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Suppose we have a Lagrangian that with fields that are acted on by a symmetry group, e.g. $$\mathcal{L} = \partial_{\mu}\phi \partial^{\mu}\phi^* - m^2 \phi \phi^*$$ with $G=U(1)$ (i.e. $\phi \to e^{i \alpha}\phi$). Then this symmetry group has a representation acting on the physical Hilbert space - to find these operators we can use Noether's first theorem to find a conserved current and integrate to get a conserved charge operator $\hat{Q}$ and then a representation of $U(1)$ by acting on the Hilbert space with $\hat{U} = e^{i \theta \hat{Q}}$.

My question now is what happens if we now have a gauge symmetry $G$ - how do we find the Hilbert space operator corresponding to gauge symmetries $G$? We can no longer use Noether's first theorem. (Of course we expect that the subspace of physical states will transform as a singlet under the Hilbert space operators corresponding to elements of $G$.)

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Not sure if this answers your question, but Noether's second theorem is the analog of Noether's ("first") theorem for gauge symmetries. For a field $A_\mu$ with a gauge symmetry $A_\mu \to A_\mu + \partial_\mu \lambda$, the conserved current (which is also the generator of gauge transformations) is given by $J^\mu = \frac{\partial \mathcal{L}}{\partial A_\mu}|_{A_\mu = 0}$. In the case of QED, $J_\mu = e \bar{\Psi} \gamma_\mu \Psi$, where the $\Psi$ are the electron Dirac fermion operators.

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  • $\begingroup$ Note that in a sense this is more of a math identity than a physical identity, in the sense that unlike the Noether current for a global symmetry, the conservation of a gauge current is purely an operator identity, and does not require making any assumptions about whether or not the action is extremized. $\endgroup$ – tparker Oct 20 '16 at 8:17
  • $\begingroup$ Thanks for your answer - I'm aware of Noether's second theorem but I'm still not entirely sure how we can then get an operator on the Hilbert space corresponding to a gauge transformation? $\endgroup$ – Wooster Oct 20 '16 at 8:41
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According to the Dirac conjecture, in the Hamiltonian and canonical formulation of gauge theory, the gauge transformations are generated by first class constraints. In the case of QED, the gauge generator is the Gauss law operator ${\bf \nabla}\cdot {\bf E}-\frac{\rho}{\epsilon_0}$.

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  • $\begingroup$ Could you be a bit more explicit about what exactly you mean by "the Gauss law operator?" $\endgroup$ – tparker Oct 20 '16 at 8:55
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Oct 20 '16 at 9:00
  • $\begingroup$ Thanks for your answer. I understand that to impose gauge symmetry by Dirac's conjecture we insist the physical states are invariant under Gauss' law - however I'm really wondering about the more general case of how do we construct the Hilbert space representation of a spacetime dependent operator - I'd like to see understand how this works even when the group $G$ isn't a symmetry and then specialise to the case where it is and see that states are invariant. Every group action should have a corresponding action on the Hilbert space? $\endgroup$ – Wooster Oct 20 '16 at 9:12

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