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For massless spin-1/2 fermions say $N$ I am using the spinors as given here say - http://theory.fnal.gov/people/ellis/Calctools/spinor.pdf - the $u_{+}(k)$ and $u_{-}(k)$ on page 2. So the $u_{+}(k)$ are the eigenspinors of the right chirality projector ($P_R = diag (I,0)$) and the $u_{-}(k)$ are the eigenspinors of the left chirality projector, ($P_L = diag(0, I )$).

Now say I have the Lagrangian which interacts between a positively charged scalar $S^{+}$, a massles spin-1/2 fermion $N$ and another massive fermion $B$ as,

$L = (\partial_\mu S^{+} ) \bar{N} \gamma^\mu P_L B + (\partial_\mu S^{-} )\bar{B}\gamma^\mu P_L N$

The two terms are Hermitian conjugates of each other.

  • Now I want to know as to from knowing this Lagrangian how (or does it?) follow that this reaction is possible of ($\bar{B} \rightarrow S^{+} + \bar{N} $) ?

  • Why I am not sure of the above is because of this - lets say that the $N$ are the known left-handed neutrinos (so we need to represent them as the $u_{-}(k)$) - then it should follow that the anti-neutrinos are right-handed and hence they need to be represented as the $u_{+}(k)$ spinors - right?

Now a matrix element of the above reaction, $\bar{B} (p) \rightarrow S^{+}(k) + \bar{N}(q)$ will look like, $v_B(p)\gamma^\mu k_\mu P_L u_{+}(q)$ -But isn't this term starightaway $0$ since $P_L$ annihilates $u_{+}$?

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