# Is the following decay process possible?

## Homework Statement

An antimuon and electron may bind together via Coulomb attraction and then decay, but is the following process possible? (µ+e-) → νe + νµ_bar

*The νµ_bar is the antiparticle of the muon neutrino - the antimuon neutrino

More than one answer (below) may be correct.

a) This decay mode can proceed naturally.

b) This decay would violate electron-lepton number. 0 + 1e → 1e + 0

c) This decay would violate muon-lepton number. -1μ + 0 → 0 - 1μ

d) This decay would violate tau-lepton number. 0 + 0 → 0 + 0

e) This decay would violate charge conservation. +1 - 1 → 0 + 0

## Homework Equations

Lepton numbers, charge, conservation

## The Attempt at a Solution

+e-) → νe + νµ_bar

µ+: lepton number: -1
e-: lepton number: +1
νe: lepton number: +1
νµ_bar: lepton number: -1

(-1µ++1e-) → +1νe + 1νµ_bar

electron-lepton number and muon-lepton number is conserved. There are no tau's in the equation. Therefore, b),c),d) are incorrect.

µ+ charge = +1
e- charge = -1
νe charge = 0
νµ_bar charge = 0

(+1µ+-1e-) → 0νe + 0νµ_bar

The net charge on both sides of the equation equals zero. Therefore, e) is incorrect.

So, by that logic I'd say the answer is a) This decay mode can proceed naturally.

Last edited:

haruspex
Homework Helper
Gold Member
Not an area I know anything about, but I note that b), c), d) concern lepton flavor conservation
https://en.wikipedia.org/wiki/Lepton_number
You only seem to have addressed flavor-independent lepton number conservation.

The electron number Le and the antimuon number Lµ+ are the same on both sides. (antimuon and antimuon neutrino both have a lepton number of -1)

haruspex
Homework Helper
Gold Member
The electron number Le and the antimuon number Lµ+ are the same on both sides. (antimuon and antimuon neutrino both have a lepton number of -1)
You are missing the point.
As well as overall lepton number conservation, there is also conservation of the three flavors of lepton number separately. Those flavor conservations do not always apply - e.g. random switching of neutrinos between flavors - but do apply in most interactions.

I've re-read the link and understand that flavor conservation does not always apply with neutrinos. I don't think I understand the point you're trying to make.

haruspex
Homework Helper
Gold Member
I've re-read the link and understand that flavor conservation does not always apply with neutrinos.
But do you understand that it is conserved in most interactions?
Is the "lepton electron number" conserved in the process mentioned in post #1?

Yes it is. The electron and electron neutrino both belong to the Le flavor

haruspex
Homework Helper
Gold Member
Yes it is. The electron and electron neutrino both belong to the Le flavor
Just noticed you have edited post #1 to clarify that. Originally you had only discussed total lepton number, so your reasoning was incomplete.

and lepton muon number is also conserved

haruspex