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

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    $\begingroup$ Related: physics.stackexchange.com/q/21801/2451 , physics.stackexchange.com/q/22449/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Feb 25, 2014 at 16:58
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    $\begingroup$ 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 $\endgroup$
    – John
    Commented Feb 25, 2014 at 17:54

1 Answer 1


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
    Commented Feb 26, 2014 at 8:03
  • $\begingroup$ Yes, I'm making the text clearer. $\endgroup$ Commented Feb 26, 2014 at 8:04
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    $\begingroup$ 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. $\endgroup$ Commented Feb 26, 2014 at 8:06

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