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.


1 Answer 1


(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. $$ \widehat{J^{\mu \nu} \phi}(x) \equiv [\hat J^{\mu \nu}, \hat \phi(x)] = S^{\mu \nu} \hat \phi(x) - i x^{\mu} (\partial^{\nu}\hat \phi)(x) + ix^{\nu} (\partial^{\mu}\hat \phi)(x). $$

  • $\begingroup$ 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? $\endgroup$
    – user984949
    Nov 5, 2021 at 17:26
  • $\begingroup$ 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? $\endgroup$
    – user984949
    Nov 5, 2021 at 17:32
  • 1
    $\begingroup$ To the first comment. I think you are basically correct, however be careful with this analogy, since here it acts on fields and not on wavefunctions, which are fundamentally different things. The second comment seems to me as a good viewpoint. $\endgroup$
    – jkb1603
    Nov 6, 2021 at 9:55
  • $\begingroup$ Great, thanks :) $\endgroup$
    – user984949
    Nov 6, 2021 at 10:24

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