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


1 Answer 1


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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.