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

+ 4 like - 0 dislike

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.


This post has been migrated from (A51.SE)
asked Mar 21, 2012 in Theoretical Physics by Henry (115 points) [ no revision ]

2 Answers

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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 $(\tfrac 3 2, 0) \oplus (0, \tfrac 3 2)$.

This post has been migrated from (A51.SE)
answered Mar 21, 2012 by Sidious Lord (160 points) [ no revision ]
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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 post has been migrated from (A51.SE)
answered Mar 21, 2012 by josh (205 points) [ no revision ]

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