Neutral current: terminology In particle physics, where does the term 'neutral current' originate? An example would be an electron exchanging a Z boson with another electron. I understand that the Z boson itself is neutral, but surely 'current' refers to electron and its associated amplitude in this case?
 A: The neutral current is electrically neutral. To see why, one must first understand what the current is. It is a composite field or (in the quantum theory) an operator, something like
$$ J^{\mathrm{(NC)}\mu}(f) = \bar{u}_{f}\gamma^{\mu}\frac{1}{2}\left(g^{f}_{V}-g^{f}_{A}\gamma^{5}\right)u_{f}, $$
where $u_f$ is the Dirac field for the fermion $f$. Note that the current is a product of the charged field and its complex conjugate, along with some coefficients, contractions, and gamma matrices, so the electric charge cancels between $\bar u_f$ and $u_f$. Equivalently, the current is an operator that is creating a particle and an antiparticle at the same moment (or creates+destroys a particle; or creates+destroys an antiparticle) and this particle pair is electrically neutral.
The field $u_f$ itself is charged but the current isn't just $u_f$.
Of course, one may see more easily why the neutral current has to be neutral. It's because the Lagrangian contains terms like $Z_\mu J^\mu$ and because $Z_\mu$ is an electrically neutral field, $J^\mu$ must be electrically neutral as well for the Lagrangian to preserve the electric charge.
The neutral currents should be contrasted with the previously identified charged currents which are schematically $\bar u_e \cdots   u_\nu $ which is a product of an electron or positron field and the neutrino field so that the charges don't cancel. Equivalently, the operator of the charged current may be seen to carry the charge $\pm e$ because the Lagrangian contains products like $J_{\rm charged}^\mu W^\pm_\mu$ and the charge has to be compensated.
In all cases, the charge of an operator is determined from the phase by which the operator transforms under gauge transformations – it adds an extra $\exp(iQ\lambda)$ – or, equivalently, from the commutator $[Q,L]$ of the charge $Q$ with the operator $L$ that is equal to the charge times $L$.
