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Using the gauge principle we can derive the QED lagrangian from the dirac lagrangian, $$ \mathcal{L}_{Dirac}=\bar{\Psi}(i\gamma^{\mu} \partial_{\mu}-m\hat{I})\Psi $$ $$ \mathcal{L}_{QED}=\bar{\Psi}(i\gamma^{\mu} \partial_{\mu}-g\gamma^{\mu}A_{\mu}-m\hat{I})\Psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}. $$

If we apply Noethers first theorem to the Dirac lagrangian we find that it is invariant under global $U(1)$ transformations, with associated Noether's current: $$ \psi \rightarrow e^{iq\theta}\psi $$ $$ J^{\mu}=q\bar{\psi}\gamma^{\mu}\psi $$ This can be taken as a statement of charge conservation. Is the statement of conservation of charge via $J^{\mu}$ referring to global charge conservation (since we consider global transformations) or local charge conservation? Further, since the Dirac lagrangian represents the charged fermion field does $q$ here represent the total charge of the field or that of an individual fermion?

Qmechanic
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Adrien
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    Related: https://physics.stackexchange.com/q/159273/2451 , https://physics.stackexchange.com/q/48305/2451 and links therein. – Qmechanic Feb 02 '22 at 10:40
  • current is local, so $q$, I suppose, it's the charge of the single particle. Noether theorem is about global transformations: you see it by checking at how it is demonstrated. – Rob Tan Feb 02 '22 at 12:52
  • $\psi$ represents a field of charged fermions, why should charge be that of the individual fermion and not the field – Adrien Feb 02 '22 at 13:31
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    What do you call a local charge? You derived the condition already, $\partial_\mu J^\mu = 0$. What else is there to say? If you want a constraint on what usually is called charge, i.e. $Q = \int d^{d-1}x J^0$, you can get it by integrating both side of the current conservation equation. By the way, there are no currents associated to gauge transformations. – nwolijin Feb 02 '22 at 14:40

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