Sign of momentum in fermion propagator Thinking of a process like Compton scattering, where we have an electron as a propagator, I would typically write down the propagator as
$$i \frac{\not q+m}{q^2-m^2}.$$
If I were to replace the electron with a positron, does the propagator become:
$$i\frac{-\not q+m}{q^2-m^2}~?$$
Sort of a silly question, but I'm finding it hard to narrow down the answer with different sign conventions out there.
 A: For reference, the fermion propagator is
$$ \left\langle 0 \right| T\psi(x)\overline\psi(y) \left|0\right \rangle= S(x-y) = \int \frac{d^4k}{(2\pi)^4} \frac{i}{\not k-m}e^{-ik\cdot(x-y)}$$
Depending on the time ordering, this describes a particle moving from $y$ to $x$, or an antiparticle moving from $x$ to $y$.

Now, consider a one-loop diagram in which an external photon decays into a particle and an antiparticle at point $y$, which subsequently annihilate to produce a photon at $x$. Let the momentum appearing in the propagators be $p$ for the photon, $l$ for the particle, and $k$ for the antiparticle.
This diagram has an integral over the positions of the vertices $x,y$; the particle's propagator $S(x-y)$, and the antiparticle's propagator $S(y-x)$; and an exponent $e^{-ip\cdot(y-x)}$ coming from the photon wavefunction. Ignoring vertex factors and polarization vectors, this diagram is
$$ \int d^4x \int d^4y \int \frac{d^4l}{(2\pi)^4} \frac{i}{\not l-m}e^{-il\cdot(x-y)}\int \frac{d^4k}{(2\pi)^4} \frac{i}{\not k-m}e^{-ik\cdot(y-x)}e^{-ip\cdot(y-x)} $$
Performing the position integrals gives us two identical delta functions $ \delta^4(l-k-p)$, which tell us to set $k=l-p$ in the calculation and drop the $k$ integral, leaving
$$ \int \frac{d^4l}{(2\pi)^4} \frac{i}{\not l-m}\frac{i}{(\not l-\not p)-m} $$

In fact wherever there is a vertex, the lines going into the point $x$ contribute a $e^{-ip_1\cdot x -...}$, and the lines going out contribute a $e^{+ik_1\cdot x+...}$, giving us a delta function $\delta^4(p_1+...-k_1...)$. 
Ingoing particles have lines that point into the vertex, and outgoing particles have lines that point out of the vertex, so if a particle $p_1$ scatters off of an ingoing photon $p_2$, then we have a delta function $\delta^4(p_1+p_2-k_1)$, the usual momentum conservation relation that we expect: $k_1=p_1+p_2$.
But ingoing antiparticles have lines that point away from the vertex, and outgoing antiparticles have lines that point into the vertex, so they always contribute minus their momentum. The same process as above gives a delta function $\delta^4(-p_1+p_2+k_1)$, giving $-k_1=-p_1+p_2$. Let us rewrite this,
\begin{align}
&q_1=-p_1, \quad q_2=-k_1 \\&
q_2=q_1+p_2
\end{align}
Now it resembles the normal momentum conservation equation, and we can relate back to the propagator momenta with the above relations.
A: If q is the momentum for positron, then the propagator for it is still $i\frac{\not{q}+m}{\not{q}^2 - m^2 + i \epsilon}$.
