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The $U(1)$ global gauge symmetry of electromagnetism implies - via Noethers theorem - that electric charge is conserved. Actually, it implies a continuity equation:

$$ \psi \rightarrow e^{i\theta}\psi \quad\Rightarrow\quad \partial_\mu J^\mu = \frac{\partial}{\partial t} \rho + \vec\nabla\cdot\vec j = 0 $$

Is it possible to turn this around, and starting from $\partial_\mu J^\mu = 0$ (maybe $J^\mu$ interpreted as classical field), show that the responsible symmetry is $U(1)$? I tried to write down Noethers theorem for an appropriate Lagrangian, and compare terms, but it led to nothing. I'd greatly appreciate a simple derivation. Literature tips and solution hints would also be helpful.

Note this is not homework, but rather a case of yak shaving. The question is a corollary to my other question "Understanding the argument that local U(1) leads to coupling of EM and matter" (renamed to make the difference clear). I'm writing a piece and trying to justify local gauge theories from simple physics (like classical electromechanics), rather than letting them fall from the sky. The part that's missing is where the $e^{i\theta}$ comes from.

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What is classical about the continuity equation here? Both $\rho$ and $\vec{j}$ are operators built from fields. The classical continuity equation appears when you take expectation values of the quantum version.

Here is a sketch of an answer (I'll let you dot the i's and cross the t's): your conservation equation means that the total charge $Q \equiv \int dx j^0$ is constant, i.e. it commutes with the Hamiltonian. A hermitian operator that commutes with the Hamiltonian generates a symmetry because it connects states with the same energy. Hence $U(\alpha) = \exp(i \alpha Q)$ is a unitary symmetry transformation. You don't need to postulate any specific field content for this.

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D'oh, that looks simple, and I don't even have to invoke Noethers theorem. I'll think about it, thanks! Regarding the continuity equation: aren't $\rho$ and $\vec j$ just classical charge and current density? As in classical, non-QM, Maxwellian electrodynamics? Of course they can be quantized, but my intent was to use a well-known classical observation as a starting point. – jdm Jul 23 '13 at 15:28
@jdm Ah ok. You can do a completely classical treatment using Poisson brackets instead of commutators. – Michael Brown Jul 23 '13 at 15:44

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