# Is the electric current in QED uncharged under the gauge field?

If I understand correctly, in gauge theory a field is defined to be "charged" under the gauge field if it transforms nontrivially under gauge transformations. (For example, in the abelian case a field $\Psi$ is defined to have charge $q$ if, under a gauge transformation in which $A^\mu \to A^\mu - \partial^\mu \Gamma$, it goes to $\exp(-i q \Gamma) \Psi$.) This then implies that any gauge-invariant quantity cannot be charged under the gauge field. For example, the electric currents $J^\mu := e \bar{\Psi} \gamma^\mu \Psi$ in spinor QED and $J^\mu := -i(\varphi^\dagger D^\mu \varphi - \varphi (D^\mu \varphi)^\dagger)$ in scalar QED are both uncharged under this definition.

Is this the standard definition? It seems funny to me to say that electric current is uncharged, even if it's made up of fundamental fields that are charged.

• I'm not sure what exactly you expect as an answer here - yes, if you define "charged" as "transforms non-trivially under the gauge group", then the QED electric currents are uncharged. And yes, that definition clashes with our classical notion of calling $j^0$ the charge density. Asking whether certain terminology is "standard" is off-topic as primarily opinion-based - is there something else you want to know about this? – ACuriousMind Jul 11 '17 at 12:15
• @ACuriousMind "A question of the form What does this notation/terminology mean? is on-topic if it cannot immediately be answered by a simple Google search/Wikipedia lookup." My question is "What does the terminology 'charged under the gauge field' mean?" – tparker Jul 11 '17 at 15:22

In general, a composite field expression has a well-defined total charge $Q$ if it is a linear combination of products of field operators, and if for each term in the linear combination, the elementary fields in that term have charges that add up to the same value $Q$. In particular, this applies to the current, and gives charge zero.