It's very common in quantum field theory (QFT) to say things like: The mass (charge, etc.) is protected by the symmetry. But I never quite understood the notion. What do we mean by that?


We mean that particular operators are forbidden by the symmetry or result in particular forms of radiative correction.

For example, a chiral symmetry $$\psi\to e^{i\theta \gamma_5}\psi$$ would forbid fermion masses at tree-level. Even if a fermion mass were present, breaking the symmetry, the symmetry in the $m\to0$ limit would guarantee that radiative corrections (and beta-functions) were proportional to the mass, $$\Delta m\propto m.$$ Thus the mass is protected from corrections from other mass scales in the theory, $$\Delta m \not\propto M $$ This is desirable as we need e.g. electrons to be much lighter than the Planck scale and don't want $\Delta m_e \propto M_P $.

Scalars, such as the Higgs, aren't protected by any syymmetries, leading to $$\Delta m_h^2 \sim M_P^2$$ This is the notorious hierarchy problem.

  • $\begingroup$ It is often said that the bare Majorana mass of the $SU(2)_L$ singlet right-handed sterile neutrino (in type-I seesaw extension of Standard model) is not protected by symmetry, and could take large values as large as $M_{pl}$. Is it due to the same reason you explained? What do you think of my answer to OP's question? @innisfree $\endgroup$
    – SRS
    May 22 '17 at 6:21

The Dirac mass of the fermions and massive gauge bosons cannot take arbitrary values in the Standard model because any Dirac mass is projected by $SU(2)_L\times U(1)_Y$ symmetry. It is only the breakdown of this symmetry which impart mass to these particles. All particle masses are determined by the symmetry breaking scale.

This is not the case, for example, right-chiral Majorana mass for the neutrinos $M_\nu \overline{(\nu_R)^c}\nu_R+h.c.$(Note that, unlike any Dirac mass, this term is a gauge singlet).

The right-handed Majorana mass term is not protected by any symmetry i.e., this mass term doesn't break $SU(2)_L\times U(1)_Y$ symmetry. Therefore, a bare Majorana mass term with arbitrarily large value can be written down even in the tree-level Lagrangian (for Standard model augmented with right-handed neutrinos).


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