Given an arbitrary quantum field theory, can I always write it in terms of another (different) quantum field theory containing only operators with 3 fields? (i.e. vertexes with 3 legs)

I guess that should be possible introducing arbitrary additional fields that once integrated away bring to the original operators. Is that so?

Can it be done generally without having additional degrees of freedom in the new (3 vertexes) theory? I am thinking of something with maybe ghost fields.

Or maybe there is some other way?

I heard this has been done with General Relativity (in Hidden Simplicity of the Gravity Action), but I really did not check in detail the content.

Maybe the simplest case to think of is that of a theory containing only one interaction vertex (let's say 4 legs, $\phi^4$). We replace the theory with a theory with only the three leg vertex ($\phi^3$) and if we quotient the space of asymptotic states removing those with any odd number of external legs (you cannot have those in $\phi^4$), then we should obtain the starting theory. This said I don't see how to upgrade this trick in cases where the starting lagrangian contains more interactions, hence the question.


1 Answer 1


Yes you can! I'm sure you can do this all sorts of ways, but here is a nice systematic one. Suppose you have a term in the lagrangian $\prod_{i=1}^{k} \phi_i\subseteq L$. We can start by introducing two new fields, $\lambda_1$ and $\sigma_1$. Add the term $i\lambda_1(\sigma_1-\phi_1*\phi_2)$. If we first integrate out the $\lambda$ variable, we get a delta function. Integrating out the $\sigma$ gives us unity. Notice that the terms added to the lagrangian are at most cubic in the vertices. We can then introduce $lambda_2, \sigma_2$, and add in the term $i\lambda_2(\sigma_2-\sigma_1*\phi_3)$. If we keep going, we get that the path integral contains a bunch of delta functions forcing $\sigma_{k-1}$ to be our our term in the lagrangian. So we can just remove the offending term by replacing it with a $\sigma_{k-1}$, and our higher-order term was replaced with a family of cubics.

You can show that this all works out for vector fields, and even fermionic fields. If you have multiple terms, just repeat the procedure multiple times. In this way, we can force every lagrangian to be cubic.

  • $\begingroup$ Neat! Thank you! Do you know where can I find something like that in the literature? $\endgroup$
    – AoZora
    May 26, 2019 at 13:00
  • $\begingroup$ Not exactly the same, but the idea of using a Lagrange multiplier and auxiliary fields to rewrite the action is pretty common with the SYK model. $\endgroup$
    – Mikewins
    May 26, 2019 at 17:25

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