# Chirality when moving around legs in Feynman diagrams

Assuming one has the following term in a Lagrangian:

$$g (\overline{A_R} B_L)(\overline{C_R}D_L)$$

where A,B,C,D correspond to spin 1/2 Dirac particles and the subscripts $R$ and $L$ denote left- and right-handed particles. From what I remember does this correspond e.g. to the scattering process

$$\tilde{A_R} + B_L \rightarrow \tilde{C_R} + D_L$$

where $\tilde{A_R}$ denotes the anti-particle of $A_R$ etc.

When 'moving around' the $B_L$ leg from the incoming to the outgoing part, one has to replace $B$ by its anti-particle $\tilde{B}$ if I remember correctly:

$$\tilde{A_R} \rightarrow \tilde{B} + \tilde{C_R} + D_L$$

However, I wonder what the chirality of B is ?

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