- #1

VintageGuy

- 3

- 0

## Homework Statement

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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]).

## Homework 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.

## The Attempt at a Solution

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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?