The Klein Gordon propagator is given (I use Peskin and Schroeder's conventions, if it matters...),

$$\frac{ i }{ p ^2 - m ^2 + i \epsilon }$$ The photon propagator (using Feynman gauge) is $$\frac{ - i \eta^{\mu\nu}}{ k ^2 + i \epsilon }$$ The time-like component of the photon field propagates with a different sign then the scalar field and the spatial components propagate with the same sign.

Is there a physical significance to the difference in sign between the two or it is just a consequence of our conventions?

Yes, there is a physical significance. The longitudinal mode $A^0$ is pure gauge, it does not propagate (in other words, the equation of motion for $A^0$ is a constraint [Gauss Law], not an equation of motion and it's canonical momenta is identically 0 , meaning we cannot impose canonical commutation relations on it). Some of the spatial modes do propagate, so they should have the same sign propagator as the scalars. This is really the easiest way to remember the sign of the photon propagator. The wrong sign in the $A^0$ propagator is at the heart of many problems in QFT. It marks the conflict between the positive definite norm of quantum mechanics and the indefinite norm on Minkowski space that we are forced to deal with because we want a manifestly local Lorentz invariant formulation of the theory.
• Fascinating! Thanks for the answer. How come $A^0$ doesn't propagate? Is this always true or does it depend the gauge that you work in? Feb 24, 2014 at 1:30
• could you elaborate how exactly the sign difference implies the equation for $A^0$ is a constraint? Feb 24, 2014 at 1:40
• @Dan: Could you kindly answer my question in the comment? I'm aware that in a vector theory, one can derive(from lagrangian) that equation for $A^0$ contains no time derivative, one can also derive there is a sign change in the propagator. I've always treated these two facts as two separate consequences of the vector theory, so I'd like to know how can you derive one from the other. Feb 26, 2014 at 11:56