11.2 Particle conservation laws  (Page 2/6)

Lepton number conservation

Lepton number conservation states that the sum of lepton numbers before and after the interaction must be the same. There are three different lepton number     s : the electron-lepton number $L_e,$ the muon-lepton number $L_{\mu },$ and the tau-lepton number $L_{\tau }.$ In any interaction, each of these quantities must be conserved separately . For electrons and electron neutrinos, $L_e=1;$ for their antiparticles, $L_e=\mathrm{-1};$ all other particles have $L_e=0.$ Similarly, $L_{\mu }=1$ for muons and muon neutrinos, $L_{\mu }=\mathrm{-1}$ for their antiparticles, and $L_{\mu }=0$ for all other particles. Finally, $L_{\tau }=1,\mathrm{-1}$ , or 0, depending on whether we have a tau or tau neutrino, their antiparticles, or any other particle, respectively. Lepton number conservation guarantees that the number of electrons and positrons in the universe stays relatively constant. ( Note: The total lepton number is, as far as we know, conserved in nature. However, observations have shown variations of family lepton number (for example, $L_e)$ in a phenomenon called neutrino oscillations. )

To illustrate the lepton number conservation law, consider the following known two-step decay process:

In the first decay, all of the lepton numbers for ${\pi }^{+}$ are 0. For the products of this decay, $L_{\mu }=\mathrm{-1}$ for ${\mu }^{+}$ and $L_{\mu }=1$ for ${\nu }_{\mu }.$ Therefore, muon-lepton number is conserved. Neither electrons nor tau are involved in this decay, so $L_{\text{e}}=0$ and $L_{\tau }=0$ for the initial particle and all decay products. Thus, electron-lepton and tau-lepton numbers are also conserved. In the second decay, ${\mu }^{+}$ has a muon-lepton number $L_{\mu }=\mathrm{-1},$ whereas the net muon-lepton number of the decay products is $0+0+\left(\mathrm{-1}\right)=–1$ . Thus, the muon-lepton number is conserved. Electron-lepton number is also conserved, as $L_{\text{e}}=0$ for ${\mu }^{+}$ , whereas the net electron-lepton number of the decay products is $\left(\mathrm{-1}\right)+1+0=0$ . Finally, since no taus or tau-neutrons are involved in this decay, the tau-lepton number is also conserved.