All irreducible finite dimensional complex representations of the Lorentz group can be specified by two positive half-integers, i.e. $(j_1, j_2)$. The $(0,0)$ representation is the trivial scalar representation, $(\tfrac{1}{2}, 0)$ is the left handed Weyl spinor representation, $(0, \tfrac{1}{2})$ is the right handed Weyl spinor representation, and $(\tfrac{1}{2}, \tfrac{1}{2})$ is the (complex) vector representation. Most QFT textbooks talk about these representations. And then they stop. What about the all the other options, like $(0,1)$, $(1, \tfrac{1}{2})$, $(\tfrac{3}{2}, 1)$, etc? Do these have any relevance at all? Have people ever speculated that such fields exist?

This also gets at a bigger question. Certainly, many different $(j_1, j_2)$ representations will have the same $SO(3)$ spin. It seems to me like there should be many interesting ways to make a "spin $\tfrac{3}{2}$" particle, for example, each behaving differently under parity.

  • 4
    $\begingroup$ Possible duplicate here. $\endgroup$
    – knzhou
    Sep 26, 2018 at 21:08
  • $\begingroup$ Of course people have speculated, ad nauseam. Comfortable with the canonical table? $\endgroup$ Sep 26, 2018 at 22:33

1 Answer 1


The $(0,1)$ rep is an antisymmetric two-tensor that is either self-dual or anti-self-dual. The field strength $F_{\mu \nu}$ in the Maxwell theory is a sum of both reps $(1,0) + (0,1)$. Likewise the (in)famous Rarita-Schwinger fermion transforms in the $(1,1/2) + (1/2,1)$ representation.

In general you'll find theorems that only a finite number of representations of the Lorentz group appear, because higher-spin fields behave pathologically. This isn't quite true, in the sense that starting with a boson $\phi$, a Dirac fermion $\Psi$ and a gauge field $A_\mu$ you can easily build composite operators that transform in any representation you want. Those composite operators don't have their own dynamics though: their behavior is completely governed by the Lagrangians of the fundamental fields they're built out of.


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