Dirac operator Feynman propagator Is it true that the following identity holds for the Feynman prescription Dirac propagator:
$$ S_F(x) \stackrel{?}{=} \gamma^0[S_F(-x)]^\dagger\gamma^0 $$ 
where $S_F$ is defined as the Green's function:
$$ (i\gamma^\mu\partial_\mu-m)S_F(x-y)=i\delta(x-y) $$
This is somewhat related to a previous question of mine: Green's function for adjoint Dirac Equation
If the statement is true, how do you prove it?
 A: I guess, it's true, although I never heard of it. Let's denote $S_F^{-1}(x)=(i\gamma_\mu\partial^\mu-m)(x)$ First let's compute $[S_F^{-1}(−x)]^\dagger$. Assuming that $(\partial_\mu)^\dagger=\partial_\mu$ we get
$$[S_F^{-1}(−x)]^\dagger=(-i\gamma_\mu^\dagger\partial^\mu-m)(-x)=(i\gamma_\mu^\dagger\partial^\mu-m)(x)$$
Using $\gamma^0\gamma^0=1$ and $\gamma^0\gamma_\mu\gamma^0=\gamma_\mu^\dagger$ we get
$$\gamma^0[S_F^{-1}(−x)]^\dagger\gamma^0=(i\gamma^0\gamma_\mu^\dagger\gamma^0\partial^\mu-m\gamma^0\gamma^0)(x)=(i\gamma_\mu\partial^\mu-m)(x)=S_F^{-1}(x)$$
But how do we prove, that it also holds for $S_F(x)$?
Let's apply this transformation on the defining equation $S_F^{-1}(x)S_F(x)=\delta(x)$. Then we get:
$$\gamma^0\left(S_F^{-1}(-x)S_F(-x)\right)^\dagger\gamma^0=\gamma^0\delta(-x)^\dagger\gamma^0$$
The l.h.s. gives: $$\gamma^0\left(S_F^{-1}(-x)S_F(-x)\right)^\dagger\gamma^0=\gamma^0S_F^\dagger(-x)[\overleftarrow{S_F^{-1}}(-x)]^\dagger\gamma^0=\gamma^0S_F^\dagger(-x)\gamma^0\gamma^0[\overleftarrow{S_F^{-1}}(-x)]^\dagger\gamma^0=\gamma^0S_F^\dagger(-x)\gamma^0\overleftarrow{S_F^{-1}}(x)$$
where in the last step we used the relation for $S_F^{-1}$ derived in the beginning.
The r.h.s. simply gives $\gamma^0\delta(-x)^\dagger\gamma^0=\delta(x)$ since the delta function is symmetric. And yeah, we need to assume that it is self-adjoint.
From this we see that $$\gamma^0S_F^\dagger(-x)\gamma^0\overleftarrow{S_F^{-1}}(x)=\delta(x)$$ and therefore $\gamma^0S_F^\dagger(-x)\gamma^0$ is the Green's function of $S_F^{-1}$ which is unique. Therefore it must hold $$\gamma^0S_F^\dagger(-x)\gamma^0=S_F(x)$$
A: It looks correct: $$S_F (x)=\intop\frac{\text d ^4 p}{(2\pi)^4}e^{-ipx}\dfrac{ \not p +m}{p^2-m^2+i\varepsilon},\\ S_F ^\dagger(-x)=\intop\frac{\text d ^4 p}{(2\pi)^4}e^{-ipx}\dfrac{\not p^\dagger +m}{p^2-m^2+i\varepsilon}=\gamma^0 S_F(x)\gamma ^0.$$ The last equality follows from $\gamma^0\gamma^\mu\gamma^0=(\gamma^\mu)^\dagger$
