# Doubt in Weinberg's book on Quantum Field Theory

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.
• 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. – Sounak Sinha Dec 12 '18 at 9:21