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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.