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?


2 Answers 2


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, 2017 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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