# Irreducible representations and elementary particles

## Main Question or Discussion Point

Hi,
I had a question about irreducible representations and elementary particles....
Namely, I've been told by teachers and read in a few texts that particles ARE irreducible representations, and I have never been able to wrap my mind around what that means.
Please keep in mind that I am no theorist or mathematician, so I may be a little loose in how I use my math jargon...

Correct me if I am wrong, but I thought particles were represented by a state vector in Hilbert space, i.e. in some particular basis an electron with a certain Hamiltonian is given by the state vector psi=(c1,c2,c3,.....) where c1, c2, c3, .... are complex.

Now, I thought that the representations of groups over some vector space were actual matrices. So if I take the group SU(2), then representations of that group would be some linear combination of the pauli matrices, since they form an irreducible representation of the group (kinda like a basis for a vector space, but instead we have matrices). I know that fermions are spin 1/2, and the spin operators in each direction (essentially the pauli matrices) are basically defined by the fact that the eigenvalues must be to +1/2, or -1/2.

This is where I get confused. I never thought of the matrices (or operators) as corresponding to particles, but instead correspond to the properties of the particles you can measure, such as the spin, energy, etc., and what I thought of the particles were the the state vectors or at least represented by them, NOT the matrices.

What do the matrices represent? There are 3 irreducible representations for SU(2), but not all three correspond to a different particle, since you can only have two, lets say an up electron or a down electron..... so does each matrix correspond to a different particle? Or is it the entire group that represents one spin half particle? Please, any clarity on the subject would be great. Thanks.

Christopher Betancourt

## Answers and Replies

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You should think of infinite dimensional unitary representations of the Poincare group extended by "internal symmetries". Poincare group gives you mass and spin (or helicity), internal symmetry groups give you additional quantum numbers.

Fredrik
Staff Emeritus
Gold Member
A representation isn't a matrix. It's a group homomorphism into the group of invertible linear operators on a vector space. Suppose that F:G→GL(V) is a representation. A subspace U of V is said to be invariant if for each g in G, the range of F(g) is a subset of U. A representation F:G→GL(V) is irreducible if V doesn't have any invariant subspaces.

In QM, we're interested in unitary representations, i.e. group homomorphisms into the subgroup U(V) of GL(V) that consists of all the unitary operators. The vector space is taken to be a complex infinite-dimensional separable Hilbert space, so I'll call it H instead of V from now on. The group we're interested in (for very technical reasons) is the universal covering group of the connected part of the Poincaré group. It doesn't have a standard name, so I'll just call it G.

If F:G→U(H) is an irreducible representation (with the specific G I just mentioned), H is the Hilbert space of one-particle states of a particular species of particle.

Chapter 2 of Weinberg's QFT book is an OK place to read about this stuff. (I don't know a great place to read about it).

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So representations are maps from the group into their unitary representation? So the representation (say a fermion) would take say infinitesimal rotations and represent them as pauli matrices (or more precisely U=exp[-i*sigma/2]) over some space H where only fermions live? Are you saying the particles are the actual mapping of of a group into a unitary operator acting on the Hilbert space of that species of particle?

I'm sorry if these questions sound dumb and elementary, but this is not easy stuff to first learn

Fredrik
Staff Emeritus
Gold Member
So representations are maps from the group into their unitary representation?
Yes, I suppose you could say that, but it sounds confusing when you call the range of that map a "representation". It's the map that should be called a representation, not its range.

So the representation (say a fermion) would take say infinitesimal rotations and represent them as pauli matrices (or more precisely U=exp[-i*sigma/2]) over some space H where only fermions live?
Not infinitesimal. A unitary representation of SO(3) would take a member R of SO(3) to a unitary operator U. That unitary operator would take a state vector to the state vector that an observer rotated by R would use. When R is a rotation around an axis, let's say the z axis, by an angle θ, the unitary operator would be expressed as exp(iθJz) with Jz self-adjoint. This isn't a derived result. It is the definition of the operator Jz, the z component of "spin". Yes, this is how "spin" is defined in QM.

The Pauli matrix σz enters the picture when you consider the matrix of components of the linear operator Jz in a basis for the 2-dimensional Hilbert space that the state vectors belong to. It's 2-dimensional if we're considering an irreducible representation of SO(3) with spin 1/2.

If we instead consider an irreducible representation of (the universal covering group of the connected part of) the Poincaré group, the Hilbert space is infinite-dimensional. One of the most interesting facts about the irreducible representations is that every vector in the associated Hilbert space has the same eigenvalue of

$$P^2=\eta_{\mu\nu}P^\mu P^\nu=-(P^0)^2+\vec P^2=-H^2+\vec P^2$$

where P is the four-momentum and H is the Hamiltonian. The eigenvalue is written as -m2, and m is called the "mass" of the representation. Yes, this is how "mass" is defined in QM.

Are you saying the particles are the actual mapping of of a group into a unitary operator acting on the Hilbert space of that species of particle?
I would say that the particle species (e.g. "electron") is the real-world concept associated with a specific irreducible representation. The actual mapping is just a mapping. The unitary operator is a mapping too, but it too represents a real-world concept: The state of motion of the labratory (or equivalently, the physical system). What I mean is that the unitary operator that corresponds to a member of the universal covering group of the connected part of the Poincaré group is the mathematical representation of the physical process of rotating your lab, boosting it to another velocity, moving it to another location, and doing the experiment at a different time.

The self-adjoint operators that appear in the exponents are mathematical representations of "observables", i.e. equivalence classes of measuring devices in the real world.

You seem to be wondering what, if anything, in this mess is the mathematical representation of a specific particle. I'd say that nothing represents the actual particle, but the one-dimensional subspaces (rays) of the Hilbert space are as always mathematical representations of the real-world concept of "states", i.e. equivalence classes of preparation procedures.

...this is not easy stuff to first learn
I agree. I had to read that chapter in Weinberg's book many times, and a few other things, just to get to this level of understanding, and there are still many things I don't fully understand. (Wigner's theorem, why exactly we use the universal covering group, etc.)

So the particle fields are the irreducible reps?

Fredrik
Staff Emeritus
Gold Member
You can construct an irreducible representation by quantization of a classical field theory without interaction terms (Klein-Gordon, Dirac, Maxwell). Noether's theorem gives you a bunch of conserved quantities, and quantization turns them into self-adjoint operators, which can be used to define unitary operators. (If A is self-adjoint and s real, exp(isA) is unitary).

The representation is the function that takes members of the (universal covering group of the connected part of) the symmetry group to these unitary operators.

You can construct an irreducible representation by quantization of a classical field theory without interaction terms (Klein-Gordon, Dirac, Maxwell). Noether's theorem gives you a bunch of conserved quantities, and quantization turns them into self-adjoint operators, which can be used to define unitary operators. (If A is self-adjoint and s real, exp(isA) is unitary).

The representation is the function that takes members of the (universal covering group of the connected part of) the symmetry group to these unitary operators.
OK, so let me piece it all together....
The quantized free fields of the particle species in question ARE the irreducible representation. The irreducible reps, or fields, takes the Poincare group (Lorentz + translations?) extended by internal symmetries (gauge groups?) and maps it into unitary operators through exp(isA), where A is a hermitian operator (or self adjoint). These unitary operators in turn only act on the Hilbert space of that particular particle species whole field we are using. And those hermitian operators (A in this case) are just the operators corresponding to observable conserved quantities that we got from taking Noether's theorem and applying the symmetry operations from the Poincare group + internal symmetries to the field in question.

So to summarize,
the fields (plus Noether) gives you a prescription to turn symmetries of the extended poincare group into hermitian operators (which one can then define unitary operators from) acting on the Hilbert space of that particular species... So each Field (irr. rep.) will have a different realization of the extended poincare group (hermitian operators acting on different Hilbert spaces). And in the end we just take the different Hilbert spaces, paste them together with a direct product to construct the bigger Hilbert space (Fock space?) where the sate vector of a physical system (where many fields exist) lives.

So different particle species are just different realizations of the underlying symmetry group that governs our universe?

Is this more or less the idea? Or am I missing something?

Quantized fields are not irreducible representations. But one-particle subspaces, for stable particles, carry nearly irreducible representations of the Poincare group. Why only "nearly"? Because we have to leave the room for parities, charges, other internal degrees of freedom. So, irreducible representations of the Poincare group enter with some (usually finite) multiplicity.

Fredrik
Staff Emeritus
Gold Member
The quantized free fields of the particle species in question ARE the irreducible representation.
I wouldn't say that, especially not with emphasis on the word "are". A quantized free field can be used to explicitly construct an irreducible representation, but technically they're two very different things. A representation is a group homomorphism into the group of linear operators on a vector space. A quantized free (scalar) field is an operator-valued distribution on Minkowski spacetime.

The rest of what you said sounds good enough. I don't know how to handle gauge groups, but it looks like arkajad just told us an important part of that.

I wouldn't say that, especially not with emphasis on the word "are". A quantized free field can be used to explicitly construct an irreducible representation, but technically they're two very different things.
OK, so there are the fields, and from the fields you can construct the irreducible reps? Is there a one-to-one correspondence between the fields and the irreducible reps? Because earlier it was said that the species of a particle is its irreducible rep, but I think of a species of a particle as really just being the underlying field it comes from, so I imagine that while they are different objects, one can go back and forth between when talking about particle species.

So what would be the irreducible rep constructed from a Dirac Field (not just the spinor part)? Could you construct that mathematical object we call the irr. rep. explicitly from the field? Or can you show me a reference online that does it (Weinberg's QFT book is on its way, but now here yet)?

Fredrik
Staff Emeritus
Gold Member
Is there a one-to-one correspondence between the fields and the irreducible reps?
Probably yes. I'm not sure though.

So what would be the irreducible rep constructed from a Dirac Field (not just the spinor part)? Could you construct that mathematical object we call the irr. rep. explicitly from the field? Or can you show me a reference online that does it (Weinberg's QFT book is on its way, but now here yet)?
I don't know all the details, but ever since I learned about how particle species correspond to different irreps, I've felt that the point of quantum fields is that they give us a way to explicitly construct the irreps. But I don't know a book that actually says that, or that goes through the details of the construction. There are however many books that describe how to find the conserved quantities associated with the one-parameter subgroups, and it seems fairly straightforward to construct Lorentz transformation operators from them. (I don't have a book that does this really well, so I don't know what to recommend).

Weinberg doesn't talk much about representations after chapter 2. In his approach, the "point" of quantum fields is that if we express the interaction Hamiltonian as a combination of quantum fields, we automatically get a Lorentz invariant S-matrix.

A. Neumaier
2019 Award
OK, so there are the fields, and from the fields you can construct the irreducible reps?
No. It goes the other way: Elementary particles correspond to certain irreducible representations of the symmetry group of the universe (Poincare group times inner symmetry group - not only the gauge group but also the flavor group). These are characterized by an irreducible rep of the proper connected Poincare group,
with nonnegative mass and finite spin or helicity together with certain additional
quantum numbers.

From these irreducible reps, one can construct a corresponding free field in the way discussed in Weinberg's book. This free field classically parameterizes the possible in- or out-states of a single particle of the type described by the representation.
The corresponding quantum field describes all particles of the given kind, including their antiparticles, as long as they can be regarded as free (i.e., asymptotically, in scattering experiments).

Is there a one-to-one correspondence between the fields and the irreducible reps?
No. Only particular combinations of mass, spin/helicity, and other quantum numbers appear in Nature.

Could you construct that mathematical object we call the irr. rep. explicitly from the field? Or can you show me a reference online that does it (Weinberg's QFT book is on its way, but now here yet)?
If you have the (free) field Phi(x), with components indexed by s, say, you get the irreducible representation by taking the vacuum state |>, define |x,s>=Phi_s(x)|>,
and have the group act naturally on the 1-particle states

$$|psi> := \int dx \sum_s psi(x,s)|x,s>$$

An explicit formula for translations generators in terms of free fields can be found for instance http://arxiv.org/pdf/1008.0244" - formula (21).
Expressions for Lorentz transformations are of a similar nature. Once you have generators, you restrict them to one-particle subspace. You calculate Casimirs there and you can find mass and spin this way.

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If you have the (free) field Phi(x), with components indexed by s, say, you get the irreducible representation by taking the vacuum state |>, define |x,s>=Phi_s(x)|>,
and have the group act naturally on the 1-particle states

$$|psi> := \int dx \sum_s psi(x,s)|x,s>$$
So the irr. rep |psi> is the sum of all the possible combinations of states the particle can be in? This might sounds dumb then, but does that mean the irr. rep. is the Hilbert space of a particular species attached with the appropriate operators (spin, charge, etc.) corresponding to the observables of that particle species? Sorry if I am struggling, but this seems like probably the most epic thing I am learning ever, and I really want a good grasp on it (or at least SOME grasp on it).

An explicit formula for translations generators in terms of free fields can be found for instance http://arxiv.org/pdf/1008.0244" - formula (21).
Expressions for Lorentz transformations are of a similar nature. Once you have generators, you restrict them to one-particle subspace. You calculate Casimirs there and you can find mass and spin this way.
Thanks for the link, I'm reading it now, but it is a little dense.... I'll probably end up back here with more questions after reading it (once, twice, or however many times it takes).

Thanks all for your patients

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A. Neumaier