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I'm a PhD student in mathematics and I have no problem in understanding what irreducible representation are. I mean that the mathematical side is not a particular problem. Nevertheless I have some problems in understanding why and in which sense these irreducible representations are considered as particles. What does this mean that if I have two electrons I have two irreducible representations of some group? And in this case of which dimension? And if they collide what do I get? Another representation? Can I get the equation of motion from this view? etc...

I know that this question should be already answered here somewhere, but I cannot find it anywhere explicitly stated in plain English. Can someone explain to me in plain English what's the point here?

Ruslan
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As you probably know, the Lie group of physical transformations of a quantum system acts on the Hilbert space of states of the system by means of a (strongly-continuous projective-) unitary representation of the group. $G \ni g \mapsto U_g$. This action is effective also on the observables of the system, represented by self-adjoint operators: The action of $g$ on the observables $A$ is $U_gAU^*_g$. The latter represents the observable $A$ after the action of the transformation $g$ on the physical system. This transformation has a twice intepretation. We can imagine that either it acts on the system or on the reference frame, our choice does not matter in this discussion.

Now let us focus on physics. There are natural elementary systems, called elementary particles. These systems are completely determined by fixing some real numbers corresponding to the values of some observables. Within the most elementary version of the story, these numbers are the mass $m$ which may attain a few positive numbers experimentally observed ad recorded, and the spin $s$ which may attain any number in $\{1/2, 1, 3/2,...\}$. Different values of the pair $(m,s)$ mean different particles.

These numbers have the property that they are invariant under the action of the most general symmetry group, I mean the (proper orthochronous) Poincaré group. A type of particle has the same fixed numbers $m$ and $s$ independently from the reference frame we use to describe it and the various reference frames are connected by the transformation of Poincaré group.

Passing to the theoretical quantum description of an elementary particle, in view of my initial remark, we are committed to suppose that its Hilbert space supports a representation of Poincaré group ${\cal P} \ni g \mapsto U_g$ (I omit technical details). Moreover there must be observables representing the mass $M$ and the spin $S$ that, on the one hand they must be invariant under the action of the group, i.e., $U_gM U_g^* =M$ and $U_gS U_g^* =S$ for every $g \in \cal P$. On the other hand they must assume fixed values $M=mI$ and $S=sI$.

Wigner noticed that a sufficient condition to assure the validity of these constraints is that ${\cal P} \ni g \mapsto U_g$ is irreducible.

Indeed, $M$ and $S$ can be defined using the self-adjoint generators of the representation, since they are elements of the universal enveloping algebra of the representation of the Lie algebra of $\cal P$ induced by the one of $\cal P$ itself. As expected, one finds $U_gM U_g^* =M$ and $U_gS U_g^* =S$ for every $g \in \cal P$. But, if $U$ is also irreducible, re-writing the identities above as $U_gM =M U_g$ and $U_gS =S U_g$ for every $g \in \cal P$, Schur's lemma entails that $M=mI$ and $S=sI$ for some real numbers $s,m$.

To corroborate Wigner's idea it turns out that the two constants $m$ and $s$ are really sufficient to bijectively classify all possible strongly-continuous unitary irreducible representations of $\cal P$ with "positive energy" (the only relevant in physics).

The mathematical theory of representations of ${\cal P}$ autonomously fixes the possible values of $s$ and they just coincide with the observed ones. The values of $m$ are not fixed by the theory of representations where any value $m\geq 0$ would be possible in principle, though not all $m \geq 0$ correspond to the masses of observed elementary particles.

If you have many elementary particles, the Hilbert space of the system is the tensor product of the Hilbert spaces of the elementary particles and there is a corresponding unitary representation of Poincaré group given by the tensor product of the single irreducible representations. Obviously, the overall representation is not irreducible.

ADDENDUM. I would like to specify that the irreducible representations of the group of Poincaré I discussed above are the faithful ones whose squared mass is non-negative. Moreover, there is another parameter which classifies the irreducible representations of Poincaré group. It is a sign corresponding to the sign of energy. Finally not all particles fit into Wigner's picture.

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    I'm a bit late here but which particles don't fit into Wigner's picture and what is the "enlarged picture" we adopt to accommodate them? – Kai Nov 18 '19 at 23:05
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    Neutrinos have ``oscillating'' mass so that they cannot be accomodated into Wigner's picture since it requires a fixed value of mass. Furthermore there are other particles which physically speaking should be cosidered elementary, for instance quarks, which have internal symmetries described by compact Lie groups ($SU(3)$ for quarks). In this case the Poincar'e group is not enough to define them and the gauge group must be included in their description. – Valter Moretti Jun 12 '20 at 13:19
  • @ValterMoretti Wait a minute, if we have a reducible representation we turn it into direct sum of irreducible representations on respective invariant subspaces. Something fishy here. Multiparticle systems are tensor product but this is direct sum. – MrPajeet Aug 17 '21 at 02:31
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    I am not sure to understand. However tensor products of irreducible reps are direct sums of irreducible representations, e.g., $1/2\otimes 1/2 = 0 \oplus 1$ for $SU(2)$. – Valter Moretti Aug 17 '21 at 19:57
  • @ValterMoretti I meant to say, suppose we have, say n systems, each is a 1 particle system. Then we have n Hilbert spaces $\mathcal{H}_i$ together with an irreducible representation $\pi_i$ of the symmetry group. Then when we form the combined system it should be the tensor product. Then the tensor product of these representations will be $\otimes_i \mathcal{H}_i$ and the representation will be $\pi=\pi_1\otimes I\otimes\cdots\otimes I+I\otimes\pi_2\otimes I\otimes\cdots\otimes I+...$.

    Shouldn't this correspond to n times reducible representation?

    I am a bit confused

    – MrPajeet Aug 17 '21 at 23:07
  • I cannot see what the point is. What it is true is that irreducible representations of the Poincaré group describe elementary particles. This does not imply a particular statements about the interplay of multiparticle systems and reducible representations. Mathematically speaking, reducible representations can be constructed as a sum of irreducible representations and physically speaking multiparticle systems are obtained as tensor products of elementary representations. These facts are compatible. – Valter Moretti Aug 18 '21 at 06:41
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    It is nowhere required that multiparticle systems are direct sum of irreducible representations. Wigner's idea regards the interplay of elementary systems and irreducible representations. – Valter Moretti Aug 18 '21 at 06:46
  • @ValterMoretti Ah I see! Thanks a lot I was confused about this for a long time. It makes sense now – MrPajeet Aug 18 '21 at 07:08
  • @ValterMoretti In this answer by Josh, he states something like what I was trying to say. Maybe this is where my confusion started.

    https://physics.stackexchange.com/questions/73593/why-do-we-say-that-irreducible-representation-of-poincare-group-represents-the-o

    – MrPajeet Aug 18 '21 at 07:13
  • I see, that answer deals with the structure of Fock space describing the space of the states of a system with an indefinite number of identical particles. That space is a direct sum of tensor products of irreducible representations. The first two are the trivial representation and the, say, "fundamental" one-particle representation (of the Poincaré group), next we add all possible (suitably symmetrized) tensor products of the latter. This is a fact arising from free QFT, independent but compatible with Wigner's idea of elementary particle. – Valter Moretti Aug 18 '21 at 08:13
  • @ValterMoretti So here is what's happening I think
    1. One particle system correspond to irreducible representation
    2. Fock space should represent many particle system (symmetric tensor product if bosonic, skew if fermionic)
    3. Representation on the Fock space is direct sum of representation of the form, $0\oplus \pi \oplus (\pi\otimes 1 + 1\otimes \pi)\oplus \cdots$.
    4. Now this is reducible to each tensor component

    My problem is then the following, if we just have some Hilbert space with 2 invariant subspaces. Is there any physical meaning to these invariant subspaces?

    – MrPajeet Aug 18 '21 at 15:24
  • For example if $(\mathcal{H},\pi)$ is an irreducible representation, does $(\mathcal{H}\oplus\mathcal{H},\pi\oplus\pi)$ have any physical meaning? – MrPajeet Aug 18 '21 at 15:33
  • @Boltzee It depends on the context and on the physical meaning of the mathematical objects. In your case, for instance, you may add a new generator to your group of symmetry (commuting with the initial group) with the meaning of "charge", and taking eigenvalues $1$ and $-1$ in each of the two subspaces. The whole representation is again irreducible (for the enlarged group) and the two subspaces describe the two states of charge of an elementary object. – Valter Moretti Aug 19 '21 at 18:20
  • @ValterMoretti Thank you very much! That was very helpful. The new generator will be projection onto the first subspace direct sum -ve of projection onto the second subspace no? – MrPajeet Aug 20 '21 at 00:00
  • @Boltzee Yes, the generator has the form you said. – Valter Moretti Aug 20 '21 at 06:54
  • "Wigner noticed that a sufficient condition to assure the validity of these constraints is that [the representation] is irreducible." Is this also a necessary condition? I.e., if $M$ and $S$ have the above properties, must the representation be irreducible? – WillG Jul 29 '22 at 22:46
  • No, it is not necessarily irreducible. – Valter Moretti Jul 30 '22 at 06:38