# Effective theories and dimension six operators

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.

• 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 – Qmechanic Apr 20 '14 at 5:03

## 1 Answer

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.

• @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? – Melquíades Apr 20 '14 at 16:22
• 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. – Matt Reece Apr 20 '14 at 20:36