Only because
Rep is unitary, so saves positive-definite norm (for possibility density),
Casimir operators of the group have eigenvalues $m^{2}$ and $m^2s(s + 1)$, so characterizes mass and spin, and
It is the representation of the global group of relativistic symmetry,
yes?
Answer
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:
- The restriction maps $\mathcal H_1$ into itself.
- 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.
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