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 $$\gamma^0S_F^\dagger(-x)\gamma^0=S_F(x)$$

This post imported from StackExchange Physics at 2014-04-13 11:29 (UCT), posted by SE-user Photon