What Is Helicity Flip Suppression in Particle Physics?

  • Thread starter Thread starter electroweak
  • Start date Start date
  • Tags Tags
    Helicity
electroweak
Messages
43
Reaction score
1
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.
 
Physics news on Phys.org
B -> W-> e nu:

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

<br /> 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:
<br /> \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}<br />

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

<br /> \bar{e} (\not p_e + \not p_{nu}) (1-\gamma_5) \nu<br />where
<br /> \bar{e} \not p_e = m_e \bar{e}<br />
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:

<br /> \bar{u} (\not p_e + \not p_{\nu})( 1 - \gamma_5 ) \nu <br />

<br /> \bar{u} (m_e + \not p_{\nu})( 1 - \gamma_5 ) \nu <br />

<br /> \bar{u} (m_e ( 1 - \gamma_5 ) +(1+\gamma_5) \not p_{\nu} )\nu <br />

<br /> \bar{u} (m_e ( 1 - \gamma_5 ) - m_{\nu} (1+\gamma_5) )\nu <br />

<br /> \bar{u} ((m_e - m_{\nu}) - (m_e + m_{\nu}) \gamma_5 )\nu <br />

.
 
Last edited:
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