2
$\begingroup$

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?

$\endgroup$
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.