# 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. Commented Jun 27, 2019 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? Commented Jun 27, 2019 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. Commented Jun 27, 2019 at 4:51
• @annav I think the OP is considering a $t$-channel process.
– JEB
Commented Jun 27, 2019 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. Commented Jun 27, 2019 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.