I am reading Quantum Field Theory for The Gifted Amateur by Lancaster and Blundell. In the book, it states that quantum fields transform under spacetime translations according to the rule $$\hat U(a_{\mu})\Psi(x_{\mu})\hat U^{\dagger}(a_{\mu})=\Psi(x_{\mu}+a_{\mu})$$ where$$\hat U(a_\mu)=\exp(ip^\mu a_\mu),\qquad p^\mu=(E,p).$$ However, when I tried to see if infinitesimal space or time translations leave the scalar field lagrangian $$\mathcal L=\frac 12(\partial_{\mu}\Psi)^2-\frac {m^2}2\Psi^2$$ invariant, I get confused as to how to proceed. Can I just taylor expand $\Psi(x_\mu+a_\mu)$ and insert it into the lagrangian? That is how I would proceed classically, but the field is an operator, so would I have to use the Baker-Hausdorff Identity since $\Psi$ transforms according to the above rule? Please explain how I should proceed and why the method works.
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$\begingroup$ The lagrangian is a functional of the classical fields, not the operators. $\endgroup$– Davide MorganteCommented Jan 25, 2021 at 20:38
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$\begingroup$ So its only the Hamiltonian that's a functional of the operators? $\endgroup$– Daniel WatersCommented Jan 25, 2021 at 20:42
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$\begingroup$ Assuming canonical quantization, of course $\endgroup$– Daniel WatersCommented Jan 25, 2021 at 20:48
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$\begingroup$ Even the Hamiltonian is a functional of the classical fields, whenever using second quantization the Hamiltonian is to be written in terms of creation and annihilation operators. $\endgroup$– Davide MorganteCommented Jan 25, 2021 at 20:55
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$\begingroup$ @DavideMorgante, okay, so canonical quantization is a step ahead of evaluating Noether currents. Thanks for helping! $\endgroup$– Daniel WatersCommented Jan 25, 2021 at 21:05
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Noether's theorem is for classical field theory with a classical action $S =\int \mathrm{d}^4x \mathcal{L}$, then we define the Noether's current from it as you mentioned. In order to quantize a theory (in canonical quantization), we replace the classical commutation relations, for instance here for the Noether's current $J^{\mu}$ with non-commutating operators $\hat{J}^{\mu}$.
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$\begingroup$ So we evaluate Noether currents before canonical quantization? $\endgroup$ Commented Jan 25, 2021 at 20:54
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$\begingroup$ @DanielWaters In reality, canonical quantization does not go very well with symmetries and Nöether's theorem, and in general it comes with all sorts of problems. A formal way of going to the quantum formalism is through functional (i.e. path integral) quantization. $\endgroup$ Commented Jan 25, 2021 at 20:57
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$\begingroup$ @Daniel Waters-Yes. Now for a quantum version of Noether's theorem, as far as I am aware, you can check Ward–Takahashi identity. $\endgroup$ Commented Jan 25, 2021 at 20:57
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$\begingroup$ @DavideMorgante Yeah I thought that canonical quantization doesn't really work for interacting theories as well, but I might be wrong on that. $\endgroup$ Commented Jan 25, 2021 at 21:01