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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.

Qmechanic
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UrsaCalli79
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  • Related: https://physics.stackexchange.com/q/73593/2451 , https://physics.stackexchange.com/q/21801/2451 and links therein. – Qmechanic Oct 12 '20 at 02:45
  • The question and answers here might be enlightening: https://physics.stackexchange.com/questions/191010/rest-mass-and-wigners-classification – WillO Oct 12 '20 at 06:06

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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.

knzhou
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    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. – Charlie Oct 11 '20 at 22:42
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    @Charlie Yes. Just saying particles "are" irreducible representations is concise, but really confusing. – knzhou Oct 11 '20 at 22:42
  • Ok I see, ty, nice answer. – Charlie Oct 11 '20 at 22:43
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    the representation might be indecomposable rather than irreducible and you might still reach all states from some states. – ZeroTheHero Oct 11 '20 at 22:43
  • @ZeroTheHero True, but does that distinction ever matter in QM? – knzhou Oct 11 '20 at 22:46
  • Thanks for the clarification. I guess a follow up question would be, how is this used to differentiate between different species of particles? – UrsaCalli79 Oct 11 '20 at 22:51
  • Yes if you want a theory with particle decay. See Dirac, P.A.M., 1984. The future of atomic physics. International journal of theoretical physics, 23(8), pp.677-681. (Dirac’s last journal paper incidentally.) – ZeroTheHero Oct 11 '20 at 22:51
  • @UrsaCalli79 The different kinds of irreducible representation tell you about different types of particles, i.e. information about their masses and spins. – knzhou Oct 11 '20 at 23:11
  • I am willing to believe that this is the best answer to the question as stated, but mathematically something is wrong (this relates to ZeroTheHero's comment): suppose you pick a vector $v$ is some irreducible representation and a vector $w$ in a different irreducible representation (not just a different copy, but truly different, i.e. not isomorphic as representations) and then let your statevector be $v + w$. Then the span of all the images of $v + w$ under all the group elements is some subrepresentation of the big Hilbert space (by definition, as you write), but not irreducible. – Vincent Oct 12 '20 at 12:54
  • I don't know what the difference between general and irreducible representations (or between state vectors generating irreducible and non-irreducible representations under rotations and boosts) looks like 'physically', but something needs to be added to the answer, it seems – Vincent Oct 12 '20 at 12:55
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    @Vincent Thanks for the comment, I was definitely oversimplifying. I tried to make things a bit more precise. – knzhou Oct 12 '20 at 17:51
  • Nice work. I upvoted – Vincent Oct 12 '20 at 20:19
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    @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. – Jan Hudec Oct 13 '20 at 18:25
  • Sorry to be dense, but as far as I understand, group representations aren't vector spaces at all, but are group homomorphisms (to $\mathrm{GL}(n)$). So what does your first sentence mean? – tparker Oct 15 '20 at 03:17
  • @tparker This is the usual issue where some people say a representation means a group homomorphism to $GL(n)$, some people say it means the vector space it acts on, and yet others take it to be an ordered pair containing both. In particle physics we always mean the second, because we think of fixing the action of the group elements first, then carving up the Hilbert space into irreducible pieces. It's very useful to give those pieces a snappy name, since they represent important things like particles or multiplets, so we call them representations. – knzhou Oct 15 '20 at 04:10
  • @knzhou So you're essentially identifying representations of the Poincare group and Hilbert subspaces corresponding to equivalence classes of physical states that are related by Poincare transformations? Wouldn't the latter subspaces all be isomorphic to the full Poincare group? – tparker Oct 16 '20 at 23:23
  • @knzhou And is there something special about single-particle states here? Wouldn't a state consisting of many interacting particles also fit into some representation under this definition? This concept of a particle "being" an irrep of the Poincare group has always frustrated me, because I don't think it's actually saying anything conceptually very complicated, but I've never seen any clear explanation of that statement so I've never had any idea what it means. – tparker Oct 16 '20 at 23:25
  • @tparker "Wouldn't the latter subspaces all be isomorphic to the full Poincare group?" I don't understand what you're asking here; how is a subspace isomorphic to a group? – knzhou Oct 16 '20 at 23:46
  • @tparker If you took the full interacting Hilbert space and decomposed it into Poincare irreps, some would contain single-particle states and some would contain multi-particle states. The only reason we're usually talking about the former is because we usually apply this reasoning to the "in" and "out" states related by the S-matrix. These correspond to infinitely separated particles, so we can just consider one at a time. – knzhou Oct 16 '20 at 23:48
  • @knzhou Start with some reference state - say, an electron wave packet at a certain position with a certain momentum (with appropriate uncertainties, etc.). Each element of the Poincare group is represented by some unitary operator, which maps that state to a new state. It seems to me that generically, each of these uncountably many new states would be linearly independent. That's what I mean by the subspace spanned by all of these translated/rotated/boosted states being isomorphic to the Poincare group (once you pick a reference state to correspond to the identity element). – tparker Oct 17 '20 at 02:04
  • @knzhou I guess strictly speaking, it's just the basis that's isomorphic to the Poincare group, not the full subspace. But it seems that under your definition, any Poincare irrep would be an infinite-dimensional subspace, which isn't actually the case. – tparker Oct 17 '20 at 02:05
  • @tparker All Poincare irreps that corresponds to elementary particles are infinite-dimensional. Finite-dimensional Poincare irreps show up all the time but don't represent the same thing; for example, the vacuum state is typically in the trivial irrep. – knzhou Oct 17 '20 at 02:59
  • I see. This is very clarifying, thank you. From what I've seen, this topic is not well-explained in textbooks. – tparker Oct 17 '20 at 15:07
  • It seems to me that another reason why the statement "elementary particles are Poincare irreps" is misleading (beyond the clear lack of semantic clarity around the word "are") is that apparently many interacting multiparticle states are also Poincare irreps. It seems circular to say that "We can identify individual particles with Poincare irreps because it's implicitly understood that we're only considering individual particles" (i.e. far-separated S-matrix states). In order to say whether the particles in an incoming state are "far-separated", you need to already know what the particles are – tparker Oct 17 '20 at 15:16
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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)
daydreamer
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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).

Yasha Karant
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  • 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. – iSeeker Oct 13 '20 at 17:49