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Advanced Dirac field propagator (spacelike separation)

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data

    I'm supposed to calculate the advanced propagator for the Dirac field, and I have no problem with that. Then I'm supposed to show it vanishes for spacelike separation (that is [itex](x-y)^2<0[/itex]).

    2. Relevant equations

    For the advanced propagator I get something like:
    [tex]
    S_A = \frac{i}{(2 \pi)^3}\int \theta (x^0 - y^0) [\frac{d^3p}{2 \omega_{0_+}}e^{-ip_+ (x-y)}(\not{p}_+ + m) + \frac{d^3p}{2 \omega_{0_-}}e^{-ip_- (x-y)}(\not{p}_- + m)] [/tex]

    where [itex]\omega_{0_{\pm}} = \pm \sqrt{\vec{p}^2 + m^2}[/itex] are the poles of the beforehand calculated energy integral, and [itex]p_{\pm}=(\omega_{0_{\pm}},\vec{p})[/itex] is the 4-impulse.

    3. The attempt at a solution

    Now, I know propagators are supposed to be Lorentz invariant, and for that purpose I have written the integral this way. But the term [itex](\not{p}+m)[/itex] keeps bugging me. I can prove that the integral is 0 if the entire thing is Lorentz invariant, that is to say, if I am allowed to stand in the system for which [itex]x^0 = y^0[/itex]. I know when checking for the invariance of Dirac equation one comes up with the condition for the unitary operators that operate in bispinor representation space that looks something like:
    [tex]S^{-1}(\lambda)\gamma ^{\mu} S(\lambda) = \lambda ^{\mu}_{\,\,\nu} \gamma^{\nu}[/tex]

    But I doubt this implies that [itex]\not{p'}=\not{p}[/itex], where [itex]p'[/itex] is the transformed 4-impulse... What am I doing wrong, how can I see the integral doesn't change when switching to another coordinate system?
     
  2. jcsd
  3. May 16, 2015 #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?
     
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