advanced37 min studyLesson 602 in Physics

Count final states with an invariant measure

Derive the positive-energy on-shell measure and reduce two-body phase space while keeping all factors of two pi and identical-particle conventions explicit.

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What you’ll understand

  • Explain the origin of the one-particle measure d³p/(2E).
  • Reduce a two-body final-state measure using energy and momentum conservation.
  • Separate phase-space normalization, matrix-element weighting, and identical-particle counting.

Momentum conservation specifies which events are possible, but a decay rate also needs a rule for integrating over them. The Lorentz-invariant phase-space measure supplies that rule. It is a geometrical weight on on-shell final states, not a probability distribution until an interaction amplitude and normalization are supplied. Keeping the measure explicit prevents a uniform grid of momenta from being mistaken for a uniform physical decay sample.

Restrict four-dimensional momentum space to the mass shell

Use the module's ℏ=c=1\hbar=c=1 and (+,−,−,−)(+,-,-,-) conventions. The four-dimensional momentum volume is invariant under proper Lorentz transformations. Impose the mass shell with a delta function and retain only positive energy. Integrating over energy uses the derivative of E2−p2−m2E^2-\mathbf p^2-m^2, which gives the factor 2E2E. This produces d3p/(2E)d^3\mathbf p/(2E) rather than the ordinary three-dimensional volume. The positive-energy restriction is preserved by proper orthochronous transformations, the frame changes relevant here.

Relativistically normalized one-particle states introduce additional factors of two pi. Different references place the overall four-momentum delta-function factor inside or outside their definition of phase space. Both conventions can be consistent, but a rate formula and a phase-space formula from different conventions cannot be combined without translation. We include the full four-dimensional factor in our definition below. The positive-energy on-shell energy is always the energy of that same particle; substituting its mass for E in a moving frame destroys invariance.

d4p δ(p2−m2)Θ(p0)=d3p2E,dΦn=(2π)4δ(4) ⁣(P−∑ipi)∏id3pi(2π)3 2Eid^4p\,\delta(p^2-m^2)\Theta(p^0)=\frac{d^3\mathbf p}{2E},\qquad d\Phi_n=(2\pi)^4\delta^{(4)}\!\left(P-\sum_i p_i\right)\prod_i\frac{d^3\mathbf p_i}{(2\pi)^3\,2E_i}
The equality denotes integrating over the positive-energy shell. Our Φn includes the (2π)⁴ conservation factor; some references place it in the rate instead. With natural units, Φn has energy dimension 2n−4 before integration over any external invariant mass.↔ Scroll wide equations horizontally.

Reduce two-body phase space in the parent rest frame

For total timelike momentum P with invariant mass M, choose its rest frame. The spatial delta function sets the daughter momenta equal and opposite. The remaining energy delta function imposes E1(p)+E2(p)=ME_1(p)+E_2(p)=M. Its derivative with respect to p is pM/(E1E2)pM/(E_1E_2) at the allowed root. Combining that Jacobian with the radial momentum volume leaves one factor of p and an angular measure. Invariance makes the final expression valid for computing the integral regardless of the original laboratory boost.

The allowed momentum is determined by the Källén function. Factoring it as [M2−(m1+m2)2][M2−(m1−m2)2][M^2-(m_1+m_2)^2][M^2-(m_1-m_2)^2] makes the ordinary positive-energy threshold visible. At M=m1+m2M=m_1+m_2 the two-body measure collapses as the momentum vanishes. For two massless daughters, p equals M/2 and the fully integrated measure is 1/(8π)1/(8\pi). That dimensionless result is a useful normalization check, but it is not itself a decay width or an angular probability.

p∗=λ(M2,m12,m22)2M,λ(x,y,z)=x2+y2+z2−2xy−2xz−2yz,dΦ2=p∗16π2M dΩp^*=\frac{\sqrt{\lambda(M^2,m_1^2,m_2^2)}}{2M},\quad \lambda(x,y,z)=x^2+y^2+z^2-2xy-2xz-2yz,\qquad d\Phi_2=\frac{p^*}{16\pi^2M}\,d\Omega
M, daughter masses, and p* are in the same energy unit, with M ≥ m1 + m2. The angular integral is 4π for fully labeled distinguishable daughters. Φ2 is dimensionless in this convention.↔ Scroll wide equations horizontally.

Add dynamics and counting only once

For a decay, multiplying our measure by ∣M∣2/(2M)|\mathcal M|^2/(2M) gives the rest-frame differential width, with appropriate spin treatment. The amplitude can weight different angles or invariant masses unequally; phase space alone does not determine that physics. For n identical final particles, integration over the complete labeled phase space counts each physical configuration n factorial times, so a corresponding symmetry factor is needed. Alternatively, an explicitly ordered region can count each configuration once. Using both remedies would undercount the rate.

A simulation also needs to distinguish its sampling coordinates from the desired measure. Drawing momentum components uniformly generally introduces a nonuniform Jacobian on the constrained surface. A generator can sample conveniently and attach weights, or design coordinates with a simpler measure, but the weighting must be documented. The same principle applies when comparing data with a phase-space shape: finite detector acceptance and momentum resolution distort the observed population. A discrepancy from an uncorrected uniform plot is not by itself a measurement of a new interaction amplitude.

A massive daughter and a massless daughter

A system of invariant mass M=5 GeVM=5\,\mathrm{GeV} produces distinguishable daughters of masses 3 GeV3\,\mathrm{GeV} and zero. Find the common rest-frame momentum and the fully integrated two-body phase space in our convention.

  1. The Källén function factors as (25−9)(25−9)=256 GeV4(25-9)(25-9)=256\,\mathrm{GeV^4}. Its positive square root is 16 GeV216\,\mathrm{GeV^2}, giving p∗=16/(2×5)=1.6 GeVp^*=16/(2\times5)=1.6\,\mathrm{GeV}.

  2. Check the energy constraint: the massive daughter has energy 32+1.62=3.4 GeV\sqrt{3^2+1.6^2}=3.4\,\mathrm{GeV}, and the massless daughter's energy is 1.6 GeV1.6\,\mathrm{GeV}. Their sum is exactly M.

  3. Integrate dΩd\Omega over 4π4\pi: Φ2=p∗/(4πM)=1.6/(20π)=0.08/π≈0.02546\Phi_2=p^*/(4\pi M)=1.6/(20\pi)=0.08/\pi\approx0.02546. There is no identical-particle division because the daughters are distinguishable.

The result

The momentum is 1.6 GeV1.6\,\mathrm{GeV} and integrated phase space is 0.08/π0.08/\pi, dimensionless. A width still requires the amplitude and the factor 1/(2M)1/(2M).

Explore the model

Interactive lab

Interactive lab · vectors

Inspect angular projections without inventing a phase-space generator

Reset and set angle to 0 degrees and magnitude to 1.6; predict x = 1.6 and y = 0. Compare with a fresh Reset, angle 90 degrees and magnitude 1.6, giving x = 0 and y = 1.6. This shows one fixed-magnitude planar vector. The lab does not draw both daughters, impose a mass shell, sample solid angle, or integrate phase space; dΩ belongs to a three-dimensional angular integration derived in the lesson.

Inspect angular projections without inventing a phase-space generatorPlayback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Current readings: x component: 1.393; y component: 0.975; Magnitude: 1.7. All parameters are adjustable with the labeled controls below.−7.46−3.7303.737.46−3.5−1.7501.753.5xyvvₓ = 1.39
Vectorx / y projectionsTrace marker
x component
1.393
y component
0.975
Magnitude
1.7
t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA fixed momentum magnitude leaves directions to integrate, with the correct geometrical measure.

Check your understanding

Why is the on-shell one-particle invariant measure proportional to d³p/(2E) rather than just d³p?

Try it yourself

A system of invariant mass 10 GeV10\,\mathrm{GeV} produces two distinguishable particles each of mass 3 GeV3\,\mathrm{GeV}. Compute their common momentum and integrated phase space. What changes in the counting if they are instead identical and the full labeled angular region is retained?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Each daughter has energy five GeV in the rest frame. Apply the identical-particle correction after computing the labeled measure.

Compare with the solution

Each momentum magnitude is 52−32=4 GeV\sqrt{5^2-3^2}=4\,\mathrm{GeV}. The full labeled integral is Φ2=4/(4π×10)=0.1/π≈0.03183\Phi_2=4/(4\pi\times10)=0.1/\pi\approx0.03183. For two identical particles, the fully labeled rate requires a factor 1/2!1/2!, giving an effective measure 0.05/π0.05/\pi for that rate. An equivalent calculation integrates an appropriate nonduplicating region. This counting correction does not replace the exchange-symmetrized or antisymmetrized amplitude required by quantum statistics.

Take it with you

  • The mass-shell Jacobian makes d³p/(2E) Lorentz invariant.
  • Declare where the overall (2π)⁴ factor sits before using a rate formula.
  • Phase space, dynamics, and identical-particle counting are separate ingredients.

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