# Charge in field theory [duplicate]

Why is it that a real scalar field is charge neutral whereas a complex scalar field (or a spinor field like Dirac theory) describe charged fields?

## 1 Answer

Charge is a measure of how strongly a field couples to the electromagnetic field. The electromagnetic field is a gauge field generated by $U(1)$ gauge symmetry1, so to couple to it the field needs to be a non-trivial representation of the $U(1)$ group. $U(1)$ is the symmetry group of points on a circle, so we need to look for mathematical objects that have a similar structure.

The trivial representation is the singlet representation, naively speaking a "point", a real scalar field. This is a singlet, meaning that it doesn't change as you act on it with $U(1)$ operators (there is no way to tell afterwards whether the point has been rotated around itself or not2). Since it does not respond to the $U(1)$ field, a real scalar field does not couple to the electromagnetic field: it is chargeless.

The simplest non-trivial representation is a complex field. This should be clear if you imagine the complex phases $e^{i\theta}$ on the complex plane, which have the same topology as points on a circle. The complex phase does change as you act on it with $U(1)$ operators: the phase angle changes. So this couples to the electromagnetic field: it is charged.

The conserved current you talk about is a "current" in a more general sense than electrical current (probability current etc.). To get the electrical current you would need to multiply the electric charge to this "probability current", which would give you zero for a real scalar field, since as I mentioned, the real scalar field has charge zero.

1 Think of this as the definition of the electromagnetic field. This will give you Maxwell's equations.

2 Using the "value" of the field at the point to tell how much it has been rotated does not work, because it doesn't have the topology of a circle: you could tell the difference between, say $0$ and $2\pi$. A real value by itself is not a representation of the $U(1)$ group.