Weak force coupling to left-handed particles

[JoePhysicsNut]

The W couples to left-handed particles only. What about the Z? Is it the same?

Thanks in advance!

 

[Bill_K]

The Z couples to both. Quoting from Halzen and Martin, the W coupling is -i(g/√2)½γ^μ(1 - γ⁵), which is left-hand only. By contrast, the Z coupling is -i(g/cos θ_W)½γ^μ(c_V - c_Aγ⁵) where c_V = T³ - 2 sin²θ_WQ and c_A = T³.
T³ is the weak isospin of the fermion in question and Q is its charge.
For e^-, μ^- the values are c_A = -1/2, c_V = -0.03.
For u, c quarks, c_A = 1/2, c_V = 0.19.
For d, s quarks, c_A = -1/2, c_V = -0.34.

 

[JoePhysicsNut]

The Z couples to both. Quoting from Halzen and Martin, the W coupling is -i(g/√2)½γ^μ(1 - γ⁵), which is left-hand only. By contrast, the Z coupling is -i(g/cos θ_W)½γ^μ(c_V - c_Aγ⁵) where c_V = T³ - 2 sin²θ_WQ and c_A = T³.
T³ is the weak isospin of the fermion in question and Q is its charge.
For e^-, μ^- the values are c_A = -1/2, c_V = -0.03.
For u, c quarks, c_A = 1/2, c_V = 0.19.
For d, s quarks, c_A = -1/2, c_V = -0.34.

Thanks for a thorough answer!