# Is this feynman diagram possible? $e^- + e^+ \to \nu_\mu + \bar{\nu}_\mu$

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? 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.)

• 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? – Luismi98 Dec 22 '18 at 14:48
• 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. – Cosmas Zachos Dec 22 '18 at 14:53

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 $$\nu_{\mu}+\bar{\nu}_{\mu}$$.

• Why this one not allowed? – Luismi98 Dec 22 '18 at 12:43
• If it is allowed, then there must be a similar diagram with a $W^+$ boson. – Vladimir Kalitvianski Dec 22 '18 at 12:46
• Yes there is, simply by changing reference frame. How is that related to it being forbidden? – Luismi98 Dec 22 '18 at 12:47