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In this post regarding quantum corrections to a massless fermion field, the answerer stated that quantum corrections to the mass will always be proportional to the mass (at least in QED). This point leads to an important claim that a massless fermion remains massless under quantum corrections in the Standard Model.

I can show that this is true to 1 loop order in QED, but is this always the case? If so then why?

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It can be seen to follow from a more general statement, namely: "if a parameter in the theory is such that the symmetry gets enhanced when it vanishes, then at every order in perturbation theory the corrections to this parameter will be proportional to its bare value".

This is because perturbation theory respects the symmetry of the classical theory. If the bare parameter is zero, there is an enhanced symmetry that is respected order by order in perturbation theory, and therefore the symmetry breaking parameter never gets generated. It follows that every correction in the theory with bare parameter $\neq 0$ must be a function of the parameter that vanishes when evaluated in $0$.

In the case of the fermion mass, the enhanced symmetry is the chiral $U(1)$.

This general statement is the reason why a parameter can be "naturally small" whenever setting it to zero results in enhanced symmetry.

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  • $\begingroup$ Interesting, thanks! While this makes sense is there a way to formulate this mathematically? In particular is it obvious that if the bare parameter is zero a symmetry breaking parameter won't be broken at loop order? $\endgroup$
    – JeffDror
    Commented Feb 17, 2014 at 16:48
  • $\begingroup$ I'll think about a reference to a mathematical formulation. But if you think about it, the idea is straightforward: you want to compute the correction to a parameter, call it $C$, that breaks a certain symmetry. The Feynman diagrams that give such correction cannot contain only vertices that respect the symmetry. However, if there is only $C$ that violate the symmetry, this imply that you must have at least an insertion of the vertex proportional to the coupling $C$. Therefore, the correction to $C$ will be proportional to $C$ itself $\endgroup$ Commented Feb 17, 2014 at 18:30
  • $\begingroup$ @Morrisey87: I see what you are saying, but it still sounds plausible to me that Feynman diagrams made of vertices that respect the symmetry will produce a broken symmetry. I will try to find time to read up on anomalies in Peskin. I suspect it will go over this idea. Thanks again! $\endgroup$
    – JeffDror
    Commented Feb 19, 2014 at 0:16
  • $\begingroup$ @Morrissey87 I'm thinking of the chiral anomaly whereby in a theory of massless fermions, chiral symmetry is broken by quantum effects, does this not violate your rule of thumb? Or does your statement only hold perturbatively? $\endgroup$ Commented Jul 23, 2017 at 11:29

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