University Quantum Field Theory I · Feynman Diagrams for Scalar Field Theories · 3.8
1PI diagrams, the self-energy and the full propagator
The Lagrangian hands you a parameter m₀; an experiment hands you the position of a pole. This topic joins the two: sort the two-point diagrams into irreducible blobs, sum the chain they form, and read the physical mass off the zero of p² − m₀² − Σ(p²), a moved pole that no truncated sum of diagrams shows.
Build the model
Connect the measurement to the mechanism.
The φ⁴ Lagrangian, ½(∂φ)² − ½m₀²φ² − (λ/4!)φ⁴ in natural units ħ = c = 1 with metric (+, −, −, −), contains a parameter m₀, but no experiment measures it: what is measured is where the two-point function has its pole. To find that pole, sort the connected two-point diagrams by whether they survive the cut of any single internal line. Those that do are one-particle irreducible, and with their external legs removed they sum to −iΣ(p²).
Every connected diagram is then uniquely a chain of these blobs joined by free propagators that carry the full external momentum, so the two-point function is a geometric series in Σ/(p² − m₀²), and summing it gives i/(p² − m₀² − Σ(p²) + iε). The pole has moved to the root of p² − m₀² − Σ(p²) = 0, and that root is the physical mass. No finite order shows it: every truncated sum keeps its only pole at m₀², and the series diverges wherever |Σ| ≥ |p² − m₀²|, a band whose edge is the shifted pole.
The resummation reorganises diagrams and adds no new ones, so it is only as good as the order to which Σ is known. At order λ, Σ is the momentum-independent tadpole, which moves the pole and leaves its residue at 1; the p² dependence that changes the residue arrives at λ², and above the three-particle threshold Σ turns complex.
- Simple definition
- A Feynman diagram is one-particle irreducible if cutting any single internal line leaves it connected; the 1PI two-point diagrams, legs removed, sum to −iΣ(p²), and chaining them gives the full propagator i/(p² − m₀² − Σ(p²) + iε).
- Example
- Through order λ² in φ⁴, the tadpole (symmetry factor 2), the sunset (6) and the tadpole on a tadpole (4) are all 1PI. Two tadpoles in series (4) fall apart when the one line between them is cut, so they belong to the chain, not to Σ.
A definition. With one momentum, Σ can depend on p² alone, and a two-point vertex read off the Lagrangian is 1PI by itself, which is how counterterms will enter.
Σ and p² in GeV², mass dimension 2; p is the four-momentum through the blob; λ and every loop integral sit inside Σ
Derived: the chain with n blobs is [i/(p² − m₀²)](n+1)(−iΣ)ⁿ, a geometric series with ratio Σ/(p² − m₀²). Set Σ = 0 and the free propagator returns.
D in GeV⁻²; m₀ the Lagrangian mass parameter in GeV; iε the Feynman prescription of topic 1.06
Derived from the rules of topics 3.01 and 3.05. The external momentum never enters the loop, so Σ₁ is a constant: it moves the mass and leaves the residue at 1.
λ dimensionless; ℓ the loop momentum in GeV; the 1/2 is the symmetry factor; the integral is ⟨0|φ²|0⟩ of the free field
Derived: the zero of the resummed denominator. For small Σ it gives m ≈ m₀ + Σ/(2 m₀); a constant Σ = δm² gives exactly the free propagator of mass² m₀² + δm².
m the pole mass and m₀ the Lagrangian parameter, both in GeV; Σ evaluated on the pole, at p² = m²
Previewed by a one-line Taylor expansion about the pole; topic 4.03 proves 0 ≤ Z ≤ 1. In φ⁴ the first p²-dependent insertion is the sunset, at order λ².
Z and dΣ/dp² dimensionless; Z = 1 exactly while Σ is constant, as at order λ in φ⁴
Derived: a geometric series diverges once |r| ≥ 1. The physical pole sits at r = 1, on the band's edge, so no finite order finds it; the closed form continues the sum there.
both sides in GeV²; inside this band the chain's ratio r = Σ/(p² − m₀²) has |r| ≥ 1, and N insertions leave a factor 1 − r(N+1)
Cut one internal line: the test that sorts the diagrams
Take a connected two-point diagram and cut one internal line. If it falls into two pieces, that line was a bridge; if no single cut disconnects it, the diagram is one-particle irreducible, 1PI. In φ⁴ through order λ² there are four connected two-point diagrams beyond the free line, with the symmetry factors topic 3.02 found. The tadpole (one vertex, one loop, symmetry factor 2), the sunset (two vertices joined by three lines, factor 6) and the tadpole on a tadpole (factor 4) are 1PI: cut any one of their lines and they hang together. Two tadpoles in series (factor 4) are not, because the single line between the vertices is a bridge. Momentum conservation fixes what a bridge carries. The whole external momentum p enters at one end, so the one line that joins the two halves must carry exactly p and no loop momentum: its factor is i/(p² − m₀² + iε), standing outside every integral. That is why a reducible diagram factorises into its irreducible pieces.
Remove the legs, and what is left is −iΣ(p²)
Define −iΣ(p²) as the sum of all 1PI two-point diagrams with their two external propagators stripped off. Three facts fix its form. It depends on p only through p², because it is a Lorentz scalar built from a single four-momentum. Its mass dimension is 2, the same as p² and m₀²: λ is dimensionless in four dimensions and the tadpole is λ times an integral of the form ∫d⁴ℓ/ℓ², so Σ can stand beside p² − m₀² in one denominator and is quoted in GeV². And it collects every two-point vertex the Lagrangian contains. By topic 3.07's rule a term −½bφ² is the vertex −ib and ½a(∂φ)² is the vertex +iap²; each is 1PI trivially, since it has no internal line to cut, so together they add b − ap² to Σ. Topic 3.07 read those vertices off; here they become insertions on the same footing as loops, which is exactly how counterterms will enter in Unit 6.
Sum the chain: a geometric series in Σ/(p² − m₀²)
Every connected two-point diagram is a unique chain: free propagator, 1PI blob, free propagator, blob, and so on, ending on a propagator. Uniqueness means nothing is counted twice. The weights also factorise. The external points x and y fix the ends of the chain, so no automorphism can swap two blobs, and the chain's symmetry factor is the product of its blobs' factors: two tadpoles in series have 2 × 2 = 4. So the chain with n blobs contributes [i/(p² − m₀²)](n+1)(−iΣ)ⁿ, with no extra 1/n!. Each added blob multiplies by (−iΣ) × i/(p² − m₀²) = Σ/(p² − m₀²), and the sum is i/(p² − m₀²) × 1/(1 − Σ/(p² − m₀²)) = i/(p² − m₀² − Σ(p²) + iε). This is derived, not assumed. The two factors of i multiply to +1, so Σ enters the denominator with a minus sign: a positive Σ raises the pole.
At order λ the tadpole only shifts the mass
The order-λ insertion is the tadpole: one vertex −iλ, symmetry factor 2, and a loop closing on that vertex, so −iΣ₁ = (−iλ/2)∫d⁴ℓ/(2π)⁴ i/(ℓ² − m₀² + iε). The external momentum enters and leaves at the same point and never reaches the loop, so Σ₁ is a constant. It equals (λ/2)⟨0|φ²|0⟩, which a mean-field count confirms: replace two of the four fields in (λ/4!)φ⁴ by their vacuum average, in 4!/(2! 2!) = 6 ways, and the interaction becomes (λ/4)⟨φ²⟩φ² = ½(λ⟨φ²⟩/2)φ², a mass term. After the Wick rotation the integrand is 1/(ℓE² + m₀²), positive everywhere, so with any cutoff Σ₁ is positive; the elementary radial integral up to a Euclidean cutoff Λ gives Σ₁ = λ[Λ² − m₀² ln(1 + Λ²/m₀²)]/(32π²), and λ = 0.10 with Λ = 10 m₀ gives Σ₁ = 0.030 m₀². Topic 6.05 treats the divergence properly. A constant insertion is the cleanest check of the resummation: Σ = δm² turns i/(p² − m₀²) into i/(p² − m₀² − δm²), exactly the free propagator of mass² m₀² + δm², with residue 1.
The pole, not m₀, is the mass you measure
The physical mass m is the zero of the denominator, m² − m₀² − Σ(m²) = 0, and for small Σ it sits at m ≈ m₀ + Σ/(2 m₀). Why the pole rather than m₀? A one-particle state of mass m puts a pole at p² = m² in the propagator (topic 1.06 for the free field; topic 4.02 proves it in general), and the pole sets every long-distance effect: the exchange potential of topic 5.06 falls as e(−mr)/r at large r with m the pole, because the nearest singularity in |q| fixes the decay. The residue is 1 when Σ is constant. When it is not, expanding the denominator about the pole gives residue 1/(1 − dΣ/dp²) at p² = m², the Z that topic 4.03 bounds between 0 and 1; in φ⁴ the first p²-dependent insertion is the sunset, so Z = 1 + O(λ²). A lepton mass in the Particle Data Group tables, such as the electron's 0.51099895 MeV, is a pole mass of this kind; no table lists the m₀ of a Lagrangian.
Why resum: no finite order moves the pole
Stop the chain after N insertions and the sum is (1 − r(N+1))/(p² − m₀² − Σ), times i, with r = Σ/(p² − m₀²). Wherever |r| is below 1 this approaches the resummed result as N grows. Inside the band |p² − m₀²| ≤ |Σ| around the bare mass it does not, and every truncation still has its only pole at p² = m₀², now of order N + 1. The physical pole sits at r = 1, on the edge of that band, exactly where the series stops converging. So the closed form is not the sum of the series at the pole; it is the unique analytic continuation from where the series converges, and it is what perturbation theory means there. Two limits remain. The resummation adds every chain of the blobs you computed but no new blobs: with Σ known to order λ, the sunset and the tadpole on a tadpole are still missing at λ². And above the three-particle threshold, p² = 9 m² in φ⁴, the sunset gives Σ an imaginary part. The φ⁴ pole lies below that threshold, so this particle stays stable; when a pole lies above a threshold its particle can decay, and topic 5.09 turns Im Σ there into a width.
Change one variable at a time
Make the relationship visible.
Keep Σ = 0.3 m₀² and raise N from 0 to 6: at the probe p² = 1.65 m₀², where r = 0.46, the dashed value closes on the solid one. Move the probe to 1.15 m₀², inside the band, where r = 2 and each extra insertion makes it worse. Set Σ to 0 for the free limit.
PHYSICAL MASS m/m₀1.140
CHAIN RATIO r AT PROBE0.462
RESUMMED m₀²D/i AT PROBE2.857
TRUNCATED AT N, AT PROBE2.576
Live interpretationPHYSICAL MASS m/m₀: 1.140. CHAIN RATIO r AT PROBE: 0.462. RESUMMED m₀²D/i AT PROBE: 2.857. TRUNCATED AT N, AT PROBE: 2.576
Catch the common trap
Explain before calculating.
In φ⁴ theory at order λ the only one-particle-irreducible two-point diagram is the tadpole, which, cut off at a large Euclidean momentum, gives a constant, positive Σ₁. After the chain of tadpole insertions is summed, what does the full propagator look like near p² = m₀²?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA φ⁴ field has Lagrangian mass m₀ = 0.600 GeV, and its order-λ tadpole gives Σ₁ = 0.0900 GeV². Find the physical mass from the resummed propagator, compare it with the first-order estimate m₀ + Σ₁/(2 m₀), and restore ħ and c to find how the range of a force carried by this particle changes.
- Σ₁ is a constant, so the resummed propagator is i/(p² − m₀² − Σ₁), a free propagator whose pole sits at m² = m₀² + Σ₁ = 0.3600 + 0.0900 = 0.4500 GeV². Dimensions agree: Σ₁ and m₀² are both GeV².
- m = √0.4500 GeV = 0.6708 GeV. The residue is exactly 1, because dΣ₁/dp² = 0.
- First order: m ≈ m₀ + Σ₁/(2 m₀) = 0.600 + 0.0900/1.200 = 0.6750 GeV, high by 0.6%. The next term, −Σ₁²/(8 m₀³) = −0.0047 GeV, closes most of the gap; Σ₁/m₀² = 0.25 is not small.
- Limiting case: as Σ₁ → 0 the pole returns to m₀ and the free field is recovered.
- Restore units: an exchange force falls as e(−r/R) with R = ħc/(mc²), fixed by the pole. For m₀, R = 197.327 MeV fm / 600 MeV = 0.329 fm; for the physical mass, R = 197.327 MeV fm / 670.8 MeV = 0.294 fm, 11% shorter.
Answerm = 0.671 GeV with residue 1, against 0.675 GeV at first order; the range is ħ/(mc) = 0.294 fm, not the 0.329 fm that m₀ would give.
MediumExpand the resummed propagator with Σ = Σ₁, the order-λ tadpole, to order λ². Show that its λ² term is the two-tadpole chain with the right weight by counting the Wick contractions of ⟨0|T φ(x)φ(y) (1/2!)[−i(λ/4!)∫d⁴z φ⁴(z)][−i(λ/4!)∫d⁴w φ⁴(w)]|0⟩ that produce it. Then, for Σ₁ = 0.030 m₀², find the chain's size relative to the free propagator at p² = 2 m₀², 1.05 m₀² and 1.02 m₀².
- Expanding i/(p² − m₀² − Σ₁) in r = Σ₁/(p² − m₀²) gives D₀ + D₀(−iΣ₁)D₀ + D₀(−iΣ₁)D₀(−iΣ₁)D₀ + …, with D₀ = i/(p² − m₀²). Each −iΣ₁ carries the tadpole's 1/2, so the λ² term has weight (1/2)² = 1/4.
- Count contractions. Choose the vertex joined to x: 2 ways. Pick its field: 4. Join y to one of the other vertex's four fields: 4. Of the first vertex's three remaining fields, choose the one that crosses: 3; it meets one of the second vertex's three: 3. The last pairs close on themselves, 1 way each.
- That is 2 × 4 × 4 × 3 × 3 = 288 contractions against the prefactor 1/(2! × 4! × 4!) = 1/1152, a weight 288/1152 = 1/4, so the symmetry factor is 4.
- Automorphisms agree: each loop can be flipped, 2 × 2 = 4, and the two tadpoles cannot be swapped because x and y are fixed. The series reproduces the diagram exactly, so nothing reducible may sit in Σ.
- Size: the chain is r² times the free propagator. At p² = 2 m₀², r = 0.030 and r² = 9.0 × 10⁻⁴. At 1.05 m₀², r = 0.60 and r² = 0.36. At 1.02 m₀², r = 1.5 and r² = 2.25: the λ² chain outweighs the free term, the series diverges, and yet the resummed m₀²D/i = 1/(0.020 − 0.030) = −100 is finite.
AnswerWeight 288/1152 = 1/4 from contractions and 2 × 2 = 4 from automorphisms, matching (1/2)² from the series; r² = 9.0 × 10⁻⁴, 0.36 and 2.25, and at 1.02 m₀² the series diverges while the resummed D/i = −100/m₀².
HardA φ⁴ Lagrangian carries two extra quadratic terms, ½a(∂φ)² − ½bφ², with a = 0.050 and b = 0.0200 GeV², treated as two-point vertices beside m₀ = 0.500 GeV and a tadpole Σ₁ = 0.0125 GeV². Find Σ(p²), the physical mass and the residue of the resummed propagator, check both against an exact rescaling of the field at λ = 0, and find where the chain stops converging.
- Read as vertices (topic 3.07), ½a(∂φ)² gives iap² and −½bφ² gives −ib. A single vertex has no internal line to cut, so each is 1PI: Σ(p²) = Σ₁ + b − ap² = 0.0325 GeV² − 0.050p².
- Resum: i/(p² − m₀² − Σ₁ − b + ap²) = [i/(1 + a)]/[p² − (m₀² + Σ₁ + b)/(1 + a)]. a is dimensionless and every other term is in GeV², as the denominator requires.
- Pole: m² = (0.2500 + 0.0125 + 0.0200)/1.050 = 0.2825/1.050 = 0.2690 GeV², so m = 0.5187 GeV. Without the derivative term the mass would be √0.2825 GeV = 0.5315 GeV; a = 0.050 lowers it by 2.4%.
- Residue: 1/(1 + a) = 0.952, and the general rule agrees, 1/(1 − dΣ/dp²) = 1/(1 + a). A constant Σ cannot change the residue; a p²-dependent one does.
- Exact check at λ = 0: φ′ = √(1 + a) φ is a canonical field of mass² (m₀² + b)/(1 + a) = 0.2700/1.050 = 0.2571 GeV², and ⟨φφ⟩ = ⟨φ′φ′⟩/(1 + a). The resummed pole and residue match this exactly; no finite order in a does.
- Convergence: r = Σ(p²)/(p² − m₀²) is −0.13 at p² = 0, but |r| ≥ 1 for 0.2289 GeV² ≤ p² ≤ 0.2690 GeV², a band whose upper edge is the pole itself.
AnswerΣ = 0.0325 GeV² − 0.050p²; m = 0.519 GeV with residue 0.952; at λ = 0 both match the rescaled free field, 0.2571 GeV² and 1/1.050; the chain diverges for 0.229 GeV² ≤ p² ≤ 0.269 GeV².