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In the weak interaction the gauge bosons couple to left handed chiral states only.

When the two interactions are unified do the bosons couple to both left and right handed chiral states?

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Absolutely (yes)!

While the charged current weak interactions, mediated by the W bosons, only coupled to left-chiral fermions (and right chiral anti fermions), both the EM and the neutral current weak interactions couple to right handed fermions. (In general: neutrinos are an exception). See formulas (10.2), (10.5a,b) of the canonical PDG review, required reading.

That is to say, even though $W^3$ couples to left-chiral fermions only, like its charged brothers, the hypercharge gauge field $B$ couples to both left and right-handed fermions in its inimitable "inelegant" way that puzzles students when they first learn about the SM. (Feynman used to call these couplings "cockeyed".)

After EW mixing, the EM interactions couples equally to left and right chiral fermions ("vectorlike" coupling), whereas the Z boson couples to special combinations of left- and right-chiral fermions, with the exception of right-neutrinos, with are thus "sterile".

The most memorable "hybrid language" formulation of the neutral current couplings is $$ \mathcal{L}_{\rm NC} = \frac g{\cos\theta_{\rm W}}(J_\mu^3-\sin^2\theta_{\rm W}J_\mu^{\rm em})Z^\mu. $$ The $J^3_\mu$ current involves strictly left-chiral fermions, as indicated, while the e.m. current which also couples to photons involves equal-strength couplings to left- and right-chiral fields. You thus see that for all charged particles, leptons and quarks, there are Z couplings, but since changeless neutrinos do not participate in the e.m. current, right-neutrinos have no way to couple to the Z s.

Geeky note: For the charged lepton sector, the "accidental" value $\sin^2 \theta_W\sim 1/4$ dictates net almost pure A couplings to Z, !, a favorite starting exercise in appreciating the devilishness of the SM .

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  • $\begingroup$ Thats interesting thanks! So since the $W^{3}$ only couples to left handed states is it fair to say that this is still a parity violation? I've heard people say that parity violation in the weak sector is the result of EW symmetry breaking, however if ive understood correctly, this isn't the case for charged interactions with W bosons. $\endgroup$ Commented Apr 18, 2022 at 16:16

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