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

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


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.

  • $\begingroup$ Which book? Which page? $\endgroup$ – Qmechanic Jul 31 '18 at 0:19
  • 2
    $\begingroup$ It means the inverse power of a derivative, i.e. an antiderivative. $\endgroup$ – Qmechanic Jul 31 '18 at 0:23
  • 2
    $\begingroup$ In general, the function of a differential operator is defined using the Fourier transform: $f(\partial)g(x)\equiv F^{-1}[f(p)F[g](p)](x)$. $\endgroup$ – AccidentalFourierTransform Jul 31 '18 at 0:24
  • $\begingroup$ @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 $A^+=0$ $\endgroup$ – Silas K Grossberndt Jul 31 '18 at 0:29
  • $\begingroup$ Is this notation conventional? I haven't ever seen it before $\endgroup$ – 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 $(\partial_+)^{-1} \sim - i/p_-$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.