Understanding attractive/repulsive boson-mediated forces based on interference of free-space fields

Question

I just worked through the derivations of the Yukawa interaction for scalar and spin one particles (i.e. Peskin and Schroeder, end of chapter 4, which covers the tree-level Feynman diagram). It's very satisfying to see that the sign of the interaction is uniquely determined, but I don't feel like doing this calculation gave me much physical insight into why even-spin forces are only attractive, but odd-spin forces like QED can be attractive or repulsive. There are so many minus signs to keep track of, but one of them gives us a very fundamental fact about nature.

A nice piece of intuition from classical electrodynamics is that the attractive/repulsive nature of the force comes down to how electric fields interfere constructively or destructively in space. As particles are brought closer together, the energy of the surrounding field changes, giving rise to an attractive or repulsive potential depending on the relative sign of the charges. This gives a nice visual explanation for why classical electromagnetism can yield repulsion or attraction, by thinking about how vector fields interfere in free space. See note below.

My question is, can we come up with a similar simple picture when thinking of force-carrying bosons? Can one easily see how the spin of the force-carrying boson gives different results depending on whether the spin is even or odd? I suppose this question is closely related to how we take the classical limit of the free-space field.

To clarify the second paragraph, the fields generated by two positively charged particles predominantly constructively interfere throughout space. This means that the energy stored in the electromagnetic field increases as the particles are brought together, explaining the repulsive interaction. The opposite is true for a positively and a negatively charged particle brought together.

Answer 1

The crucial difference is that in the Yukawa interaction we have no $\eta_{\mu\nu}$ in the scalar propagator $\frac{\operatorname{i}}{p^2-m^2+\operatorname{i}\epsilon}$ and the vertex. We therefore have always an attractive potential (since the potential is the Fourier transform).

Answer 2

By the CPT theorem, antiparticles can be thought of as normal particles travelling backwards in time. A scalar mediator doesn't see the difference between the two because it doesn't have any way to detect directions in spacetime, so it will always treat particles/antiparticles the same. On the other hand, a vector mediator can see the difference, and so it does treat them differently. Let's see this in slightly more detail.

For a scalar mediator, the amplitude for an electron/positron scattering is just proportional to $i\mathcal{M} \propto (\bar{u} u )(\bar{v} v)$. These are both positive, and there's nothing here that explicitly depends on the "direction in spacetime" the particles are moving.

On the other hand, because the propagator if a vector boson has a factor of $\eta_{\mu\nu}$ in it, and the vertices each have a $\gamma^\mu$, you can write the tree level amplitude for scattering between an electron and a positron as being proportional to $i\mathcal{M} \propto (J_-)_\mu (J_+)^\mu$, where each $J_\mu$ is the electric current for the electron/positron, respectively. If you think of $ J_-^\mu \sim -e u^\mu$, then you could heuristically either think of $J_+^\mu$ as coming from the shift $ e \to -e$ or $u^\mu \to -u^\mu$. If you take the second perspective, the different treatment of particles/antiparticles becomes more clear: although $J_\pm^\mu$ are both timelike, their dot product can still be any sign because they could be aligned or anti-aligned. This difference in sign is what leads to a repulsive/attractive force between them.