Which type of states carry the irreducible unitary representations of the Poincare group? Multi-particle states or Single-particle states?


Essentially by definition (due to Wigner), one-particle Hilbert spaces of elementary particles support unitary strongly continuous irreducible representations of Poincaré group.

Conversely, any multi-particle Hilbert space, with either fixed or undefined number of particles either identical or distinguishable, cannot be irreducible under the action of the associated representation of Poincaré group.

Proof. A multi-particle representation is the tensor product of the representations in each factor one-particle subspace. If $P_\mu$ denotes the total four-momentum operator of the system of particles, the bounded unitary operator $e^{i a P_\mu P^\mu}$ ($a\in \mathbb R$) commutes with all the unitary operators of the tensor representations and it is not proportional to the identity operator (as it instead happens for a one-particle space). In view of Schur's lemma the representation cannot be irreducible.

An invariant closed subspace is, evidently, the subspace of state vectors where the squared mass $M^2= P_\mu P^\mu$ assumes values (in the sense of spectral decomposition) inside a fixed interval $[a,b]$.

  • $\begingroup$ Thanks for your answer. By "The said representation", do you mean the multiparticle Hilbert space? Furthermore, is $P^\mu P_\mu$ a Casimir operator? $\endgroup$ – Hunter Feb 26 '14 at 8:03
  • $\begingroup$ Yes, I'm making the text clearer. $\endgroup$ – Valter Moretti Feb 26 '14 at 8:04
  • 1
    $\begingroup$ Yes, it is a Casimir operator in the sense that it commutes with all the generators of the representation, but it is not a constant, as instead it turns out for one-particle reprs. $\endgroup$ – Valter Moretti Feb 26 '14 at 8:06

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.