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So the Feynman propagator for a Klein Gordon is manifestly Lorentz invariant clearly by looking at the momentum space representation written in terms of Lorentz scalars.

But in the case of the Dirac/ fermion propagator one can express the denominator as a Lorentz invariant, yet the numerator appears Lorentz covariant because of the $\gamma \cdot p$ term. Thus is this indicative of a Lorentz covariance ie under Lorentz transformation there is a change in helicity/ polarisation for Dirac spinors?

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The Dirac propagator $\Delta_{ab}$ is defined to be the solution of the following differential equation

$$ \left(\gamma. \partial + m \right)^{\phantom{a}c}_{a} \Delta_{cb}(x-y) = \delta_{b}^{a} \delta^{(4)}(x-y) $$

subjected to Feynman boundary conditions. The $a,b$ and $c$ are four-dimensional spinor indices.

This $\Delta_{ab}$ is the two-point function of the free theory

$$ \langle \bar{\psi}_{a} (x) \psi_{b} (y) \rangle_{\text{free}} = \Delta_{ab} (x - y) $$

so you are right: due to the two-point function interpretation, you must have Lorentz covariance.

Compare this with a scalar field:

$$ \langle \phi (x) \phi(y) \rangle_{\text{free}} = \Delta(x - y). $$

You can see manifest Lorentz invariance since $\phi$ is a scalar.

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  • $\begingroup$ How do we reconcile the fact that it is a lorentz covariance quantity? does it physically mean that when I measure the propagation of a fermion in two coordinate frames, spin/helicity/something else changes according to two observers? $\endgroup$ – MKF Sep 13 '18 at 10:06
  • $\begingroup$ For instance, send $p\rightarrow \Lambda p$, then $\Delta_{fermion} \propto \gamma\cdot (\Lambda^{-1} p) + m$, with the denominator unchanged (in Minkowski spacetime/local inertial frame). So inherently the different coordinate system changed the observed helicity? $\endgroup$ – MKF Sep 13 '18 at 10:09
  • $\begingroup$ How do you reconcile what with what? "does it physically mean that when I measure the propagation of a fermion in two coordinate frames, spin/helicity/something else changes according to two observers?" Yes. If you are worried about this, think a little bit about what the propagator means. $\endgroup$ – OkThen Sep 13 '18 at 18:19
  • $\begingroup$ My intuition tells me because spinors are eigenstates of spin and thus helicity, if I transform the fields, I automatically attain a different projection of helicity along the propagation direction $\endgroup$ – MKF Sep 13 '18 at 18:47
  • $\begingroup$ Only massless spinors are eigenstates of helicity. For massive spinors helicity does not have to be conserved. Now, for massless spinors, take the massless propagator and boost it, then show that the helicity projection remains the same. $\endgroup$ – OkThen Sep 13 '18 at 19:41

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