University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.7

Mandelstam variables and crossing symmetry

Every two-to-two amplitude in the rest of this course is written in s, t and u, and this is where that habit is earned. We fix the kinematics once, read a scattering angle off t, get a second process from the first without drawing a new diagram, and learn to see in each pole either a force or a particle.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A spinless two-to-two amplitude is a Lorentz scalar built from four on-shell momenta tied by one conservation law, so it can depend on only two numbers. The Mandelstam variables s, t and u, whose sum the masses fix, are the natural choice, because each is the squared centre-of-mass energy of the reaction read in a different direction. LSZ then gives more than a labelling.

It builds every S-matrix element from one correlation function, and in that function one field stands for both an incoming particle and an outgoing antiparticle with its momentum reversed. So a single ℳ(s, t, u) holds ψψ → ψψ in one region of the s-t plane and ψψ̄ → ψψ̄ in two others, the regions disjoint and joined only by analytic continuation. Read this way, every pole of ℳ is a particle: a pole in t or u lies outside the physical region, and its distance from it sets the range of a force, while a pole in s is a particle the colliding pair can become, produced for real once the pole crosses threshold.

At tree level ℳ is a rational function, so the continuation is trivial. Beyond it the picture needs care: loops add branch cuts to steer around, an unstable particle's pole leaves the real axis, a massless exchange puts its pole on the physical boundary t = 0, and fermions bring spinors and a crossing sign of their own.

Simple definition: The Mandelstam variables are the invariants s = (p₁ + p₂)², t = (p₁ − p₃)² and u = (p₁ − p₄)² of a two-to-two process, and crossing symmetry says one function of them, continued between regions, gives every process made by moving a particle across the reaction as its antiparticle.
Example: Two ψ of mass M = 1.00 GeV meet head-on with |p| = 1.00 GeV and scatter through 90°: s = 4(|p|² + M²) = 8.00 GeV², t = u = −2|p|² = −2.00 GeV², and s + t + u = 4.00 GeV² = 4M².

The Mandelstam invariantss = (p₁ + p₂)², t = (p₁ − p₃)², u = (p₁ − p₄)²

√s is the total centre-of-mass energy; t and u are the squared four-momentum transfers from particle 1 to particles 3 and 4. A spinless ℳ can depend on nothing else.

p₁, p₂ incoming and p₃, p₄ outgoing four-momenta in GeV; s, t, u in GeV², unchanged by any Lorentz transformation

Only two are independents + t + u = m₁² + m₂² + m₃² + m₄²

ℳ is a function of two variables. Four equal masses M give 4M², and the massless limit gives s + t + u = 0.

the four masses in GeV; derived from p₁ + p₂ = p₃ + p₄ and pᵢ² = mᵢ²

Equal-mass elastic scattering, centre of masst = −2|p|²(1 − cos θ) = −4|p|² sin²(θ/2)

θ from 0 to π takes t from 0 to −(s − 4M²), so ψψ → ψψ is physical only for s ≥ 4M², t ≤ 0, u ≤ 0.

|p| the CM momentum in GeV, θ the CM angle; s = 4(|p|² + M²) and u = −2|p|²(1 + cos θ), all derived

The crossing ruleincoming ψ, momentum p ⇔ outgoing ψ̄, momentum k = −p

One correlation function, so one ℳ. The crossed process sits where p⁰ = −k⁰ is negative, which only analytic continuation reaches.

LSZ gives both from the field ψ†, with e^(−ip⋅x) for the incoming ψ and e^(+ik⋅x) for the outgoing ψ̄; derived from topic 4.04

ψψ → ψψ crossed into ψψ̄ → ψψ̄ℳ_ψψ̄(s, t, u) = ℳ_ψψ(u, t, s) = g²[1/(m² − t) + 1/(m² − s)]

The u-exchange of ψψ → ψψ becomes annihilation into φ. Its pole s = m² enters the physical region s ≥ 4M² only if m exceeds 2M.

derived by crossing legs 2 and 4, which swaps s and u; g in GeV, ℳ dimensionless; agrees with topic 4.06's diagrams

A t-channel pole as a forceV(r) = −(g²/4M²) e^(−mr)/(4πr)

Derived. The pole lies m² beyond the physical edge t = 0, and that distance sets the range ħ/(mc): 1.41 fm for a 139.57 MeV pion.

from g²/(m² − t) at t = −|q|², non-relativistic, matched to the Born amplitude; r in fm after ħc = 197.327 MeV fm

Three invariants, and why only two count

Work in ħ = c = 1 with metric (+, −, −, −), and write ℳ for the invariant amplitude, M for the ψ mass and m for the φ mass of the scalar Yukawa model. For 1 + 2 → 3 + 4 define s = (p₁ + p₂)², t = (p₁ − p₃)² and u = (p₁ − p₄)². Each is the square of a four-vector, hence Lorentz invariant, and topic 4.01 showed a spinless ℳ is a Lorentz scalar, so it can depend on the momenta only through such products. Expanding, s + t + u = 3m₁² + m₂² + m₃² + m₄² + 2p₁⋅(p₂ − p₃ − p₄), and conservation turns p₂ − p₃ − p₄ into −p₁, which cancels 2m₁²: s + t + u = m₁² + m₂² + m₃² + m₄², derived. Of the six products pᵢ⋅pⱼ, conservation and the mass shells leave two free, so ℳ is a function of two variables. Dimensions check every step: s, t and u are GeV², and ℳ for four external particles has mass dimension 4 − 4 = 0, so each tree term must be g², of dimension 2 in this model, divided by a squared mass.

In the centre-of-mass frame t and u measure the angle

For ψψ → ψψ take p₁ = (E, p), p₂ = (E, −p), p₃ = (E, p′) and p₄ = (E, −p′), with |p′| = |p| because the masses are equal, and θ the angle between p and p′. Then s = (2E)² = 4(|p|² + M²). The difference p₁ − p₃ = (0, p − p′) has no time component, so t = −|p − p′|² = −2|p|²(1 − cos θ), and likewise u = −|p + p′|² = −2|p|²(1 + cos θ), all derived; the three sum to 4M². As θ runs from 0 to π, t falls from 0 to −(s − 4M²) while u rises to 0, so the physical region is s ≥ 4M², t ≤ 0, u ≤ 0, a wedge bounded by the lines t = 0 (forward) and u = 0 (backward). Swapping the identical final particles sends θ to π − θ, which is t ↔ u, so their ℳ must be symmetric in t and u. With particle 2 at rest in the lab, s = 2M² + 2ME_lab. Two limits check this: for |p| ≪ M, √s − 2M ≈ |p|²/M, the pair's non-relativistic kinetic energy, and t = −|q|² with q = p − p′, the momentum transfer of the Born approximation; for M → 0, t = −(s/2)(1 − cos θ) and s + t + u = 0.

Crossing comes from LSZ: one correlation function

Topic 4.04 builds each S-matrix element from the Fourier transform of a correlation function, here ⟨Ω|T ψ ψ† ψ ψ†|Ω⟩. In that recipe the field ψ† does two jobs: it creates an incoming ψ of momentum p in the far past, entering with ∫d⁴x e^(−ip⋅x), and it annihilates an outgoing ψ̄ of momentum k in the far future, entering with ∫d⁴x e^(+ik⋅x), while ψ does the same for an outgoing ψ and an incoming ψ̄. These are the same integral at p = −k. So the amplitude with an incoming ψ and the amplitude with an outgoing ψ̄ in its place are one function of the external momenta, evaluated at p and at −k: crossing, derived. The catch is where −k sits. A physical ψ̄ has positive energy k⁰, so p⁰ = −k⁰ is negative, a point no physical ψ reaches, and ℳ, known on its own physical region, must be continued there. At tree level ℳ is a rational function of s, t and u, so the continuation is unique and trivial. With loops ℳ has branch cuts and the path must go around them through complex values; that this works for 2 → 2 in any theory with a mass gap was proved by Bros, Epstein and Glaser in 1965, quoted.

One function, three channels

Start from topic 4.06's ψ(p₁)ψ(p₂) → ψ(p₃)ψ(p₄), ℳ = g²[1/(m² − t) + 1/(m² − u)], derived there. Cross particle 2 to an outgoing ψ̄ of momentum k₄ = −p₂ and particle 4 to an incoming ψ̄ of momentum k₂ = −p₄. The new process ψ(p₁)ψ̄(k₂) → ψ(p₃)ψ̄(k₄) has its own invariants s′ = (p₁ + k₂)² = u, t′ = (p₁ − p₃)² = t and u′ = (p₁ − k₄)² = (p₁ + p₂)² = s. Substituting gives ℳ′ = g²[1/(m² − t′) + 1/(m² − s′)], the t-exchange plus the s-channel annihilation that topic 4.06 found by drawing the diagrams afresh, so crossing passes a real test. The physical regions are disjoint: ψψ → ψψ at s ≥ 4M² with t, u ≤ 0; ψψ̄ → ψψ̄ at u ≥ 4M² with s, t ≤ 0; and, at t ≥ 4M² with s, u ≤ 0, the same ψψ̄ process with its final pair labelled the other way round, which is why one t ↔ u symmetric function serves both. Between them lies the Mandelstam triangle, 0 ≤ s, t, u ≤ 4M², where nothing is physical. Scalars cross with no sign; fermions add one, topic 9.08.

A pole in t or u is an exchange, and it sets a range

The term g²/(m² − t) has its pole at the positive value t = m², and physical ψψ → ψψ has t ≤ 0, so the pole is never reached: it sits m² beyond the forward edge t = 0, and that distance sets the range of a force. In the centre-of-mass frame t = −|q|², so the term is g²/(m² + |q|²). Dividing out the relativistic normalisation, √(2M) per external leg, and matching to the Born amplitude of UQMII-15 gives Ṽ(q) = −ℳ/(4M²), so V(r) = −(g²/4M²) e^(−mr)/(4πr), an attractive Yukawa potential of range 1/m, derived but for the Fourier transform, which the Born treatment supplies. With units restored the range is ħc/(mc²): 197.327 MeV fm ÷ 139.57 MeV = 1.414 fm for the measured pion mass, the reach of the nuclear force's long tail from which Yukawa predicted the meson in 1935. The pole also shapes the angular distribution: the t-term halves at −t = m², that is at 1 − cos θ = m²/(2|p|²), so a light exchange gives a forward peak about m/|p| wide. As m → 0 the pole reaches the edge and the t-term becomes −g²/t = g²/(4|p|² sin²(θ/2)), whose square is Rutherford's 1/sin⁴(θ/2), dominant at forward angles while the u-term mirrors it backward: infinite range.

A pole in s is a particle the pair can become

In ψψ̄ → ψψ̄ the crossed u-term has become g²/(m² − s), and a pole in s means something new: the colliding pair can merge into one φ carrying their total momentum, so s = m² is that φ on its mass shell. Whether it is reached depends on m² against the threshold 4M². For m below 2M the pole lies below the physical region, the φ cannot decay to ψψ̄, and the s-term is negative wherever a pair can collide, eating into the attractive t-exchange. For m above 2M the pole moves inside: at √s = m the pair makes a real φ, the tree amplitude diverges there, and topic 5.09's width Γ cures it by turning m² − s into m² − s − imΓ, quoted, so an experiment sees a resonance peak, as e⁺e⁻ collisions see the Z at √s = 91.19 GeV. ψψ → ψψ has no s-pole at all, because no single particle carries two units of ψ number. In every channel the poles name exactly the particles with that channel's quantum numbers, and crossing turns an exchange in one channel into a production in another.

Change one variable at a time

Make the relationship visible.

Interactive model

Exchanged mass m0.6 M

Centre-of-mass energy squared s8.0 M²

Centre-of-mass angle θ60 °

Sweep θ from 0° to 180° and the dot runs from the t = 0 edge to the u = 0 edge without touching a pole. Then raise m past 2M: the dashed poles leave the triangle and cut the two ψψ̄ wedges, where a real φ is made, but never enter ψψ → ψψ.

t AT THE DOT-1.00 M²

u AT THE DOT-3.00 M²

ℳ(ψψ → ψψ) / g²1.033 M⁻²

ψψ̄ POLE − THRESHOLD-3.64 M²

Live interpretationt AT THE DOT: −1.00 M². u AT THE DOT: −3.00 M². ℳ(ψψ → ψψ) / g²: 1.033 M⁻². ψψ̄ POLE − THRESHOLD: −3.64 M²

Catch the common trap

Explain before calculating.

Misconception check

t is not a positive squared momentum transfer that grows until it hits m². Students who picture t as |q|² expect the exchange term g²/(m² − t) to blow up at the angle where the momentum transfer equals the exchanged mass, and look for a peak there. In the metric (+, −, −, −) a purely spatial four-vector has a negative square, and in elastic centre-of-mass scattering p₁ − p₃ carries no energy, so t = −|q|² ≤ 0 at every angle. The pole at t = m² corresponds to |q|² = −m², an imaginary momentum transfer no collision supplies. That is not a defect but the content of the pole: its place at |q| = im is exactly what makes the Fourier transform fall as e^(−mr), so the unreachable pole sets the range. A t-pole can be reached only in a crossed channel, where t is a squared energy, and there only once m exceeds 2M.

Topic 4.06 gives ψ(p₁)ψ(p₂) → ψ(p₃)ψ(p₄) as ℳ = g²[1/(m² − t) + 1/(m² − u)]. Crossing particle 2 into an outgoing ψ̄ and particle 4 into an incoming ψ̄ gives ψ(p₁)ψ̄(−p₄) → ψ(p₃)ψ̄(−p₂). Written in this new process's own Mandelstam variables s, t and u, what is its amplitude?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo ψ particles of mass M = 1.00 GeV collide head-on in their centre-of-mass frame, each with momentum |p| = 0.750 GeV, and scatter through θ = 60.0°. Find s, t and u, check the sum rule, and find the beam energy that gives the same s on a ψ at rest.

E = √(|p|² + M²) = √(0.5625 + 1.0000) GeV = 1.250 GeV for each particle, so s = (2E)² = 6.250 GeV² and √s = 2.500 GeV.
t = −2|p|²(1 − cos θ) = −2 × 0.5625 GeV² × (1 − 0.500) = −0.5625 GeV².
u = −2|p|²(1 + cos θ) = −2 × 0.5625 GeV² × 1.500 = −1.6875 GeV².
Check: s + t + u = 6.250 − 0.5625 − 1.6875 = 4.000 GeV² = 4M², and t and u are both negative, as the physical region of ψψ → ψψ requires.
Fixed target: s = (p₁ + p₂)² = 2M² + 2ME_lab, so E_lab = (s − 2M²)/(2M) = (6.250 − 2.000)/2.000 GeV = 2.125 GeV, a kinetic energy of 1.125 GeV against 2 × 0.250 = 0.500 GeV shared in the centre-of-mass frame.

Answers = 6.25 GeV², t = −0.563 GeV², u = −1.69 GeV², summing to 4.00 GeV² = 4M²; a ψ beam on a ψ at rest needs E_lab = 2.13 GeV for the same s.

MediumIn the scalar Yukawa model take M = 1.00 GeV and m = 0.500 GeV. At √s = 2.50 GeV and θ = 90° in the centre-of-mass frame, evaluate ℳ/g² for ψψ → ψψ and, by crossing, for ψψ̄ → ψψ̄. Which term makes the difference, and by what factor do the two |ℳ|² differ?

Both processes have four particles of mass M, so the kinematics is shared: |p|² = s/4 − M² = 1.5625 − 1.0000 = 0.5625 GeV², and at 90° t = u = −2|p|² = −1.125 GeV².
ψψ → ψψ: ℳ/g² = 1/(m² − t) + 1/(m² − u) = 2/(0.250 + 1.125) GeV⁻² = 1.455 GeV⁻².
Crossing legs 2 and 4 swaps s and u, so ψψ̄ → ψψ̄ has ℳ/g² = 1/(m² − t) + 1/(m² − s) = 1/1.375 + 1/(0.250 − 6.250) = 0.7273 − 0.1667 = 0.5606 GeV⁻².
The difference is one term: the u-exchange, +0.727 GeV⁻² for ψψ, has become the s-channel annihilation, −0.167 GeV⁻². Its pole at s = m² = 0.250 GeV² lies far below the threshold 4M² = 4.00 GeV², so in the physical region it is small and negative.
Ratio: |ℳ_ψψ|²/|ℳ_ψψ̄|² = (1.455/0.5606)² = 2.595² = 6.73. Dimensions: ℳ/g² in GeV⁻² times g² in GeV² is dimensionless, as a four-particle amplitude must be.

Answerℳ_ψψ = 1.45 g² GeV⁻² and ℳ_ψψ̄ = 0.561 g² GeV⁻², so |ℳ|² differs by a factor of 6.73, all of it from the u-exchange term turning into a negative s-channel term.

HardTopic 4.06 gives ψ(p₁)ψ̄(p₂) → φ(p₃)φ(p₄) as ℳ = g²[1/(M² − t) + 1/(M² − u)]. Cross it into ψφ → ψφ, locate the new amplitude's poles relative to its physical region, and evaluate ℳ/g² at centre-of-mass momentum |p| = 0.300 GeV and θ = 90°, for M = 1.00 GeV and m = 0.400 GeV.

Move ψ̄(p₂) to an outgoing ψ of momentum −p₂ and φ(p₃) to an incoming φ of momentum −p₃, φ being its own antiparticle. For ψ(p₁)φ(−p₃) → ψ(−p₂)φ(p₄): s′ = (p₁ − p₃)² = t, t′ = (p₁ + p₂)² = s, u′ = u, so ℳ = g²[1/(M² − s′) + 1/(M² − u′)].
Kinematics: E_ψ = √(1.00 + 0.09) GeV = 1.04403 GeV and E_φ = √(0.16 + 0.09) GeV = 0.500 GeV, so s = (1.54403 GeV)² = 2.3840 GeV², above the threshold (M + m)² = 1.96 GeV².
The s-pole at M² = 1.00 GeV² lies 0.96 GeV² below threshold, outside the physical region: it is the ψ itself, which ψφ can form virtually but never as a real final state, because M is less than M + m, just as the nucleon pole sits below threshold in pion-nucleon scattering.
For the u-pole: u = M² + m² − 2(E_ψE_φ − |p|²) − 2|p|²(1 + cos θ), and (E_ψE_φ − |p|²)² − M²m² = |p|²(E_ψ − E_φ)² ≥ 0, so u ≤ (M − m)² = 0.36 GeV². The pole at u = M² stays at least 0.64 GeV² outside: an exchanged ψ.
At 90°: t′ = −2|p|² = −0.180 GeV², and u′ = 2(M² + m²) − s′ − t′ = 2.320 − 2.3840 + 0.180 = 0.1160 GeV². Then ℳ/g² = 1/(1.00 − 2.3840) + 1/(1.00 − 0.1160) = −0.7225 + 1.1312 = 0.4087 GeV⁻².
Checks: s′ + t′ + u′ = 2.320 GeV² = 2(M² + m²); at threshold, s′ = 1.96 and u′ = 0.36 GeV² give ℳ/g² = −1.0417 + 1.5625 = 0.521 GeV⁻², finite as it must be with both poles outside the physical region.

Answerℳ(ψφ → ψφ) = g²[1/(M² − s) + 1/(M² − u)]; its s-pole lies 0.96 GeV² below threshold and its u-pole at least 0.64 GeV² beyond u ≤ (M − m)²; at the given point ℳ/g² = 0.409 GeV⁻².