# Notation: Dividing by a partial differential

## Setup

Hi, I am working on a DIS scattering problem in the Light Cone Gauge, and this has me needing to calculate the currents. In doing this, I have come across the following equation

$$t^a \partial_+ \left( \frac{-2}{\partial^+} \partial_{\perp}^j \cdot A_j^{\perp} +\frac{2g}{(\partial^+)^2} \left\{ \left[i \partial^+ A_{\perp}^j, A^j_{\perp}\right] +2 q^{\dagger}_+ t^b q_+ t^b \right\} \right)$$

## Question

My question is how does one interpret $\frac{2}{\partial^+}$?

Does it cancel?

## Further Context

I know this book is using Lapage and Brodsky's notation from their 1980 paper, but I could not find a good explanation for this.

• Which book? Which page? – Qmechanic Jul 31 '18 at 0:19
• It means the inverse power of a derivative, i.e. an antiderivative. – Qmechanic Jul 31 '18 at 0:23
• In general, the function of a differential operator is defined using the Fourier transform: $f(\partial)g(x)\equiv F^{-1}[f(p)F[g](p)](x)$. – AccidentalFourierTransform Jul 31 '18 at 0:24
• @Qmechanic it is from Quantum Chromodynamics at High Energies by Kovghegov and Levin page 9 equation 1.41 combined with equation 1.4 and choice of $A^+=0$ – Silas K Grossberndt Jul 31 '18 at 0:29
• Is this notation conventional? I haven't ever seen it before – Silas K Grossberndt Jul 31 '18 at 3:14

It's defined as the inverse operator by convolution, ie., the inverse in the Fourier space. Say for example that $\psi(x^+) = \int dp \, e^{i x^+ p_-}$, then $(\partial_+)\sim i p_-$ and $(\partial_+)^{-1} \sim - i/p_-$.