# How to derive operator form of the parton distribution function

A similar question is found here in Stackexchange a year ago without any response.

I am following the formulation of the parton densities from the handbag diagram in Collins Handbook of Perturbative QCD, Chapter 6. I know that the quark density can be written as (left as an exercise 6.2)

$f_{j/h}(\xi) = \int\frac{dw^-}{2\pi}e^{-i\xi P^+w^-} <P|\bar{\psi}_j(0, w^-,\textbf{0}_T)\frac{\gamma^+}{2}\psi_j(0)|P>$ (6.31).

I am having difficulty figuring out where $\gamma^+$ comes from.

Also in Chapter 7, 7.5.4 gauge-invariant quark density in QCD, they make make the (bare) quark density function gauge invariant by inserting a Wilson line in between the two fermion fields:

$f_{(0)f/h}(\xi)=\int\frac{dw^-}{2\pi}e^{-i\xi P^+w^-} <P|\bar{\psi}_j(0, w^-,\textbf{0}_T)\frac{\gamma^+}{2}W(w^-,0)\psi_j(0)|P>$ (7.40).

Where exactly does the Wilson Line come from? I feel uncomfortable just saying that the Wilson Line is put there by hand to make that matrix element gauge-invariant, and would like to see a derivation.

• Actually, the full formula comes with a $\gamma^0$ but in the consideration of the higher twist suppressed terms, one can sudakov decompose the gamma matrix and pick out the subleading term $\sim \gamma^+$. I can elaborate later. – CAF Aug 7 '18 at 20:07

From a physically motivated perspective within the framework of collinear factorisation, Schwartz finds $$f(\xi) = \int \frac{\text{d}\lambda}{2 \pi} e^{- \mathrm{i} \lambda \xi n \cdot P} \langle P | \bar{\psi}(\lambda n) \gamma^0 \psi(0) | P\rangle$$
One can write a generic $4$ dimensional vector in a Sudakov basis spanned by vectors e.g $n \sim (\Lambda,0,0,\Lambda)$ and $\bar n \sim (\Lambda,0,0,-\Lambda)$ such that $$A^{\mu} = (n \cdot A) \bar n^{\mu} + (\bar n \cdot A) n^{\mu} + A^{\mu}_{\perp}.$$ Setting $\Lambda$ as unity we make the identification of the Sudakov basis with the light cone basis as used in Collins’ formulation. That is to say, $A^+ \equiv (n \cdot A)$ and $A^- \equiv (\bar n \cdot A)$ so that $$A^{\mu} = A^+ \bar n^{\mu} + A^- n^{\mu} + A^{\mu}_{\perp}.$$
In an analogous fashion, one can do the same with a gamma matrix (i.e send $A \rightarrow \gamma$) and with an appropriate normalisation of the Sudakov vectors may identify $$\gamma^0 = \frac{1}{\sqrt{2}} \left(\gamma^+ + \gamma^-\right).$$ This of course also follows from the canonical light cone coordinate definitions: $$\begin{cases} \gamma^+ = \frac{1}{\sqrt{2}} \left(\gamma^0 + \gamma^3 \right) \\ \gamma^- = \frac{1}{\sqrt{2}} \left(\gamma^0 - \gamma^3\right) \end{cases}$$
As per Schwartz, if the proton is going mostly in the $\bar n$ direction, then the projection of $\psi$ in the $n$ direction is suppressed (contributing at higher twist outwith the famous hard and soft sector factorisation). This all means to say $(\bar n \cdot \gamma) \psi = \gamma^- \psi \approx 0$ and we obtain $$\gamma^0 \psi \approx \gamma^+ \psi$$