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  Elementary particles as irreducible representations

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It is known that, in Wigner's view, free elementary particles are nothing but irreducible representations of Poincaré group. However, Arnold Neumaier recently suggested to me that this picture is not appropriate,  in fact it is misleading,  due to what  nowadays we know  about (physically) elementary particles: Both quarks and neutrinos are not described in terms of such representations because the mass is not a Casimir operator of the representation as the mass of these particles is not a constant observable, but is described by the Cabibbo-Kobayashi-Maskawa Quark Mass Matrix  and a Neutrino Mass Matrix. He also pointed out to me that these particles actually are irreducible representation of the global symmetry group.

Could you show me the irreducibility of these representations of the global symmetry group or provide me some technical reference where an explicit  proof appears?

asked Aug 11, 2014 in Theoretical Physics by Valter Moretti (2,085 points) [ revision history ]

1 Answer

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The particles of the standard model fall into irreducible representation of the direct product of the Poincare group ISO(1,3), the gauge group S(U(3) x U(2)) - often written as SU(3) x SU(2) x U(1) -, and a group U(3)  interchanging the particle generations participating in the weak interaction. This defines the symmetry group of the standard model. The unbroken symmetry group is only SU(3) x U(1) x U(1)^3; the remaining part of the symmetry group is broken in Nature: The Poincare symmetry through the nonuniform small-scale distribution of matter, the SU(2) x U(1) symmetry through the different masses of proton and neutron, and the generation SU(3) through the different masses of the electron generations.

There are two irreducible fermionic representation: the quarks form one, the leptons form the other. There are three irreducible bosonic representation: the  gluons form one, photon, Z-boson and W-boson another, and the Higgs bosons the third.

Each of these representations splits into multiple irreducible representations under the Poincare group. Since it is customary to give names to the particles in each of these representations, we have 6 quarks, 6 leptons (3 electrons and 3 neutrinos), 8 gluons, 1 photon, 1 Z-boson, 2 W-boson, and several Higgs bosons, depending on the precise structure of the Higgs sector. Each of these particles has a fixed mass, the mass being a Casimir operator of the Poincare Lie algebra. The different particles in this sense may be viewed as the different components of the various fields appearing in the standard model Lagrangian if the (additional) spin index is suppressed.

If one splits instead into irreducible representations of Poincare x Isospin SU(2), one gets 3 quark generations, 3 lepton generations (each consisting of one electron and one neutrino), 8 gluons, 1 photon, 1 Z-boson, 1 W-boson generation, and in the minimal case one Higgs generation. The generation U(3) interchanges the three generations of quarks, and the three generations of leptons.

Quark can appear in a superposition of the mass states; this is due to the existence of a quark mass mixing matrix: http://en.wikipedia.org/wiki/Cabibbo–Kobayashi–Maskawa_matrix. Similarly, neutrinos can appear in a superposition of the mass states; this is due to the existence of a neutrino mass mixing matrix: http://en.wikipedia.org/wiki/Pontecorvo–Maki–Nakagawa–Sakata_matrix. Though the latter define by convention the standard quarks and neutrinos (as irreducible representations of the Poincare group), they appear in the form of superpositions in the weak interaction. For neutrinos, this explains the neutrino oscillations. There is no corresponding mixing matrix for the three generations of electrons since these are subject to the charge superselection rule, which forbids superpositions of different eigenstates of charges. 

However, because of the mass mixing, the irreducible representations of the Poincare group do not form superselection sectors. As a consequence, there is no superselection rule for the mass, and the mass is a nontrivial operator (3 times 3) on the single particle sector of quarks (and of neutrinos).This contrasts with nonrelativistic mechanics, where the structure of the Galilean group forces such a (Bargmann) superselection rule, and hence every single-particle sector is characterized by a numerical mass.

answered Aug 11, 2014 by Arnold Neumaier (15,787 points) [ revision history ]
edited Aug 11, 2014 by Arnold Neumaier

(Galileian vs Galilean: Galilean arises from the name, Galileian from the family name. I have learned it from the translator of a  book of mine. http://en.wiktionary.org/wiki/Galileian ) Regarding the question. I do not understand if your answered. So neutrinos and quarks are irreducible representations? If yes, of which global symmetry group? 

The spelling tradition is deeply rooted and shouldn't be changed: 

Galilean 2.4 million hits on Google; Galileian 21.000 hits on Google. 

Hermitian 1.5 millions hits; Hermitean 51.000 hits

I'll edit my answer to address the other issue.

There is a further statement I do not understand well in your answer. "Thus the irreducible representations of the Poincare group do not form superselection sectors. As a consequence, there is no superselection rule for the mass".  Well, irreducible representations of the Poincaré group do not form superselection sectors even in Wigner's picture for "naive" free  elementary particles, since  coherent superpositions of states with different masses (different numbers of particles of the same kind) are allowed in relativistic physics, differently from classical physics, due to Bargmann superselection rule.  So, I am not sure I understand well the meaning of that sentence in the context of your answer. Please clarify it when editing your answer. 

I completely rewrote my answer, hoping to satisfy you.

Thanks! I have to read it carefully...

Let me understand. Are you saying that these particles are irreducible representations of the remaining unbroken symmetry? Could you explain better how Poincaré symmetry breaking occurs? (This is very interesting! You know I am now a mathematician, I studied some of these things many many years ago).

If you only consider the unbroken symmetry group SU(3) x U(1) x U(1)^3, it has very many different irreps. This makes it meaningless for classification purposes.

Think of the different particles (i.e., irreps of Poincare) as indices of fields (without the spin>0 indices). Isospin SU(2) groups the particle with nonzero isospin into pairs. Generation U(3) groups the particles with nontrivial generation into triples. Thus the six quarks give three isospin triples and a single irrep under SU(2)_iso x U(3)_gen.

Poincare-symmetry means translation symmetry, rotation symmetry and boost symmetry. Space translation and rotation symmetry is broken since we see around us objects that are neither translation invariant nor rotation invariant.  Time translation symmetry is broken since we observe change. Boost symmetry is broken since an observer (e.g., you, me, or a camera) distinguishes a particular future-directed world line. (The empty universe is translation invariant and even diffeomorphism invariant, but as soon as you put in matter, it breaks symmetry unless the matter is homogeneous in space and time.) 

In QFT, the vacuum is the only Poincare invariant state. The theory is Poincare invariant since one considers all states simultaneously. But reality is uninteresting without preferred states.

Note that the generation U(3) is usually not considered to be a symmetry, as it is explicitly broken in the standard model Lagrangian to U(1) x U(1) x U(1). But it is naturally associated with the mixing matrices. Dropping it from the description doesn't change the main conclusion that even for single particle systems, mass may be a non-numerical operator.

It's interesting to compare "Galileian,Galilean" using the Google Ngram Viewer for the English corpus and then for the Italian corpus, which shows something of why an Italian reader might have a different take on this (or both compared together). For this kind of comparison I'd generally choose to use the Google Ngram Viewer, which is based on published books and gives some insight into how usage has changed over the last 200 years, instead of using Google on the web, as it is today only.

There is a point I do not understand Arnold. You say that Poincaré invariance is broken "in practice", but what about  the Lagrangian of neutrinos including the matrix of masses? I suspect that it is not broken there. My idea is that the one-particle Hilbert space of the "final" theory is a product $H\otimes H_{int}$ where $H$ is the Hilbert space of the external parameters, momenta and spin, $H_{int}$ is finite dimensional and includes all (surviving) internal symmetries. I guess that this gobal Hilbert space  space is an irreducible rep of $SO(3,1)_+ \times G_{int}$, where $G_{int}$ is the group of internal symmetries. Now the mass operator is a matrix in $H_{int}$ function of the ($n\geq 2$) Casimir operators of $G_{int}$. That matrix, at last for neutrinos, does not commute with the Hamiltonian operator.

Very interesting Peter...Indeed, as a representative of Italian community, I am always confused when I have to decide between Galilean and Galileian.

There is a similar problem/confusion in Italian: "gruppo di Galileo" (name of the famous scientist)  or "gruppo di Galilei" (family name of the famous scientist).

''neutrinos including the matrix of masses'' - this only breaks the generation symmetry, which is already broken because of the different masses of the electron and its heavier generation partners.

The 1-particle Hilbert space is the space of linear mappings from $R^3$ (coordinatizing the spatial momentum) - with time component of $p$ determined by the masses (in the mass eigenstates) - to a finite-dimensional complex representation space which is the direct sum of the representation spaces of the four irreps of the biggest symmetry group mentioned at the beginning of my answer. Thus it is not irreducible. The smaller one makes the group the more irreducible components one gets. (Various grand unification schemes therefore try to work with larger groups, where all particles fit into a single irrep, but there is no experimental evidence for that, as it invariably means that many additional particles must also exist, and none of them turned up so far.)

It is not irreducible, you argued that it was irreducible. I quote from you comment to my answer in PSE:

"In today's theories, elementary particles are irreducible representations of the full symmetry group, which is not just the Poincare group. If the mass transforms nontrivially under the internal symmetry group, it is a matrix rather than a number."

What did you mean there so?

My quoted remark referred to each class of elementary particles separately. 

The representation of all elementary particles is not irreducible. But this is because elementary particles with different spin cannot be in a common finite-dimensional irreducible representation of a group. But the discussion on PSE was not about all elementary particles but about neutrinos (or quarks), and these are in a single representation of Poincare x SU(2)_iso x U(3)_gen. All components of this representation space are mixed since there is no corresponding superselection rule. The representation splits into 3 (neutrinos) resp. 6 (quarks)  irreducible representations of the Poincare group. These representation are mixed by the interactions, hence neutrino and quark masses are mixed as well. 

In fact, this implies that there is even a simpler example of mass mixing (which I didn't think of before) - the nucleon. It is the basic particle in nuclear physics and appears in their models as superpositions of proton, neutron and their antiparticles. The charge superselection rule (for the U(1) charge of electromagnetism) doesn't apply on the nuclear level as nuclear physics is blind to electromagnetic interactions. 

Thank you for posting this discussion.  (1)  Can you be more explicit about how U(3)_gen acts?  (2)  In the high energy limit where the SM gauge group is unbroken, shouldn't fermions be massless, and hence LH and and RH fermions in distinct irreps?

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