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.

(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. $$\widehat{J^{\mu \nu} \phi}(x) \equiv [\hat J^{\mu \nu}, \hat \phi(x)] = S^{\mu \nu} \hat \phi(x) - i x^{\mu} (\partial^{\nu}\hat \phi)(x) + ix^{\nu} (\partial^{\mu}\hat \phi)(x).$$
• This is very helpful, many thanks. I guess when I said "$\mathcal J$ are operators acting on the Hilbert space of fields" I was a bit sloppy, since the space of fields is not the Hilbert space of states. I guess this is the same distinction as in standard quantum mechanics, where strictly speaking $\hat P$ is the momentum operator acting on the Hilbert space (consisting of kets $| \psi\rangle$) while its differential realisation $\mathcal P=-i\nabla$ acts on wavefunctions, $\langle x|\psi\rangle$. Would you agree? Nov 5, 2021 at 17:26
• So just to confirm, we start with the usual poincare generators $J$, which act on spacetime. We promote these to operators $\hat J$ acting on the Hilbert/Fock space. We may then write e.g. $[\hat J,\hat \phi(0)]=-S\phi(0)$ (where I omitted indices, and $S$ is what i previously labelled $\mathcal J$), where hatted operators act on the Hilbert space, while S is just an operator acting on fields (or just a matrix at the end of the day). Is this correct? Nov 5, 2021 at 17:32