Advanced Dirac field propagator (spacelike separation)

[VintageGuy]

Homework Statement

[/B]
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 $(x-y)^2<0$).

Homework 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.

The Attempt at a Solution

[/B]
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?

 

[mark726]

First of all, it is important to note that the integral you have written is not the full advanced propagator. It is only the part that corresponds to the positive energy solutions of the Dirac equation. The full advanced propagator also includes the contribution from the negative energy solutions.

To show that the advanced propagator vanishes for spacelike separation, you need to use the fact that the energy-momentum conservation law holds for all particles and fields in nature. This means that the 4-impulse of a particle or field can only change due to interactions with other particles or fields. In the absence of interactions, the 4-impulse of a particle or field remains constant.

Now, let us consider the integral you have written. The term (\not{p}+m) appears because it corresponds to the positive energy solutions of the Dirac equation. This means that it describes the propagation of a particle with positive energy and momentum. However, for spacelike separation, the 4-momentum of the particle is no longer conserved. This means that the particle can no longer have a definite energy and momentum, and the term (\not{p}+m) becomes meaningless.

To see this more clearly, let us consider the Lorentz transformation of the 4-impulse. Under a Lorentz transformation, the 4-impulse transforms as p'^{\mu} = \Lambda^{\mu}_{\,\,\nu} p^{\nu}, where \Lambda^{\mu}_{\,\,\nu} is the Lorentz transformation matrix. Now, for spacelike separation, the 4-momentum does not transform in a way that preserves its magnitude, i.e. p'^2 \neq p^2. This means that the term (\not{p}+m) does not transform in a Lorentz invariant manner, and hence cannot be part of the advanced propagator for spacelike separation.

To summarize, the fact that the advanced propagator vanishes for spacelike separation is a consequence of the energy-momentum conservation law and the non-Lorentz invariance of the term (\not{p}+m) for spacelike separation. This is not a problem, but rather a manifestation of the fundamental principles of nature.