What is the importance of dimension six operators in the study of physics beyond the Standard Model? Are these operators more relevant than dimension five operators like $HHFF$ or operators with derivative couplings?

I often see lagrangians with dimension six operators in effective studies of the standard model, but I do not understand this choice. An example is the paper arXiv:1304.1151, where they have defined: $$\mathcal{L}_{\rm eff}= \sum_n \frac{g_n}{\Lambda^2}\mathcal{O_n},$$ whit $g_n$ being the corresponding couplings and $\mathcal{O}_n$ the dimension six operators.

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    $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1304.1151 $\endgroup$ – Qmechanic Apr 20 '14 at 5:03

The only dimension-five operators allowed by the SM are neutrino masses, $(HL_i)(HL_j)$. So we mostly talk about dimension-six operators because for almost any question they're the first higher-dimension operators that can appear.

  • $\begingroup$ @Matt_Reece, thank you for your answer! Do you know why dimension-six operators with derivative couplings are never considered? I also have a by-product question: we can give mass to neutrinos in the SM using dimension-4 terms. So, why is it interesting to consider dimension-5 (non-renormalizable) operators to give them mass? $\endgroup$ – Melquíades Apr 20 '14 at 16:22
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    $\begingroup$ Operators with derivative couplings are considered frequently. And giving mass to neutrinos with dimension 4 operators requires the introduction of a field that isn't present in the Standard Model (a right-handed neutrino). In the Standard Model as an effective theory with fixed field content, the dimension-5 Majorana masses are the only possible neutrino mass. Which doesn't mean it's how neutrino mass works in the real world; we don't know yet. $\endgroup$ – Matt Reece Apr 20 '14 at 20:36

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