9

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

DanielC
  • 4,333
SRS
  • 26,333
  • 2
    Related: http://physics.stackexchange.com/q/21801/2451 , http://physics.stackexchange.com/q/22449/2451 and links therein. – Qmechanic Feb 25 '14 at 16:58
  • 4
    by definition, single particle states are irreps of Poincare or any other group of interest. Multiparticle states belong to symemtrized (bosons) or antisymmetrized (fermions) products thereof – John Feb 25 '14 at 17:54

1 Answers1

10

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 $[a,b]$.

  • Thanks for your answer. By "The said representation", do you mean the multiparticle Hilbert space? Furthermore, is $P^\mu P_\mu$ a Casimir operator? – Hunter Feb 26 '14 at 08:03
  • Yes, I'm making the text clearer. – Valter Moretti Feb 26 '14 at 08:04
  • 1
    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. – Valter Moretti Feb 26 '14 at 08:06