I believe (but please correct me if I'm wrong) that I understand the basic philosophy and most of the mathematics involved in Wigner's classification of particles via group representations. But I'm missing one key bit of physical intuition that I'm hoping someone can provide.

Here's the part I think I understand (feel free to skip this paragraph and the next one if you just want to get to the question): The state space of a physical system is the projectivization of a Hilbert space $H$. An element of the Poincare group moves this system to a new location in spacetime and changes its state in some way. We believe that this should not affect the outcomes of experiments, so the inner product of two state vectors should be unaffected. Thus the Poincare group acts on the projectivization of $H$ in a way that preserves inner products. This action lifts to an action of the universal cover of the Poincare group on $H$ itself. Wigner proves (more or less) that this action must be unitary. Also, if the physical system is a single particle, it makes sense that the representation should be irreducible. So to classify particles, we should classify irreducible unitary representations of the universal cover of the Poincare group.

I also (mostly) understand how the classification goes: We let $SL_2({\mathbb C})$ act on Minkowski space, and find representations of the isotropy groups. Thus each particle is associated with some orbit, and the orbit is characterized (more or less) by its (Lorentzian) distance from the orign $m$. For each such orbit, we get a discrete set of irreducible representations (i.e. particles) coming from representations of the isotropy group of a representative point.

Here's what I don't get: We now identify the (continuous) parameter $m$ with the particle's rest mass, and the remaining discrete parameter with the particle's spin. My question is:

Why do we identify $m$ with the rest mass as opposed to some other property of the particle?

Perhaps the answer is just that $m$ seems to be telling us something about the particle and that rest mass is the only natural candidate that comes to mind. But I suspect there's a deeper reason, and that I'm missing it.

One could ask an analogous question about the discrete parameter and spin, but mass is easier for me to think about, and at least (in this context) equally mysterious to me, so I'd like to understand that first.

  • $\begingroup$ What are particles? No such thing exists in nature. Are you talking about the metastable states of a quantum field and associated (near) conservation laws? $\endgroup$
    – CuriousOne
    Commented Jun 24, 2015 at 3:16
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    $\begingroup$ @CuriousOne: I sympathize with your insistence that there are no such things as particles in modern quantum field theory, but I do believe (and once again, please correct me if I'm wrong) that Wigner thought in terms of particles, and I'm currently trying to understand things at that level. Eventually, I'm sure I'll want to transfer this understanding to a fields-only model, but for now, I think I'll get there fastest if I can understand what Wigner was thinking. $\endgroup$
    – WillO
    Commented Jun 24, 2015 at 3:25
  • $\begingroup$ What Wigner thought makes no more difference to nature than what you or I think, so, again, I have to ask you where in nature have you seen particles? I have only been counting quanta with my photomultiplier tubes and silicon strip detectors, particles never got caught in there, for sure. Keep chasing those particles, though, it ought to be fun once you get to the level of actual experimental data and you realize that they are not in there. $\endgroup$
    – CuriousOne
    Commented Jun 24, 2015 at 3:32
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    $\begingroup$ There is in principle a definition of particles independent of nature. What Wigner was calling a particle was simply an "irreducible representation of the Poincare Algebra". Thus, the notion of a particle in this sense is simply a mathematical construct. It is in this sense that @WillO is asking his question. Whether this notion of a particle has anything to do with the usual ones encountered in nature or QFT is a completely separate question. $\endgroup$
    – Prahar
    Commented Jun 24, 2015 at 9:51
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    $\begingroup$ @Prahar: I'm sorry that you had trouble following the question, but it still seems to me that the question is clear as written: We have a number $m$ that shows up in a purely mathematical context and then we interpret it as having a physical meaning. The question is: What is the intuition that connects the two? $\endgroup$
    – WillO
    Commented Jun 24, 2015 at 17:56

2 Answers 2


We have a number m that shows up in a purely mathematical context and then we interpret it as having a physical meaning. The question is: What is the intuition that connects the two?

The intuition connecting the two is essentially identical with the reason why Wigner connects the purely physical notion of an elementary particle with the purely mathematical notion of an irreducible unitary representation of the Poincare group.

For any relativistic quantum system, the total 4-momentum is a 4-vector $P$ whose components are commuting operators. In a common eigenstate of these operators, $P$ becomes a numerical vector, from which one can calculate the total mass $m$ through the formula $(mc)^2=P^2$ (in the +--- metric). In a rest frame, the spatial components vanish, so that $P^2=P_0^2=(E/c)^2$, where $E$ ist the rest energy, giving $E=mc^2$. This is why $m$ is called the rest mass.

Noether's theorem now says that the 4-vector $P$ is the infinitesimal generator of the group of translation symmetries, which is a subgroup of the Poincare group. Therefore the Poincare group acts in some unitary representation, and $P$ represents the image of the translation group generator in this representation.

For an elementary system (usually referred to as an - elementary - particle) one makes the extra requirement that there is no Poincare invariant subsystem, which translates to irreducibility of the representation. Since $P^2$ is a Casimir operator, it has in any irreducible representation a constant numerical value $M^2$. If this irreducible representation is the representation of an elementary system, we therefore have $M^2=(mc)$^2, which gives $M=mc$. Now in the present context the units are typically chosen such that the speed of light takes the value 1.

Therefore we find that the number $M$ arising in a purely mathematical context has the physical interpretation of a rest mass whenever this representation belongs to that of a particle.

The same kind of reasoning is applied more generally - whenever a mathematical concept gets a physical label.

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    $\begingroup$ This is exactly what I was looking for, and in fact had pretty much figured it out long after I posted the question (but long before you posted this answer). It is great to have my tentative understanding confirmed. Thanks very much indeed. $\endgroup$
    – WillO
    Commented Jul 28, 2015 at 15:51

Usually the first step in deriving the reps of Poincaire is to go to the rest frame of the particle. This amounts to choosing a basis where $P^0$ acting on the state is nonzero, and where the eigenvalue a of $P^i$ are zero. We can do this if the momentum is timelike, that is if the eigenvalue of $P_\mu P^\mu$ is negative (in -+++ signature). Furthermore the sign of the eigenvalue of $P^0$ matters, we'll focus on the positive case. (Note that the irreducible reps are classified but the value of $P_\mu P^\mu$, as well as the Pauli Lubanski operator, because these are casimir operators (commute with all the group generators of Poincaire)--these operators will have the same eigenvalue acting on any state in an irreducible rep).

Long story short, we've picked a basis/frame where the energy P^0 is positive and the momentum P^i is zero. The energy in the rest frame is the mass.

Also notice that $P_\mu P^\mu = - m^2$ (eigenvalue of the operator $P^2$ in this irrep) is the on shell condition for a massive particle.

  • $\begingroup$ Thanks for this answer. I'm still trying to digest it. My first problem is that I'm not sure what the $P^\mu$ are. My guess is that $P^\mu$ is defined by something like $\rho(G^\mu)=e^{iP^\mu}$ where $\rho$ is a given group representation and $G^\mu$ is a group element representing translation of spacetime in the $\mu$ direction. Do I have this right? $\endgroup$
    – WillO
    Commented Jun 26, 2015 at 2:54
  • $\begingroup$ Yes exactly right. Although, I would write it more like $G(x^\mu) =e^{i P_\mu x^\mu)$. The x are the parameters of the transformation (equivalently the coordinates on the Lie group manifold), the P_\mu are a particular representation for the Lie algebra elements / group generators, and G is a group element in that representation. The analog to normal quantum mechanics is, for example, time translations are given by (for a time independent hamiltonian H) $U(t)=e^{I H t}$, or space translations are given by $G(x) =e^{iPx}$, where P is the momentum operator. $\endgroup$
    – Andrew
    Commented Jun 26, 2015 at 4:12
  • $\begingroup$ Andrew: Thank you. This now allows me to get to the heart of my confusion. The self-adjoint operators $P^\mu$ act on some Hilbert space $H$, which is associated with our arbitrary group representation. Energy and momentum are self-adjoint operators (call them $Q^\mu$) acting on some Hilbert space $H'$ which is the state space of our particle. There is, as you say, some analogy between the $P^\mu$ and the $Q^\mu$. But why do we identify the $P\mu$ with the $Q^\mu$? [CONTINUED...] $\endgroup$
    – WillO
    Commented Jun 26, 2015 at 4:21
  • $\begingroup$ [CONTINUED] There is, after all, also an analogy between the $P^\mu$ and the observables that represent location (that is, both transform similarly under the Lorentz group). So why do we identify them with energy/momentum as opposed to, say, time/location or some other observables? $\endgroup$
    – WillO
    Commented Jun 26, 2015 at 4:22
  • $\begingroup$ PS---I feel sure that whatever I'm failing to understand is something entirely basic and completely obvious to everyone but me. I'm pretty confident that I understand the hard parts but there's one simple thing somewhere I'm overlooking. $\endgroup$
    – WillO
    Commented Jun 26, 2015 at 4:36

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