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  What is meant by the phrase "this operator does not renormalize this other operator", and how can one understand it using diagrammatic arguments?

+ 2 like - 0 dislike

I am trying to understand some sentences in a paper http://arxiv.org/pdf/1412.7151v2.pdf. In section two  the following theory of a (complex) massless scalar coupled to a $U(1)$ gauge boson is introduced
where $\Lambda$ is an energy scale suppresing the dimension 6 operators and
What I want to understand is what is meant in the last paragraph of the same page 

>"Many of the one-loop non-renormalization results that we discuss can be understood from arguments based on the Lorentz structure of the vertices involved. Take for instance thenon-renormalization of $\cal{O}_{FF}$ by $\cal{O}_R$."

my first question is, what is exactly meant when they say that an operator doesn't renormalize the other? ( I somehow suspect this has something to do with the renormalization group but since my knowledge on this matter is very recent I would like as explicit an explanation as possible)

the paragraph continues

>"Integrating by parts and using the EOM, we can eliminate $\cal{O}_r$ in favor of $\cal{O}'_r=(\phi{}D_{\mu}\phi^*)^2+h.c..$ Now it is apparent that $\cal{O}'_r$ cannot renormalize $\cal{O}_{FF}$ because either $\phi{}D_{\mu}\phi^*$ or $\phi^*{}D_{\mu}\phi$ is external in all one-loop diagrams, and these Lorentz structures cannot be completed to form $\cal{O}_{FF}$."

This whole part confuses me. I want to know how do these diagrammatic arguments arise in this context and how can I learn to use them (it would be nice also if someone pointed out which are "all one-loop diagrams" that are mentioned").

asked May 19, 2015 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ revision history ]
edited May 20, 2015 by Dmitry hand me the Kalashnikov

This paper: http://arxiv.org/abs/1505.01844, is more worthy of study on these issues.

The paper cited in the comment of anonymous cites the paper in the OP with the comment "Interestingly, the superfield formalism offers an enlightening albeit partial explanation of these cancellations [16]".

The complete reference to the previous work is in fact this:

"Interestingly, the superfield formalism offers an enlightening albeit partial explanation of these cancellations [16] as well as analogous effects in chiral perturbation theory [17]. These results are clearly connected to our own via the well-known “effective” supersymmetry of tree-level QCD [18], and so merits further study."

As Savas Dimopoulos likes to say: "don't be a hater".

For a successful technology, reality must take precedence over public relations, for Nature cannot be fooled. --

--Richard P. Feynman

http://arxiv.org/abs/1505.01844  has the title "Non-renormalization Theorems without Supersymmetry"

Precisely so: cancellations without supersymmetric spectrum. That's the point. And still supersymmetry helps your understanding, same way it helped with spinor-helicity cancellations in non-supersymmetric QCD. Facts, not PR.

Certainly there is no PR here. None whatsoever.

Another point of view is that supersymmetry supplies a way to misunderstand the actual physics at work quite profoundly.

Here is another fact. A relevant one. The word "helicity" never appears in http://arxiv.org/pdf/1412.7151.pdf.

Answers turned into comments since they dont answer the question.

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