Does the $U(1)$ charge of a scalar particle flip under charge conjugation?

Question

Consider a complex scalar particle $\phi$ coupled to an electromagnetic field. The Lagrangian is given by

$$ \mathcal{L} =(D_\mu \phi)^* D^\mu \phi - m^2 \phi^2 - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}$$

where $D_\mu = \partial_\mu - ie A_\mu $ and $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. This Lagrangian has global $U(1)$ symmetry under $\phi \rightarrow e^{i \alpha} \phi$, $\phi^* \rightarrow e^{-i\alpha } \phi$. The corresponding Noether current is given by

$$ j^\mu(\phi) = -i[\phi^* D^\mu \phi - (D^\mu \phi)^* \phi] = -i(\phi^* \partial_\mu \phi -\phi \partial_\mu \phi^* - 2ieA_\mu|\phi|^2)$$

and is interpreted as the electric current, as discussed in this question

One would expect that under charge conjugation $\phi \rightarrow \phi^*$ that the electric current would change sign. If I replace $\phi $ with $\phi^*$ in the current above, I find

$$j^\mu(\phi^*)= -i(\phi \partial_\mu \phi^*-\phi^* \partial_\mu \phi - 2ie A_\mu |\phi|^2) \neq -j^\mu(\phi) $$

So it hasn't flipped sign due to the $|\phi|^2$ term. What is going on?

Answer 1

Note that charge conjugation flips the sign of the vector potential $A_\mu$. You know this even from freshman physics: the potential of a positive charge is positive, while the potential of a negative charge is negative.

Answer 2

To identify which (if any) transformation should be called "charge conjugation," follow this process:

1. List the theory's discrete symmetries,

2. Discard the ones that affect spacetime (like parity and time-reversal),

3. If the ones that remain include one that has the same effect on the lagrangian as flipping the sign of the charge, then that one deserves to be called charge conjugation.

The transformation $\phi\to\phi^*$ is not a symmetry of the theory shown in the question, because it doesn't preserve the kinetic term: $$ (D_\mu^*\phi)(D^\mu\phi^*)\neq (D_\mu\phi)^*(D^\mu\phi). $$ To preserve the kinetic term, you also need to flip $A\to-A$, as indicated in knzhou's answer: the transformation $(\phi,A)\to (\phi^*,-A)$ is a symmetry.