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Field redefinitions and new counterterms

+ 6 like - 0 dislike
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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?

This post imported from StackExchange Physics at 2014-06-19 11:27 (UCT), posted by SE-user JeffDror
asked Jun 17, 2014 in Theoretical Physics by JeffDror (650 points) [ no revision ]
retagged Jun 19, 2014
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.

This post imported from StackExchange Physics at 2014-06-19 11:27 (UCT), posted by SE-user Adam

Even without Lagrangian, it is clear that the new $\phi$ is expressed via the old $\phi_0$ and $\lambda$ in a complicated way. If you want to develop a series in powers of $\lambda$, try it first in this variable change to make sure it is sensible. Only one root out of two makes it possible.

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