Why the Pauli-Lubansky and momentum operator build an irreducible representation of Poincarè group?

We know that particle states in QFT are identified with irreducible representation of Poincarè group, in particular they can be identified using Pauli-Lubansky and (squared) Momentum operator (wich are Casimir) so they can be labeled by their mass and Spin. My question is: how can we prove that this representation is irreducible? I think I'm missing the link between the Casimir operators and the irreducible nature of the representation.

• A pure math question. In short, the 2 operators from the universal enveloping algebra of the Poincare algebra are Casimirs, therefore they commute with all generators. This means that in a basis of their own eigenvectors they are proportional to the unit operator > by Schur's lemma a representation of the algebra built with a basis of Casimir's eigenvectors is irreducible. Commented Feb 15, 2023 at 11:30
• @DanielC you should post this as an answer. Commented Feb 15, 2023 at 12:10
• More on representations of the Poincare group. Commented Feb 15, 2023 at 12:15

Let $$G$$ be a group and $$\rho:G\to{\rm GL}(V)$$ be a representation. We say that the representation is irreducible when there are no non-trivial invariant subspaces. A subspace $$W\subset V$$ is said to be invariant when $$\rho(g)\cdot W \subset W$$ for all $$g\in G$$. The trivial invariant subspaces are then $$V$$ and $$\emptyset$$.

Now let us consider a unitary representation of Poincaré (more precisely of its universal cover). Now consider the mass Casimir $$P^2$$, we may slice the representation space $${\cal H}$$ by the value it attains on the states. In other words, we define $${\cal H}_{m^2}\subset{\cal H}$$ to be the subspace of all states $$|\psi\rangle$$ such that $$P^2|\psi\rangle=-m^2|\psi\rangle$$. Likewise, we can do the same for the Pauli-Lubanski Casimir $$W^2$$, and further slice the Hilbert space by the values of both operators, defining subspaces $${\cal H}_{m^2,\sigma}\subset{\cal H}$$.

What happens then is that you may show that $$P^2$$ and $$W^2$$ commute with the action of the Poincare group, because they are invariants. As such let $$|\psi\rangle\in{\cal H}_{m^2,\sigma}$$,

$$P^2U(\Lambda,a)|\psi\rangle = U(\Lambda,a)P^2|\psi\rangle=-m^2U(\Lambda,a)|\psi\rangle,$$

with the analogous computation for $$W^2$$, showing that $$U(\Lambda,a)|\psi\rangle\in{\cal H}_{m^2,\sigma}$$. In other words, the subspaces $${\cal H}_{m^2,\sigma}$$ are invariant.

If they are invariant and the representation is to be irreducible, they must be trivial. So they are either $$\emptyset$$, in which case they are not really there, or they are the whole space $${\cal H}$$, in which case there is really just a single one of them and the representation gets indexed by the values of $$m^2$$ and $$\sigma$$.

Showing that $$P^2$$ and $$W^2$$ commute with all Poincaré transformations can be done explicitly by first showing that if $$C^\mu$$ is an operator that transforms as a four-vector, namely,

$$U(\Lambda,a)C^\mu U^{-1}(\Lambda,a)=\Lambda^\mu_{\phantom\mu\nu}C^\nu,$$

then $$C^2 = C_\mu C^\mu$$ is a scalar

$$U(\Lambda,a)C^2 U^{-1}(\Lambda,a) = C^2.$$

Now one just needs to show that $$P^\mu$$ and $$W^\mu$$ transforms as four-vectors. In particular, this is accomplished using

$$\begin{eqnarray} U(\Lambda,a)J^{\rho\sigma}U^{-1}(\Lambda,a) &=& \Lambda^{\rho}_{\phantom \rho\mu}\Lambda^{ \sigma}_{\phantom \sigma\nu}(J^{\mu\nu}-a^\mu P^\nu + a^\nu P^\mu),\\ U(\Lambda,a)P^\rho U^{-1}(\Lambda,a) &=& \Lambda^{\rho}_{\phantom \rho\mu}P^\mu. \end{eqnarray}$$

• How can you show that Casimir operators commute with every U(\Lambda,a)?Are there some good references? Thank you for the answer btw Commented Feb 15, 2023 at 15:02
• In greater generality, Casimir operators lie in the center of the universal enveloping algebra of the Lie algebra of the group you are working with. As a result, they commute with all generators and all powers of all generators and in particular, they will commute with the finite transformations. We can nevertheless show explicitly that they commute with any finite Poincaré transformation. I've added an edit about it, see if it helps you.
– Gold
Commented Feb 15, 2023 at 15:31
• Thank you so much, I think you missed a $C^{2}$ in the right hand side of $C^{2}$ transformation, but now it's clear! Commented Feb 15, 2023 at 15:53
• You're welcome! Thanks for letting me know, I indeed missed a $C^2$ on the RHS of that equation. Just fixed it.
– Gold
Commented Feb 15, 2023 at 16:06

There are some theorems in mathematics that basically help us ascertain that:

• The (universal enveloping algebra of the) Lie algebra of the Poincare group has two Casimir operators (both quadratic), one the total momentum squared, the other the squared Pauli-Lubanski 4-vector.
• The Casimirs commute with all group generators (members of the algebra), therefore, if one considers a basis made up of eigenvectors of the Casimirs, this basis can be taken to span the representation space.
• The representation is irreducible by Schur's lemma applied to the two Casimirs.

More details here: https://people.fjfi.cvut.cz/snobllib/Casimirs_Lesna.pdf

There is, actually, no way to prove that, because this is a qution of the definition of an elementary particle (it is not so for system of particles). It can be only motivated. Why do we say that, say, positron and proton are two different particles, but two protons with different momenta are differen state of the same kind of particle (let us forget about weak and strong interactions for simplicity)? Because in the second case one state can be transformed to another by a change of the reference frame. In the quantum case it is more difficult (because a state is a more complicated thing), but if you think about it, the idea of irreducibility does the job. If the statespace consists of two subspaces, invariant under the action of the symmetry group, it means that it describes more than one type of particle. The connextion between Casimirs and irreducibility is explained in the Gold's answer. But I want to underline that irreducibility implies that Casimir operators are represented by numbers, not the other way around.