1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz Transformation in Bjorken & Drell QFT

  1. Nov 15, 2014 #1
    1. The problem statement, all variables and given/known data

    I'm trying to derive (14.25) in B&J QFT. This is

    ##U(\epsilon)A^\mu(x)U^{-1}(\epsilon) = A^\mu(x') - \epsilon^{\mu\nu}A_\nu(x') + \frac{\partial \lambda(x',\epsilon)}{\partial x'_\mu}##, where ##\lambda(x',\epsilon)## is an operator gauge function.

    This is all being done in the radiation gauge, i.e. ##A_0 = 0## and ##\partial_i A^i=0##, with ##i \in {1,2,3}##.

    ##\epsilon## is an infinitesimal parameter of a Lorentz transformation ##\Lambda##.

    2. Relevant equations


    3. The attempt at a solution

    ##\epsilon## is an infinitesimal parameter of a Lorentz transformation ##\Lambda##.

    Under this transformation, ##A^\mu(x) \rightarrow A'^\mu(x')=U(\epsilon)A^\mu(x)U^{-1}(\epsilon)##.

    The unitary operator ##U## which generates the infinitesimal Lorentz transformation

    ##x^{\mu} \rightarrow x'^{\mu} = x^{\mu} + \epsilon^{\mu}_{\nu}x^{\nu}## is

    ##U(\epsilon)=1 - \frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}##

    where ##M## are the generators of Lorentz transformations. (I guess really I should have ##M^{\mu\nu}=(M^{\rho\sigma})^{\mu\nu}##. M is a hermitian operator, so

    ##U^{-1}(\epsilon)=1 + \frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}##

    Now I tried writing out ##U(\epsilon)A^\mu(x)U^{-1}(\epsilon)## explicitly but it didn't really get me anywhere. The answer is supposed to have ##x'## as the argument of ##A^\mu## on the RHS but I only get ##x##. I'm not sure how to Lorentz transform the function and the argument at the same time.



    Underneath the formula in B&J it says the gauge term is necessary because ##UA_0U^{-1}=0## since ##A_0=0##. I don't see why this warrants the need of a gauge term.

    Edit: Oh wait, it's needed because otherwise there will be no conjugate momenta for the ##A_0##. Okay I get that, but still don't understand where the initial formula comes from.
     
    Last edited: Nov 15, 2014
  2. jcsd
  3. Nov 20, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Dec 20, 2014 #3
    bump
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Lorentz Transformation in Bjorken & Drell QFT
  1. Lorentz transformation (Replies: 1)

  2. Lorentz Transformations (Replies: 29)

Loading...