# Why does the chances of particular Feynman diagram occurring reduces by 1% at each photon-electron interaction?

I saw a youtube video regarding Quantum Electrodynamics which explained how one can eliminate the Feynman diagrams with complex photon-electron interactions or loops. The guy explained that each time a photon interacts with the electron the chances of that Feynman diagram reduces by 1%. Thus, the diagram involving least photon-electron interaction (or the "vertices" as he called them) is more likely to happen in an electron scattering event. My question is that - Is there any possible reason for this probability reducing?

• Feynman diagrams do not "happen". They represent mathematical terms in an expansion, not distinct physical processes. – fqq Apr 19 at 15:44
• What do those mathematical terms represent? – Ajinkya Naik Apr 19 at 16:17
• They represent contributions to the complex-valued probability amplitude that the initial state in the diagram transitions into the final state in the diagram. What happens “in between” cannot be described classically in terms of particle trajectories. – G. Smith Apr 19 at 17:41

$$\alpha=\frac{1}{4\pi\epsilon_0}\frac{e^2}{\hbar c}\approx\frac{1}{137}$$
Another way say it is that it is due the fact that the coupling constant between the electron-positron field and the photon field is $$e$$. In quantum field theory, the electric charge of any particle is a measure of how strongly its field couples to the electromagnetic (i.e., photon) field. Each vertex in a diagram contributes a factor of $$e$$ to the probability amplitude represented by that diagram, and thus a factor of $$e^2$$ to the probability itself.