in Weinberg Vol. I section 5.9 (in particular p. 251 and surrounding discussion), it is explained that the smallest-dimension field operator for a massless particle of spin-1 takes the form of a field strength, $F_{\mu\nu}$. This is because a massless vector field does not transform properly under Lorentz transformations (Weinberg, Vol. I eq. 5.9.22) and requires additional gauge symmetry/redundancy to 'mod out' unphysical degrees of freedom.

One of the questions in Banks' Modern QFT textbook (problem 2.10) asks to repeat this general analysis for spin-3/2 and spin-2 fields, for which we know the main examples are the Rarita-Schwinger field and the graviton. For spin-3/2, however, I am confused as to why the Rarita-Schwinger field is described by an action which looks very similar to the Dirac action. How is it that the massless Rarita-Schwinger field, which carries a vector index as well as an additional spin-1/2 index, is able to avoid the ill-behavior under certain Lorentz transformations (which Weinberg calls $S(\alpha,\beta)$) that is exhibited in the massless vector?

Naively I would have wanted to write a spin-3/2 field as a $$(1,1/2) = (1/2,1/2)\oplus(1/2,0)$$ representation of the Lorentz group so that the vector index can be treated independently of the spinor index. Shouldn't the argument for massless photons carry over and imply that I should have some kind of `field strength' for the spin-3/2 field?

I suspect that what I'm missing is something trivial and silly.


1 Answer 1


I think you're a bit confused with the Lorentz representation. The field is written with a 4-vector and Dirac spinor index. So naively it transforms as $[(0,\tfrac{1}{2})\oplus(\tfrac{1}{2},0)]\otimes(\tfrac{1}{2},\tfrac{1}{2})$. But this has two subduced irreducible representations under the subgroup of spatial rotations, the spin-$\tfrac{3}{2}$ Rarita-Schwinger field you want, and a spin-$\tfrac{1}{2}$ you are going to project out. This is your big clue that the spinor and vector indices are not independent. They must be "coordinated" to keep only the Rarita-Schwinger spin-$\tfrac{3}{2}$ field. Its representation under the full Lorentz group is $(1,\tfrac{1}{2})\oplus(\tfrac{1}{2},1)$.

Alternatively, looking at the answer (i.e. the equation for the Rarita-Schwinger field) reveals that the two kinds of indices talk to the same terms and aren't independent of one another.


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