Is the multiplet given by $$\left( -\frac12,0,0,\frac12 \right)$$ self-CPT conjugate?

There seems to be no common agreement upon that:

  • Weinberg (QFT 3, page 47) and many others claim it is not, basing on $SU(2)$ $R$-symmetry reasoning

  • Terning (modern susy, page 12) and some others claim it is, provided the fermions transform in a real representation of the gauge group, regardless of $R$-symmetry considerations.

  • $\begingroup$ Let me add that also Wess-Bagger (page 18) say it is CPT complete. $\endgroup$ – jj_p Apr 24 '13 at 7:11

It looks like Terning and Weinberg are talking about answering slightly different question. Terning is first answering the question, is there an N=2 hypermultiplet that is CPT self conjugate? The answer is yes, if the fermions transform in a real representation of the gauge group. If the gauge group representation is complex we have to add the CPT conjugate of the hypermultiplet to get a self CPT conjugate multiplet (the group representation gets complex conjugated and for a complex representation by definition this representation is not equivalent to the original), so the hypermultiplet was not self-CPT conjugate.

Weinberg does not consider a real gauge group representation. For physical reasons he considers a complex representation of the gauge group. Once again, the hypermultiplet is then not self conjugate, so he adds its CPT conjugate, but then the fermions transform in a real representation of the gauge group. For physical reasons we do not want fermions to transform in a real representation of a gauge group, so the N=2 hypermultiplet, either by itself or by with its complex conjugate, is unappealing.

Just as an addendum, the fermions should transform under a complex representation of the gauge group because nature is chiral. Particles of opposite helicity are treated differently. A spin 1/2 particle should transform differently under a gauge transformation compared to a spin -1/2 particle. They transform under representations which are complex conjugate to one another, so having fermions transform in a real representation would ruin chirality.

Edit: I apologize, I did not see where Weinberg had taken into account the R-symmetry originally. There does some appear to be some disagreement within the literature, but the way around Weinberg's argument is if the scalars transform in a psuedoreal representation of the gauge group. An argument is illustrated in http://arxiv.org/abs/1107.0973

  • $\begingroup$ Hi, thanks for your answer; can you be more precise about the last point, i.e. the way around Weinberg? $\endgroup$ – jj_p Jun 13 '13 at 8:26
  • $\begingroup$ Well Weinbergs argument rests on the fact that two real scalar fields cannot form a doublet of SU(2). The doublet of SU(2) is the usual fundamental representation given by the Pauli matrices, which is pseudoreal. That is it is equivalent to its complex conjugate representation, but it can not be written in terms of real matrices. It seems then if the gauge group rep is psuedoreal also, we can still maintain CPT invariance (because its psuedoreal) and have a doublet representation of SU(2) (because its only "psuedo" real). $\endgroup$ – David M Jun 13 '13 at 15:50
  • $\begingroup$ As another reference I would look at page 18 of this paper: arxiv.org/pdf/1204.2141.pdf They cite some of the original literature which will probably give a better and more accurate explanation, but unfortunately I can't access it now. Maybe you'll have better luck! $\endgroup$ – David M Jun 13 '13 at 15:58

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