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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?

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  • $\begingroup$ Just wanted to say, I am not a degree level student. So I would appreciate if someone could simplify things a bit. $\endgroup$
    – user63248
    Oct 31, 2014 at 16:56
  • $\begingroup$ Simply put, some confusion seems to arise from your statement that "a neutron can change into a proton when it interacts with a neutrino [...] because $W^-$ bosons are exchanged". Why would having 2 interacting particles in the initial state (neutron and neutrino) demand the exchange of a boson? $\endgroup$
    – J-T
    Oct 31, 2014 at 18:19
  • $\begingroup$ That assumption was made because of my book saying: "A neturino can interact with a neutron and make it change into a proton. An electron is created and emitted as a result of the change". $\endgroup$
    – user63248
    Oct 31, 2014 at 19:09
  • $\begingroup$ How would that alone imply that a boson is exchanged? $\endgroup$
    – J-T
    Oct 31, 2014 at 19:15
  • $\begingroup$ Because it continues with "These interactions are due to the exchange of particles referred to as W bosons." $\endgroup$
    – user63248
    Oct 31, 2014 at 20:47

3 Answers 3

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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: 4-point contact interaction 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:

  • For low enough energies, the 4-point interaction description is accurate and still useful;
  • In the Standard Model, the W boson couples to the quarks out of which the proton and neutron are made of.
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  • $\begingroup$ Very detailed response, beacuse of the level of Physics we are doing I can't understand fully what you said. It's interesting to see though two different way to represent these reactions depending on the level of energy involved. $\endgroup$
    – user63248
    Oct 31, 2014 at 16:55
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    $\begingroup$ @daljit97 It is important to realise that if you consider the 'true' UV complete theory, these 4-point diagrams are just an effective, low energy representation of the reaction. In this effective representation one has 'integrated out' the presence of the W bosons. These bosons are nonetheless part of our theory (and are, as mentioned, indispensable for computations at high energy). $\endgroup$
    – J-T
    Oct 31, 2014 at 17:43
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Here are the neutron decay Feynman diagram :

neutron decay

A free neutron will decay by emitting a W-, which produces an electron and an antineutrino.

and the diagram for neutrino neutron scattering :

neutrino neutron scattering

This interaction is the same as the one at top since a W+ going right to left is equivalent to a W- going left to right.

In the quantum mechanical framework, if a state is at a higher energy level than a lower energy level state that is allowed by quantum number conservations, then a decay will occur.

The neutron has a higher mass, ~939MeV than the proton's ~938, and charge and lepton number can be conserved with the decay shown, therefore it happens with a calculable and measurable probability. These are the dominant diagrams for these reactions , that give measurable probabilities for the reactions to go.

In the interaction case energy has to be supplied in order to turn a neutron into a proton, (and not wait for nature to have its way by the lifetime decay) and it is supplied by the incoming neutrino. In both cases, the W which has a mass about 100GeV is not on its mass shell. It is a virtual exchange carrying the quantum numbers of the W but the four vector exchanged does not have the mass of the W.

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The boson serves as an "interaction propagator" particle. Namely it is short lived and serves as the "carrier" of the weak force that governs the beta decay. Similar cases are in the strong interactions where qluons play this role or in electromagnetism, where the electromagetic force id mediated by photons. You dont have to study quantum field theory to gain an insight in these matters. Instead a good reading on Yukawa's theory and virtual particles would be much helpful

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