In page number 59 of his book on QFT, Weinberg mentions that for the operator $U$, defined for infinitesimal parameters $\omega$ and $\epsilon$ as: \begin{equation} U(1+\omega,\epsilon)=1+\dfrac{1}{2}i\omega_{\rho\sigma}J^{\rho\sigma}-\dfrac{1}{2}i\epsilon_{\rho}P^{\rho}+...\tag{2.4.3} \end{equation} (eq.(2.4.3)) to be unitary, the operators $J$ and $P$ must be Hermitian. But the Poincaré group is not compact and hence should have no non-trivial unitary representations of finite dimension. Doesn't Weinberg's statement violate this as he is assuming U to be unitary, even though $1+\omega$ and $\epsilon$ both belong to the Poincaré group?


You are right that non compact groups should not have finite dimensional unitary representations. But the generators $J^{\rho\sigma}$ and $P^\rho$ do not act on a finite dimensional vector space here. Using Weinberg notation they act on states $|p,\sigma, n, \ldots\rangle$ and, in particular, $p$ does not take values in a bounded set, the energies are arbitrarily high (indeed $J^{\rho\sigma}$ can boost). Equivalently you can think of these operators in the differential form $P^\rho = -\mathrm{i}\partial^\rho$ and $J^{\rho\sigma} = -\mathrm{i}(x^\rho \partial^\sigma - x^\sigma\partial^\rho)$. In any case they would act on $C^\infty$ functions which live in an infinite dimensional vector space.

Your point might be raised later on when he discusses massless representations. In fact the little group for a massless representation is $\mathrm{ISO}(2)$ (like Poincaré but in two dimensions). Since it is non compact we expect unitary representations to be infinite dimensional, nevertheless it is not the case: the photon and the graviton have two helicities! This is correct because such representations trivialize the action of the translation generators in $\mathrm{ISO}(2)$ (which have nothing to do with actual four dimensional translations by the way), thus effectively reducing it to $\mathrm{SO}(2)$. There is also the possibility of keeping such two dimensional translations and, sure enough, we end up with infinite dimensional representations, which are called continuous spin representations.

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  • $\begingroup$ Thanks a lot! I checked the book once again and realized that Weinberg had defined the $U(\Lambda)$ operators to act on state vectors $\Psi$ which belong to a $C^\infty$ dimensional Hilbert space with the inner product defined to be $\int \Psi^* \Psi dx$ which naturally makes U a unitary operator. $\endgroup$ – Sounak Sinha Dec 12 '18 at 9:21

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