# Does the W boson necessarily change an anti-fermion's flavor to its anti-neutrino counterpart?

I'm writing the diagrams for the following process in Standard Model: $$\nu_e + e^+\rightarrow \mu^++\nu_\mu+\gamma$$ I want to know if the W boson changes the flavor of $$\mu^+$$, for instance, necessarily to its anti-neutrino counterpart $$\bar{\nu_\mu}$$ or it can also change into $$\nu_\mu$$.

• Muon and electrons are different flavors, not muons and muon neutrinos. As for what neutrino appears with a muon depends on muon number conservation. – Triatticus Jun 27 '19 at 1:49
• I'm confused. I thought that within the leptons category we could distinguish between 6 different flavors: electron, muon, tau, electron neutrino, muon neutrino and tau neutrino. Am I wrong? – RicardoP Jun 27 '19 at 2:03
• you can distinguish, but each lepton number is conserved in their own right. The reaction you write annihilates the flavor number so just energy and charge remains, an annihilation, ( like e+e-, anything can happen). It is conservation of charge on the right that asks for a W+, , which then has to conserve whatever lepton it creates with antilepton. – anna v Jun 27 '19 at 4:51
• @annav I think the OP is considering a $t$-channel process. – JEB Jun 27 '19 at 5:43
• @JEB depending on charge conservation in the specific process. anything that is not forbidden by conservation laws can happen, though it might be of very low probability. – anna v Jun 27 '19 at 6:00

I want to know if the W boson changes the flavor of μ+, for instance, necessarily to its anti-neutrino counterpart $$\bar ν_μ$$, or it can also change into $$ν_μ$$.
No it does not, in the Standard Model. All vertices there conserve lepton number, so the virtual $$W^+$$ involved will decay to just $$\mu^+ ~ ν_μ$$. To instead go to $$\bar ν_μ$$ would violate lepton number by 2. The photon in the reaction will attach to any charged line in your diagram.
Equivalently, only $$\bar ν_μ$$ can go to $$W^- ~ \mu^+$$ while $$ν_μ$$ can never do this.