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I am working with an action of this form:

$$ S=\int d^4 x [ g^{\mu\nu}\partial_\mu \phi \partial_\nu \phi +\lambda \phi^4 ] $$

Typically, an action would contain a mass term $m^2 \phi ^2$. But here the term is missing. Also typically the $\phi^4$ represents a quartic interaction.

Here however, we have no mass but we do have the quartic interaction. Is the interpretation still the same, or does the quartic term acquire the missing role of the mass somehow? Is a theory without mass but with interaction terms possible?

Qmechanic
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Anon21
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  • As a supplement, my answer to this question uses lattice qft, where everything is finite, to give some intuition about why the bare $m^2$ needs to be negative (not zero!) in order to achieve masslessness. This is true already on a lattice, whether or not any nontrivial continuum limit exists. – Chiral Anomaly Nov 04 '21 at 21:45

1 Answers1

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It has been rigorously established that in 3d there are massless interacting $\phi^4$ theories. See for example Section 9 (4) of this article by Brydges, Fröhlich and Sokal.

Note that the parameters $\lambda$ and eventually $m^2$ in $S$ don't really mean anything. These are bare couplings which can very well become $\pm\infty$. One usually introduces a UV cutoff $\Lambda$ and lets the bare couplings $\lambda_\Lambda$ and $m^2_{\Lambda}$ depend on the cutoff. In 3d for instance it is possible to get the massless (but not conformally invariant) theory with $\lambda_{\Lambda}$ fixed and $m^2_\Lambda\rightarrow -\infty$.

Abdelmalek Abdesselam
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