University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.5

Three-body phase space and the Dalitz plot

A two-body decay at rest has no spectrum to measure: every event is the same up to a rotation. A three-body decay has two free variables, and this lesson chooses them so that phase space contributes nothing but a constant. Then a scatter of events is a picture of |ℳ|², and its edge is pure kinematics.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Three final momenta carry nine components and four-momentum conservation removes four, leaving five. Three of the five are the Euler angles that orient the decay plane, and for a spinless or unpolarised parent |ℳ|² cannot depend on them, because nothing in the initial state picks out a direction; integrating them gives a bare 8π² and leaves two variables. Chosen as two product energies in the rest frame, the measure is flat, dΠ₃ = dE₁ dE₂/(32π³), because each d³p/(2E) supplies p dE and the energy δ function, spent on the opening angle, returns E₃/(p₁p₂) and cancels the rest.

Since m₂₃² = m² + m₁² − 2mE₁ is linear in E₁, the measure stays flat in the pair masses, dΓ = |ℳ|² dm₁₂² dm₂₃²/(256π³m³), and that is the whole content of a Dalitz plot: its edge, where the three momenta turn collinear, depends only on the four masses, and its density is |ℳ|², so a uniform plot means a constant matrix element and every band or empty edge is dynamics. The picture stops where its assumptions do. A polarised parent's angular information is integrated away, products must be stable and sharp in mass, a resonance in one pair needs its width (topic 5.09), identical products need 1/n!, and four or more bodies leave five or more variables, so no single plane displays |ℳ|².

Simple definition: A Dalitz plot is the scatter of three-body decays in the plane of two pair masses squared, such as m₁₂² and m₂₃², chosen because phase space is uniform there, so the density of events is proportional to the spin-averaged |ℳ|².
Example: A 1 GeV parent decaying to three massless products fills the triangle m₁₂², m₂₃² ≥ 0, m₁₂² + m₂₃² ≤ 1 GeV², of area 0.5 GeV⁴. With constant |ℳ| = 1 its width is 1 GeV/(512π³) = 6.30 × 10⁻⁵ GeV, a lifetime of 1.04 × 10⁻²⁰ s.

Three-body phase space, orientation integrateddΠ₃ = dE₁ dE₂/(32π³)

Derived. The delta functions remove p₃ and the opening angle, the decay plane's three Euler angles give 8π², and the three 1/(2E) factors cancel, leaving a flat measure.

E₁, E₂ two product energies in the parent rest frame, GeV; dΠ₃ in GeV², the GeV⁽²ⁿ⁻⁴⁾ of topic 5.02 at n = 3

Dalitz form of the decay ratedΓ = |ℳ|² dm₁₂² dm₂₃²/(256π³ m³)

Derived from dΓ = |ℳ|² dΠ₃/(2m) with dm₂₃² = 2m dE₁. Flat in the pair masses, so the density of events on the plot is |ℳ|² and nothing else.

m the parent mass in GeV; mᵢⱼ² = (pᵢ + pⱼ)² in GeV²; ℳ dimensionless for 1 → 3, so Γ comes out in GeV

The boundary at fixed m₁₂²(m₂₃²)± = (E₂* + E₃*)² − (p₂* ∓ p₃*)²

Derived: the extremes are p₂* and p₃* parallel and antiparallel. Only the four masses enter, so the edge of the plot is the same whatever |ℳ|² is.

E₂* = (m₁₂² − m₁² + m₂²)/(2m₁₂), E₃* = (m² − m₁₂² − m₃²)/(2m₁₂) in the 1-2 rest frame; pᵢ* = √(Eᵢ*² − mᵢ²); GeV

Projection onto m₁₂² for constant |ℳ|dΓ/dm₁₂² = |ℳ|² λ¹⁄²(m₁₂², m₁², m₂²) λ¹⁄²(m², m₁₂², m₃²)/(256π³ m³ m₁₂²)

A flat plot does not give a flat mass spectrum: the projection is the chord length, which falls to zero at the top of the range, and at the bottom too unless m₁ = m₂ = 0.

λ the Källén function of topic 5.02; λ¹⁄²λ¹⁄²/m₁₂² = 4p₂*p₃*, the plot's chord at that m₁₂², in GeV²

Constant |ℳ|, three massless productsΓ = |ℳ|² m/(512π³)

Computed. Against two massless bodies the integrated phase space shrinks by m²/(32π²), about m²/316: a third body is expensive.

the triangle m₁₂², m₂₃² ≥ 0, m₁₂² + m₂₃² ≤ m² has area m⁴/2; divide by 3! if the products are identical

Muon decay through the Dalitz integralΓ(μ → eν̄ₑν_μ) = G_F² m_μ⁵/(192π³)

Phase space derived, |ℳ|² quoted: τ = 2.187 μs against the measured 2.197 μs, the 0.44% gap the quoted radiative and mₑ corrections.

G_F = 1.1664 × 10⁻⁵ GeV⁻², m_μ = 0.105658 GeV, mₑ neglected; |ℳ|² quoted from Fermi's contact interaction

Five variables, and three of them are angles

Three final momenta have nine components and conservation removes four, so a three-body decay has five kinematic variables where a two-body decay had two. In natural units ħ = c = 1 with the metric (+, −, −, −), kept throughout, work at rest and spend δ³ on p₃. Write d³p₁ = p₁E₁ dE₁ dΩ₁, using p dp = E dE, and measure p₂ from the direction of p₁: d³p₂ = p₂E₂ dE₂ d(cos θ₁₂) dφ. Since E₃² = p₁² + p₂² + 2p₁p₂ cos θ₁₂ + m₃², the energy δ function fixes the opening angle with Jacobian E₃/(p₁p₂), and the product E₁E₂E₃ from the three 1/(2E) factors cancels exactly: dΠ₃ = dE₁ dE₂ dΩ₁ dφ/(256π⁵). Ω₁ and φ are the three Euler angles of the decay plane. A spinless parent, or an unpolarised one once its spin is averaged, offers |ℳ|² no direction to depend on, so they integrate to 4π × 2π = 8π² and dΠ₃ = dE₁ dE₂/(32π³), derived. Its dimension, GeV², is the GeV⁽²ⁿ⁻⁴⁾ of topic 5.02 at n = 3.

Why pair masses, and why the plot is flat

Energies change from frame to frame; pair masses do not. At rest, m₂₃² = (P − p₁)² = m² + m₁² − 2mE₁, so dm₂₃² = 2m dE₁, and likewise dm₁₃² = 2m dE₂. The Jacobian is the constant 1/(4m²), and with topic 5.03's dΓ = |ℳ|² dΠ₃/(2m) the rate becomes dΓ = |ℳ|² dm₁₃² dm₂₃²/(256π³m³), derived. The sum rule m₁₂² + m₂₃² + m₁₃² = m² + m₁² + m₂² + m₃² leaves only two pair masses independent, and swapping which two you plot has unit Jacobian, so every choice of axes is flat. Flatness is the point: the number of events in any cell is |ℳ|² times the cell's area times a constant, so the plot is a picture of the matrix element. It can be filled from parents in flight, because each mᵢⱼ² is computed from measured four-momenta in whatever frame they were measured. Dimensions: with four external legs ℳ has mass dimension 4 − 4 = 0, and GeV⁴/GeV³ leaves Γ in GeV.

The boundary is kinematics alone

Fix m₁₂ and go to the rest frame of the 1-2 pair. There particles 2 and 3 have fixed energies, E₂* = (m₁₂² − m₁² + m₂²)/(2m₁₂) and E₃* = (m² − m₁₂² − m₃²)/(2m₁₂), and only the angle between p₂* and p₃* is free. So m₂₃² = (p₂ + p₃)² runs between its values for parallel and antiparallel momenta, (E₂* + E₃*)² − (p₂* ∓ p₃*)², a chord of length 4p₂*p₃*. The chord closes at m₁₂ = m₁ + m₂, where 1 and 2 move together, unless both are massless, and at m₁₂ = m − m₃, where 3 is at rest; everywhere on the edge the three momenta are collinear. Three massless products give the triangle m₁₂², m₂₃² ≥ 0, m₁₂² + m₂₃² ≤ m². A 1 GeV parent decaying to two 0.300 GeV products and a massless third has a curved region of 0.112 GeV⁴, 22% of that triangle, computed numerically. The massive region need not sit inside the massless one: since m₁₂² + m₂₃² = m² + m₁² + m₂² + m₃² − m₁₃² and m₁₃² ≥ (m₁ + m₃)², it can reach past the line m₁₂² + m₂₃² = m² by up to m₂² − 2m₁m₃. |ℳ|² never enters the edge.

Reading |ℳ|² off the density

A uniform plot means constant |ℳ|²; anything else is dynamics. A resonance R decaying to 2 + 3 lays a band along m₂₃² ≈ m_R², about 2m_RΓ_R wide (topic 5.09). The decay ω → π⁺π⁻π⁰ has |ℳ|² proportional to |p₊ × p₋|² in the ω frame, quoted, so its plot is empty along the whole edge, where the momenta are collinear. Identical products make the plot symmetric, six-fold for three identical particles, so either divide the full integral by 3! or integrate one sextant. Summed over the plot, Γ = ⟨|ℳ|²⟩A/(256π³m³), with A the plot's area and ⟨|ℳ|²⟩ the average over it. Dalitz drew the first such plot in 1953 for the τ meson, today's K⁺, decaying to π⁺π⁺π⁻; its uniformity pointed to spin-parity 0⁻, one half of the τ–θ puzzle that ended in parity violation (history, quoted).

Massless products, and what a third body costs

With constant |ℳ| and massless products the triangle's area is m⁴/2, so Γ = |ℳ|²(m⁴/2)/(256π³m³) = |ℳ|² m/(512π³), computed. Three checks. Dimensions: ℳ is dimensionless for 1 → 3, so Γ ∝ m is forced. The massless limit of the boundary formula: with every mass zero, E₂* = p₂* = m₁₂/2 and the chord at m₁₂² is m² − m₁₂², the triangle again. Two bodies: the integrated phase space is 1/(8π) for massless products and m²/(256π³) here, a ratio m²/(32π²) ≈ m²/316. The three-body ℳ carries one more coupling and one less power of mass than the two-body one, so for a dimensionless coupling a third body typically costs a factor of order (coupling)²/(32π²) in the rate. Near threshold the scaling changes as well: two-body phase space opens as √Q in the energy release Q, three-body as Q², because the non-relativistic region is an ellipse whose two axes both grow in proportion to Q (the Hard example derives it).

Muon decay: a test of the 192π³

For μ⁻ → e⁻ν̄ₑν_μ, Fermi's contact interaction, whose weak-interaction theory is left to a later course, gives the spin-averaged, spin-summed |ℳ|² = 64G_F²(p_μ⋅p_ν̄ₑ)(pₑ⋅p_νμ), quoted. With mₑ neglected it depends on one pair mass, s = (pₑ + p_νμ)²: |ℳ|² = 16G_F² s(m_μ² − s). The chord at fixed s is m_μ² − s, so Γ = 16G_F² ∫ s(m_μ² − s)² ds/(256π³m_μ³) from 0 to m_μ², which is G_F² m_μ⁵/(192π³), derived from the quoted |ℳ|². With G_F = 1.1664 × 10⁻⁵ GeV⁻² and m_μ = 0.105658 GeV, Γ = 3.009 × 10⁻¹⁹ GeV and τ = ħ/Γ = 2.187 μs, against the measured 2.197 μs. The 0.44% gap is the QED correction 1 − (α/2π)(π² − 25/4) = 1 − 0.0042 in the rate, plus 0.02% from mₑ, both quoted. G_F is extracted from this lifetime, so the check tests the phase-space coefficient: a lost 2 or π would miss by a factor. The m⁵ itself is tested by the τ lepton, m_τ = 1.777 GeV, whose electron branch B = 17.82% predicts B τ_μ (m_μ/m_τ)⁵ = 291 fs for its lifetime, against 290.3 fs measured (inputs quoted).

Change one variable at a time

Make the relationship visible.

Interactive model

Mass of products 1 and 20.30 GeV

Mass of product 30.00 GeV

Probe: 1-2 pair mass squared0.64 GeV²

The defaults are the Easy example: a chord of 0.238 GeV² at m₁₂² = 0.64 GeV². Set both masses to zero and the region fills the dashed triangle while the projection becomes the line 1 − m₁₂²; push both to 0.32 GeV and only a small threshold ellipse is left.

ENERGY RELEASE Q400 MeV

CHORD AT THE PROBE0.238 GeV²

SPECTRUM PER UNIT |ℳ|²3.00 10⁻⁵ GeV⁻¹

CHORD ÷ MASSLESS CHORD0.66

Live interpretationENERGY RELEASE Q: 400 MeV. CHORD AT THE PROBE: 0.238 GeV². SPECTRUM PER UNIT |ℳ|²: 3.00 10⁻⁵ GeV⁻¹. CHORD ÷ MASSLESS CHORD: 0.66

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that a flat Dalitz plot makes every distribution flat, so a peak or slope in a one-dimensional spectrum must be dynamics. Neither follows. Phase space is uniform over a plane, two squared pair masses or equivalently two rest-frame energies taken together; it is not uniform in any single variable, not in m₁₂ rather than m₁₂², and not in any projection. Project a flat plot onto m₁₂² and you get the chord length. For three massless products with constant |ℳ|, one product's energy spectrum is dΓ/dE₁ ∝ E₁, rising linearly to its endpoint m/2, though nothing in |ℳ| prefers fast particles. Replot the same events against m₁₂ and the density picks up a factor 2m₁₂. Only in the full plane of squared masses is a non-uniform density a statement about |ℳ|²; every one-dimensional spectrum has to be compared with its phase-space shape before structure is claimed.

A spinless 2.0 GeV parent at rest decays to three identical massless scalars through a contact coupling with constant ℳ = 3.0. What is its width?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spinless particle X of mass 1.00 GeV decays at rest to ψψ̄χ, with ψ and ψ̄ of mass 0.300 GeV and χ massless; label them 1, 2 and 3. Find the full ranges of m₁₂² and m₂₃² on the Dalitz plot, then the range of m₂₃² allowed when m₁₂² = 0.640 GeV².

m₁₂ runs from m₁ + m₂ = 0.600 GeV, with ψ and ψ̄ at rest relative to each other, to m − m₃ = 1.000 GeV, with χ at rest: m₁₂² from 0.360 to 1.000 GeV².
Likewise m₂₃ runs from m₂ + m₃ = 0.300 GeV to m − m₁ = 0.700 GeV: m₂₃² from 0.090 to 0.490 GeV².
At m₁₂ = 0.800 GeV, in the ψψ̄ rest frame: E₂* = (0.640 − 0.090 + 0.090)/1.600 = 0.400 GeV and p₂* = √(0.160 − 0.090) = 0.2646 GeV; E₃* = (1.000 − 0.640 − 0)/1.600 = 0.225 GeV = p₃*.
(m₂₃²)± = (E₂* + E₃*)² − (p₂* ∓ p₃*)² = 0.3906 − (0.0396)² and 0.3906 − (0.4896)², so m₂₃² runs from 0.151 to 0.389 GeV².
Checks: the chord 0.238 GeV² equals 4p₂*p₃* = 4 × 0.2646 × 0.225 GeV², and its midpoint 0.270 GeV² equals (m² + m₁² + m₂² + m₃² − m₁₂²)/2 = (1.180 − 0.640)/2 GeV², the point where m₂₃² = m₁₃² and ψ and ψ̄ are interchangeable.

Answerm₁₂² from 0.360 to 1.000 GeV² and m₂₃² from 0.090 to 0.490 GeV²; at m₁₂² = 0.640 GeV², m₂₃² runs from 0.151 to 0.389 GeV², a chord of 0.238 GeV².

MediumA spinless parent of mass 2.00 GeV decays at rest to three identical massless scalars. Bose symmetry makes |ℳ|² symmetric in the three pair masses; take |ℳ|² = A(m₁₂⁴ + m₂₃⁴ + m₁₃⁴)/m⁴ with A = 50. Find the average of |ℳ|² over the Dalitz plot, the width and the lifetime, and say where the events crowd.

Use x = m₁₂²/m² and y = m₂₃²/m², so m₁₃²/m² = 1 − x − y and the plot is the triangle x, y ≥ 0, x + y ≤ 1, of area 1/2 in these units and m⁴/2 = 8.00 GeV⁴ in real ones.
Over the triangle, ∫x² = ∫ x²(1 − x) dx from 0 to 1 = 1/3 − 1/4 = 1/12, and by the triangle's symmetry y² and (1 − x − y)² give the same. The sum is 1/4, so ⟨|ℳ|²⟩ = A(1/4)/(1/2) = A/2 = 25.
Γ = (1/3!) ⟨|ℳ|²⟩ (m⁴/2)/(256π³m³) = A m/(6144π³) = (100/190503) GeV = 5.25 × 10⁻⁴ GeV, the 1/3! because the full plot counts each event six times.
τ = ħ/Γ = 6.582 × 10⁻²⁵ GeV s/(5.25 × 10⁻⁴ GeV) = 1.25 × 10⁻²¹ s.
|ℳ|² is A/3 at the centre, where every pair mass is m²/3 and the momenta form a symmetric star, A/2 at the midpoint of each edge, and A at each corner, where one scalar is soft and the other two fly back to back with m/2 each: the events crowd into the three corners.
Check: a constant |ℳ|² = 25 would give the same Γ, which is what the average means; ℳ is dimensionless and Γ ∝ m, as it must be.

Answer⟨|ℳ|²⟩ = 25; Γ = 5.25 × 10⁻⁴ GeV and τ = 1.25 × 10⁻²¹ s. The density rises from A/3 at the centre to A at the corners, where one scalar is soft.

HardA spinless parent of mass 1.000 GeV decays at rest to three distinct spinless particles, each of mass 0.320 GeV, so the energy release is Q = 40 MeV, with constant |ℳ| = 0.50. Treating the products non-relativistically, find the allowed region in their kinetic energies, its area, the width and the lifetime. Compare with the exact relativistic region and state how Γ scales with Q near threshold.

Non-relativistically Tᵢ = pᵢ²/(2M), with M = 0.320 GeV and T₁ + T₂ + T₃ = Q. The momenta close into a triangle only if |p₁ − p₂| ≤ p₃ ≤ p₁ + p₂; squaring, with p₃² = 2M(Q − T₁ − T₂), gives (Q − 2T₁ − 2T₂)² ≤ 4T₁T₂.
With u = T₁ + T₂ and v = T₁ − T₂, 4T₁T₂ = u² − v² and the condition becomes v² + 3(u − 2Q/3)² ≤ Q²/3: an ellipse of semi-axes Q/√3 in v and Q/3 in u, centred where all three share Q/3.
Its area in (u, v) is π(Q/√3)(Q/3), and dT₁ dT₂ = du dv/2, so the area in (E₁, E₂) is πQ²/(6√3) = π(0.040 GeV)²/10.39 = 4.837 × 10⁻⁴ GeV².
With A_E that area, Γ = |ℳ|² A_E/(64π³m) = 0.25 × 4.837 × 10⁻⁴ GeV²/(1984.4 × 1.000 GeV) = 6.09 × 10⁻⁸ GeV, so τ = ħ/Γ = 1.08 × 10⁻¹⁷ s.
Integrating the exact chord 4p₂*p₃* over m₁₂² numerically gives 1.896 × 10⁻³ GeV⁴, or 4.740 × 10⁻⁴ GeV² after dividing by 4m², 2.0% below the ellipse, so Γ = 5.97 × 10⁻⁸ GeV and τ = 1.10 × 10⁻¹⁷ s. At Q = 10 MeV the gap is 0.5%: it shrinks in proportion to Q, as a relativistic correction of order Q/M should.
The area grows as Q², so Γ ∝ Q² near threshold, against √Q for two bodies (topic 5.02): halving the energy release quarters the width. Against the massless triangle, m²/8 = 0.125 GeV², the masses cost a factor 3.8 × 10⁻³, or 3.9 × 10⁻³ from the ellipse.

AnswerThe region is the ellipse (Q − 2T₁ − 2T₂)² ≤ 4T₁T₂ of area πQ²/(6√3) = 4.84 × 10⁻⁴ GeV²; Γ = 6.09 × 10⁻⁸ GeV and τ = 1.08 × 10⁻¹⁷ s, 2.0% above the exact 5.97 × 10⁻⁸ GeV; Γ ∝ Q² near threshold.