9
$\begingroup$

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 ?

$\endgroup$
1
  • 3
    $\begingroup$ 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. :) $\endgroup$
    – 299792458
    Jun 26, 2014 at 12:20

2 Answers 2

4
$\begingroup$

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).

Now introduce a small mass term, treat it as an perturbation that softly breaks the formerly exact chiral symmetry. "Technical Naturalness" is the term you're looking for: it states, that quantum corrections to this parameter can only be of the order of magnitude of the symmetry breaking term itself. This also holds for massive gauge bosons, whose mass term explicitly breaks gauge invariance, and whose mass is kept small due to the softly broken gauge symmetry: as you see, this is not an ambiguity of the chiral symmetry.

I might also add that the mass of the Higgs is not protected by any "custodial" symmetry and therefore would diverge quadratically with some cut-off scale. This is one of the reasons to introduce supersymmetry.

$\endgroup$
0
$\begingroup$

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 $m_e \to 0$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.