Fermi and Gamow-Teller beta decays

[JesseC]

Am I right in thinking Fermi type beta decay is helicity suppressed, due to the necessity of having a right handed electron? This was implied but not explicitly stated in my lectures. And is this why Gamow-Teller type beta decay is the generally favoured mode?

 

[Bill_K]

No, the reason that Fermi transitions are suppressed has to do with the matrix element, because the nuclear initial and final states will have a smaller overlap.

As far as the spins go, note that the electrons produced by a given decay have a wide range of energies, and while they may be relativistic they are not always so. For a nonrelativistic electron the helicity and the handedness are not the same thing.

Secondly, there's the momentum vectors to consider. The spins for Fermi transitions are antiparallel but the particles may come off at any angle, so antiparallel spins does not necessarily mean opposite helicities. It does mean the angular distribution between the two particles will be different for Fermi and Gamow-Teller.

 

[JesseC]

Thanks for that explanation Bill. I can see where I was getting it wrong, not considering the momentum of the daughter and the range of angles for the neutrino and electron. Shame my course doesn't go into more detail on these sort of things.