# In relativistic QFT, is it ever possible that the bare mass be finite and equal to the physical mass?

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)?

• How about the photon in the real world. The bare mass is the same as the physical mass and remarkably, it is even equal to the finite number zero. – Oбжорoв Jun 22 at 12:09
• @Oбжорoв probably not the best example, due to ultraviolet inconsistencies of QED. But Yang-Mills for a compact group (read – group with a compact algebra; and $\mathfrak{u}_1$ is not compact) is asymptotically safe, so it is a better example. – Prof. Legolasov Jun 22 at 14:58
• @Solenodon Paradoxus The OP wasn't asking for a UV consistent theory. Strange that a theory that describes the real world, whether QED or the SM, disqualifies on the basis that there may be inconsistencies at an energy scale where the theory is not supposed to be correct ... – Oбжорoв Jun 23 at 7:52

$$S[\phi] = \int d^4 x \left( \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{m^2}{2} \phi^2 \right).$$
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.
• Well, they usually call the level a topological mass. The propagator has a pole at $p^2=g^2k$, I think, where $g$ is the coupling constant. But I guess the pole does get renormalised, through $g$. So never mind my comment – AccidentalFourierTransform Jun 21 at 23:44