It is known that

  • Finite dimensional irreducible representations of Lorentz group can be indexed by two half integers $(s_1,s_2)$ and the sum $s_1+s_2$ is called the spin.
  • Infinite dimensional unitary irreducible representations of (the universal cover of) Poincare group can be indexed by two numbers $m$ and $s$, while the $s$ is also called spin.
  • We can construct representations of Poincare group from those of Lorentz group.

My question is:

  1. Are all infinite-dimensional unitary irrps of Poincare group are constructed from finite-dimensional irreps of Lorentz group?
  2. Does the spin $s$ of the representation of Poincare group equal to $s_1+s_2$ if it is constructed from a $(s_1,s_2)$ irrp of Lorentz group?

For example:

  • left-handed electron is a $(1/2,0)$ Lorentz representation. Does it induces a spin $1/2$ Poincare representation?
  • Is a $(1/2,1/2)$ Lorentz representation induces a spin 1 Poincare representation?

1 Answer 1


First, note that while you can induce unitary representations of the Poincaré group from representations of the Lorentz group, they are not in general irreducible (although sometimes they are), and the various physical field equations (Dirac equation, Lorentz gauge condition) actually serve as projectors onto the irreps.

The irreducible unitary representations are actually constructed via Wigner's little group method. The construction is rather technical; you can read the full details in Figueroa O'Farril's (incomplete) notes which you can find here, which require familiarity with principal bundles. So to answer your questions:

  1. In the notes, it is stated on page 18 without proof that all unitary irreps are constructed in this manner. I unfortunately do not know the proof.

  2. First note that to obtain half-integer spin representations, you have to consider the double cover of the Poincaré group $\mathbb{R}^{(1,3)}\rtimes \text{SL}(2,\mathbb{C})$. The constructed representations will also be representations of $\text{SL}(2,\mathbb{C})$ as it is contained as a subgroup, so yes, the definition of spin will be the same. (What I'm saying is that the $(\frac{1}{2},0)$ representation is not actually an honest-to-goodness representation of the Lorentz group, but rather a projective one; but it is a representation of its double cover. This is the same situation as in the non-relativistic case with $\text{SU}(2)$ and $\text{SO}(3)$). A few comments:

    • The $(0,0)$, $\left(\frac{1}{2},0\right)$ and $\left(0,\frac{1}{2}\right)$ representations induced by (the double cover of) the Lorentz group are irreducible, that is to say, they are also induced by the little group method.
    • The spin $1$ massive $\left(\frac{1}{2},\frac{1}{2}\right)$ induced representation decomposes as a $3$-dimensional vector irrep $\oplus$ $1$-dimensional scalar irrep, both constructed under the little group method; the Lorentz gauge-like condition $\partial_\mu A^\mu=0$ serves as the projector onto the first one, while the second is just the $(0,0)$ irrep. The massless "spin" $1$ particles (i.e. photons) cannot be embedded in this representation (see below).
    • The $\left(0,\frac{1}{2}\right)\oplus\left(\frac{1}{2},0\right)$ representation is obviously reducible by projecting onto the chiral subspaces, but it can also be shown that the Dirac equation is actually also projector onto a spin $\frac{1}{2}$ irrep that mixes both chiralities.

One further thing to note, however, is the mass $m$ that you mentioned, which also determines the representation; in the massless case it will not be appropriate to speak of "spin".

When the mass is nonzero, the little group in question is $\text{SU}(2)$, and the little group method consists of taking unitary finite-dimensional representations of it, so we can speak of spin as usual. It can also be seen that the mass induces the mass-shell condition upon choosing a rest frame for the representation, which acts as the Klein-Gordon equation (which is why it occurs in all massive field theories). The fact that in this case the double cover of $\text{SO}_3$ is the little group is no coincidence; it is exactly the group of symmetries of the rest frame.

However, in the massless case, the little group is given by $\mathbb{R}^2\rtimes\text{Spin}_2$, the double cover of the Euclidean group. So its finite-dimensional representations are more accurately labeled by the name helicity. There is no rest frame for the particle, so its symmetry group is given by rotations around its axis of motion*: (the double cover of) $\text{SO}(2)$. The presence of the $\text{Spin}_2$ group severely restricts the possible Lorentz representations it can be embedded in; we cannot, for example, describe a particle of helicity $\pm1$ with the $\left(\frac{1}{2},\frac{1}{2}\right)$ representation, and we must resort to $(0,1)$ and/or $(1,0)$ (the self-dual and anti-self-dual fields respectively).

* The plane part $\mathbb{R}^2$ acts trivially, as we look for finite-dimensional unitary representations of the little group and $\mathbb{R}^2$ is non-compact.

  • $\begingroup$ great answer, thanks! $\endgroup$
    – HanXu
    Commented Oct 24, 2020 at 15:08

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