Consider a theory of a free massless complex scalar $\phi$ which undergoes global $U(1)$ transformations. The conserved current associated to this symmetry is the usual scalar current

$$ J^\mu = i\left(\phi^\dagger \partial^\mu \phi - \partial^\mu\phi^\dagger \phi\right) \tag{1} $$

which is divergence-less on-shell: $$\partial_\mu J^\mu = \phi^\dagger \square \phi -h.c = 0 \tag{2}$$ since $$\square\phi=0.\tag{3}$$

When we gauge the $U(1)$, we expect that gauge symmetriy should not spoil global current conservation. The Lagrangian is now

$$ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - (D_\mu\phi)^\dagger(D_\mu\phi) = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}- (\partial_\mu\phi)^\dagger(\partial_\mu\phi) - A_\mu J^\mu - A_\mu^2\phi^\dagger\phi \tag{4} $$ where $$D_\mu\phi = \partial_\mu \phi -i A_\mu \phi. \tag{5}$$

The equation of motions for the photon are

$$ \partial_\mu F^{\mu\nu} = J^\mu + 2A_\mu\phi^\dagger\phi \tag{6} $$

which seems to imply that the global current is not even conserved, that is

$$ \partial_\mu J^\mu = -2\partial_\mu\left[A^\mu \phi^\dagger \phi\right]. \tag{7} $$

Is this result wrong? It is against the expectations. The theory is still invariant under global $U(1)$ transformations, so the global current should be conserved.


1 Answer 1


No, the gauging just changes the partial derivatives $\partial_{\mu}$ in the current (1) to covariant derivatives $D_{\mu}$. The global $U(1)$ Noether current for the gauged theory [i.e. the right-hand side of OP's eq. (6)] is still conserved on-shell, cf. e.g. this Phys.SE post.

  • $\begingroup$ Yes. But, what about the Noether current still associated to U(1) global transformations? Is it conserved? $\endgroup$
    – apt45
    Nov 11, 2018 at 20:16
  • $\begingroup$ There's only one global $U(1)$ symmetry in the gauged theory, so there's only one Noether current. And it's conserved on-shell due to Noether's theorem. $\endgroup$
    – Qmechanic
    Nov 11, 2018 at 20:22
  • $\begingroup$ I know. The current of the global $U(1)$ symmetry is Eq.(1). In the gauged theory it is, or not, conserved? $\endgroup$
    – apt45
    Nov 11, 2018 at 20:25
  • 1
    $\begingroup$ @newUser The current of U(1) is not (1), but the r.h.s. of (6). $\endgroup$ Nov 11, 2018 at 22:30

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