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 (or is it?). In this sense, the smallness of electron mass it 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?


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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    $\begingroup$ Instanton contributions can break the Z2 and U(1) symmetry, see journals.aps.org/prl/abstract/10.1103/PhysRevLett.61.794 (although not in QED, since the gauge group is abelian). $\endgroup$ – thedoctar Apr 17 at 12:33
  • $\begingroup$ @thedoctar I am aware that discrete symmetries can have anomalies, but as you agree this one doesn't. $\endgroup$ – user178876 Apr 17 at 13:51
  • $\begingroup$ So how do I see that the $Z_2$ axial symmetry is anomaly free? $\endgroup$ – Thomas Apr 18 at 18:18
  • $\begingroup$ @Thomas AFAIK the first paper on this is by Krauss & Wilczek, then there were papers by Ibañez and Ross, by Banks and Dine and subsequently several others who derived the anomaly constraints for discrete symmetries. The arguably cleanest derivation is based on the path integral method, but this has been done much later. Here it is sort of pointless since the U(1)$_\mathrm{em}$ does not have instantons. $\endgroup$ – user178876 Apr 18 at 20:36

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

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    $\begingroup$ I think the OP has a legitimate question: How can chiral symmetry protect the electron mass if there is no chiral symmetry? It does not really matter that there is a finite number of primitive diagrams that reflect the anomaly. I can always insert these primitive vertices into the electron self energy. I think the point is that the RHS of the anomaly really is a total derivative and QED (as opposed to QCD) does not have topological sectors, theta dependence etc, $\endgroup$ – Thomas Apr 15 at 13:14
  • $\begingroup$ @Thomas The electron mass is protected by chiral symmetry, because the chiral current is conserved in the absence of an external field. Otherwise, write down a Feynman diagram with 1 ingoing eL and 1 outgoing eR! Topological considerations are non-perturbative so we can't think it terms of diagrams anymore. You're right, that the chiral current can be violated if the gauge field has non-trivial topology. This is because the integral of the divergence of the chiral current is a Chern number. But since U(1) is abelian, the photon doesn't have a topological winding number. $\endgroup$ – thedoctar Apr 16 at 10:38
  • $\begingroup$ @Thomas Yes you're right, for some reason I forgot all about that topological stuff! These effects are suppressed by the instanton action exp(-S_inst) = exp(-8 pi/g) (for n=1). One has to use the saddle-point expansion -- it's not as simple as using the triangle diagram as an effective vertex. I've included a link in my answer where you can learn about this technique. $\endgroup$ – thedoctar Apr 16 at 12:39

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 $m_d$.

  • $\begingroup$ Your answer invokes confinement. Without confinement, instanton contributions can (in principle) induce a mass term in QCD. $\endgroup$ – thedoctar Apr 17 at 12:34
  • $\begingroup$ @thedoctor Are you thinking of the $\eta'$ mass? -- or is there a machanism that gives a mass directly to the fermions? $\endgroup$ – mike stone Apr 17 at 12:40
  • $\begingroup$ see journals.aps.org/prl/abstract/10.1103/PhysRevLett.61.794 -- they generate an up mass from massless, but they need massive down & strange quark. Still there is a U(1) symmetry. The comment was more in general, that instanton contributions can change the picture. $\endgroup$ – thedoctar Apr 18 at 6:36

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