My question was motivated by my attempt to answer this question. Suppose we are given an action and we make a change of variables such that the theory is non-renormalizable. Does the new theory then require an infinite number of counterterms?

As an explicit example lets consider the situation brought up in the linked question (though I change notation for my convenience). We start with the Lagrangian, $${\cal L}= \frac{1}{2}\partial^\mu\phi_0\partial_\mu\phi_0-\frac{1}{2}m^2\phi_0^2$$

Then we make the substitution, If I make $\phi_0=\phi+\frac{\lambda}{M} \phi^2$ such that $\lambda$ is dimensionless and $M$ is some mass scale. Then the Lagrangian is $${\cal L}= \frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}m^2\phi^2+2\frac{\lambda}{M}\phi\partial^\mu\phi\partial_\mu\phi-\frac{\lambda}{M} m^2\phi^3 + 2\frac{\lambda^2}{M^2}\phi^2\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}\frac{\lambda^2 }{M^2}m^2\phi^4$$

Now originally we could have found all the counterterms with calculating a few simple diagrams. On the one hand I'd think that since we still have a single coupling, $\lambda$, we should still have the same number of counterterms in the new theory. On the other hand I've learned that operators get renormalized, and not couplings, so since we have more operators we also need more counterterms. How many counterterms does this new theory actually need?

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    $\begingroup$ Haven't you forget a Jacobian with this change of variable? I think that they usually do the job to make the theory after change of variable ok. $\endgroup$
    – Adam
    Jun 17, 2014 at 13:30

1 Answer 1


The general recipe for the renormalization of a renormalizable QFT produces (in perturbation theory) an finite-dimensional space of possibilities, described by a finite, typically small number of renormalization parameters.

The general recipe for the renormalization of an ''unrenormalizable'' QFT produces (in perturbation theory) an infinite-dimensional space of possibilities, described by a countably infinite number of renormalization parameters.

If you make an invertible nonlinear transform of a renormalizable QFT with p renormalization parameters (and include the logarithm of the transformation Jacobian, so that the actions are classically equivalent) you get a p-dimensional submanifold of the infinite-dimensional manifold of QFTs corresponding to the transformed Lagrangian. This means that the infinitude of renormalization parameters for the latter are parameterized in terms of the p original renormalization parameters. However, this parameterization is not easily recoverable if you start from the transformed Lagrangian without any knowledge of the transformation. Moreover, it doesn't mean that other choices from the infinite-dimensional manifold are not allowed, just that they don't come from a transformation of the original family of QFTs.


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