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I am starting to learn about Feynman diagrams and I have been told that flavour has to be conserved at vertices in the case that the exchange particle was neutral. Since here the exchange particle is a charged boson, my guess is that this diagram is possible. Is it true?

enter image description here

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I don't see why you "guessed" this. It is impossible by inspection.

Go to your SM Lagrangian and observe that the relevant charged current coupling, the coupling of leptons to the Ws is, broadly, of the form $$ \propto W^+_\kappa \bar {\nu}_e \gamma^\kappa (1-\gamma_5) e +\mathrm{h.c.} ~, $$ which tells you the Feynman diagram vertex must be of the form "e goes in, and W- and $\nu_e$ go out".

The wrong fictitious superposition of neutrino mass states you are writing, $\nu_\mu$ , is the one coupling to μs in such vertices; in fact, that is how it (neutrino flavor) is defined!

The convenience superposition of neutrino mass eigenstates coupling to electrons is defined to be $\nu_e$.

(Reviews, such as the PDG (10.4) skip the neutrino flavor label, since it is self-explanatory, and might unnecessarily enhance the illusion these states are real particles.)

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  • $\begingroup$ I am starting to learn about Feynman diagrams so I can't understanding this with the level I have. Is there a way to simplify why it is forbidden without having to go into current couplings? $\endgroup$
    – Luismi98
    Dec 22, 2018 at 14:48
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    $\begingroup$ It is a "just so" statement: Whenever you write a vertex with a W and an electron, the neutrino coming out must be a $\nu_e$; alternatively, $\nu_e$ is defined as the neutrino occurring in this type of vertex. If you write a Feynman diagram, it corresponds to a given Lagrangian, in this case that of the SM, where you read such vertices off. $\endgroup$ Dec 22, 2018 at 14:53
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Even though this question is more than a year old, here is another explanation:

We know that at each vertex, the individual lepton numbers must be conserved, i.e. $L_e$, $L_{\mu}$ and $L_{\tau}$ are constants.

In the diagram you show, the lepton numbers - both for $L_{e}$ and $L_{\mu}$ - are not conserved. Hence, your diagram cannot be valid.

Does this make sense to you?

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The process $e^-+e^+\to \nu_{\mu}+\bar{\nu}_{\mu}$ is possible, but via a neutral Z-boson that decays into a truly neutral couple $ \nu_{\mu}+\bar{\nu}_{\mu}$.

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  • $\begingroup$ Why this one not allowed? $\endgroup$
    – Luismi98
    Dec 22, 2018 at 12:43
  • $\begingroup$ If it is allowed, then there must be a similar diagram with a $W^+$ boson. $\endgroup$ Dec 22, 2018 at 12:46
  • $\begingroup$ Yes there is, simply by changing reference frame. How is that related to it being forbidden? $\endgroup$
    – Luismi98
    Dec 22, 2018 at 12:47

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