In Schwartz's book, 'Quantum Field Theory and the Standard Model' P.$696$, he starts to derive an expression for a parton distribution function in terms of matrix elements evaluated on the lightcone. Most of the derivation is clear to me, except a couple of things at the start and midway. The derivation begins by saying that the probability for finding a quark within a proton with a given momentum fraction (=PDF) is given by $$f(\zeta) = \sum_X \int \text{d}\Pi_X | \langle X|\psi|P \rangle |^2 \delta( \zeta n \cdot P - n \cdot p),$$ with $P = p + p_X$. Why is this the correct mathematical representation for a PDF? I understand the delta function constrains the parton to take a fraction $\zeta$ of the proton's momentum in the $n$ direction and that probabilities are squared matrix elements but what does the sum over $X$ mean here (we have no scattering taking place) and what is the meaning of $\langle X | \psi | P \rangle$?

The other thing was he states that since the proton moves mostly in the $\bar n = (1,0,0,-1) $ direction, $\not \bar n \psi \approx 0$. Why does this approximation hold?


The amplitude to pluck out a parton at spacetime location $0$, say, from the lightcone directed proton $P$ momentum is $\langle X | \psi(0) | P \rangle. $ The probability to pick out a parton with a particular momentum fraction $\xi$ of the mother hadron is dependent on the resolution scale with which the hadron is probed. Typically, in DIS, this probe is a high virtuality photon and for large $\gamma^* p$ c.m energies (small Bjorken $x$ or the Regge limit) the gluon density is dominant. Regardless, at a particular scale, we have to sum over all permissible constituents and their phase space configurations inside the proton.

In the premise of collinear factorisation, the chosen parton is aligned with some arbitrary $n$ direction (in accordance with the direction $n$ of the mother hadron, which we may as well realign with the plus lightcone direction) up to corrections from contributions in the transverse and opposite directions pertaining $\sim 1/Q^2$ corrections so suppresssed at higher twist. The delta function $\delta(\xi n \cdot P - n \cdot p)$ facilitates this in the explicit summation. Putting all this together yields the desired equality.

Analogously to how one can make a Sudakov decomposition of any $d$ dimensional vector $k^{\mu}$, we may do the same for a gamma matrix such that $\gamma^{\mu} = (n \cdot \gamma) \bar n^{\mu} + (\bar n \cdot \gamma) n^{\mu} + \gamma^{\mu}_{\perp}$ with the usual relations amongst the chosen basis vectors. Now, for the proton going mainly in the $ \bar n$ direction, say, the dominant contribution in the matrix element is due to the coefficient $n \cdot \gamma$ and so the projection of $\psi$ in the direction $n$ is approximately zero up to subleading terms. This is to say, equivalently, that $(\bar n \cdot \gamma) \psi \approx 0$.


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