the pseudoscalar coupling (ψ̄γ⁵ψ)† = −ψ̄γ⁵ψ, so Hᵢₙₜ = igφψ̄γ⁵ψ is Hermitian and its vertex is −i(igγ⁵) = +gγ⁵, derived; written with the opposite sign it is −gγ⁵.

The propagator has a numerator and a direction

Contracting ψ(x) with ψ̄(y) gives topic 7.01's S_F(x − y), quoted, with its row index at x and its column index at y: it carries fermion number from y to x, along the arrow. In momentum space it is i(p̸ + m)/(p² − m² + iε) with p along the arrow. For the complex scalar a momentum's direction never mattered, since p² is even. Here the p̸ in the numerator is odd, so a momentum routed against the arrow brings i(−p̸ + m)/(p² − m² + iε). The numerator also says what propagates. Where p = (E, 0), p̸ + m = Eγ⁰ + m, and with P± = (1 ± γ⁰)/2 the propagator splits into i[P₊/(E − m) − P₋/(E + m)]: the pole at E = m sits only in the upper, particle block. At E = 1.10m the upper block is 1/(0.10m) and the lower −1/(2.10m), twenty-one times smaller but not zero. On shell p̸ + m = Σₛ u(p, s)ū(p, s), UAQM-04.08's spin sum, so near its pole an internal fermion is summed over its two spin states.

A line is read from its head back to its tail

At each vertex ψ̄ stands left of ψ, and contracting a vertex's ψ with the ψ̄ of the vertex before it along the arrow gives S_F, its row index at the later vertex. At the ends topic 7.03's LSZ factors take over, quoted: an incoming fermion gives u(p, s), an outgoing one ū(p, s), an incoming antifermion v̄(p, s), an outgoing one v(p, s). So a line is written from where its arrow ends back to where it begins. For ψφ → ψφ the s-channel chain is ū(p′)(−ig)S_F(p + k)(−ig)u(p) and the u-channel chain ū(p′)(−ig)S_F(p − k′)(−ig)u(p); for ψψ̄ → φφ the arrow enters with the fermion and leaves through the incoming antifermion, giving v̄(p₂)(−ig)S_F(p₁ − k₁)(−ig)u(p₁). A row, square matrices and a column multiply to a number; reversed, u … ū would be a 4 × 4 matrix. Dimensions agree: [u] = 1/2 and [S_F] = −1, so ū S_F u is dimensionless, as a 2 → 2 amplitude with dimensionless g must be. Order matters once vertices differ, since γ⁵ anticommutes with p̸.

Closing a loop costs one anticommutation too many

Put n vertices on a loop, ψ̄₁ψ₁ ψ̄₂ψ₂ ⋯ ψ̄ₙψₙ, and contract ψ(xᵢ) with ψ̄(xᵢ₊₁) and ψ(xₙ) with ψ̄(x₁). Topic 7.02's rule weights a term by the sign of the permutation that makes its pairs adjacent in the order ψψ̄, and every pair but the last already is. The last ψ stands at the far right and its partner ψ̄₁ at the far left, so ψₙ must pass the 2n − 2 fields between them and then ψ̄₁ itself, 2n − 1 in all: an odd number for every n, hence exactly one −1. S_F(xₙ − x₁) then links the last vertex's matrix to the first one's, and the product closes into a trace read against the arrow, −Tr[V S_F V S_F ⋯ V S_F], derived. Two checks. A contraction with c closed loops carries (−1)^c, and a permutation of n vertices with c cycles has sign (−1)^{(n − c)}, so for index-free fermions the sum is (−1)ⁿ det[S(xᵢ − xⱼ)], topic 7.02's determinant. And no convention removes the sign: the self-energy goes as g², blind to the sign of g, and topic 7.08 finds a negative width without it.

What the numerator does at threshold

Fermion-scalar scattering needs no relative fermion sign: one open line, an s-channel chain through S_F(p + k) and a u-channel chain through S_F(p − k′). At threshold everything is at rest, so the external spinors have upper components only and both propagators split into blocks as above. A scalar vertex keeps the spinors upper, so only the particle block contributes, 1/m_φ from the s-channel and −1/m_φ from the u-channel, and M = 0 exactly, for any masses, derived. In the Dirac representation γ⁵ swaps upper and lower components, so with gγ⁵ vertices the same chains see only the antiparticle block, −1/(2m + m_φ) and −1/(2m − m_φ), which add: M = −8g²m²/(4m² − m_φ²), derived. The algebra behind both is γ⁵(q̸ + m)γ⁵ = −q̸ + m, which turns the mass terms in each numerator from adding into cancelling. Far above threshold the mass stops mattering and both vertices give the forward amplitude −2g², the massless limit in which the chiral rotation ψ → exp(iπγ⁵/4)ψ turns one coupling into the other. The figure follows the two chains between these limits.

Change one variable at a time

Make the relationship visible.

Interactive model

Mass ratio m_φ/m0.50

Vertex: 0 is −ig, 1 is gγ⁵0

Marker energy ω/m_φ2.0

With the scalar vertex the dashed and faint curves start as mirror images and the sum starts at zero, for every mass ratio. Switch to gγ⁵: the s-chain shrinks, the u-chain turns negative and they add. Push ω up and both sums settle towards −2, where the mass no longer tells the two vertices apart.

s-CHANNEL Mₛ/g²-2.67

u-CHANNEL Mᵤ/g²1.14

SUM M/g² AT ω-1.52

SUM AT THRESHOLD0.00

Live interpretations-CHANNEL Mₛ/g²: −2.67. u-CHANNEL Mᵤ/g²: 1.14. SUM M/g² AT ω: −1.52. SUM AT THRESHOLD: 0.00

Catch the common trap

Explain before calculating.

In Yukawa theory with Hᵢₙₜ = gφψ̄ψ, a φ of momentum p enters a fermion loop at one vertex and leaves at the other. With q the momentum along the arrow on one internal line and q + p along the arrow on the other, which expression is −iΣ(p²)?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA Dirac fermion of mass m = 0.50 GeV and a scalar of mass m_φ = 0.20 GeV, coupled by Hᵢₙₜ = gφψ̄ψ with g = 0.80, scatter at threshold, both at rest in the centre-of-mass frame. Write the two tree chains in the right order, evaluate each, and find M.

One open line, its arrow running from the incoming to the outgoing fermion. Read back from its head: the s-channel chain is ū(p′)(−ig)S_F(P)(−ig)u(p) with P = p + k, the u-channel chain the same with P′ = p − k′. The prefactor (−ig)²i = −ig² gives M = −g²[ū(P̸ + m)u/(P² − m²) + ū(P̸′ + m)u/(P′² − m²)].
At rest p = p′ = (m, 0), so the spinors have upper components only: ūu = 2m and ūγ⁰u = u†u = 2m, while ūγⁱu = 0. With P = (m + m_φ, 0) = (0.70 GeV, 0) and P′ = (m − m_φ, 0) = (0.30 GeV, 0), each numerator is 2m(P⁰ + m).
s-channel: 2(0.50)(0.70 + 0.50) = 1.20 GeV² over P² − m² = 0.49 − 0.25 = 0.24 GeV², a ratio of 5.0, so Mₛ = −0.64 × 5.0 = −3.2.
u-channel: 2(0.50)(0.30 + 0.50) = 0.80 GeV² over 0.09 − 0.25 = −0.16 GeV², a ratio of −5.0, so Mᵤ = +3.2. In general the ratios are ±2m/m_φ: the particle block of S_F at energies m ± m_φ.
Checks: GeV² over GeV² leaves M dimensionless, as a 2 → 2 amplitude must be, and the chain is diagonal in spin, so nothing flips at threshold. There is no measurement for this toy; the next example changes the vertex.

AnswerMₛ = −3.2 and Mᵤ = +3.2, so M = 0 at threshold: with the scalar vertex the two chains cancel exactly, for any m and m_φ.

MediumSwitch to Hᵢₙₜ = igφψ̄γ⁵ψ and use it as a crude model of π⁰p scattering: m = 0.938 GeV, m_φ = 0.135 GeV, g = 13.1, so g²/4π = 13.7. Find the threshold amplitude, the scattering length |f| = |M|/(8π√s) and the threshold cross-section, and compare with experiment.

Each vertex is now gγ⁵, so iM = g²i[ūγ⁵(P̸ + m)γ⁵u/(P² − m²) + (P → P′)]. As γ⁵ anticommutes with every γ^μ and squares to 1, γ⁵(P̸ + m)γ⁵ = −P̸ + m, and M = g²[ū(−P̸ + m)u/(P² − m²) + ū(−P̸′ + m)u/(P′² − m²)].
At rest each numerator is 2m(m − P⁰). s-channel: −2m m_φ = −0.2533 GeV² over m_φ(2m + m_φ) = 0.2715 GeV², ratio −0.933. u-channel: +2m m_φ = 0.2533 GeV² over −m_φ(2m − m_φ) = −0.2350 GeV², ratio −1.078. The mass terms cancel and the two chains add.
Together −2m[1/(2m + m_φ) + 1/(2m − m_φ)] = −8m²/(4m² − m_φ²) = −2.010, so M = 171.6 × (−2.010) = −345, close to −2g² because m_φ ≪ 2m.
With √s = m + m_φ = 1.073 GeV, |f| = 345/(8π × 1.073 GeV) = 12.8 GeV⁻¹, and restoring ħc = 0.1973 GeV fm gives 2.52 fm, or 1.73 m_π⁻¹ with ħc/m_π = 1.462 fm. The threshold cross-section 4π|f|² is 80 fm² = 8.0 × 10² mb.
Checks: M is dimensionless and |f| a length. Experiment, quoted: pionic hydrogen and deuterium put the isospin-even πN scattering length, the π⁰p one in the isospin limit, at about 0.01 m_π⁻¹ or below, so the model is some 170 times too large. The antiparticle block that γ⁵ selects is the culprit; chiral symmetry, through a pseudovector coupling or an added scalar exchange, cancels it, stated, not derived.

AnswerM ≈ −345, |f| ≈ 2.5 fm ≈ 1.7 m_π⁻¹ and σ ≈ 8.0 × 10² mb at threshold, more than a hundred times the measured scattering length.

HardIn the scalar Yukawa theory with g = 1.0 and m = 0.40 GeV, six φ lines carrying zero four-momentum attach to a single fermion loop. Count the contractions, fix the sign, take the trace and the loop integral, and give M₆, the one-loop amputated six-point function at zero external momenta, in GeV⁻².

The 1/6! of the Dyson series cancels against attaching six distinct external φ's to six vertices. The ψ's must then form one cycle through six labelled vertices: (6 − 1)! = 120 directed cycles, the two orientations of each ordering counted separately and no symmetry factor. Each is a closed loop, −Tr[((−ig)S_F(q))⁶], every line carrying q at zero external momentum.
Constants: (−ig × i)⁶ = g⁶. Since q̸q̸ = q², odd powers of q̸ are traceless and Tr[(q̸ + m)⁶] = 4(m⁶ + 15m⁴q² + 15m²q⁴ + q⁶). So iM₆ = −120g⁶ ∫ d⁴q/(2π)⁴ 4(m⁶ + 15m⁴q² + 15m²q⁴ + q⁶)/(q² − m² + iε)⁶.
Wick-rotate as in Unit 6: q⁰ = iq_E⁰, d⁴q = i d⁴q_E and q² = −Q with Q = q_E², while d⁴q_E/(2π)⁴ = Q dQ/(16π²). With Q = m²t the integral becomes (i/(16π²m²)) × 4J, where J = ∫ t(1 − 15t + 15t² − t³)(1 + t)⁻⁶ dt, taken from t = 0 to ∞.
Beta functions, B(a, b) = ∫ t^(a−1)(1 + t)^(−a−b) dt from 0 to ∞, give J = B(2, 4) − 15B(3, 3) + 15B(4, 2) − B(5, 1) = 1/20 − 1/2 + 3/4 − 1/5 = 1/10. So iM₆ = −120g⁶ × 4i/(160π²m²), and M₆ = −3g⁶/(π²m²).
Numbers: M₆ = −3/(9.870 × 0.160 GeV²) = −1.90 GeV⁻². Checks: a six-point amplitude has mass dimension 4 − 6 = −2; topic 6.01's counting, one power of q per fermion line, gives D = 4 − 6 = −2, so the loop is finite; and the heavy fermion decouples as 1/m².
A second route: the one-loop fermion potential −m_eff⁴ ln(m_eff²/μ²)/(16π²) plus a quartic, with m_eff = m + gφ, quoted rather than derived here, has sixth derivative +3g⁶/(π²m²), and an effective φ⁶ term in −V gives the vertex −iV⁽⁶⁾, so M₆ = −3g⁶/(π²m²) again. Drop the loop's −1 and the sign of M₆ reverses, disagreeing with it.

AnswerM₆ = −3g⁶/(π²m²) = −1.90 GeV⁻², finite, and negative because of the loop's −1; this toy model has no measurement to meet.