PDFs, sum rules, and flavour structure
Definitions, evolution basis, sum rules derived, proton/deuteron flavour separation.
The parton model handed us a set of functions $f_i(x)$ and a formula weighting them by squared charge. This section makes those functions first-class citizens: what a PDF actually is, the exact constraints (sum rules) that any proton must satisfy, the flavour anatomy of $F_2^p$ and $F_2^n$, and the change of basis in which fits and evolution codes actually work. It ends with the question that shapes this lab's results pages: which flavour combinations can electromagnetic $F_2$ determine at all — and which are flat directions until other data arrives.
What a PDF is — and is not
Operationally, $f_i(x)\,dx$ is the expected number of partons of species $i$ carrying a fraction of the proton's light-cone momentum between $x$ and $x + dx$. It is a number density, not a probability: its integral $\int_0^1 f_i\,dx$ counts partons (and diverges for sea quarks and gluons as $x \to 0$ — there are infinitely many soft ones), while $\int_0^1 x f_i\,dx$ is the momentum share, which is what must add to one. Quarks and antiquarks get separate densities, $q(x)$ and $\bar q(x)$.
Field theory sharpens this into an operator definition. In light-cone coordinates ($P^+$ the big momentum component), the quark density is the Fourier transform of a correlator of quark fields separated along the light cone:
Read qualitatively: $\gamma^+$ picks out the "good" light-cone component that counts quanta rather than interaction energy; the Wilson line $\mathcal{W}$ (a path-ordered gluon exponential) makes the separated fields gauge-invariant; and antiquarks appear as the $x \to -x$ continuation, $\bar q(x) = -q(-x)$. Two consequences matter for everything downstream: PDFs are universal proton matrix elements — defined without reference to any particular scattering process — and beyond leading order they acquire a scale and scheme label, $q(x, \mu^2)$ in $\overline{\rm MS}$, which is Topic 2's story. Our fits parametrize them at $Q_0 = 1.65$ GeV and evolve upward.
Flavour anatomy of $F_2^p$ and $F_2^n$
Write out the master formula with the actual charges $e_u = +\tfrac23$, $e_d = e_s = -\tfrac13$ (charges enter squared, so signs drop):
All PDFs here are proton PDFs. The neutron needs no new functions: strong isospin — the near-exact symmetry under $u \leftrightarrow d$ — maps proton to neutron, so the neutron's up distribution is the proton's down distribution:
(Charge-symmetry violation is a sub-percent effect, ignored at our precision.) Split each flavour into valence plus sea, $q = q_v + \bar q$ with $q_v \equiv q - \bar q$, and the $p$–$n$ difference becomes a scalpel:
Subtract the two decompositions; strange and heavier flavours cancel identically:
$$ F_2^p - F_2^n = x\Big(\tfrac{4}{9} - \tfrac{1}{9}\Big)\Big[(u + \bar u) - (d + \bar d)\Big] = \frac{x}{3}\Big[(u + \bar u) - (d + \bar d)\Big] $$
Now insert $u = u_v + \bar u$, $d = d_v + \bar d$, so $u + \bar u = u_v + 2\bar u$ and likewise for $d$:
$$ F_2^p - F_2^n \;=\; \frac{x}{3}\Big[u_v - d_v\Big] \;+\; \frac{2x}{3}\Big[\bar u - \bar d\Big] $$
Valence difference plus sea asymmetry — nothing else survives.
Divide by $x$ and integrate. The valence integrals are fixed by counting ($\int u_v = 2$, $\int d_v = 1$, derived below), giving
$$ S_G \;\equiv\; \int_0^1 \frac{dx}{x}\Big[F_2^p - F_2^n\Big] = \frac{1}{3}\big(2 - 1\big) + \frac{2}{3}\int_0^1\!\big(\bar u - \bar d\big)\,dx = \frac{1}{3} + \frac{2}{3}\int_0^1\!\big(\bar u - \bar d\big)\,dx $$
A flavour-symmetric sea ($\bar u = \bar d$, the natural guess since gluons split democratically into light $q\bar q$ pairs) predicts $S_G = 1/3$. NMC measured (1991–94, muon DIS on $p$ and $d$):
$$ S_G = 0.235 \pm 0.026 \qquad\Longrightarrow\qquad \int_0^1\big(\bar d - \bar u\big)\,dx = \frac{3}{2}\Big(\frac{1}{3} - S_G\Big) = 0.148 \pm 0.039 $$
The proton's sea is not flavour-symmetric: it holds measurably more $\bar d$ than $\bar u$. Drell–Yan ratio experiments (E866, later SeaQuest) confirmed the excess and mapped its $x$-shape. The standard physical reading is the pion cloud: the fluctuation $p \to n\pi^+$ is easier than $p \to \Delta^{++}\pi^-$, and a $\pi^+ = u\bar d$ carries exactly the surplus $\bar d$ observed. A "sum rule" that fails is often worth more than one that holds — this failure is a permanent, structural fact about the proton that any fit's sea parametrization must be flexible enough to express.
Deuterons: the poor man's neutron
No free neutron target exists, so $F_2^n$ is reached through deuterium: to first approximation the deuteron is a loosely bound $p + n$ and, per nucleon,
The corrections are nuclear physics: Fermi motion smears the steeply falling large-$x$ tail (a large relative
effect for $x \gtrsim 0.6$), binding and off-shell effects shift the mid-$x$ region at the percent level, and
shadowing depletes $x \lesssim 0.1$ slightly. None are known to the accuracy of the DIS data itself. That is why
the SLAC and BCDMS deuteron tables enter our fits as the deweighted _dw variants — carrying an
inflated, nuclear-model-aware uncertainty — so that deuteron modelling error cannot masquerade as neutron
structure. Every isospin separation in what follows silently leans on this caveat.
Define the valence counts $N_q \equiv \int_0^1\big[q(x) - \bar q(x)\big]dx$. These are net charges of conserved currents, so they are fixed by the proton's quantum numbers — and (Topic 2) DGLAP evolution preserves them exactly, since gluons only ever add $q\bar q$ pairs. Impose baryon number, electric charge, and strangeness:
$$ B = \tfrac{1}{3}\big(N_u + N_d + N_s\big) = 1, \qquad Q = \tfrac{2}{3}N_u - \tfrac{1}{3}N_d - \tfrac{1}{3}N_s = 1, \qquad S = -N_s = 0 $$
Three equations, three unknowns: $N_s = 0$, then $N_u + N_d = 3$ and $2N_u - N_d = 3$, so
$$ \int_0^1\big(u - \bar u\big)dx = 2, \qquad \int_0^1\big(d - \bar d\big)dx = 1, \qquad \int_0^1\big(s - \bar s\big)dx = 0 $$
These are exact, valid at every scale, and our fits impose them analytically on the parametrization normalizations (the Euler-Beta machinery of Topic 3, §3.2; see sum rules).
The proton's momentum is the expectation of the energy-momentum tensor, $\langle P|T^{++}|P\rangle = 2(P^+)^2$. In light-cone quantization $T^{++}$ splits into a sum of quark and gluon pieces whose forward matrix elements are precisely the second moments of the PDFs. Dividing out the normalization:
$$ \sum_q \int_0^1 x\big[q(x) + \bar q(x)\big]dx \;+\; \int_0^1 x\,g(x)\,dx \;=\; 1 $$
Note what is not conserved: each individual term. §1.3's audit found the quark share $\approx 0.54$ at $Q^2 \sim 10\ \text{GeV}^2$; evolution slowly pumps momentum from quarks to gluons and sea as $Q^2$ grows, but the total stays pinned at 1. The gluon PDF in an $F_2$-only fit is constrained mostly by this rule plus the $Q^2$-slopes — a point our results return to repeatedly.
The evolution basis: coordinates that diagonalize the physics
Fits could parametrize $\{u, \bar u, d, \bar d, s, \bar s, g\}$ directly, but DGLAP evolution (Topic 2) mixes them in a very structured way: gluons talk only to the total quark singlet, and flavour differences evolve autonomously. The natural coordinates split sums from differences. With $q^\pm \equiv q \pm \bar q$:
For the light sector this is an invertible linear map — written out once, explicitly:
(the same matrix maps $\{u^-, d^-, s^-\}$ to $\{V, V_3, V_8\}$). Why bother? Because in this basis DGLAP is block-diagonal: $\Sigma$ couples to $g$ in a 2×2 system, and every $T_k$, $V_k$ evolves by itself with a non-singlet kernel (the kernels differ among $T$, $V$ only beyond leading order). The sum rules become one-liners: $\int V = 3$, $\int V_3 = 1$, $\int V_8 = 3$, $\int x(\Sigma + g) = 1$, and the Gottfried asymmetry is $\int(\bar u - \bar d) = \tfrac12\int(T_3 - V_3)$. One practical wrinkle at our $Q_0 = 1.65$ GeV: if charm is generated purely by evolution then $c^+(Q_0) = 0$, i.e. $T_{15}(Q_0) = \Sigma(Q_0)$ — one fewer function to parametrize. This basis — see evolution basis — is the one our parametrizations and the EKO evolution pipeline march in.
What electromagnetic $F_2$ can and cannot see
Now the payoff question. Express $F_2$ in the evolution basis using the inverse map above (three light flavours; a two-line exercise):
$$ 9\,F_2^p/x = 4u^+ + d^+ + s^+ = 2\Sigma + \tfrac{3}{2}T_3 + \tfrac{1}{2}T_8 $$
(check the coefficients: $u^+$ alone gives $\Sigma = T_3 = T_8 = 1 \Rightarrow 2 + \tfrac32 + \tfrac12 = 4$ ✓; $d^+$ gives $2 - \tfrac32 + \tfrac12 = 1$ ✓; $s^+$ gives $2 - 1 = 1$ ✓). Isospin flips the sign of $T_3$:
$$ 9\,F_2^n/x = 2\Sigma - \tfrac{3}{2}T_3 + \tfrac{1}{2}T_8, \qquad\qquad 9\,F_2^d/x = 2\Sigma + \tfrac{1}{2}T_8 \;\;(\text{per nucleon}) $$
So a proton–deuteron dataset measures exactly two linear combinations at each $x$: $T_3$ (from the $p$–$d$ difference) and $2\Sigma + \tfrac12 T_8$ (from the deuteron). Everything else is invisible at fixed $Q^2$: the photon never couples to $V$, $V_3$, $V_8$ (charge-squared weighting is blind to quark-vs-antiquark), cannot split $\Sigma$ from $T_8$ (i.e. cannot isolate strangeness), and sees the gluon not at all — only through the $Q^2$-evolution slopes and the momentum sum rule.
This is not an academic worry; it is the geometry of this lab's likelihood surface. Our fits to SLAC + BCDMS $F_2^{p,d}$ (648 points, NNLO, $Q_0 = 1.65$ GeV, MSHT20-style parametrization fitted Bayesianly — PPDF-10 onward) exhibit exactly the predicted flat directions: strangeness and sea-decomposition parameters ride their priors, and the posterior stretches along the combinations Derivation 5 says are unmeasured — documented in PPDF-4 and, for the MSHT20 basis, PPDF-15. The cures are other probes with other weightings: charged-current neutrino DIS, whose $xF_3 \approx x(u_v + d_v + \cdots)$ (on an isoscalar target) measures the valence sector directly; dimuon production off strange quarks; $W^\pm$ charge asymmetries at hadron colliders, which key on $u\bar d$ vs $d\bar u$; and Drell–Yan $pd/pp$ ratios, which map $\bar d/\bar u$. A global fit is precisely the enterprise of stacking enough distinct projections to invert the whole flavour matrix.
No experiment measures "the strange PDF". Each observable is one fixed linear (or, for hadron colliders, quadratic) functional of the PDFs, with weights set by the probe — $e_i^2$ for photons, weak couplings for $W/Z$. A dataset determines the components of the PDF vector along its own weight directions and says nothing about the orthogonal complement, no matter how many points or how small the errors. That is why posterior widths in a Bayesian fit can be data-dominated in one direction and pure prior in another at the same $x$ — and why adding a kind of data beats adding statistics. Keep this picture; it is the single most useful lens on every results card in this lab.
Toy set at $x = 0.3$ (values chosen near global-fit magnitudes at $Q^2 \sim 10\ \text{GeV}^2$): $xu_v = 0.55$, $xd_v = 0.22$, $x\bar u = x\bar d = 0.018$, $xs = x\bar s = 0.009$.
Step 1 — flavour sums: $x(u + \bar u) = 0.55 + 2(0.018) = 0.586$; $x(d + \bar d) = 0.22 + 0.036 = 0.256$; $x(s + \bar s) = 0.018$.
Step 2 — structure functions:
$$ F_2^p = \tfrac{1}{9}\big[4(0.586) + 0.256 + 0.018\big] = \tfrac{2.618}{9} = 0.291, \qquad F_2^n = \tfrac{1}{9}\big[4(0.256) + 0.586 + 0.018\big] = \tfrac{1.628}{9} = 0.181 $$
$$ F_2^d = \tfrac{1}{2}(0.291 + 0.181) = 0.236 \qquad\Longrightarrow\qquad F_2^p/F_2^d = 1.23, \quad F_2^n/F_2^p = 0.62 $$
Step 3 — invert, as a fit would: from "measured" $F_2^p$ and $F_2^d$, reconstruct $F_2^n = 2F_2^d - F_2^p = 0.181$ and then $x\,T_3 = 3(F_2^p - F_2^n) = 3(0.110) = 0.330$ — which indeed equals the input $x(u^+ - d^+) = 0.586 - 0.256 = 0.330$ ✓. The $u$–$d$ difference is recovered exactly; the strange fraction, by Derivation 5, could have been anything.
Model: $\bar d(x) - \bar u(x) = A\,(1 - x)^7$ — a soft, sea-like shape with one normalization.
Step 1 — the integral: $\int_0^1 A(1-x)^7\,dx = A/8$.
Step 2 — match NMC: $S_G = \tfrac13 - \tfrac23\cdot\tfrac{A}{8} = 0.235$ requires $A = 12\,(\tfrac13 - 0.235) = 1.18$, i.e. $\int(\bar d - \bar u)\,dx = 0.148$ — the NMC central value by construction.
Step 3 — predict the shape and test it elsewhere: the momentum-weighted asymmetry $x(\bar d - \bar u) = 1.18\,x(1-x)^7$ peaks where $\tfrac{d}{dx}[x(1-x)^7] = 0 \Rightarrow x = \tfrac18 = 0.125$, with peak value $1.18 \times 0.125 \times (0.875)^7 = 0.058$. Compare E866 Drell–Yan: peak $x(\bar d - \bar u) \approx 0.06$ near $x \approx 0.1\text{–}0.2$ — a one-parameter toy, fed only the NMC integral, lands on the independent Drell–Yan shape within $\sim\!10\%$. This is the sum-rule game at its best: an integral constraint from one process, a differential test in another.