We define this operator : $$M^{\mu\nu} = \int d^3x~(x^{\mu}T^{0\nu} - x^{\nu}T^{0\mu})$$ where $T_{\mu\nu}$ is the energy momentum tensor (see e.g. Energy momentum tensor from Noether's theorem) for a scalar field. Concentrate on the boost operator $M^{0i}$ (where $x^0$ appears explicitly). We can easily prove that its derivative with respect to $x^0$ is zero using integration by parts and the local conservation of $T_{\mu\nu}$. Now using the Heisenberg picture, we can take this operator and evolute it with $$U=e^{-iHt}$$ and $U^\dagger$. Taking $t$ small, you develop, keep first order in t and get (where the term in the parenthesis goes for $x^0$) $$M^{\mu\nu}(t) = M^{\mu\nu}(0) + it[H,M^{\mu\nu}(0)].$$ But, $M^{\mu\nu}$ is a generator of the Poincaré group and its commutation relations must satisfy in particular : $$[M^{0i},H]=iP^i$$ (it's not zero and serves to make the good translation in the first term of $M^{0i}$). Where is the blunder ? On one side, you have a conservation law, on the other, a consistent evolution.
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The quantum version of Noether's theorem are the Ward-Takahashi identities.
Classical conservation laws only hold "inside expectation values", so you should not expect a classically conserved operator to have no time evolution quantumly.
Additionally, switching to the Hamiltonian formalism breaks manifest Lorentz invariance, so you should also not expect anything to behave in an evidently covariant way here - the final S-matrix elements/expectation values will be Poincare invariant, but steps on the way may well look as if they violate Poincare symmetry/some conservation law derived from it.
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