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The lagrangian $$\mathcal{L} = \frac{1}{2}\partial_\mu\phi\partial^\mu\phi - \frac{1}{2}m^2\phi^2$$ is invariant under spatial rotations, and one can calculate the conserved Noether charge for it. The conserved charge is the angular momentum, and it can be written as $$\textbf{L} = - \int \mathbf{d^3x\Pi}(\textbf{x})(\textbf{x} \times \nabla) \phi(\textbf{x}),$$ where $\mathbf{\Pi}(\textbf{x})$ is the canonical conjugate momentum of the field.

One can prove that $\textbf{L}$ is the generator of rotations at a quantum level too: $$[\hat{\phi}(\textbf{x}), \mathbf{\hat{L}}] = -i (\textbf{x} \times \mathbf{\nabla}_x) \hat{\phi}(\textbf{x}), $$ where the commutator is evaluated at equal times.

There is a question in an exam of a previous semester that I can't figure out. Why can that commutator be evaluated at equal times? The way the question is asked makes me believe this is not a trivial question.

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  • $\begingroup$ Well, does $\boldsymbol L$ depend on time? $\endgroup$ – AccidentalFourierTransform May 29 '18 at 3:17
  • $\begingroup$ It does. The question implies that there could be a case when you could not evaluate said equal-time commutator, but in this case you can. $\endgroup$ – Luke May 29 '18 at 3:25
  • $\begingroup$ ...but does it? Are you sure? $\endgroup$ – AccidentalFourierTransform May 29 '18 at 3:29
  • $\begingroup$ Is $\phi(x)$ not a Heisenberg operator? $\textbf{L}$ should depend on time if it is. $\endgroup$ – Luke May 29 '18 at 4:47
  • $\begingroup$ Have you tried computing the time derivative of L, just to make sure? $\endgroup$ – knzhou May 29 '18 at 7:31

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