University Quantum Field Theory I · The S-Matrix from Correlation Functions: Kallen-Lehmann and LSZ · 4.1

In and out states, S = 1 + iT and the invariant amplitude M

Experiments never see a field. They prepare particles far apart and catch them far apart, and this topic turns that into mathematics: states named by their past or their future, a unitary S that overlaps them, and one function per process, M, from which every later rate, cross-section and loop in the course is built.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

An interacting theory has no free particles at any finite time, but a collider asks about only two regimes: the far past, when packets were prepared far apart, and the far future, when they are counted far apart. So label states by their asymptotics. An in state |α in⟩ is an exact eigenstate of the full Hamiltonian whose wave packets, followed back to t → −∞, become separated, freely moving particles of physical mass with momenta α; an out state looks like that as t → +∞. Both are Heisenberg-picture vectors in one Hilbert space, and S_βα = ⟨β out|α in⟩ is the overlap of two bases.

It is unitary because each basis is complete, and Poincaré invariant because separated packets in the far past mean the same in every frame. Translations force a factor δ⁴(p_α − p_β), and writing S = 1 + iT, the 1 for particles that miss each other, leaves iM: for spinless particles with invariant normalisation, a Lorentz scalar of mass dimension 4 − n. All of that is derived. It does not say how to compute M.

Topic 2.09's recipe, free states of the Lagrangian either side of U(+∞, −∞), silently takes the asymptotic particles to be free quanta of mass m₀ with unit overlap with φ; interactions move the mass and shrink the overlap to √Z, which the rest of this unit repairs. And the construction needs stable particles and a mass gap: unstable states and massless photons strain it, as topic 4.09 shows.

Simple definition: The invariant amplitude M is what an S-matrix element leaves once the no-scattering term and the momentum-conserving factor are removed: ⟨β|S − 1|α⟩ = (2π)⁴ δ⁴(p_α − p_β) iM(α → β).
Example: At lowest order in φ⁴ theory, φφ → φφ has M = −λ, a pure number as 4 − n = 0 requires. K_S → π⁺π⁻, with n = 3, has |M| = 0.39 keV from its measured lifetime: a mass, and the same in every frame.

In and out states: the asymptotic conditione^(−iHt) ∫dα g(α)|α in⟩ → e^(−iH₀t) ∫dα g(α)|α⟩ as t → −∞

Defined: |α in⟩ is an exact eigenstate of H, energy Σ√(pᵢ² + mᵢ²), that looks like free separated packets in the far past. Derived from it: ⟨β in|α in⟩ = ⟨β|α⟩.

g(α): smooth packet profile; H: full Hamiltonian; H₀: free particles of physical mass; t in GeV⁻¹; out states take t → +∞

The S-matrix as an overlapS_βα = ⟨β out|α in⟩, with ⟨β|S|α⟩ ≡ S_βα on free-particle states

Defined: the amplitude that a system prepared as α in the far past is counted as β in the far future. One Heisenberg vector, two bases; nothing evolves.

α, β: lists of physical-particle momenta; S_βα has mass dimension −n for n external particles in all

Unitarity from completenessΣ_γ S*_γβ S_γα = Σ_γ ⟨β in|γ out⟩⟨γ out|α in⟩ = δ(β − α)

Derived: S†S = 1, and completeness of the in states gives SS† = 1. With S = 1 + iT it forces −i(T − T†) = T†T, so T is Hermitian at lowest order.

Sum over a complete set of out states; δ(β − α) is the free normalisation, 2E(2π)³δ³ for each particle

Poincaré invariance of SU₀(Λ, a) S U₀(Λ, a)⁻¹ = S

Derived from the covariance of the asymptotic condition. Translations force ⟨β|S|α⟩ ∝ δ⁴(p_α − p_β); boosts give M(Λp₁, …, Λpₙ) = M(p₁, …, pₙ) for spin 0.

U₀: topic 1.03's representation on free Fock states, U₀(Λ)|p⟩ = |Λp⟩ for spin 0; Λ a Lorentz matrix, a a translation in GeV⁻¹

Defining the invariant amplitude MS = 1 + iT⟨β|iT|α⟩ = (2π)⁴ δ⁴(p_α − p_β) iM(α → β)

Defined: the 1 is particles that miss each other, a δ³ for every particle; iT carries one overall δ⁴, and M is the smooth function multiplying it.

p_α, p_β: total four-momenta in GeV; M is read from the fully connected part of T; the (2π)⁴ is matched in phase space

Mass dimension of M[M] = 4 − n

Derived: φφ → φφ in φ⁴ has M = −λ, a pure number; a two-body decay amplitude is a mass (K_S → π⁺π⁻: 0.39 keV); a 2 → 4 amplitude goes as GeV⁻².

n: external particles in all; ⟨p|q⟩ = 2Eₚ(2π)³δ³(p − q) has dimension −2, so each ket has −1, and [δ⁴] = −4

An in state is named by its past, not dated by it

Natural units, ħ = c = 1, and the metric (+, −, −, −) hold throughout. In the Heisenberg picture no state moves. |α in⟩, with α a list of momenta of physical particles, is one fixed vector: an exact eigenstate of the full H with energy E_α = Σ√(pᵢ² + mᵢ²), physical masses mᵢ, whose packets e^(−iHt)∫dα g(α)|α in⟩ approach free, separated packets as t → −∞. |α out⟩ does the same as t → +∞. Because e^(−iHt) is unitary, ⟨β in|α in⟩ may be evaluated in the far past, where it is the free product ⟨β|α⟩: in states obey topic 1.03's normalisation, derived. The idealisation is generous. Strong forces reach about ħc/(m_π c²) = 1.41 fm, crossed in 4.7 × 10⁻²⁴ s, while a detector 1 m away is reached after 3.3 ns, some 7 × 10¹⁴ times longer. Followed forward, the same vector contains the collision and all it produces.

S overlaps two complete bases, so it is unitary

Define S_βα = ⟨β out|α in⟩: prepare α in the far past, ask for β in the far future. If the out states are complete, Σ_γ |γ out⟩⟨γ out| = 1, then Σ_γ S*_γβ S_γα = Σ_γ ⟨β in|γ out⟩⟨γ out|α in⟩ = ⟨β in|α in⟩ = δ(β − α), so S†S = 1, derived; completeness of the in states gives SS† = 1 the same way. Completeness is physics, not algebra: every state of the theory, bound states counted as particles, must be a superposition of in states, which topic 4.09 examines. The vacuum and a single stable particle cannot scatter, so |Ω in⟩ = |Ω out⟩ and |p in⟩ = |p out⟩, and S acts on them as 1. Writing S = 1 + iT, (1 − iT†)(1 + iT) = 1 gives −i(T − T†) = T†T, derived. At first order in a coupling the right side is second order, so the leading T is Hermitian, M(α → β) = M(β → α)*, and the forward amplitude gains an imaginary part only at the next order, the seed of topic 5.07's optical theorem.

Translations give the δ⁴, boosts make M a scalar

Separated packets of physical particles in the far past mean the same thing in every inertial frame, so the exact Poincaré operators map in states to in states with transformed labels, U(Λ)|p₁…pₙ in⟩ = |Λp₁…Λpₙ in⟩ for spin 0, and out states likewise. The S operator therefore commutes with topic 1.03's free-particle generators P^μ and J^μν, derived. Translations first: in and out states are eigenstates of the exact P^μ, so ⟨β out|α in⟩ = ⟨β out|e^(−iP⋅a)e^(iP⋅a)|α in⟩ = exp[i(p_α − p_β)⋅a] ⟨β out|α in⟩ for every four-vector a, possible only if the element vanishes unless p_β = p_α: it is proportional to δ⁴(p_α − p_β). Boosts next: δ⁴ is invariant since det Λ = 1, and 2Eₚ δ³(p − q) is invariant since d³p/2Eₚ is, so the coefficient of δ⁴ is unchanged when every momentum is boosted. Normalise instead to δ³(p − q) and each leg brings a factor √(E(Λp)/E(p)), so the coefficient changes from frame to frame.

Peeling off the 1 and the δ⁴

Most of S does nothing. For two identical particles, ⟨k₁k₂|p₁p₂⟩ = 2E₁2E₂(2π)⁶[δ³(k₁ − p₁)δ³(k₂ − p₂) + δ³(k₁ − p₂)δ³(k₂ − p₁)]: packets that miss each other, a δ³ fixing every particle. Write S = 1 + iT so that this sits in the 1. What remains is constrained only by conservation of the total four-momentum, so pull that factor out, by definition: ⟨k₁k₂|iT|p₁p₂⟩ = (2π)⁴δ⁴(p₁ + p₂ − k₁ − k₂) iM(p₁p₂ → k₁k₂). M is then an ordinary function of the momenta, smooth apart from poles, which topics 4.06 and 4.07 locate, and the branch cuts that loops add. When three or more particles come in, iT also holds pieces in which a spectator passes untouched, a δ³ times a lower amplitude; M is read from the fully connected part. Limits: switch the coupling off and T = 0, S = 1 and every |α in⟩ = |α out⟩. At first order, contracting fields with external states as topic 2.09 did gives M = −g for φ → ψψ̄ in the scalar Yukawa model and M = −λ for φφ → φφ in φ⁴, at every angle and energy, both quoted.

Dimension 4 − n, and three checks

⟨p|q⟩ = 2Eₚ(2π)³δ³(p − q) has mass dimension 1 − 3 = −2, so each one-particle ket has −1 and ⟨β|S|α⟩ has −n for n external particles in all. The δ⁴ carries −4, so [M] = 4 − n, derived. Couplings: M = −λ is a pure number for n = 4, and M = −g has dimension 1 for n = 3, since the scalar Yukawa g does; one tree diagram for φφ → 4φ is −λ²/(P² − m²), dimension −2 = 4 − 6. Non-relativistic limit: for slow, distinguishable particles the invariant normalisation puts 2m₁ × 2m₂ into M, and topic 5.06 shows Ṽ(q) = −M/(4m₁m₂), quoted. Ṽ(q) = ∫d³r e^(−iq⋅r)V(r) is an energy times a volume, dimension 1 − 3 = −2, exactly that of M/(4m₁m₂). Measurement: K_S → π⁺π⁻ has spin 0 throughout, so M is a single number of dimension 1. The measured mean life 0.8954 × 10⁻¹⁰ s and branching fraction 69.20% give |M| = 0.39 keV through Unit 5's rate formula, quoted: 8 × 10⁻⁷ of m_K, the weakness of the weak force in one number.

What topic 2.09's shortcut assumed

Topic 2.09 sandwiched U(+∞, −∞), the time-ordered exponential of −i∫d⁴x H_I, between free states a†(p)…|0⟩ of the field with the Lagrangian's mass m₀, and called the result S. That takes the asymptotic particles to be those free quanta: mass m₀, and ⟨0|φ(0)|p⟩ = 1. Interactions spoil both, stated here and derived in topics 4.02 and 4.03. Topic 3.08's resummed propagator i/(p² − m₀² − Σ(p²)), quoted, has its pole where m² = m₀² + Σ(m²), not at m₀², and the field reaches the one-particle state only with amplitude ⟨Ω|φ(0)|p⟩ = √Z, where 0 ≤ Z ≤ 1 and in general Z is strictly below 1. A free state of the wrong mass never settles into an in state: for a particle at rest, a 1 MeV mismatch turns the relative phase through 2π every 2πħ/(1 MeV) = 4.1 × 10⁻²¹ s, so the t → −∞ limit does not exist. In diagrams the same disease is a self-energy on an external leg multiplying 1/(p² − m₀²), infinite at p² = m₀². At leading order nothing is lost, which is why M = −g stands; beyond it, LSZ (topic 4.04) takes over, with physical m and √Z per leg.

Change one variable at a time

Make the relationship visible.

Interactive model

CM rapidities ±Y of the pair1.4

Observer's boost rapidity η−1.4

The opening η = −Y is the laboratory, particle 2 at rest: the curve sits at cosh²Y/cosh 2Y of its CM value while M has not moved. Raise Y and that ratio tends to one half; drag η towards ±3, where both particles are fast, and the δ³-normalised amplitude dies away.

√s / m, EVERY FRAME4.30

E₁ / m IN THIS FRAME8.25

E₂ / m IN THIS FRAME1.00

δ³-NORMALISED ÷ CM VALUE0.561

Live interpretation√s / m, EVERY FRAME: 4.30. E₁ / m IN THIS FRAME: 8.25. E₂ / m IN THIS FRAME: 1.00. δ³-NORMALISED ÷ CM VALUE: 0.561

Catch the common trap

Explain before calculating.

Misconception check

The tempting picture is that |α in⟩ is the system's state at t = −∞, |β out⟩ a state at t = +∞, and S the time evolution carrying one into the other, so that in states are free-particle states and |α in⟩ and |α out⟩ are one state seen at two times. Heisenberg-picture states do not evolve: |α in⟩ is one time-independent vector, an exact eigenstate of the full H, whose history starts as separated packets and later contains the collision and everything it produces; in the out basis it reads Σ_β |β out⟩ S_βα. So |α in⟩ and |α out⟩, with the same labels, are different vectors whose overlap is the forward element S_αα, and they coincide only for the vacuum and a single stable particle, where nothing can happen. Nor are they free states: they carry the physical masses, which the Lagrangian's m₀ is not.

In φ⁴ theory the lowest-order amplitude for φφ → φφφφ comes from two vertices joined by one propagator. With the invariant normalisation ⟨p|q⟩ = 2Eₚ(2π)³δ³(p − q), what are the mass dimensions of ⟨k₁k₂k₃k₄|iT|p₁p₂⟩ and of M?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA π⁻ beam of momentum 2.00 GeV strikes protons at rest, and the detector sits 1.50 m downstream. Take the strong force's range as the pion's reduced Compton wavelength ħ/(m_π c), with m_π c² = 139.57 MeV. Compare the time the pion spends within range of a proton with its flight time to the detector, then find how fast a free comparison state whose pion mass is wrong by 1.0 MeV dephases from the true one.

Range: R = 1/m_π = 1/(0.13957 GeV) = 7.165 GeV⁻¹; restoring ħc = 0.197327 GeV fm gives R = 1.414 fm.
Speed: E = √(2.000² + 0.13957²) GeV = 2.0049 GeV, so β = p/E = 0.99757. Crossing a region of size 2R takes 2.828 fm ÷ (0.99757 × 2.998 × 10²³ fm s⁻¹) = 9.45 × 10⁻²⁴ s.
Flight: 1.50 m ÷ (0.99757 × 2.998 × 10⁸ m s⁻¹) = 5.02 × 10⁻⁹ s, so the pion spends 5.3 × 10¹⁴ times longer as a free packet than interacting. The limits t → ±∞ are an idealisation real experiments satisfy with vast room.
Wrong mass: at fixed momentum ∂E/∂m = m/E, so ΔE = (0.13957/2.0049) × 1.0 MeV = 0.0696 MeV, and the relative phase grows at ΔE/ħ = 0.0696 MeV ÷ 6.582 × 10⁻²² MeV s = 1.06 × 10²⁰ rad s⁻¹.
Over the flight that is 5.3 × 10¹¹ rad, still growing: a comparison state with the wrong mass never approaches the true one, so the asymptotic condition has no limit unless H₀ carries the physical mass.

AnswerThe pion is within range for 9.5 × 10⁻²⁴ s of a 5.0 ns flight, a ratio of 5.3 × 10¹⁴; a 1.0 MeV mass error dephases at 1.1 × 10²⁰ rad s⁻¹, 5.3 × 10¹¹ rad before the detector.

MediumTwo identical spinless particles of mass 0.500 GeV scatter forward, k₁ = p₁ and k₂ = p₂, at lowest order in φ⁴ theory with λ = 0.80. In the laboratory particle 1 has momentum 4.00 GeV and particle 2 is at rest. Find s, M, and the amplitude M′ = M/(2E₁ × 2E₂) that states normalised to ⟨p|q⟩ = δ³(p − q) would give, in the laboratory and in the centre-of-momentum frame.

Laboratory: E₁ = √(4.00² + 0.500²) GeV = 4.0311 GeV, so s = (E₁ + m)² − p₁² = 4.5311² − 16.00 = 4.531 GeV², equal to 2m² + 2mE₁ as it must be.
Centre of momentum: the same s, so each particle has energy √s/2 = 1.0643 GeV. Forward scattering has t = 0, so s is the only invariant in play.
M: one vertex gives iM = −iλ, so M = −0.80 in both frames, dimensionless as 4 − n = 0 requires. Here it does not even depend on s; in general it depends on s and t, and is blind to the frame.
δ³ normalisation rescales each leg by 1/√(2E), four legs in all, so M′ = M/(2E₁ × 2E₂). Laboratory: 8.062 × 1.000 GeV² = 8.062 GeV², so M′ = −0.80/8.062 GeV⁻² = −0.0992 GeV⁻². Centre of momentum: (2 × 1.0643)² GeV² = s = 4.531 GeV², so M′ = −0.1766 GeV⁻².
The two M′ differ by the factor 4.531/8.062 = 0.562, which is cosh²Y/cosh 2Y for the centre-of-momentum rapidity Y = 1.388, the curve in the figure. M′ has dimension −2: four kets at −3/2 each give −6, and removing the δ⁴'s −4 leaves −2.

Answers = 4.53 GeV²; M = −0.80 in every frame; M′ = −0.0992 GeV⁻² in the laboratory and −0.177 GeV⁻² in the centre-of-momentum frame, a factor 0.562 apart.

HardK_S is spinless and decays to π⁺π⁻ with branching fraction 69.20%; its measured mean life is 0.8954 × 10⁻¹⁰ s, with m_K = 497.611 MeV and m_π = 139.570 MeV. Using Unit 5's two-body rate Γ = |M|²p*/(8π m_K²), quoted here, extract |M| and check its dimension. Then find the mean decay length of a K_S of laboratory momentum 10.0 GeV, saying what in the rate changes between frames and what does not.

Form: n = 3, so [M] = 4 − 3 = 1. Every momentum is on its mass shell and p_K = p₊ + p₋, so every invariant, p₊⋅p₋ = m_K²/2 − m_π² for one, is fixed by the masses: M is one constant with the dimension of mass, the same in every frame.
The quoted rate is Γ = (1/(2m_K))∫dΠ₂|M|² with ∫dΠ₂ = p*/(4π m_K) for distinguishable π⁺ and π⁻. Dimensions: [|M|²] = 2 and [p*/m_K²] = −1, so [Γ] = 1, an energy, as a rate must be in natural units.
Measured partial width: Γ = 0.6920 ħ/τ = 0.6920 × 6.5821 × 10⁻²² MeV s ÷ 0.8954 × 10⁻¹⁰ s = 5.087 × 10⁻¹² MeV.
Pion momentum in the rest frame: p* = √(m_K²/4 − m_π²) = √(61904 − 19480) MeV = 206.0 MeV. Then |M|² = 8π m_K² Γ/p* = 8π × (497.611 MeV)² × 5.087 × 10⁻¹² MeV ÷ 206.0 MeV = 1.537 × 10⁻⁷ MeV², so |M| = 3.92 × 10⁻⁴ MeV, or 7.9 × 10⁻⁷ m_K.
Limit: at fixed |M|, Γ vanishes as p* → 0 at threshold, a phase-space effect, not a dynamical one; the small |M| is dynamics, the weak interaction.
Moving K_S: M and the invariant phase space are unchanged; only the 1/(2E_K) from the decaying state's normalisation changes, so Γ_lab = Γ m_K/E_K, time dilation. With p_K/m_K = 10.0/0.4976 = 20.10 and cτ = 2.684 cm, the mean decay length is 20.10 × 2.684 cm = 53.9 cm.

Answer|M| = 3.92 × 10⁻⁴ MeV = 0.39 keV, of mass dimension 1 as 4 − n requires; a 10.0 GeV K_S travels 53.9 cm on average, the frame entering only through 1/(2E_K), never through M.