What Is Helicity Flip Suppression in Particle Physics?

[electroweak]

Can someone please explain what is meant by "helicity flip suppression" and how this mechanism operates? (I'd like to see an explicit amplitude and/or cross section if possible.)

I've been reading papers in which a Majorana fermion self-annihilates into some resonance (say, a Z). Fermi-Dirac statistics at the annihilation vertex require these incoming fermions to have opposite spins, and conservation of angular momentum requires the final state fermion and antifermion (coming from the Z decay) to have opposite spins as well. Somehow this restriction yields a suppression factor equal to the final state fermion mass. I'd like to see exactly how this effect plays out.

 

[Hepth]

B -> W-> e nu:

Have a spin 0 -> two fermions, requires spin flip on the electron:

$$M \approx (...) V_{ub} \langle 0 | \bar{u} \gamma^{\mu} (1-\gamma_5) b | B \rangle \bar{e} \gamma_{\mu} (1 - \gamma_5) \nu$$
Where, using the definition of the decay constant:
$$\langle 0 | \bar{u} \gamma^{\mu} (1-\gamma_5) b | B \rangle = i f_B p_B^{\mu} = i f_B (p_e + p_{\nu}) ^{\mu}$$

Plug in, and the momenta get dotted into the gamma in the leptonic current.

$$\bar{e} (\not p_e + \not p_{nu}) (1-\gamma_5) \nu$$where
$$\bar{e} \not p_e = m_e \bar{e}$$
Hence, helicity suppression.

So because the decaying particle is a pseudoscalar (spin 0) the only 4-vector that it can be represented by its decay is its momentum. The e-nu vertex is a left handed, V-A current in this weak decay. The momentum gets dotted into the gammas, and if you use conservation of momentum you get that it is the sum of the two final state particles, and then just get the mass of both the e and the nu.

Notice, if the neutrino had a mass, the slash on the p_nu:

$$\bar{u} (\not p_e + \not p_{\nu})( 1 - \gamma_5 ) \nu$$

$$\bar{u} (m_e + \not p_{\nu})( 1 - \gamma_5 ) \nu$$

$$\bar{u} (m_e ( 1 - \gamma_5 ) +(1+\gamma_5) \not p_{\nu} )\nu$$

$$\bar{u} (m_e ( 1 - \gamma_5 ) - m_{\nu} (1+\gamma_5) )\nu$$

$$\bar{u} ((m_e - m_{\nu}) - (m_e + m_{\nu}) \gamma_5 )\nu$$

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