I'm trying to figure out why a left-handed Majorana mass term is mathematically forbidden by the $SU(2)_L$ symmetry in the context of the seesaw model.

As far as I get it, it is because the left handed neutrino in the Majorana mass term is part of a $SU(2)$ doublet, such that it would be changed in a $SU(2)$ symmetry transformation, e.g. $v_L => e_L$. And for the right handed Majorana mass term it would be allowed, because the right handed neutrino is part of a singlet and not a doublet. Can anybody show me the explicit gauge transformation for that term that breaks symmetry?

Somewhere I've also read that $v_L$ possesses non-zero isospin and hypercharge and that's why the mass term is actually forbidden. That's basically the same like in my interpretation from above, right?

I just want to be sure that I don't get anything wrong here.


1 Answer 1


The left-handed neutrino is a 2-spinor field $\eta_A$, $A=0,1$, and the Majorana mass term is a bilinear, $\Delta L = \pm 2$ term without the complex conjugation in each term, $$ m\cdot \eta_A \eta_B \cdot \epsilon^{AB} + \text{Hermitian conjugate}$$ Note that this Majorana term is the only bilinear term without derivatives that one may construct from a single 2-component spinor (even if the complex conjugate is allowed as well).

However, in the $SU(2)$ gauge-invariant theory, $\eta_A$ is a component of a doublet, so it carries an extra index $i=1,2$ where $i=1$ is the neutrino and $i=2$ is the left-handed electron. So the Majorana term would have to be $$ m\cdot \eta_{Aj} \eta_{Bk} \cdot \epsilon^{AB}t^{jk} + \text{Hermitian conjugate}$$ However, the only $SU(2)$-covariant tensor $t^{jk}$ with two indices of the same kind is $\epsilon^{jk}$ which is however antisymmetric, so with this choice of $t$, the term would vanish. Note that $t^{jk}=\delta^{jk}$ is not possible because $\delta$ must have one upper index and one lower index, not two indices of the same kind.

Equivalently, if there were an $SU(2)$-invariant mass term "neutrino . neutrino", by the neutrino-electron symmetry which is an element of $SU(2)$, there would also have to be a term "electron . electron" which would however violate the charge by $\Delta Q = 2$, so it clearly cannot be gauge-invariant.

One may show that the Majorana term would violate the hypercharge, too. Clearly, the neutrino left-handed field has a nonzero hypercharge $Y$, so the bilinear term $\eta\eta$ has a nonzero (double) value of the hypercharge, too, so such a term cannot be gauge-invariant – a faster way to show the same result.

  • $\begingroup$ Thank you very much! I only have a few issues with the notation as I've never seen that particular tensor notation in my books (e.g. QFT & SM by Schwartz, or I forgot it). So I guess the epsilon tensor with indices A,B is like a basis? E.g. like the minkowski tensor for a 4-vector (do you have some introductional literature to that?)? Your solution shows that the mass term vanishes, but is it also possible to apply the SU(2) gauge transformation and show that it is clearly not invariant? $\endgroup$
    – 0vbb
    May 27, 2016 at 3:31
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    $\begingroup$ Dear Michael, 2-component spinors in relativity are the "same" as the non-relativistic 2-component spinors e.g. from the "Pauli equation", one just allows the action of $\sigma^0\sim 1$, the time component, as well. ... A zero term is always invariant under everything. What is non-invariant would be a particular nonzero term. For example, if you choose my 2nd Ansatz and only choose the component $t_{\nu,\nu}$ nonzero, then you reproduce the original Majorana term for neutrinos and nothing else, but a tensor 2x2 whose only component is $t_{22}$ is clearly not invariant under $SU(2)$, right? $\endgroup$ May 27, 2016 at 3:42
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    $\begingroup$ But there is no "the" term that we should consider, the $SU(2)$ completion of the neutrino Majorana term could hypothetically be different than one I just described. That's why one must consider all alternatives, which is what I did, and prove that the only solution is zero, which is what I did. What you propose now is simply not a general way to deal with the problem because you haven't specified the "right and only candidate" that should be tested on $SU(2)$ invariance. $\endgroup$ May 27, 2016 at 3:44

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