# Protection of the electron mass by chiral symmetry

In many textbooks it is said that mass renormalization of the electron mass is only logarithmic $$\delta m \sim m\, log(\Lambda/m)$$ because it is protected by the chiral symmetry. I understand that in case of massless fermions to keep them massless in the renormalization procedure it must be like this. Or differently said, the renormalization procedure respects the axial current conservation.

But is there a compulsory reason for the renormalization procedure to respect the axial current conservation ? Does every renormalization procedure respect that ? Apparently Pauli-Villars and dimensional renormalization do it, but what for other procedures ? I also know that in triangular Feynman diagrams anomalies occur which do break the axial current conversation. So why can't it happen for something simpler like the electron mass respectively self-energy ?

• Can you please add sources here, e.g. when you say ''in many textbooks'', you may post a link to at least one of them. :) Commented Jun 26, 2014 at 12:20

As far as I have understood the topic, the fact that the correction to the mass is going with the log is not the point. The point is that it is proportional to the mass of the electron itself (The log enters in this argument via the fact that it doesn't blow up as fast as e.g. quadratic divergence). You will know that the chiral symmetry is exact for $m=0$, and, in any renormalization scheme, all corrections to $m=0$ will be zero as well (just like the photon doesn't acquire mass).
The only mass scales are the mass of the electron $m_e$, and, if we consider other kind of interactions of a fermion (here the electron) with a massive gauge vector boson (ex : $Z$ for the weak interaction), or a massive scalar ( scalar electrodynamics), you would have this boson mass $m_B$ too.
However, the chiral symmetry being exact for $m_e=0$, we know that the quantum corrections $\delta m$ of the electron mass, when $m_e \to 0$, will go towards zero.
If the quantum corrections $\delta m$ were (in the limit where $m_e \to 0$) proportionnal to $m_B$, it would contradict the above zero limit, so the only possibility is that $\delta m$ is proportionnal to $m_e$ (always in the limit where $m_e \to 0$).