TITLE: In relativistic QFT, is it ever possible that the bare mass be finite and equal to the physical mass?
QUESTION [3 upvotes]: In renormalization, one follows the philosophy that the bare mass is unobservable and could be infinite, and the physical mass comes from the pole of the two-point function. Is it possible that in any case the bare mass is same as the physical mass? Do we have an immediate example (perhaps in some extensions of the Standard Model or more mundane)?

REPLY [5 votes]: Yes, it is possible.
One possible example is the non-interacting theory, for example, a free scalar (bosonic) field with action
$$ S[\phi] = \int d^4 x \left( \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{m^2}{2} \phi^2 \right). $$
If you're looking for examples of interacting theories, Yang-Mills theory for any compact group $G$ has massless gluons in both the bare and renormalized actions. This is an example of the mass term (or rather the absence of thereof) protected by the symmetry.
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