EDIT: Often in textbooks one discusses QFT with massless particles, e.g. massless fermions or scalar particles. What mass vanishes: bare or physical? or both?
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$\begingroup$ I have never seen a textbook cover QFT with massless particles at the start of it. When it is covered, it is always the physical mass that is zero; the bare mass tends to also be zero for them too. $\endgroup$– naturallyInconsistentCommented Jul 17 at 17:05
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$\begingroup$ I meant Peskin-Schroeder book. It would be helpful to have a reference for the argument that vanishing of physical mass implies vanishing of bare mass. Thank you. $\endgroup$– MKOCommented Jul 17 at 17:12
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1$\begingroup$ Pesky and Shredder is not at all a normal textbook. The front part is essentially impossible to read. $\endgroup$– naturallyInconsistentCommented Jul 17 at 17:20
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4$\begingroup$ Peskin and Schroeder briefly discuss this in chapter 10 of the first edition: "But the electron mass shift must actually be proportional to m, since chiral symmetry would forbid a mass shift if m were zero." $\endgroup$– hftCommented Jul 17 at 19:46
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When textbooks specifically mention massless particles, they always mean renormalized mass. Unless protected by some type of symmetry, if the bare mass vanishes, the renormalized mass does not (due to loop corrections).
Examples of cases where renormalized mass vanishes are:
Goldstone bosons (spin-0): Goldstone bosons arise due to spontaneous symmetry breaking. They have a non-linear symmetry transformation of the form $\phi \to \phi + a$. The mass term $m^2 \phi^2$ is not invariant under such a shift symmetry, so the renormalized mass vanishes.
Chiral fermions (spin-1/2): The fermion mass term $m {\bar \psi} \psi$ is not invariant under chiral symmetry transformations, so chiral fermions are massless. Note, however, that a chiral symmetry is often anomalous, in which case the renormalized mass is not expected to be zero.
Gauge symmetry (spin-1): The mass term $m^2 A^\mu A_\mu$ is not gauge-invariant so in this case, the renormalized mass is also zero.
Diffeomorphism symmetry (spin-2): Same as above.
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$\begingroup$ Pedantic nitpick on 1: It is possible to have SSB due to radiative corrections on a theory lacking it, so, then, it is the renormalized effective theory goldston that is massless. $\endgroup$ Commented Jul 17 at 19:31
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$\begingroup$ Thank you. What happens in $\phi^4$-theory? It has no symmetry $\phi\to \phi+a$. $\endgroup$– MKOCommented Jul 18 at 6:54
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$\begingroup$ The scalar field in $\phi^4$ theory is not a Goldstone boson. $\endgroup$– PraharCommented Jul 18 at 7:07
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$\begingroup$ @MKO ... but you can get there, through extra interactions, somehow . $\endgroup$ Commented Jul 18 at 12:54
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$\begingroup$ What is the loophole in the argument where we apply (3) in 2 dimensions where the photon becomes massive due to loop corrections? $\endgroup$– SanjanaCommented Jul 25 at 21:11