I am a beginner Physics student and I am studying the weak nuclear force and how particle interactions work. Now, from my book and the Feynman diagram, I learned that a neutron can change into a proton when it interacts with a neutrino, this happens because W$^-$ bosons are exchanged between $\nu$ and $n$; therefore $n$ is converted into a proton and $\nu$ is converted into a positron.
I am not sure if what I wrote above is correct. My real doubts come when talking about beta decay; I know that in beta decay, a neutron converts into a proton, but where does the W$^-$ boson come from? I mean the neutron doesn't interact with another particle, so why is the boson is even there?
Best Answer
At a first glance, W bosons are not needed at all in those scenarios: you could model the interaction by supposing a direct coupling exists between the 4 particles (neutron, proton, electron, neutrino) -- which is what was initially proposed by Fermi (see Fermi's 4-point Interaction).
Thus, both the $n+\nu \rightarrow p + e$ interaction and the decay $n\rightarrow p + e + \overline{\nu}$ can be diagrammatically represented by:
where time flows from left to right.
Problems, however, arise for this ansatz: as the center-of-mass energy (usually denoted $\sqrt{s}$) of the interacting neutron and neutrino in the left diagram rises, so does the cross-section $\sigma(s)$ computed in this context, in a nonsensical way (probabilities exceed 1).
One solution is to 'UV complete' the theory -- i.e. complete the theory by specifying behaviour in the ultraviolet/at high energies -- by abandoning the naïve contact interaction altogether and replacing it with the exchange of W bosons (this cannot happen just for the left diagram; one has changed the model). So, indeed, in this completed theory, even neutron decay is explained by exchange of a W boson.
Final comments: