# Why do we say that irreducible representation of Poincare group represents the one-particle state? 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^2 s(s+1), so characterizes mass and spin, and 3. It is the representation of the global group of relativistic symmetry,

wstecznyg5 2022-07-16 Answered
Why do we say that irreducible representation of Poincare group represents the one-particle state?
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 ${m}^{2}s\left(s+1\right)$, so characterizes mass and spin, and
3. It is the representation of the global group of relativistic symmetry,
yes?
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## Answers (2)

Kyan Hamilton
Answered 2022-07-17 Author has 12 answers
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.
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 $\left\{0\right\}$. 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.
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ingwadlatp
Answered 2022-07-18 Author has 2 answers
The irreducible representations of the Poincare group are labeled by mass $m$ and the spin $s$ [this corresponds to Casimir invariants ${m}^{2}$ and ${m}^{2}s\left(s+1\right)\right)$, so it corresponds naturally to $1$-particle relativist states.
The states corresponding to a representation $m,s$ are labelled $|p,\lambda ⟩$, with ${p}^{2}={m}^{2}$ and $-s\le \lambda \le s$, 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:
$|p\lambda {⟩}_{1}|{p}^{\prime }{\lambda }^{\prime }{⟩}_{2}+|{p}^{\prime }\lambda {⟩}_{1}^{\prime }|p\lambda {⟩}_{2}$
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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