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Technical or 't Hooft naturalness A parameter $\theta$ in the Lagrangian of a field theory is said to be natural, if in the limit of vanishing $\theta$, the theory has some enhanced symmetry. If this happens, the smallness of the parameter $\theta$ is said to be natural.

An example Let us consider the theory of QED with massless electrons. It can be shown that the electron mass does not receive quantum corrections and remains zero. If we repeat the same calculation with massive QED, we find that the bare electron mass $m_0$ receives a correction which itself is proportional to $m_0$ i.e. $$m_0\to m=m_0+\frac{3\alpha}{4\pi}m_0\ln\Big(\frac{\Lambda^2}{m_0^2}\Big)\tag{1}$$ where $\Lambda$ is the cut-off. Thus, massive QED reproduces the result of massless QED in the limit $m_0\to 0$. This clearly shows that if $m_0$ is zero in the classical action to start with it will remain zero; if $m_0$ is nonzero but small to start with, it will remain small. In this sense, the smallness of electron mass is technically natural.

Question But I am still uncomfortable with the role of symmetry here and cannot fully digest the idea of technical naturalness. Because, if the symmetry is anomalous in the limit $m_0\to 0$, what is it that stabilizes the electron mass against large quantum corrections? Since the symmetry is anomalous, we cannot say for sure that it is the symmetry that protects $m_0$ from receiving large correction. What is really going on at the heart of the matter?

SRS
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4 Answers4

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To forbid this mass term, we do not need a full chiral U(1) symmetry, but only a $\mathbb{Z}_2$ symmetry under which, say, the left-handed electron picks up a sign. This $\mathbb{Z}_2$ is anomaly free and can thus forbid the mass term also at the quantum level.

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In a general chiral gauge theory in D-dimensions, the anomaly only appears in the one-loop diagram with (D-1) external gauge bosons (arXiv:0802.0634).

The chiral anomaly you are talking about only appears in the triangle diagram (in 4 dimensions), i.e. 3 external photons with an electron loop. Adler, Bell and Jackiw proved this perturbatively, but a more modern treatment can be found in arXiv:0802.0634.

So the loop corrections to the electron mass don't break the chiral symmetry. Only the triangle diagram does.

The non-conservation of left- and right-handed fermions only happens with non-zero electric/magnetic fields, since the divergence of the chiral current is proportional to the field-strength tensor.

See chapter 19.2 of Peskin & Schroeder, as well as problem 19.1.

I'd also recommend following Fukikawa's analysis, where the anomaly comes from the Jacobian of the field measure in the path-integral. Also done in P&S.

EDIT: I've neglected non-perturbative effects in this answer (which is okay for the photon, since abelian gauge groups have no topological configurations). See the comments section.

If you want to learn more about calculating such effects (saddle-point expansion/semiclassical method) I'd recommend https://iopscience.iop.org/article/10.1070/PU1982v025n04ABEH004533/meta

thedoctar
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It's not just the chiral anomaly that can muddle the argument that chiral symmetry protects masslessness. If we start with massless quarks the strong interactions would still cause chiral symmetry breaking and give the quarks a "constituent" mass. Indeed in the real world the "current" masses of the the $u$ and $d$ quarks are rather small and most of the masses of the hadrons containing them comes from chiral symmetry breaking and so are proportional to $\Lambda_{\rm QCD}$ rather than $m_u$ or $m_d$.

mike stone
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The previous answers have focused on the chiral anomaly, however, I think the question is about the 'Hooft naturalness.

In this question about the fermion mass, the enhanced/restored symmetry in the limit $m_0 \to 0$ is the chiral symmetry, which is the best example for discussing the naturalness.

The essence is, the (chiral) symmetry breaking effect is proportional to $m_0$. The mass is chiral symmetry breaking interaction. Therefore, a loop correction to the mass has to break the the chiral symmetry, thus is proportional to $m_0$.

The consequences are:

  1. Every loop correction is proportional to $m_0$ thus the total loop correction is also, i.e. multiplicative. This means, if for some reason $m_0$ is small, the total correction is small.
  2. From the dimensional analysis, the fermion mass is mass-dimension one and every loop correction (without chiral symmetry) should be proportional to the momentum cutoff $\Lambda$. (To be precise, a mass scale in the UV). However, it is also proportional to $m_0$, thus the divergence is $m_0\log \Lambda$ as the question shows.

Thus the fermion mass has been known technically small.

(However, I must also add that this is not the end of the story. Since the bare mass $m_0$ should be close to $\Lambda$ to have small observed mass. I have posted the explanation at On the naturalness problem)

user112002
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