University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.6

Scalar Yukawa Feynman rules: arrows for ψ number

Real φ⁴ taught you to count identical lines. The scalar Yukawa model is the first theory whose lines are not alike, and its bookkeeping changes with them: a vertex with no factorial, an arrow that carries ψ number through every diagram, and a closed loop that drops the ½ a real field needs. QED will ask for exactly these habits.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Wick's theorem does not care which fields it contracts, but the vacuum does. With −gψ†ψφ, ψ pairs only with ψ†, since ψψ would change the vacuum's charge, and φ only with φ, since the fields are independent. The rules follow from that table.

The three fields at a vertex all differ, so each line end has one place to attach: the Dyson series' 1/n! still cancels against relabelling the vertices, nothing else needs cancelling, and the vertex is −ig, which is why the coupling carries no 1/3!. Draw each ψψ† contraction as an arrow from ψ† to ψ and every vertex has one arrow in and one out, so arrows run unbroken from an external ψ† to an external ψ or close into loops: ψ number is conserved diagram by diagram. The arrowed propagator i/(p² − M² + iε) is even in p, so the arrow is not a momentum, yet positive energy flowing against it is a ψ̄.

And the arrows tell apart lines a real field would interchange, so the ψ loop in φ's self-energy has weight 1 where a real field's has ½, and every ψ loop counts twice a real one, once for each real component inside ψ. The rules stop at correlation functions of scalar fields: external states wait for topic 4.05, Dirac fields bring spinor matrices and a −1 per loop in topic 7.04, and the loop integrals they write down are evaluated in Units 5 and 6.

Simple definition: The scalar Yukawa Feynman rules turn each Wick contraction of −gψ†ψφ theory into a diagram: −ig at every vertex where one arrow enters, one leaves and one φ line attaches, i/(p² − M² + iε) along each arrow and i/(p² − m² + iε) on each φ line.
Example: At order g² the φ two-point function has 6 full contractions. Two of them join x to y through a ψ loop, and with the 1/2! they give that loop weight 1, twice the ½ that a real field χ coupled by −(g/2)χ²φ would get.

Only ψ with ψ† and φ with φ contract⟨0|Tψ(x)ψ†(y)|0⟩ = D_F(x − yM), while ψψ, ψ†ψ† and ψφ pair to zero

ψψ would lower the vacuum's charge by 2 and ψ, φ are independent, so at order g² ⟨φφ⟩ has 6 full contractions where a real cubic theory has 105.

ψ is complex with mass M, φ real with mass m, both of mass dimension 1; D_F is in GeV², from topic 1.08.

Vertex factor, derived−ig (2π)⁴ δ⁴(Σpᵢₙ − Σpₒᵤₜ)

Each line end fits only one of ψ, ψ† and φ, so there is no 3! to cancel. Writing the coupling as −(g/3!)ψ†ψφ would give −ig/6.

g in GeV, mass dimension 1; one arrow in, one arrow out and one φ line meet here; momenta in GeV.

Arrowed ψ line, from topic 1.08i/(p² − M² + iε), with p along the arrow from ψ† to ψ

Even in p, so the label may point either way; the arrow may not. On an external line, positive energy flowing against the arrow is a ψ̄.

p is the line's four-momentum in GeV and M the ψ mass in GeV; the propagator is in GeV⁻².

Undirected φ linei/(p² − m² + iε)

At zero energy it is −i/(|q|² + m²), whose Fourier transform, quoted, has range ħ/mc: 1.414 fm for the measured pion mass, 139.57 MeV.

m is the φ mass in GeV; φ is real, so φ(x)φ(y) has no orientation; the propagator is in GeV⁻².

Closed ψ loop in the φ self-energy, weight 1−iΠ(p²) = (−ig)² ∫ d⁴l/(2π)⁴ [i/(l² − M² + iε)] [i/((l + p)² − M² + iε)]

A real χ coupled by −(g/2)χ²φ gets ½ here. ψ = (χ₁ + iχ₂)/√2 holds two such fields, and 2 × ½ = 1.

l is the loop momentum along one arrow, in GeV; Π is in GeV²; the weight 1 is derived by counting contractions.

Vertices in a connected diagramV = 2L + E_ψ + E_φ − 2

Derived by counting line ends: each loop costs g². With E_ψ = 0 every arrow must close, so connected ⟨φφφφ⟩ starts at one loop and order g⁴.

V vertices, L loops, I internal and E external lines, split into arrowed ψ and φ; from 2V = 2I_ψ + E_ψ and V = 2I_φ + E_φ.

Only ψ with ψ† and φ with φ

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. The model is ℒ = (∂_μ ψ†)(∂^μ ψ) − M²ψ†ψ + ½(∂_μ φ)(∂^μ φ) − ½m²φ² − gψ†ψφ: a complex ψ of mass M, a real φ of mass m, and a coupling g of mass dimension 1, as topic 2.01 found. With no time derivatives in the interaction, H_I = g∫d³x ψ†ψφ, so the order-n term of the Gell-Mann-Low numerator is (1/n!)(−ig)ⁿ∫d⁴z₁…d⁴zₙ times a free-vacuum expectation of the external fields and n copies of ψ†ψφ. Wick's theorem, topic 2.08, makes that a sum of full contractions, and only two kinds survive, ψ with ψ† and φ with φ: ⟨0|Tψψ|0⟩ = 0 because ψ lowers the charge by one and the vacuum has none (topic 1.08), and ⟨0|Tψφ|0⟩ = 0 because the two fields' mode operators are independent. A full contraction is therefore a pairing of ψ's with ψ†'s times a pairing of φ's: at order g², ⟨φ(x)φ(y)⟩ has 3 × 2 = 6, where a real cubic theory with eight identical fields has 7!! = 105.

Why the vertex is −ig, with no 1/3!

In −(λ/4!)φ⁴ the 4! is there to be cancelled: the four lines at a vertex attach to its four identical fields in 4! ways, leaving −iλ. At a Yukawa vertex the three fields differ, and each line end has exactly one place to go. An arrow arriving at z must contract with ψ(z), an arrow leaving z must start on ψ†(z), and the φ line takes φ(z). There is no permutation to cancel, which is why the interaction is written −gψ†ψφ with no factorial, and the vertex is −ig. The Dyson series' 1/n! still cancels against the n! relabellings of the vertices whenever those give distinct contractions, as in φ⁴. At order g the connected part of ⟨φ(w)ψ(x)ψ†(y)⟩, −ig∫d⁴z D_F(w − z; m) D_F(x − z; M) D_F(z − y; M), is a single contraction of weight 1, the same −g that topic 2.09 read as the φ → ψψ̄ amplitude. Had the coupling been written −(g/3!)ψ†ψφ, the vertex would be −ig/6 and nothing in the counting would restore the 6.

Arrows carry ψ number and never stop

Draw each contraction ψ(x)ψ†(y) as a line with an arrow from y to x, from the ψ† to the ψ: ψ†(y) raises the ψ number by one at y and ψ(x) lowers it by one at x. At a vertex z, ψ(z) receives one arrow and ψ†(z) sends one, so every vertex has exactly one arrow in, one arrow out and one φ line, and the φ line carries no ψ number. Following an arrow forward from any vertex leads to another vertex or to an external ψ, never to a dead end, so every arrowed line either runs from an external ψ† to an external ψ or closes on itself. Two consequences hold at every order, derived. A correlation function with unequal numbers of ψ and ψ† vanishes: ⟨Ω|Tψ(x₁)ψ(x₂)ψ†(y)φ(w)|Ω⟩ = 0 to all orders, since one arrow would have nowhere to start. And a diagram with any vertex whose external lines are all φ must contain a closed ψ loop, so the first correction to ⟨φφ⟩ comes at order g² and connected ⟨φφφφ⟩ begins at g⁴. This is the U(1) symmetry ψ → e^(iα)ψ, seen one diagram at a time.

Momentum along the arrow, antiparticle against it

Fourier transforming as in topic 3.05 puts i/(p² − M² + iε) on each arrowed line, i/(p² − m² + iε) on each φ line, −ig(2π)⁴δ⁴(Σp) at each vertex and ∫d⁴l/(2π)⁴ on each loop. Give each ψ line its momentum along the arrow, as topic 1.08's propagator carries it, from ψ† to ψ. The propagator is even in p, so reversing a label changes nothing in the integrand, but the arrow still decides which diagrams exist. On an external line the reading becomes physical: positive energy along the arrow is a ψ, against it a ψ̄. At the φ → ψψ̄ vertex the ψ leaves along an outgoing arrow with p₁, while the ψ̄'s arrow points into the vertex and its momentum p₂ flows out; along that arrow the momentum is −p₂, with energy −E₂, and conservation reads k = p₁ + p₂. The non-relativistic limit checks this. With p⁰ = M + E and E, |p| ≪ M, p² − M² ≈ 2M(E − |p|²/2M), so the line becomes (1/2M) i/(E − |p|²/2M + iε), QM II's retarded Schrödinger propagator: the arrow becomes the direction of time, and a closed ψ loop, which must run backward in time on one leg, vanishes if only the particle pole is kept, so it exists only through the antiparticle pole, as a virtual pair.

A closed ψ loop carries no ½

At order g², the term of ⟨φ(x)φ(y)⟩ that joins x to y through a ψ loop takes φ(x) to z₁ and φ(y) to z₂ or the reverse, two ways that cancel the 1/2!, and the ψ fields then close the loop in exactly one way, ψ(z₁) with ψ†(z₂) and ψ(z₂) with ψ†(z₁). Its weight is 1, derived. Compare a real χ of mass M coupled by −(g/2)χ²φ, whose vertex is also −ig: the two χ's at z₁ pair with the two at z₂ in 2 ways, and (1/2!)(1/2)² × 2 × 2 = ½, the symmetry factor of two interchangeable lines. In the ψ loop the lines run z₁ → z₂ and z₂ → z₁, so swapping them reverses both arrows and is no symmetry. Writing ψ = (χ₁ + iχ₂)/√2 turns −gψ†ψφ into −(g/2)(χ₁² + χ₂²)φ, two real loops of ½ each, which checks the 1. A loop through n labelled vertices closes in (n − 1)! oriented ways of weight 1, while a real field's loops total (n − 1)!/2, so the complex loop is always twice the real one. A symmetry factor survives only where relabelling vertices maps arrows onto arrows, as in the order-g² vacuum bubble, which keeps S = 2.

The rules collected, counted and checked

Collected: i/(p² − M² + iε) along each arrow, i/(p² − m² + iε) on each φ line, −ig with momentum conservation at each vertex, ∫d⁴l/(2π)⁴ for each loop, and 1/S only for symmetries that respect the arrows. A closed scalar loop takes no sign, unlike the −1 and trace topic 7.04 derives for Dirac fields. Counting line ends gives 2V = 2I_ψ + E_ψ and V = 2I_φ + E_φ, so with L = I_ψ + I_φ − V + 1 a connected diagram has V = 2L + E_ψ + E_φ − 2: each loop costs g². Dimensions check the rules: g is a mass, each propagator a mass⁻² and each loop integral a mass⁴, so −iΠ has dimension 2 − 4 + 4 = 2, as a shift of p² − m² must. As g → 0 only free lines remain. A φ line carrying no energy is −i/(|q|² + m²), whose Fourier transform e^(−mr)/(4πr), quoted from QM II, gives a force of range ħ/mc = ħc/mc²: 1.414 fm for the measured charged-pion mass, 139.57 MeV, the range of the nuclear force that Yukawa's meson theory was built to explain.

Change one variable at a time

Make the relationship visible.

Interactive model

Energy p⁰ carried along the arrow−1.75 GeV

Spatial momentum |p|0.7 GeV

ψ mass M1.6 GeV

Slide p⁰ from −1.75 to +1.75 GeV: the dashed energy arrow turns to run with the ψ arrow, and the marker moves from the ψ̄ peak to the ψ peak of one propagator. Raise M and only the solid peaks move out; the φ peaks carry no arrow, and both are the same particle.

ψ POLES AT p⁰ = ±E_ψ1.746 GeV

φ POLES AT p⁰ = ±E_φ1.221 GeV

OFF SHELL BY p² − M²0.01 GeV²

KINETIC E_ψ − M0.146 GeV

Live interpretationψ POLES AT p⁰ = ±E_ψ: 1.746 GeV. φ POLES AT p⁰ = ±E_φ: 1.221 GeV. OFF SHELL BY p² − M²: 0.01 GeV². KINETIC E_ψ − M: 0.146 GeV

Catch the common trap

Explain before calculating.

Misconception check

Treating ψ†ψ as if it were χ², the square of one real field. Students then either double the vertex to −2ig, as a −gχ²φ coupling would give, or keep the ½ of a real bubble on the closed ψ loop. Both import a symmetry that is not there. In Wick's theorem ψ and ψ† are different operators: ψ contracts only with ψ†, never with itself, so an incoming arrow can attach only to ψ(z) and an outgoing one only to ψ†(z), one way, and the vertex is −ig. In the bubble the two internal lines run in opposite directions, so interchanging them reverses both arrows and is not a symmetry of the diagram: S = 1. The check is ψ = (χ₁ + iχ₂)/√2, which makes −gψ†ψφ = −(g/2)(χ₁² + χ₂²)φ. Each real field has vertex −ig and a bubble of weight ½, and there are two of them, so the ψ loop is 1, twice a single real loop and never equal to it.

In the scalar Yukawa model, the order-g² term of ⟨Ω|Tφ(x)φ(y)|Ω⟩ that joins x to y through a closed ψ loop is c(−ig)²∫d⁴z₁ d⁴z₂ D_F(x − z₁; m) D_F(z₁ − z₂; M) D_F(z₂ − z₁; M) D_F(z₂ − y; m). What is c?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyDecide whether ⟨Ω|Tψ(x₁)ψ(x₂)ψ†(y)φ(w)|Ω⟩ can be nonzero at any order. Then, for ⟨Ω|Tψ(x₁)ψ(x₂)ψ†(y₁)ψ†(y₂)|Ω⟩ at order g², count the full contractions, find how many give connected diagrams, and state each connected diagram's weight.

Arrows run from a ψ† to a ψ, and every vertex passes one arrow in and one out, so external ψ's and ψ†'s must pair off along unbroken lines. The first function has two arrow heads and one tail: one arrow could never start, so it vanishes at every order.
At order g² the second is (1/2!)(−ig)²∫∫ times ⟨0|Tψ(x₁)ψ(x₂)ψ†(y₁)ψ†(y₂)(ψ†ψφ)(z₁)(ψ†ψφ)(z₂)|0⟩. Four ψ's pair with four ψ†'s in 4! = 24 ways and the two φ's in 1, so there are 24 full contractions.
Connected needs the φ line to join the two arrowed lines, so z₁ sits on one and z₂ on the other. The line from y₁ can end at x₁ or x₂ (2 choices) and carry z₁ or z₂ (2 choices): 4 contractions.
The other 20 are disconnected: 8 put both vertices on one line, 8 hang a closed ψ loop on one line through the φ, and 4 leave both lines bare beside a vacuum bubble.
The 4 connected ones pair under z₁ ↔ z₂, cancelling the 1/2!: two diagrams, y₁ → x₁ with y₂ → x₂ and y₁ → x₂ with y₂ → x₁, each (−ig)² times four ψ propagators and one φ propagator, weight 1.

AnswerThe first vanishes to all orders. The second has 24 full contractions at order g², 4 of them connected, forming 2 diagrams (direct and exchanged) of weight 1 each.

MediumSort all the full contractions of the order-g² Gell-Mann-Low numerator for ⟨φ(x)φ(y)⟩ into diagrams with weights. Then count the same order for a real field χ of mass M coupled by −(g/2)χ²φ, and show that the ratio of the two connected loop weights is the number of real components in ψ.

φ pairings of x, y, z₁, z₂: (xy)(z₁z₂), (xz₁)(yz₂), (xz₂)(yz₁), which is 3. ψ pairings: ψ(z₁)ψ†(z₂) with ψ(z₂)ψ†(z₁), a loop through both vertices, or ψ(z₁)ψ†(z₁) with ψ(z₂)ψ†(z₂), a closed arrow at each, which is 2. Total 3 × 2 = 6.
With the 1/2!: (xz₁)(yz₂) and (xz₂)(yz₁) with the loop give the connected bubble, weight 2 × ½ = 1; with the closed arrows they give two tadpoles, ⟨φ(x)⟩⟨φ(y)⟩ at order g, weight 1.
(xy)(z₁z₂) with the loop gives D_F(x − y; m) times the two-vertex vacuum bubble, weight ½, and with the closed arrows D_F(x − y; m) times the dumbbell, weight ½. Check: 1 + 1 + ½ + ½ = 3 = 6 × ½.
Real χ: the φ pairings are still 3, and the four χ's, two at z₁ and two at z₂, pair in 3 ways, 2 joining the vertices and 1 closing each on itself: 9 contractions, each weighted (1/2!)(1/2)² in units of (−ig)².
The connected χ bubble takes 2 φ ways × 2 χ ways = 4 contractions, weight 4 × ½ × ¼ = ½. The ψ loop's 1 is twice that.
ψ = (χ₁ + iχ₂)/√2 gives ψ†ψ = (χ₁² + χ₂²)/2, so −gψ†ψφ is two copies of the χ coupling, each with a loop of ½, and 2 × ½ = 1.

AnswerSix contractions: connected ψ bubble 1, two tadpoles 1, vacuum bubble ½ and dumbbell ½ (each times D_F(x − y; m)). The real-χ bubble is ½, from 4 of its 9 contractions; the ratio 2 counts the two real fields in ψ.

HardAt order g⁴, ⟨φ(x₁)φ(x₂)φ(x₃)φ(x₄)⟩ gets its first connected term from one closed ψ loop. Count the diagrams and their weights, and check the order with V = 2L + E_ψ + E_φ − 2. Using ∫d⁴l/(2π)⁴ (l² − M² + iε)⁻⁴ = i/(96π²M⁴), quoted here and derived by Wick rotation in Units 5 and 6, sum the loop factors at zero external momenta, read the sum as a −(λ/4!)φ⁴ coupling, and evaluate λ for g = 2.00 GeV and M = 1.00 GeV.

E_φ = 4, E_ψ = 0 and L = 1 give V = 2 + 4 − 2 = 4, so order g⁴. No tree exists: with no external arrows every arrow must close into a loop.
In (1/4!)(−ig)⁴∫…, attaching φ(x₁)…φ(x₄) to the four vertices takes 4! ways, cancelling the 1/4!. With xᵢ at zᵢ, the ψ contractions forming one loop are the 4-cycles of the vertices: (4 − 1)! = 6, each a distinct diagram of weight 1.
They pair under reversal, z₁ → z₂ → z₃ → z₄ → z₁ with z₁ → z₄ → z₃ → z₂ → z₁ and so on: 3 cyclic orders × 2 orientations. The propagator is even, so a reversed loop has the same integrand. A real χ coupled by −(g/2)χ²φ has only the 3 unoriented boxes, weight 1 each.
At zero external momenta every internal line carries l, so each diagram's loop factor, which multiplies the four external φ propagators, is (−ig)⁴ i⁴ ∫d⁴l/(2π)⁴ (l² − M² + iε)⁻⁴ = g⁴ × i/(96π²M⁴). The six sum to 6g⁴ i/(96π²M⁴) = i g⁴/(16π²M⁴).
A contact −(λ/4!)φ⁴ would give −iλ, so λ_eff = −g⁴/(16π²M⁴). Dimensions: g⁴/M⁴ is pure, as a φ⁴ coupling must be. With g/M = 2.00, λ_eff = −16/157.91 = −0.101.
Limits: λ_eff → 0 as M → ∞ at fixed g, and as g → 0. The real χ, with half the diagrams, gives −g⁴/(32π²M⁴) = −0.0507.

Answer6 oriented one-loop boxes at order g⁴, weight 1 each, summing at zero momenta to i g⁴/(16π²M⁴): λ_eff = −g⁴/(16π²M⁴) = −0.101 for g = 2.00 GeV and M = 1.00 GeV, twice the real-field −0.0507.