Pion0 Decay: C-Violation and the Search for 3-Photon Decay

  • Thread starter Thread starter ChrisVer
  • Start date Start date
  • Tags Tags
    Decay
ChrisVer
Science Advisor
Messages
3,372
Reaction score
465
The neutral pion is dominantly decaying to 2 photons via electromagnetic interaction \pi^0 \rightarrow 2 \gamma.
However if one allows for P-violation in electromagnetic interactions, he will also get C violation (due to CPT theorem). In that case a decay of the form \pi^0 \rightarrow 3 \gamma could be observable.
I read in "Discrete symmetries and CP Violation: From experiment to theory" by M.S.Sozzi that without the selection rule applied on the ##\pi^0## decay, then the rate of the three-gamma is reduced by a factor of order \mathcal{O}(\alpha) (that means ~130 times less), since we have 3 photon vertices instead of 2. Would that imply that:
\frac{Br(\pi^0 \rightarrow 3 \gamma )}{Br(\pi^0 \rightarrow 2 \gamma)} \sim 10^{-2}?
How can this be in agreement with the experimental result of <3~ 10^{-8}? It's not and that's why we say that C-invariance is there...then why looking for the 3-photon decay?
 
Physics news on Phys.org
ChrisVer said:
then why looking for the 3-photon decay?
To check if the electromagnetic interaction is really C-conserving. Smaller C-violations lead to smaller branching fractions.
 
ChrisVer said:
without the selection rule applied

This.
 
ChrisVer said:
why looking for the 3-photon decay

Good reason: to see if the EM interaction is really C-conserving.
Better reason: to see if there is some new interaction that interferes with electromagnetism, thus making it experimentally accessible.
 
  • Like
Likes ChrisVer and mfb
Yup but if we find some C-violation it won't be from EM interactions since EM interactions cannot explain the 3photon decay. It would have to be something else.
So I would go with the "better reason"...which atm I cannot challenge o0)
 
ChrisVer said:
Yup but if we find some C-violation it won't be from EM interactions since EM interactions cannot explain the 3photon decay.
It can, if it has a small C-violating term.
Other C-violating interactions are possible as well, of course (and given the precision experiments done with electromagnetism, I guess that would be a more likely explanation).
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top