University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.6
Scalar Yukawa tree amplitudes and the s-channel pole
This is where the Feynman rules first produce amplitudes worth reading. Three processes, at most two diagrams each, and every feature of the answer, its symmetry, its signs, which poles it has and where they sit, can be predicted from the arrows and checked before a single cross-section is computed.
Build the model
Connect the measurement to the mechanism.
At order g² every two-to-two process in the scalar Yukawa model is one propagator strung between two vertices, so its amplitude is a sum of simple poles, one for each way the external lines can be paired, each sitting at the squared mass of the particle that internal line carries. Which pairings exist is settled before any arithmetic by the arrows that track ψ number: ψψ → ψψ exchanges a φ in the t and u channels, ψψ̄ → ψψ̄ exchanges one in t and annihilates into one in s, ψψ̄ → φφ exchanges a ψ, and ψψ → φφ cannot happen at all. Reading the answer is then a matter of asking where each pole sits relative to the physical region.
Exchange poles never reach it in these processes: the quantum passed between the colliding particles carries momentum with q² ≤ 0, off its mass shell by at least its squared mass, so exchange terms are finite, positive and forward peaked, with a width set by the exchanged mass, and at fixed angle they fade as g²/s. The s-channel pole is different, because annihilation loads timelike momentum, s ≥ 4M², onto the φ. With m below 2M the pole stays outside and the annihilation term only competes in sign with the exchange; with m above 2M it lies inside, and at √s = m the tree amplitude is infinite.
That is where this picture stops: a φ heavy enough to be made from ψψ̄ is heavy enough to decay into it, and the cure is the width that topic 5.09 resums.
- Simple definition
- A tree amplitude is the lowest-order sum of connected, amputated diagrams without loops; in the scalar Yukawa model each two-to-two tree is a single φ or ψ propagator between two vertices, so ℳ is a sum of simple poles in s, t or u.
- Example
- At threshold, where t = u = 0, ψψ → ψψ has ℳ = 2g²/m²: with g = 0.50 GeV and m = 0.25 GeV that is 8.0, a pure number, as a four-point amplitude in natural units must be.
Two contractions, one in t and one in u, added with a plus sign, so ℳ is unchanged by t ↔ u as Bose symmetry demands. At threshold t = u = 0 and ℳ = 2g²/m².
g in GeV (mass dimension 1), m the φ mass in GeV, t and u in GeV²; ℳ is dimensionless (derived)
The s ↔ u image of the ψψ → ψψ result. With m below 2M the annihilation term is negative wherever the process is physical, and it competes with the positive exchange.
s = (p₁ + p₂)² ≥ 4M² in GeV², M the ψ mass in GeV; no u-channel term, which ψ number forbids (derived)
Now a ψ is exchanged, so the gap that keeps each term finite is M². There is no s-channel term because the Lagrangian has no φ³ vertex.
t = (p₁ − k₁)² and u = (p₁ − k₂)² in GeV², k the φ momenta; s + t + u = 2(M² + m²) (derived)
The exchanged φ is spacelike, so m² − t ≥ m²: an exchange term never diverges, and for p ≫ m it falls to half its forward value g²/m² near θ = m/p.
Elastic, equal masses; p the centre-of-mass momentum in GeV, θ the scattering angle; u has cos θ → −cos θ (derived)
A coupling with the dimension of mass fades as g²/s, as topic 2.01 predicted from counting dimensions alone.
s in GeV², g in GeV; from the first card with p² → s/4, corrections of order M²/s and m²/s (derived)
Below 2M the pole is harmless at every energy. Above it the tree diverges at √s = m until topic 5.09's width, quoted, turns m² − s into m² − s − imΓ.
m and M in GeV; the same condition that allows φ → ψψ̄, found in topic 2.01 (derived here)
Which diagrams exist: follow the arrows
Units are natural, ħ = c = 1, with metric (+, −, −, −) throughout. Quoted from topic 2.01: ℒᵢₙₜ = −gψ†ψφ, with ψ a complex scalar of mass M, φ a real scalar of mass m, and g a mass. Quoted from topic 3.06: each vertex gives −ig and joins one arrow in, one arrow out and one φ line; a ψ line gives i/(p² − M² + iε) and a φ line i/(p² − m² + iε). Topic 4.05 makes iℳ the sum of connected, amputated diagrams, with Z = 1 at tree level. At order g² a two-to-two tree has two vertices and one internal line, so listing the trees means listing the ways to pair the external lines at the vertices, and the arrows decide it. An incoming ψ and an outgoing ψ̄ each carry an arrow into the diagram, an outgoing ψ and an incoming ψ̄ an arrow out, and a vertex takes one of each. So ψψ → ψψ has no s-channel diagram, ψψ̄ → ψψ̄ no u-channel one, ψψ̄ → φφ no s-channel one for want of a φ³ vertex, and ψψ → φφ none at all: two arrows would end inside, which ψ number, an exact symmetry, forbids at every order.
Exchange between identical particles: ψψ → ψψ
Label the process ψ(p₁)ψ(p₂) → ψ(p₃)ψ(p₄). Each incoming arrow must leave through an outgoing ψ at its own vertex, and there are two ways: p₁ to p₃ and p₂ to p₄, with the φ carrying q = p₁ − p₃ so that q² = t; or p₁ to p₄ and p₂ to p₃, with q² = u. The 1/2! of second order is cancelled by the two ways of assigning the vertices, as in Unit 3, so neither diagram carries a symmetry factor. Then iℳ = (−ig)² i/(t − m²) + (−ig)² i/(u − m²), so ℳ = g²[1/(m² − t) + 1/(m² − u)], derived. Three checks follow. Dimensions: g² over a squared mass is dimensionless, as the rule 4 − n requires for n = 4 external particles. Symmetry: swapping the identical final particles swaps t and u and leaves ℳ unchanged, and the terms add with a plus sign because bosonic contractions carry no sign. Threshold: at p = 0, t = u = 0 and ℳ = 2g²/m², independent of angle, as any amplitude must become when the momentum that defines the angle vanishes.
Physical t and u are never positive
In the centre-of-mass frame of an elastic collision between equal masses each particle keeps its energy, so q = p₁ − p₃ has no time component and t = q² is minus the squared three-momentum transfer, −2p²(1 − cos θ), derived here and developed in topic 4.07. So −4p² ≤ t ≤ 0, and likewise for u: the exchanged φ has q² ≤ 0, never on shell, and m² − t ≥ m², so each exchange term is finite and positive. The pole at t = m² sits a distance m² outside the physical region, and that gap sets the shape. The t-term falls from g²/m² forward to half that where 1 − cos θ = m²/(2p²), a forward peak of width about m/p; in the massless limit the gap closes and the peak becomes Rutherford's forward divergence. At fixed angle and large s, ℳ → 4g²/(s sin²θ): the dimensionful coupling fades as g²/s, as topic 2.01 said it must. The gap also sets a range, 1/m, as topic 5.06 will show: for the charged pion, whose measured mass is 139.57 MeV, ħc/(mc²) = 197.33 MeV fm / 139.57 MeV = 1.41 fm, the reach of the longest-range part of the nuclear force, which Yukawa built this picture to explain (quoted).
Annihilation: ψψ̄ → ψψ̄ and the s channel
Label ψ(p₁)ψ̄(p₂) → ψ(p₃)ψ̄(p₄). The arrow entering at p₁ can leave through the outgoing ψ, giving the exchange term g²/(m² − t) as before, or it can turn back through the incoming ψ̄ at the same vertex, so that the pair annihilates into a φ of momentum p₁ + p₂, which then makes the final pair. That diagram is (−ig)² i/(s − m²), and ℳ = g²[1/(m² − t) + 1/(m² − s)], derived. Exchanging s and u in the ψψ → ψψ result gives the same function, the crossing that topic 4.07 explains. Here the φ is timelike, q² = s ≥ 4M². If m is below 2M, m² − s is negative throughout the physical region, so the annihilation term is finite and opposite in sign to the exchange, and the two compete. With M = 1 GeV, m = 1.6 GeV and p = 0.8 GeV, s = 6.56 GeV² and the s-term is −0.250 g² GeV⁻², while the t-term runs from 0.391 g² GeV⁻² forward to 0.195 g² GeV⁻² backward. ℳ vanishes where m² − t equals s − m², at t = −1.44 GeV², which is θ = 97.2°.
Pair annihilation into two φ: a ψ is exchanged
For ψ(p₁)ψ̄(p₂) → φ(k₁)φ(k₂) the arrow runs in at p₁, through one internal ψ line, and out at p₂, and each φ attaches at one end. Attaching φ(k₁) at the p₁ end gives the internal line momentum p₁ − k₁, the other choice p₁ − k₂, so ℳ = g²[1/(M² − t) + 1/(M² − u)] with t = (p₁ − k₁)² and u = (p₁ − k₂)², derived. Swapping the identical φs swaps t and u, a check, and s + t + u = 2(M² + m²), the sum of the four squared masses, another. Here the gap is M², and it never closes. In the centre-of-mass frame t = M² + m² − 2E² + 2|p||k| cos θ, with E = √s/2, and since 2|p||k| ≤ |p|² + |k|² = 2E² − M² − m², every physical t is at most 0, so M² − t ≥ M² and both terms are finite and positive. No s-channel diagram appears because ℒ has no φ³ vertex. The theory has no φ → −φ symmetry to protect that absence, though, and a ψ loop generates an effective φ³ coupling at order g³, so an s-channel term arrives at one loop.
When the s-channel pole becomes physical
The pole of the annihilation term at s = m² can be reached only if m² ≥ 4M², that is m ≥ 2M: exactly the condition topic 2.01 found for φ → ψψ̄ to be allowed. That is no coincidence. A pole inside the physical region means a real φ can be made from the pair, and a φ that can be made from ψψ̄ can decay back into it. At √s = m the tree amplitude is infinite, which no measured amplitude is, so here, and only here, tree level fails. Topic 5.03 computes the rate, Γ = (g²/(16πm))√(1 − 4M²/m²), and topic 5.09 shows that the self-energy's imaginary part turns m² − s into m² − s − imΓ, both quoted. The tree is therefore good wherever |s − m²| ≫ mΓ. With m below 2M the φ is stable and its pole lies below threshold, at distance 4M² − m², outside the physical region at every energy. Nature shows the same structure with spin: e⁺e⁻ → μ⁺μ⁻ peaks at √s = 91.19 GeV, the Z mass, with a full width of 2.50 GeV (quoted).
Change one variable at a time
Make the relationship visible.
At m = 1.5M the pole sits in the shaded strip, the s-channel term is negative wherever the process is physical, and at 90° the total crosses zero at √s = 2.24M. Drag m past 2M and the pole walks into the physical region, where the tree diverges. Set θ = 0° and the exchange curve flattens at g²/m².
t AT THE CURSOR-2.530 M²
EXCHANGE g²/(m² − t)0.209 g²/M²
ANNIHILATION g²/(m² − s)-0.147 g²/M²
TOTAL ℳ0.062 g²/M²
Live interpretationt AT THE CURSOR: −2.530 M². EXCHANGE g²/(m² − t): 0.209 g²/M². ANNIHILATION g²/(m² − s): −0.147 g²/M². TOTAL ℳ: 0.062 g²/M²
Catch the common trap
Explain before calculating.
In the scalar Yukawa model with m = 1.5M, ψψ̄ → ψψ̄ is computed at tree level at every physical energy and angle. Which statement is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn the scalar Yukawa model take M = 1.00 GeV, m = 0.40 GeV and g = 1.2 GeV. Two ψ particles collide with centre-of-mass momentum p = 0.50 GeV and scatter through θ = 30°. Find s, t and u and check them, evaluate the tree amplitude ℳ and each channel's share, and show that θ = 150° gives the same ℳ.
- Kinematics: s = 4(M² + p²) = 4(1.00 + 0.25) GeV² = 5.00 GeV²; t = −2p²(1 − cos θ) = −0.50 × (1 − 0.8660) GeV² = −0.0670 GeV²; u = −2p²(1 + cos θ) = −0.9330 GeV².
- Check: s + t + u = (5.00 − 0.0670 − 0.9330) GeV² = 4.00 GeV² = 4M², as four equal masses require, and both t and u are negative, as physical values must be.
- Exchange in t: g²/(m² − t) = 1.44 GeV² / (0.160 + 0.0670) GeV² = 6.34. Exchange in u: 1.44 GeV² / (0.160 + 0.933) GeV² = 1.32. Each is a squared mass over a squared mass, so ℳ is dimensionless, as a four-point amplitude must be.
- Add with a plus sign: ℳ = 6.34 + 1.32 = 7.66. The t-term dominates because the momentum transfer is small: at 30° it is still 70% of its forward value g²/m² = 9.00.
- At 150° the values of t and u trade places, so the two terms trade places and ℳ = 1.32 + 6.34 = 7.66 again: the amplitude is symmetric about 90°, as Bose symmetry requires for identical final particles.
Answers = 5.00 GeV², t = −0.0670 GeV², u = −0.933 GeV², with s + t + u = 4.00 GeV² = 4M². ℳ = 6.34 + 1.32 = 7.66, dimensionless, and the same at 150°.
MediumWith M = 1.00 GeV, m = 0.60 GeV and g = 1.0 GeV, a ψ and a ψ̄ annihilate into two φ at √s = 2.50 GeV. In the centre-of-mass frame the first φ leaves at θ = 60° to the incoming ψ. Find the momenta, t and u, the tree amplitude, and the range of values ℳ takes as θ varies.
- Every particle has E = √s/2 = 1.25 GeV, so the ψ momentum is |p| = √(E² − M²) = 0.750 GeV and the φ momentum is |k| = √(E² − m²) = √1.2025 GeV = 1.0966 GeV.
- t = (p₁ − k₁)² = M² + m² − 2E² + 2|p||k| cos θ = (1.00 + 0.36 − 3.125 + 0.8224) GeV² = −0.9426 GeV², and u, with cos θ → −cos θ, is −2.5874 GeV².
- Check: s + t + u = (6.25 − 0.9426 − 2.5874) GeV² = 2.720 GeV² = 2(M² + m²). Both t and u are negative, so M² − t and M² − u exceed M², as the bound 2|p||k| ≤ 2E² − M² − m² guarantees.
- ℳ = g²[1/(M² − t) + 1/(M² − u)] = 1.0/1.9426 + 1.0/3.5874 = 0.5148 + 0.2788 = 0.794, dimensionless.
- Range: ℳ is largest when a φ leaves along the beam, at 0° or 180°, where t = −0.1201 GeV² and ℳ = 0.8928 + 0.2268 = 1.120, and smallest at 90°, where t = u = −1.765 GeV² and ℳ = 2/2.765 = 0.723. It is symmetric about 90°, as the identical φs require.
Answer|p| = 0.750 GeV and |k| = 1.097 GeV; t = −0.943 GeV² and u = −2.587 GeV². ℳ = 0.794 at 60°, ranging from 0.723 at 90° to 1.12 along the beam.
HardTake M = 0.50 GeV, m = 1.20 GeV and g = 0.80 GeV, so φ outweighs two ψ. For ψψ̄ → ψψ̄ at θ = 90°, show the s-channel pole is physical, compare the two tree terms at √s = 1.25 GeV, then use topic 5.03's width and topic 5.09's propagator, both quoted, to say how near the pole the tree holds, how high the peak is, and how long φ lives in seconds.
- Physical s starts at 4M² = 1.00 GeV², and the pole sits at s = m² = 1.44 GeV², inside: the tree amplitude is infinite at √s = 1.20 GeV. Because m exceeds 2M, φ can also decay to ψψ̄.
- At √s = 1.25 GeV, s = 1.5625 GeV² and p² = s/4 − M² = 0.1406 GeV², so at 90° t = −2p² = −0.2813 GeV². Exchange: 0.64/(1.44 + 0.2813) = 0.372. Annihilation: 0.64/(1.44 − 1.5625) = −5.22, fourteen times larger and negative, since s is above the pole.
- Width, quoted: Γ = (g²/(16πm))√(1 − 4M²/m²). With √(1 − 1.00/1.44) = 0.5528, Γ = 0.64 × 0.5528/(16π × 1.20) GeV = 5.87 × 10⁻³ GeV = 5.87 MeV, so mΓ = 7.04 × 10⁻³ GeV².
- Quoted from topic 5.09, the annihilation term becomes g²/(m² − s − imΓ). At 1.25 GeV, |s − m²| = 0.1225 GeV² = 17.4 mΓ, so |ℳₛ| shrinks by 1/√(1 + 17.4⁻²), only 0.16%. At |√s − m| = Γ/2 = 2.9 MeV, where |s − m²| ≈ mΓ, the factor is already 1/√2: the tree fails only that close.
- At s = m² the term is ig²/(mΓ), of size g²/(mΓ) = 16π/√(1 − 4M²/m²) = 90.9. The coupling cancels, so kinematics alone fixes the peak, 236 times the exchange term there (0.386, with t = −0.220 GeV²).
- Restoring ħ: τ = ħ/Γ = 6.582 × 10⁻²⁵ GeV s / 5.865 × 10⁻³ GeV = 1.12 × 10⁻²² s, and cτ = ħc/Γ = 197.3 MeV fm / 5.865 MeV = 33.6 fm.
AnswerThe pole at √s = 1.20 GeV lies inside s ≥ 1.00 GeV². At 1.25 GeV the s-term is −5.22 against 0.372. Γ = 5.87 MeV, so the tree errs by 0.16% there; the peak is 90.9, independent of g; τ = 1.12 × 10⁻²² s.