Semileptonic Lambda baryon decay suppression

  • Context: Undergrad 
  • Thread starter Thread starter Triple_S
  • Start date Start date
  • Tags Tags
    Baryon Decay Lambda
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
Triple_S
Messages
1
Reaction score
0
Hi there,

I'm currently studying the decay modes of the Lambda baryon (uds).
According to literature, the decay mode with leptons (e.g. Lambda -> p + e- + anti ve) is suppressed against the decay modes with pions (e.g. Lambda -> p + pi-) by a factor of about 10^-4.
I was looking for the reason for this suppression, but could'nt find anything. Could anyone explain it to me? I was thinking that it might be due to parity, cause it's not conserved in weak interactions, but I'm not sure whether that's correct.

Thanks in advance.
 
Physics news on Phys.org
I don't think it's helicity suppression. If that were the case, the semimuonic decay would be much, much larger (about 7800x) than the semielectronic. It's probably 3-body vs. 2-body phase space.
 
Shouldn't the phase space favor the three-body decay? Not much energy released in the two-body decay.

The electron decay is more frequent than the muon decay by a factor of ~5, I can see the phase space factor there.
 
mfb said:
Shouldn't the phase space favor the three-body decay?

I can't do it in my head, except to note that phase space suppression goes up with increasing numbers of particles, especially when you have a W. The W propagator is trying to drive the e-nu system to high mass, and that bites into phase space. But what I am sure is that it's not helicity suppression, since the electron decay is 5x more frequent than the muon decay and not 1/7800 as frequent.