It is known that to specify a finite-dimensional irreducible representation of the Lorentz group, one needs to specify two half-integers, $(j_1,j_2)$. For instance, the left-handed and right-handed Weyl spinors are non-equivalent representations, despite both having spin half. There is the $(1/2,1/2)$ 4-vector representation, and the $(0,1)$ 2-form representation.

In QFT, in classifying the possible types of particle that can exist, we look for unitary (infinite-dimensional) irreducible representations of the Poincaré group. Here, we need to specify a mass $m$ and a single 'spin' $j$. Fixing the mass to some given number, then, it appears that there are fewer types of particle allowed than the finite-dimensional representations would suggest. For instance, there is only one type of spin half particle, $j = 1/2$. Is this a left-handed Weyl fermion or a right-handed Weyl fermion, or something else?

More generally, how should we interpret the 'particle types' given to us by the infinite-dimensional Poincaré group representations in terms of the 'particle types' given to us by the finite-dimensional Lorentz group representations?

My thoughts: one can show that the Casimirs used in classifying representations of the Lorentz group don't commute with the $P^2$ Poincaré Casimir. This might suggest that when we enlarge our group from Lorentz to Poincaré, we get new transformations that can mix states within the old representations, so that the new irreducible representations must also be enlarged. But I cannot see how a translation (followed by any combination of boosts or rotations) could turn a left-handed Weyl spinor into a right-handed one.


1 Answer 1


I think most of the confusion is due to mixing up what the Lorentz and Poincare groups are typically acting on. When we talk about Lorentz irreps, we usually mean finite-dimensional non-unitary irreps in the space of fields, while when we talk about Poincare irreps, we usually mean infinite-dimensional unitary irreps in the Hilbert space.

Starting from the Poincare group acting on the Hilbert space, we can restrict to the Lorentz group (also acting on the Hilbert space). As you pointed out, restricting the group generically makes irreps fall apart, and this is exactly what happens, because we've lost the ability to do translations; we get different irreps roughly corresponding to where the particle could be. But these are not at all related to the Lorentz irreps of fields; they are instead infinite dimensional irreps of the Lorentz group. You can read about those horrible things here and I personally learned about these things from Wu-Ki Tung's excellent group theory book.

The procedure from going from fields to particles is:

  • define a classical field transforming under a finite-dimensional non-unitary representation of the Lorentz group
  • perform canonical quantization in the usual way, associating each mode with a creation operator on a Hilbert space
  • the Poincare action on the fields induces a Poincare action on the Hilbert space; we identify particles with Poincare irreps on that Hilbert space

In particular, nothing goes 'missing', every field gives particles as we'd expect. For example:

  • the Dirac spinor field yields two kinds of particles, each with $j = 1/2$ and $m > 0$.
  • both Weyl spinor fields yield two kinds of particles, with $h = \pm 1/2$ and $m = 0$, where $h$ is the helicity which labels massless representations of the Poincare group.
  • a massive real vector field yields particles with $j = 1$ and $j = 0$, with $m > 0$. We may also impose constraints on the field to remove the $j = 0$ particle.
  • a massless real vector field yields particles with $h = \pm 1$ and $m = 0$.

Note that a Lorentz irrep on the fields, the Weyl spinor, yields a decomposable Poincare representation of particles, even though the Poincare group is bigger. Physically, this is because we have a distinct particle and antiparticle and the field can generate both.

  • $\begingroup$ Thanks for the great answer. I have one question: with regards your last comment, could we make an identical statement about a complex scalar field? That it yields a direct sum of two identical Poincaré irreps? $\endgroup$
    – gj255
    Commented Apr 16, 2018 at 15:57
  • 1
    $\begingroup$ @gj255 Yes, and an easy way to see that must be true is to split it into two real fields. $\endgroup$
    – knzhou
    Commented Apr 16, 2018 at 16:16
  • $\begingroup$ @knzhou so your line of argument suggests that the spin label used in Poincaré does not pin down the irrep labels of the finite dimensional labels of Lorentz (or alternatively the list of possible values of $j$ in a finite dimensional Lorentz rep). $\endgroup$ Commented Mar 19, 2019 at 16:59
  • $\begingroup$ @ZeroTheHero Yes, that's right. And that's possible because these irreps are for different things: Poincare for particles, and Lorentz for fields which may create these particles. The very same particles can be created by different fields. For example, $A_\mu$ creates photons, but $F_{\mu\nu}$ does too. In fact, so does $\partial_\mu \partial_\rho \partial_\nu \partial_\sigma A_\delta$, which has rather complicated Lorentz transformation properties. $\endgroup$
    – knzhou
    Commented Mar 19, 2019 at 17:02

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