= Solution
For fields $\varphi_A$ and a continuous infinitesimal internal symmetry $\delta\varphi_A=\epsilon\Delta\varphi_A$, suppose the density changes by a <divergence>, $\delta\mathcal L=\epsilon\partial_\mu K^\mu$. Varying the <action> and using the <Euler-Lagrange equations> gives the <Noether current>
$$
\boxed{j^\mu=\sum_A\frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_A)}\Delta\varphi_A-K^\mu,
\qquad\partial_\mu j^\mu=0.}
$$
The <conserved charge> is $Q=\int j^0\,d^3x$ when the flux through the boundary vanishes; more generally $dQ/dt=-\oint\mathbf j\cdot d\mathbf S$. If the <Lagrangian> is exactly invariant, $K^\mu=0$. Treat a complex field and its conjugate as independent variables during variation, imposing conjugacy on physical solutions. Time and space translations give energy and momentum through the corresponding spacetime version of the theorem. Continuous global symmetry gives an ordinary charge; local gauge redundancy requires additional care and is not an independent <conserved charge> for every arbitrary function.
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