Only because

  1. Rep is unitary, so saves positive-definite norm (for possibility density),

  2. Casimir operators of the group have eigenvalues $m^{2}$ and $m^2s(s + 1)$, so characterizes mass and spin, and

  3. It is the representation of the global group of relativistic symmetry,


  • 4
    $\begingroup$ related: physics.stackexchange.com/q/65839 $\endgroup$ Commented Aug 8, 2013 at 19:35
  • $\begingroup$ But not identical. The author of the question linked above asked about paragraph in Weinberg book. My question is more generalized. $\endgroup$
    – user8817
    Commented Aug 8, 2013 at 19:40
  • $\begingroup$ I totally agree; that's why I used the term "related" as opposed to "duplicate." $\endgroup$ Commented Aug 8, 2013 at 19:41
  • 1
    $\begingroup$ Even though I just answered this question, I think that I found a possible duplicate (even though I hope that the content of my answer is sufficiently different from the answer there to warrant a non-close) physics.stackexchange.com/q/21801 $\endgroup$ Commented Aug 8, 2013 at 20:13
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/21801/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Sep 23, 2014 at 22:31

2 Answers 2


First, note that in physics, we consider unitary representations $U$ of the Poincare group acting on the Hilbert space $\mathcal H$ of the theory because we are interested in a precise formulation of the concept of Poincare transformations acting on the quantum mechanical states of the theory as symmetries (since the laws of physics should be inertial frame-invariant); and by Wigner's theorem, we choose these symmetries to be realized by unitary operators. These observations are related to your #1 and #3 and I think they should be kept conceptually distinct from the notion of a state that represents a single particle state.

Second, since such quantum field theories are supposed to allow for the emergence of states of particles, and in particular should account for states in which there is a single elementary particle, we expect that there is some subset $\mathcal H_1$ of the Hilbert space of the theory corresponding to states "containing" a single elementary particle.

Given these observations, let's rephrase your question as follows:

What properties do we expect that the action of the representation $U$ will have when its domain is restricted to the subspace $\mathcal H_1$?

In particular, we would like to justify the following statement

The restriction of the unitary representation $U$ acting on $\mathcal H$ to the single-particle subspace $\mathcal H_1$ is an irreducible representation of the Poincare group acting on $\mathcal H_1$.

This requires justifying two things:

  1. The restriction maps $\mathcal H_1$ into itself.
  2. The restriction is irreducible.

I think that the justification of the first property is pretty intuitive. If all we are doing is applying a Poincare transformation to the state of the system, namely we are just changing frames, then the number of particles in the state should not change. It would be pretty strange if you were to, for example, boost or rotate from one inertial frame into another and find that there are suddenly more particles in our system.

The irredicibility requirement means that the only invariant subspace of the single particle subspace $\mathcal H_1$ is itself and $\{0\}$. The physical intuition here is that since we are considering a subspace of the Hilbert space in which there is a single elementary particle, expect that there is no non-trivial subspace of $\mathcal H_1$ in which vectors of this subspace are simply "rotated" into one another. If there were, then the particle would not be "elementary" in the sense that the non-trivial invariant subspace would represent the states of some "more elementary" particle. When it really comes down to it, however, I'm not sure if there is some more fundamental justification for why the restriction of $U$ to $\mathcal H_1$ is irreducible aside from the decades of experience we've now had with particle physics and quantum field theory.

  • $\begingroup$ @ joshphysics- Your intuitive explanation is very clear. However, I would like to know whether the representations labeled by Casimir operators implies that the representation is irreducible. If yes, then why? $\endgroup$
    – SRS
    Commented May 12, 2015 at 6:43
  • $\begingroup$ @SRS That is a separate mathematical question -- I think it would be best addressed as its own new question. $\endgroup$ Commented May 12, 2015 at 6:58

The irreducible representations of the Poincaré group are labeled by mass $m$ and the spin $s$ [this corresponds to Casimir invariants $m^2$ and $m^2s(s+1))$, so it corresponds naturally to $1$-particle relativist states.

The states corresponding to a representation $m, s$ are labelled $|p,\lambda \rangle$, with $p^2=m^2$ and $-s \leq \lambda \leq s$, and it corresponds here too to $1$ particle.

For multi-particle states (Fock states), we have symmetric or anti-symmetric tensorial products of these states, for instance, a $2$-particle bosonic state may be written:

$$|p \lambda \rangle_1|p' \lambda' \rangle_2 + |p' \lambda \rangle'_1|p \lambda \rangle_2$$

It is clear that these multi-particle representations are no more irreductible (because they are a sum of product of irreductible representations) .

The unitarity has no influence on this, Fock states corresponds to a unitary representation.


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