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Gauge invariance and the form of the Rarita-Schwinger action

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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. Thanks!

This post has been migrated from (A51.SE)
asked Mar 19, 2012 in Theoretical Physics by Henry (115 points) [ no revision ]

1 Answer

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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.

This post has been migrated from (A51.SE)
answered Mar 20, 2012 by josh (195 points) [ no revision ]

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