# Weinberg dimension 5 operator

How to prove that the $\Delta L=2,$ dimension=5 Weinberg operator $LLHH$ is the unique operator which violates lepton number by two units, without derivative couplings, etc.??

## 1 Answer

This $LLHH$ is not a unique operator. It just the operator with a lowest dimension using only Standard Model particles and giving Majorana neutrino masses. The way to understand this is as follows. Majorana neutrino mass must be of the form

\begin{equation} \nu _{ L }^{\,\, T } m C \nu _L \end{equation} This term breaks $SU(2) _L$ and $U(1) _Y$ and so cannot be written down in a $SU(2) _L \times U(1) _R$ invariant theory. However, It may come from a spontaneously broken term, just as the other Standard Model fermion mass terms do. In order to find what sort of form the $SU(2) _L \times U(1) _Y$ invariant term we have, we consider the charges of the low energy term. It has a $T _3$ value of $1$ and $U(1) _Y$ value of $- 2$ (this may change slightly depending on your conventions). Since the low energy term has $T _3 = 1$, it can only arise from a term that transforms like a vector under $SU(2) _L$ transformations. The lowest order such term is (you can easily check that this transforms as a vector under isospin transformations) \begin{equation} L ^T i \sigma _2 {\vec \sigma}C L \end{equation} To make an $SU(2) _L$ invariant out of this term that gives a mass we must take the scalar product with another isospin triplet that gets a VEV. Since we don't have an isospin triplet Higgs in the SM we need to use a product of Higgses. Thus the lowest order term that gives neutrino Majorana masses is given by, \begin{equation} m\left( L ^T i \sigma _2 {\vec \sigma}C L \right) \cdot \left( \phi ^T i \sigma _2 {\vec \sigma} C \phi \right) \end{equation} This term is also invariant under $U(1)_Y$. Note, that it is helpful to go ahead and write down many different terms to convince yourself this operator is truly unique at this order.