Commutators in Poincare algebra Consider the method of induced representations for the Poincare algebra, i.e. given a field $\phi$ (which need not be a scalar field despite its notation), we have the commutator $$[J^{\mu\nu},\phi(0)]=-\mathcal{J}^{\mu\nu}\phi(0)$$ where $J$ are the Lorentz generators, promoted to operators, while $\mathcal{J}$ are operators acting on the Hilbert space of fields. I want to find the commutator for non-zero $x$. My idea was to translate $\phi(0)=\mathcal{T}^{-1}(x)\phi(x)\mathcal{T}(x)$, where $\mathcal{T}(a)=e^{-ia^\mu P_\mu}$, where $P$ is the generator of translations (and so we may define an operator $\mathcal{P}_\mu=-i\partial_\mu$ on the Hilbert space).
When I do this, I find: $$[J^{\mu\nu},e^{ix^\alpha P_\alpha}\phi(x)e^{-ix^\alpha P_\alpha}]=-\mathcal{J}^{\mu\nu}e^{ix^\alpha P_\alpha}\phi(x)e^{-ix^\alpha P_\alpha}$$
My idea was to then multiply both sides by $e^{-ix^\alpha P_\alpha}$ from the left and $e^{ix^\alpha P_\alpha}$ from the right, to find $$[e^{-ix^\alpha P_\alpha}J^{\mu\nu}e^{ix^\alpha P_\alpha},\phi(x)]=-e^{ix^\alpha P_\alpha}\mathcal{J}^{\mu\nu}e^{-ix^\alpha P_\alpha}\phi(x) \tag{1}$$
Now on the left hand side i can use $$e^{-ix^\alpha P_\alpha}J^{\mu\nu}e^{ix^\alpha P_\alpha}=J^{\mu\nu}-ix^\alpha[P_\alpha,J^{\mu\nu}]=J^{\mu\nu}-x^\mu P^\nu+x^\nu P^\mu \tag{2}$$
where I used the Poincare algebra. Substituting this into (1), (I might have gotten a sign wrong somewhere) $$[J^{\mu\nu},\phi(x)]+i(x^\mu\partial^\nu-x^\nu\partial^\mu)\phi(x)=-e^{ix^\alpha P_\alpha}\mathcal{J}^{\mu\nu}e^{-ix^\alpha P_\alpha}\phi(x)$$ where I used $[P^\mu,\phi(x)]=-i\partial^\mu\phi(x)$. Now for the right hand side, I am tempted to use (in analogy with (2)) $$e^{ix^\alpha P_\alpha}\mathcal{J}^{\mu\nu}e^{-ix^\alpha P_\alpha}=\mathcal J^{\mu\nu}-ix^\alpha[P_\alpha,\mathcal J^{\mu\nu}]$$ however I am unsure as to how to proceed, because, as far as I know, the usual commutation relations hold between $P$ and $J$ (or equivalently on their representations $\mathcal P$ and $\mathcal J$), but here I have a "mixed" commutator, between P and $\mathcal J$.
I know the answer should be $$[J^{\mu\nu},\phi(x)]=-\mathcal J^{\mu\nu}\phi(x)+i(x^\mu\partial^\nu-x^\nu\partial^\mu)\phi(x)$$ so if what I wrote above is right (which it isn't, to the very least due to a sign error somewhere which I'm not too bothered about at the moment), then it must be that $[P_\alpha,\mathcal J^{\mu\nu}]=0$, which leaves my a bit perplexed.
 A: (signs might be completely wrong here) In the following I use hats on quantum Hilbert-space operators to distinguish them from the differential operators acting on fields, which have no hats.
Further I use that for any operators $\hat O(x)$ we have
$$
\hat O(x) = e^{-ix\cdot \hat P} \hat O(0) e^{i x \cdot \hat P}.
$$
This is equivalent to the statement that
$$
[\hat P^{\mu},\hat O(x)] \equiv \widehat{P^{\mu} O}(x) = -i (\partial^{\mu} \hat O)(x)
$$
where in the "field representation" we have $P^{\mu} = -i \partial^{\mu}$. Also I say that
$$
[\hat J^{\mu \nu}, \hat \phi(0)] \equiv \widehat{J^{\mu \nu} \phi}(0) = S^{\mu \nu} \hat \phi(0),
$$
where $S^{\mu \nu}$ are matrices in some internal space in which the fields live. The question is now, given that we know $\widehat{J^{\mu \nu} \phi}$ at space-time pt $x = 0$, namely $S^{\mu \nu} \hat \phi$, what is $\widehat{J^{\mu \nu} \phi}$ at arbitary pt $x$. This is of course determined by the Poincare algebra.
$$
[\hat J^{\mu \nu}, \hat \phi(x)]  = [\hat J^{\mu \nu}, e^{-i x \cdot \hat P} \hat \phi(0) e^{ix \cdot \hat P}] = e^{-i x \cdot \hat P} [e^{i x \cdot \hat P} \hat J^{\mu \nu} e^{-i x \cdot \hat P}, \hat \phi(0)] e^{i x \cdot \hat P}
\\
= e^{-i x \cdot \hat P} [\hat J^{\mu \nu} + x^{\mu} \hat P^{\nu} - x^{\nu} \hat P^{\mu}, \hat \phi(0)] e^{i x \cdot \hat P}
= e^{-i x \cdot \hat P} \Big ( [\hat J^{\mu \nu}, \hat \phi(0)] + x^{\mu} [\hat P^{\nu},\hat \phi(0)] - x^{\nu} [\hat P^{\mu},\hat \phi(0) ] \Big) e^{i x \cdot \hat P}
\\
= e^{-i x \cdot \hat P} \Big ( \widehat{ J^{\mu \nu} \phi}(0) + x^{\mu} \widehat{ P^{\nu} \phi}(0) - x^{\nu} \widehat{ P^{\mu} \phi}(0) \Big) e^{i x \cdot \hat P}
= e^{-i x \cdot \hat P} \Big ( S^{\mu \nu} \hat \phi(0) - i x^{\mu} (\partial^{\nu}\hat \phi)(0) + ix^{\nu}  (\partial^{\mu}\hat \phi)(0) \Big) e^{i x \cdot \hat P}
\\
= S^{\mu \nu} \hat \phi(x) - i x^{\mu} (\partial^{\nu}\hat \phi)(x) + ix^{\nu}  (\partial^{\mu}\hat \phi)(x).
$$
I.e.
$$
\widehat{J^{\mu \nu} \phi}(x)  \equiv [\hat J^{\mu \nu}, \hat \phi(x)] = S^{\mu \nu} \hat \phi(x) - i x^{\mu} (\partial^{\nu}\hat \phi)(x) + ix^{\nu}  (\partial^{\mu}\hat \phi)(x).
$$
