# What does it mean for particles to "be" the irreducible unitary representations of the Poincare group?

I am studying QFT. My question is as the title says. I have read Weinberg and Schwartz about this topic and I am still confused. I do understand the meanings of the words "Poincaré group", "representation", "unitary", and "irreducible", individually. But I am confused about what it means for it to "be" a particle. I'm sorry I'm not sure how to make this question less open-ended, because I don't even know where my lack of understanding lies.

Irreducible representations of the Poincare group are the smallest subspaces that are closed under the action of the Poincare group, which includes boosts, rotations, and translations. The point is that we should interpret these subspaces as the set of possible states of a particle. For example, if you start with a state representing a particle at rest, then you can boost it (so it starts moving), rotate it, translate it, and so on. But all the states you can reach represent, by definition, the same kind of particle, just in different states of motion.

The requirement that the representation be unitary just means these operations keep states normalized.

• So is it more correct to say "irreducible representations of the Poincaré group form the state spaces of elementary particles"? Rather than saying that they are the particles. Oct 11, 2020 at 22:42
• @Charlie Yes. Just saying particles "are" irreducible representations is concise, but really confusing. Oct 11, 2020 at 22:42
• the representation might be indecomposable rather than irreducible and you might still reach all states from some states. Oct 11, 2020 at 22:43
• @Vincent Thanks for the comment, I was definitely oversimplifying. I tried to make things a bit more precise. Oct 12, 2020 at 17:51
• @knzhou, whether particles “are” the irreducible representations depends on your philosophical view. If your view is that there is something in the outside world that corresponds to the term particle, then the irreducible representations just describe it, but if your view is that the outside world (if it exists at all) is a mess and particles are just a concept in our heads we use to describe it, then particles are just the model. Since physics can only give you models, it is compatible with either philosophy. Oct 13, 2020 at 18:25

This is really a deep question. I am still learning, so any feedback is more than welcome

My key takeaways and interpretations are:

1. Particles are interpreted as field excitations
2. The complexified (thanks ZeroTheHero, for clarifying) full Poincaré ISO(3,1) when studied, e.g. through the Little Group (Wigner method), turns out to have an algebra that is isomorphic to $$su (2) \oplus su(2)$$
3. That decomposition gives us the allowed subspaces (incarnated through Weyl spinors, the Electromagnetic Tensor and such) for physical theories
4. From that, irreps are found
5. Irreps are the foundational blocks to represent any group in a physical theory
6. Particles, being field excitations, have its quantum numbers (spin for instance)
7. The irreps give natural quantum numbers, which can be discriminated through the Spin-Statistics Theorem as bosons or fermions (anyons if we're working with different dimensions)
• Point 2. is not strictly true. The complexification is isomorphic to $su(2)\oplus su(2)$ but the real algebra is not. In particular, the adjoint representation over the reals is irreducible, although of course over the complex is it precisely a direct sum. Oct 12, 2020 at 2:22
• Many thanks, @ZeroTheHero Oct 12, 2020 at 3:14
• As zero the hero writes correctly in the comment, but your answer does not: it is $\oplus$ and not $\otimes$ in point 2 Oct 12, 2020 at 12:32
• Otherwise I really like the answer (+1) Oct 12, 2020 at 12:32
• My bad! Thanks a lot, Vincent! Oct 12, 2020 at 17:17

I apologize for not reading all of the answers. In "classical" QFT (e.g., QED and the current standard model that extends QED to electroweak theory -- except that real neutrinos observed by experiment flavor mix and thus effectively are not massless -- as well as QCD), the Poincaire group describes the requirement of Einstein special relativity (as contrasted with Galileo relativity of Newtonian physics). The field operators that create and destroy particle states in QFT must in terms of all observable quantities in the theory obey special relativity and thus the simplest requirement is to pick states that are irreducible for spatial coordinate system transformations (e.g., the parameters of a field space-time point -- x y z t as one representation of a space-time point) and boosts as required by special relativity -- which classically is the Poincaire group of transformations. Note that non-invariant elements may appear in the theory provided these do not make non-invariant observables (i.e., nominally what could be measured in an experiment).

• As a new contributor, Welcome. However, it’s helpful to read preceding answers before writing to see whether you have anything to add, especially on a subject like this one, about which a lot has already been written, both in the links added in comments under the Question, and in any others listed in the LINKED and RELATED columns to the right side of this screen. Oct 13, 2020 at 17:49