# Charged and neutral currents in electroweak theory

We describe electroweak theory by $$SU(2)_L\times U(1)_Y$$. This group has four generators which will be the W's, Z and $$\gamma$$ and we get 4 associated currents. We complexify the first 2 generators of SU(2) to describe the charged W bosons.

$$\sigma^\pm = \frac{1}{\sqrt{2}} (\sigma_1\pm i\sigma_2)$$ And then the charged current is $$J^\pm \propto \bar{\psi}\sigma^\pm\gamma_\mu\psi$$, which are the charged currents describing the W bosons. The remaining SU(2) generator and the U(1) generator form the neutral currents of the Z and $$\gamma$$.

Why are the complexified currents now charged currents and the remaining two said to be neutral? Why is it a charged current now that it we complexify it? and how come a minus and a plus sign for the complex number changes the charge of the current/boson

You are effectively asking what the action of the (conserved) charge operator is on the gauge bosons. First note the (weak) hypercharge action on both the Ws and the B is null (trivial), as the Ws don't couple to B, and B doesn't couple to itself being abelian (the analog of the photon having zero charge). So Y=0 for their charges, and the Gell-Mann—Nishijima formula collapses: $$Q=I_3+Y \to ~~~I_3.$$ the Ws are in the adjoint (triplet) representation of weak isospin, with hermitian generator $$Q_{ij}= i \epsilon_{3ij}, \leadsto \\ Q_{ij}W_j=i \epsilon_{3ij}W_j \leadsto \\ Q (W_1\mp iW_2)=\pm (W_1\mp iW_2).$$ This identifies then $$W^{\pm}=(W_1\mp iW_2)/\sqrt{2}$$, while the action of Q on $$W_3$$ and $$B$$ is trivial, so they are both neutral and can then weak-mix.
Correcting for the nonvanishing hypercharges, you may likewise compute the charge action on the fermions, left and right, now in the doublet/singlet weak isospin representation, and insert that in the fermion currents. For instance, for the leptons, $$Q_L= \tau^3/2+Y/2= \tau^3/2 -1/2,\\ Q_R=Y/2=-1,$$ where the weak hypercharge of hypothetical R neutrinos would be 0. (You then see your $$J^+_\mu \propto \overline{\nu_L}\gamma_\mu e_L^- +...$$ transforming with a - eigenvalue under Q (!), and coupling to $$W^+_\mu$$ to produce a neutral Lagrangian term! Many students are first confused by this notation.)