The W couples to left-handed particles only. What about the Z? Is it the same?
Thanks in advance!
The Z couples to both. Quoting from Halzen and Martin, the W coupling is -i(g/√2)½γμ(1 - γ5), which is left-hand only. By contrast, the Z coupling is -i(g/cos θW)½γμ(cV - cAγ5) where cV = T3 - 2 sin2θWQ and cA = T3.
T3 is the weak isospin of the fermion in question and Q is its charge.
For e-, μ- the values are cA = -1/2, cV = -0.03.
For u, c quarks, cA = 1/2, cV = 0.19.
For d, s quarks, cA = -1/2, cV = -0.34.
Thanks for a thorough answer!
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