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  Power divergences from loops

+ 6 like - 0 dislike
1175 views

I do not know what I should think about power divergences from loops.

Most QFT textbooks tell us how to deal with logarithmic divergences from loops $\sim\ln(\Lambda^2/\Delta)$: we can set a counterterm $\sim\ln(\Lambda^2/\mu^2)$ to cancel the divergence at some scale $\mu$, then at any other scale this stuff in terms of renormalized quantities are finite $\sim\ln(\mu^2/\Delta)$.

However, sometimes we encounter power divergences like $\sim\Lambda^2s$ in theories where we have momentum dependent interactions (such as a non-renormalizable theory with interaction $\phi^2\partial_\mu\phi\partial^\mu\phi$), where $s$ is one of the Mandelstam variables which depend on the external momenta. If we try to apply the same method to add counterterms, we may have a counterterm $\sim\Lambda^2\mu^2$ such that the power divergence is cancelled at energy scale $\mu$. But if we go to another scale with $s\neq \mu^2$, we will have $\Lambda^2(\mu^2-s)$, which is divergent and invalidates perturbation theory. This is very bad and it means we cannot cancel the infinity if we only have this type of counterterms.

Fortunately, there is another possibility: we may be able to add counterterms that gives $s\Lambda^2$ (for example, by field strength renormalization in the theory with interaction $\phi^2\partial_\mu\phi\partial^\mu\phi$), which is able to cancel the power divergences at all scales. This is nice, and it is even nicer than the case with logarithmic divergences, because we can cancel power divergences identically at all scales and we can cancel logarithmic divergences only at a particular scale that we choose to be our renormalization scale. However, my concern is we can only add terms that is compatible with the symmetries of the system, which may rule out some terms that are necessary for cancelling some power divergences.

My questions are:

1) Is this (the second type of counterterms) the only way that power divergences will be cancelled?

2) If yes, are we always able to set counterterms with respect to the symmetries of the system in this way, so that power divergences can always be cancelled at all scales (then we do not have to worry about power divergences at all!)?

3) If no, how should we cancel power divergences?

By the way, one may say if we use dimensional regularization, power divergences appear in the same way as logarithmic divergences (as $1/\epsilon$ where $\epsilon=4-d$), so we can apply the method of dealing with logarithmic divergences to them. But I think this is too tricky and I want to face this problem with other regularization schemes which have a cutoff.

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user Mr. Gentleman
asked Apr 27, 2014 in Theoretical Physics by Mr. Gentleman (270 points) [ no revision ]
retagged May 4, 2014

1 Answer

+ 0 like - 0 dislike

The interaction term which you are using is $\phi^2\partial_\mu\phi\partial^\mu\phi$ ( add a coupling constant in front of this term), this term has by power counting mass dimension 6 hence the coupling constant has mass dimension -2. This reflects that the interaction term is non-renormalizable. This tells you that this theory is not valid upto every high energy scale.

The best you can do is to make an effective field theory of it. The situation is similar to Einstein-Hilbert lagrangian, which is also power counting non-renormalizable.

None of your analysis will simply hold in QFT for this type of interaction. you can not just add any counter term you want, you must find that it's need arise as a redefinition of some quantities which in this case will not be finite in number ( like field strength, mass etc.).

This post imported from StackExchange Physics at 2014-05-04 11:35 (UCT), posted by SE-user user44895
answered Apr 28, 2014 by user44895 (35 points) [ no revision ]

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