# Majorana Mass For Neutrinos In Standard Model

Neutrinos cannot have Dirac masses because there are no $SU(2)_L$ singlet right-handed neutrinos in the Standard Model (SM). But can neutrinos be given Majorana masses in the SM? I heard that Majorana mass term cannot be included for neutrinos in the $SU(2)_L\times U(1)_Y$ SM without any modification or extension. Why is that?

• There are extensions of the SM - new Higgs fields, which generates both of Dirac and Majorana masses. For examples, one of the most popular extension is called see-saw mechanism. – Andrew McAddams Mar 17 '14 at 5:09

The important point is the fact that such a mass term breaks the gauge symmetry (Edit: I am assming that you want to build the Majorana mass term using SM available fields -- no extension considered -- of which there is only $\nu_L$). Namely, the desired term is (one generation suffices):

$$\frac{1}{2}\,M\, \nu_L^T \,\mathcal{C}^\dagger\,\nu_L\, +\, \textrm{H.c.}$$

Now, for this term to respect the U(1) part of the gauge symmetry, the total hypercharge must be zero (it is not, as Orbifold pointed out).

Moreover, as far as the SU(2) part is concerned, $\nu_L$ is part of a doublet. So, if your theory is invariant under that symmetry, you should be able to rotate the doublet $(\nu_L,\ell_L)$ under SU(2) with impunity, i.e. without changing the Lagrangian. Here, $\ell_L$ denotes a charged lepton. Clearly, a nontrivial SU(2) rotation would mix both $\nu_L$ and $\ell_L$ fields and the term/Lagrangian is not kept the same/invariant.

Recommended reading: beginning of section 6.4 of C. Giunti and C. Kim's 'Fundamentals of Neutrino Physics and Astrophysics' (Oxford, 2007).

Yes, they can via a dimension 5 operator that contains two Higgs doublets and two lepton doublets.

This is sometimes called Weinberg operator, it violates lepton number conservation and it was introduced in this work http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.43.1566 .

Since the mass comes from an higher dimensional operator (i.e. an operator whose scaling dimension is bigger than the number of space-time dimensions, also called irrelevant), by dimensional analysis it must enter the action suppressed by a scale, call it $\Lambda$. Substituting the Higgs VEV in the operator, you get that the resulting Majorana mass is of order of the (VEV)$^2$ divided by $\Lambda$.

The meaning of the scale $\Lambda$ is that the operator gets generated by some new physics which resides at that scale (think about the four-fermions weak interaction suppressed by the scale $M_W$).

Using the existing bounds to estimate the mass as 0.1 eV, and taking VEV $\approx 100\,GeV$ you will find that $\Lambda \approx 10^{14} GeV$, which is nicely close (on log scale) to the Grand Unification scale. Nevertheless you have to keep in mind that the operator could be generated with a small dimensionless coefficient, and that could change the estimate.

• Why Majorana mass term cannot be included for neutrinos in the $SU(2)_L\times U(1)_Y$ standard model without any modification? – SRS Apr 4 '14 at 12:53
• Because there is no right-handed (rh) neutrino in the SM. Take the quarks: there's an SU(2)-doublet of lh fields, and two rh SU(2)-singlets, up and down. For the leptons, instead there's the SU(2)-doublet with the l-component of the electron and the lh neutrino, and you have an SU(2) singlet right-component of the electron, but no rh neutrino. If you want you can extend the Standard Model by adding such field. It turns out it would be neutral under the whole gauge group. As a result, you can give it an arbitrarily large Majorana mass. This is one possible way to generate the Weinberg operator – Morrissey87 Apr 7 '14 at 8:09

Lets analyse the Majorana condition and the Majorana mass term.

A massive Majorana neutrino $\chi_j$ (a Majorana spin $1/2$ fermion) having mass $m_j>0$ can be described in a local quantum field theory (eg. the standard model) by a four component spin $1/2$ field $\chi_j(x)$ which satisfies the Dirac equation and the Majorana condition which reads:

$$C(\overline{\chi_j}(x))^T=\eta_k\chi_j, \text{ where }|\eta_k|^2=1$$ $C$ is the charge conjugation matrix, $C^{-1}\gamma_{\alpha}C=-(\gamma_{\alpha})^T$ $(C^T=-C,C^{-1}=C^{\dagger})$ and $\eta_k$ is a generic unphysical phase. The Majorana condition is invariant under proper Lorentz transformation. It reduces by a factor of 2 the number of independent components in $\chi_j (x)$.

Notice that the Majorana condition is invariant a global $U(1)$ transformation of the field $\chi_j(x)$ (carrying a $U(1)$ charge $Q$): $\chi_j(x)\to e^{i\alpha Q} \chi_j(x)$ iff $Q=0$. Therefore:

$\chi_j(x)$ cannot carry nonzero additive quantum numbers (e.g. lepton charge)

Given this let us now construct mass terms. In the absence of RH singlet neutrino fields in the theory (which is the case in the Standard model), the flavour neutrinos and antineutrinos $\nu_l$ and $\bar{\nu}_l,l=e,\mu,\tau$ can have a mass term of the Majorana type given by :

$$\mathcal{L}_M^{\nu}(x)=-\frac{1}{2}\overline{\nu^c_{l'R}}(x)M_{l'l}\nu_{lL}(x)+h.c., \nu^c_{l'R}\equiv C(\overline{\nu_{l'L}}(x))^T,$$ Where $M$ is a general $3\times 3$ complex matrix. Now coming to your question: Is this mass term $SU(2)_L\times U(1)_Y$ invariant? The answer is clearly NO! It is not invariant under the weak isospin symmetry, and it changes the weak hypercharge by two units (the hypercharge of all the SM neutrinos is $-1$ which implies that this mass term has a total $U(1)_Y$ charge $-2$). Therefore we need to generate this term via. a Higgs mechanism (as for the Dirac mass term for the electron) or higher loop correction to the SM Lagrangian or possibly some beyond the standard model process.

In conclusion the Standard Model (which does not include RH singlet neutrino fields and contains only the usual Higgs doublet), there is no way to give (Majorana) mass to the neutrinos if fermion (Lepton) number is conserved. Massive Majorana neutrinos appear in theories with no conserved additive quantum number, and more specifically, in which the total lepton charge L is not conserved and changes by two units.

However if you consider Standard model as an effective field theory then there is possibility to generate the above mass term via. higher dimensional operators (which is the context of the previous answer!)