Momentum conservation in the one-loop contribution of the photon propagator The lowest contribution to the photon self-energy is represented by the following diagram (Taken from F.Schwabl, Advanced quantum mechanics, p.365)::
($k$ is the momentum of the photon that decays in the positron-electron pair.)
My question concerns the labelling of the momentum: why the momentum of the lower fermion is not $k-q$ if we demand momentum conservation at each vertex?
 A: Implicitly, the image you posted is assuming that momentum flow lines up with the arrows which you usually take to distinguish "particle" and "antiparticle":

The small arrows outside the diagram indicate momentum flow and in the book they are omitted since they line up with the arrows that already make up the diagram.
Aparently, the situation you are considering in your mind is one where momentum flows from left to right. Denoting then by $q$ the momentum of the "antiparticle",  one should now draw the following diagram:

Notice the chosen direction of the momentum flow in the diagram. Since the momentum of the "particle" is chosen to flow from left to right, the "particle" has momentum $k-q$.

A third way to draw it would be to consider my first image but reversing the upper arrow, so that if $q$ flows to the left, $-q$ is what flows to the right, and one draws:

By renaming, $q\rightarrow -q$, one recovers the second drawing. The arbitrary momentum $q$ is to be integrated over. By being careful with propagators one obtains the same result for the integral independently of the convention.
(Incidentally, one should keep in mind that Feynman diagrams are a useful tool but one shouldn't ascribe too much meaning to the lines that make it up: the photon cannot be said to "decay" into a positron-electron pair.)
