# Advanced Dirac field propagator (spacelike separation)

1. May 11, 2015

### VintageGuy

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 $(x-y)^2<0$).

2. Relevant equations

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

where $\omega_{0_{\pm}} = \pm \sqrt{\vec{p}^2 + m^2}$ are the poles of the beforehand calculated energy integral, and $p_{\pm}=(\omega_{0_{\pm}},\vec{p})$ 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 $(\not{p}+m)$ 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 $x^0 = y^0$. 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:
$$S^{-1}(\lambda)\gamma ^{\mu} S(\lambda) = \lambda ^{\mu}_{\,\,\nu} \gamma^{\nu}$$

But I doubt this implies that $\not{p'}=\not{p}$, where $p'$ 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. May 16, 2015