research33 min studyLesson 625 in Physics

Evolve parton densities without inventing their initial shape

Derive convolution evolution, include virtual endpoint terms, and use moment sum rules to diagnose physically consistent PDF evolution.

On this page

What you’ll understand

  • Interpret the convolution and daughter-to-parent momentum fraction.
  • Explain why plus distributions and virtual terms are required.
  • Check valence-number and total-momentum conservation through kernel moments.

A hard probe resolves different partonic structure when its factorization scale changes. DGLAP evolution describes the logarithmic scale dependence of leading-twist collinear parton densities. It predicts how an initial distribution changes, rather than deriving that initial nonperturbative distribution from a few valence constituents. The splitting kernels are perturbatively calculable, while the boundary functions normally come from global fits or other nonperturbative information. Keeping that distinction clear prevents an evolution equation from being mistaken for a complete proton wavefunction calculation.

The parent carries more momentum than its daughter

A daughter observed at momentum fraction x may originate from a parent with fraction x divided by z, where z is the daughter’s share of the parent longitudinal momentum. Since the parent fraction cannot exceed one, the convolution runs from z equal to x to one. Its Jacobian supplies a factor one over z. The kernel index tells which parent produces which daughter, so writing the index order explicitly avoids interchanging quark-to-gluon and gluon-to-quark kernels.

At leading order the overall coefficient is alpha s divided by two pi when differentiating with respect to logarithm of factorization scale squared. The PDF’s scale is not an elapsed physical time. Radiation migrates momentum between x bins and between species as the resolution convention changes. Unpolarized PDFs have a probability-like leading-order interpretation, but renormalized distributions and subtractions beyond leading order are scheme dependent. Individual higher-order PDF values should not be treated as directly observable positive event counts.

∂fi(x,μ2)∂ln⁡μ2=αs(μ)2π∑j(Pi←j⊗fj)(x),(P⊗f)(x)=∫x1dzzP(z)f(x/z)\frac{\partial f_i(x,\mu^2)}{\partial\ln\mu^2}=\frac{\alpha_s(\mu)}{2\pi}\sum_j(P_{i\leftarrow j}\otimes f_j)(x),\qquad(P\otimes f)(x)=\int_x^1\frac{dz}{z}P(z)f(x/z)
μ is the factorization scale, commonly also chosen for αs in this LO expression. fi is a collinear PDF per unit x. P i←j produces daughter i from parent j; this is a leading-twist fixed-flavor evolution equation.↔ Scroll wide equations horizontally.

Real radiation is not the whole kernel

For a gluon producing one selected quark flavor, the leading kernel is one half times z squared plus one minus z squared. For a quark producing a gluon, the kernel has a soft singularity as the gluon share goes to zero. Quark retention has a complementary endpoint singularity as the radiated gluon becomes soft. Integrating only real emission would produce divergent probabilities and would incorrectly deplete conserved valence quantum numbers without a compensating unresolved contribution.

The full quark-retention kernel includes a plus distribution and a delta term. The plus prescription subtracts the endpoint value of a test function before integrating the singular denominator. It is a distributional statement, not an instruction to evaluate infinity at z equal to one. Writing the kernel as two over one minus z with the plus prescription, followed by regular minus one minus z terms, makes its integral especially transparent. The remaining delta coefficient restores the required zero valence moment.

Pq←g=TR[z2+(1−z)2],Pg←q=CF1+(1−z)2z,Pq←q=CF[2(1−z)+−1−z+32δ(1−z)],∫01ϕ(z) dz(1−z)+=∫01ϕ(z)−ϕ(1)1−z dzP_{q\leftarrow g}=T_R[z^2+(1-z)^2],\quad P_{g\leftarrow q}=C_F\frac{1+(1-z)^2}{z},\quad P_{q\leftarrow q}=C_F\left[\frac2{(1-z)_+}-1-z+\frac32\delta(1-z)\right],\quad\int_0^1\frac{\phi(z)\,dz}{(1-z)_+}=\int_0^1\frac{\phi(z)-\phi(1)}{1-z}\,dz
These are LO space-like kernels with z the named daughter fraction. The q←g formula is for one quark, not the full quark-plus-antiquark singlet. The last identity defines the plus distribution on smooth test functions.↔ Scroll wide equations horizontally.

Moments test the bookkeeping

Integrating a nonsinglet valence density over x gives a conserved net flavor number. The zero integral of its evolution kernel expresses that conservation. Weighting by x instead measures momentum, which can leave the quark sector and enter the gluon sector. For each parent, the momentum-weighted sum over all daughter kernels must vanish. These constraints are stringent checks on endpoint coefficients and flavor multiplicities; replacing kernels by their positive real parts violates them even if individual emissions look plausible.

Evolution requires an initial scale, boundary densities, a coupling prescription, perturbative order, and heavy-flavor matching rules. Very small x or endpoint regions may need logarithmic resummation, and low virtualities admit power corrections outside leading twist. A finite-state Markov illustration can conserve a total probability, but its two bins are not x-dependent parton species and its transition rates are not DGLAP kernels. Its useful role is to expose why apparent loss from one component must be balanced elsewhere.

∫01Pq←q(z) dz=0,∑i∫01zPi←j(z) dz=0,∫01dx x[g(x,μ2)+∑f(qf+qˉf)(x,μ2)]=1\int_0^1P_{q\leftarrow q}(z)\,dz=0,\qquad \sum_i\int_0^1zP_{i\leftarrow j}(z)\,dz=0,\qquad\int_0^1dx\,x\left[g(x,\mu^2)+\sum_f(q_f+\bar q_f)(x,\mu^2)\right]=1
The momentum sum assumes the proton’s momentum is carried only by the included QCD partons; adding QED requires a photon term. The column-wise kernel sum includes all daughter species and flavor multiplicities. Net valence and total momentum are scale invariant in consistent evolution.↔ Scroll wide equations horizontally.

Check a quark-parent momentum balance

For a quark parent, use the displayed LO kernels with CF=4/3. Compute ∫₀¹z Pq←q dz and ∫₀¹z Pg←q dz and check their sum.

  1. For the plus term, the test function is z. Thus ∫z×2/(1−z)+ dz=∫2(z−1)/(1−z)dz=−2. The regular terms integrate to−1/2−1/3=−5/6.

  2. The delta term contributes3/2. Their sum is−2−5/6+3/2=−4/3; multiplying CF gives the quark momentum moment−16/9.

  3. For the gluon daughter, multiplication by z removes the denominator. The integral of1+(1−z)² is4/3, giving+16/9 after CF. The two moments add to zero, as required for this parent.

The result

The quark and gluon daughter moments are−16/9 and+16/9, conserving the parent’s total longitudinal momentum.

Explore the model

Interactive lab

Interactive lab · markov

Track transfer while conserving a total

For fresh runs Reset and set BOTH p0.2, q0.1, then Step once; separately Reset and set p0.4, q0.1, then Step once. The exact distributions should be(0.8,0.2) and(0.6,0.4), each totaling1. This discrete two-state chain is a conservation analogy, not PDF evolution in x or a parton shower.

Track transfer while conserving a totalExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. 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 component can lose weight while the combined conserved total remains fixed.

Check your understanding

Why must DGLAP evolution be supplied with PDFs at an initial scale?

Try it yourself

For one quark flavor compute ∫₀¹z Pq←g(z)dz with TR=1/2. Then include its antiquark and nf=3 flavors. What gluon diagonal momentum moment is needed to conserve momentum?

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

Give me a hint

Integrate z[z²+(1−z)²]/2, then multiply by2nf before applying the column sum rule.

Compare with the solution

For one quark the integral is(1/2)∫(2z³−2z²+z)dz=(1/2)(1/2−2/3+1/2)=1/6. Its antiquark contributes equally. Three flavors therefore receive2×3×1/6=1 from a gluon parent in the kernel momentum sum. The diagonal gluon moment must be−1, including its virtual endpoint contribution. This is a kernel moment, not a branching probability of exactly one.

Take it with you

  • DGLAP predicts scale dependence once boundary PDFs are supplied.
  • Virtual endpoint terms are needed for finite and conserving evolution.
  • Flavor multiplicities and kernel moments provide independent consistency checks.

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