# Representing momentum exchange in Feynman diagrams

I'm a bit confused about how momentum exchange is represented in Feynman diagrams. I'm going through Tongs qft notes on interactions right now and he demonstrates how to draw Feynman diagrams with a few examples using the scalar Yukawa interaction. For nucleon-nucleon scattering the exchange particle is said to be a virtual meson. For the decay of a nucleon-antinucleon pair into two mesons the exchange particle is a virtual nucleon. For meson-meson scattering Tong states that the simplest way to represent the exchange of momentum is through a loop. I don't understand how he decides what particle(s) to use for momentum exchange. Can someone please point me in the right direction?

Each interaction is given by a term in the Lagrangian (density), which in the case of Yukawa interactions is $$\mathscr{L}_Y = g \bar{\psi}\psi \phi,$$ where $\psi$ and $\phi$ are the field operators of nucleons and mesons, each containing the corresponding creation/annihilation operators.
This term is what determines what are the possible interactions. It tells you that every vertex must join exactly two nucleons/antinucleons and one meson, so for two mesons to scatter you must use at least four such terms: at each vertex there can only be one meson $\phi$, so you need two vertices for the two incoming mesons to annihilate into nucleon/antinucleon pairs, and now you need another couple of vertices to turn the (virtual) nucleons/antinucleons back into a pair of outcoming mesons.
• That makes a lot of sense, thank you. Just to be sure that I'm clear on this, does that mean for $\phi^4$ interactions we will have four mesons at each vertex? Dec 11 '14 at 21:49