# Why aren't the spin-3/2 fields in the (3/2,0)+(0,3/2) representation?

Why is it that spin-$\frac 32$ fields are usually described to be in the $(\frac 12, \frac 12)\otimes[(\frac 12,0)\oplus(0,\frac 12)]$ representation (Rarita-Schwinger) rather than the $(\frac 32,0)\oplus(0,\frac 32)$ representation? Does the latter not describe a spin-$\frac 32$ field? Why is the gravitino given by the Rarita-Schwinger-type representation rather than the $(\frac 32,0)\oplus(0,\frac 32)$ representation?

This is related to a recent question I asked on gauge invariance of the Rarita-Schwinger field.

Thanks!

You could have asked the same question about a spin one field. Why do they transform in the $(\tfrac 1 2, \tfrac 1 2)$ representation and not in $(1,0) \oplus (0,1)$? The reason is gauge invariance; the gauge fields $A_\mu$ transform in $(\tfrac 1 2, \tfrac 1 2)$, but the gauge invariant field strength $F_{\mu \nu}$ transforms in $(1,0) \oplus (0,1)$.
The same holds for the gavitino. The Rarita-Schwinger field $\psi_{\mu \alpha}$ is like the gauge field $A_\mu$. It has a gauge transformation $\delta \psi_{\mu \alpha} = \partial_\mu \chi_\alpha$. Its gauge invariant field strength $\partial_\mu \psi_{\nu \alpha} - \partial_\nu \psi_{\mu \alpha}$ transforms as $(\tfrac 3 2, 0) \oplus (0, \tfrac 3 2)$.
Note that the field $\psi_{\mu\alpha}$ isn't all of the representation $(\tfrac{1}{2},\tfrac{1}{2})\otimes[(\tfrac{1}{2},0)\oplus(0,\tfrac{1}{2})]$ but only the part of it satisfying $\gamma^\mu \psi_{\mu\alpha}$. This selects the $(1,\tfrac{1}{2})\oplus(\tfrac{1}{2},1)$ representation. See Weinberg's QFT Sect. 5.6 for more on this.
We can expand on Sidious Lord's answer. The field $A_\mu$ transforms in a inhomogeneous way under Lorentz transformations. $$U(\Lambda)A_\mu(x)U(\Lambda)^\dagger = \Lambda_\mu{}^\nu A_\nu(x) + \partial_\mu \Omega(x,\Lambda)\,.$$ So, this field isn't technically a 4-vector representation of the Lorentz group. Weinberg treats this in section 5.9. The inhomogenous part cancels out of the field strength.
• This remark in Weinberg is one I have never been 100% comfortable with. Isn't the $A_\mu$ Weinberg talks about transforming funny because he has already constrained it in the Coulomb gauge with $A_0 = 0$? – DJBunk Feb 1 '13 at 20:52