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I would like to prove that Noether charge $Q$ is a generator of the same symmetry as the one that due to the Noether's theorem led to the current $j^\mu$ and charge $Q$ in classical field theory (I have found a lot of proves in case of classical mechanics), i.e. I would like to prove $$\delta_\epsilon\phi = \{\phi, Q_\epsilon\}$$

I would like to show it for theory with scalar fields and general transformation $\epsilon^\mu(x)$, not just constant translations.

My attempt:

Let us assume some infinitesimal transformation of the field $$\phi(x) \rightarrow \phi'(x) = \phi(x) + \epsilon^\mu(x)\partial_\mu\phi(x) = \phi(x) + \delta_\epsilon\phi(x)$$ With Noether's theorem I obtain on-shell conserved current and charge $$Q_\epsilon = \int d^{n-1}x j^0$$

From here I would like to prove this $$\delta_\epsilon\phi = \{\phi, Q_\epsilon\}$$ $$\delta_\epsilon\pi = \{\pi, Q_\epsilon\}$$

For this I have Poisson bracket as $$\{A(\textbf{x},t), B(\textbf{y},t)\} =\int d{\textbf{x'}} \frac{\delta A(\textbf{x},t)}{\delta\phi(\textbf{x'},t)}\frac{\delta B(\textbf{y},t)}{\delta\pi(\textbf{x'},t)} - \frac{\delta A(\textbf{x},t)}{\delta\pi(\textbf{x'},t)}\frac{\delta B(\textbf{y},t)}{\delta\phi(\textbf{x'},t)}$$

If I now vary the Hamiltonian action and use off-shell expression $$\{Q_\epsilon, H(t)\} + {\partial Q_\epsilon \over \partial t}=0$$ I get $$ \delta A_H[\pi,\phi] = \int dt\, \epsilon\frac{d}{dt}(\int d^{n-1}x\pi\frac{\delta Q_\epsilon}{\delta\pi}-Q_\epsilon)$$.

Here I am stuck. This closely resembles the proof for point mechanics, but for fields I would like to have total divergence. I can assume $$j^\mu_\epsilon = \frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\delta_\epsilon\phi-K^\mu_\epsilon$$ But even with that I struggle to show $$\delta A_H[\pi,\phi]=\int d^nx \partial_\mu K^\mu_\epsilon$$

Any idea what I am missing to derive that?

Qmechanic
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AccidentalThought
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    OP's field-theoretic case is a straightforward generalization of this Phys.SE post. – Qmechanic Nov 23 '20 at 10:36
  • @Qmechanic thank you for the suggested post, but I am still struggling with the generalization. I have an issue with generalizing the usage of the Noether identity in (10). I guess my main trouble is to see the connection between the total derivative with respect to time $d/dt$, which is the point of Hamiltonian description and the divergence $\partial_\mu$ in Noether's theorem in Lagrangian description in the field theoretic case. I would be grateful for any advice. – AccidentalThought Dec 19 '20 at 13:30
  • For all this construction to work, you need that the infinitesimal transformation (your starting point) is a symmetry of the action. The transformation you wrote is a symmetry only if the theory is diffeomorphism invariant, which means that you also need to transform the metric as $g_{\mu\nu}\to g_{\mu\nu}+2\partial_{(\mu}\epsilon_{\nu)}$. – Ali Seraj Dec 26 '20 at 11:50

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