How Many Counterterms Are Needed for Logarithmic Divergences?

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SUMMARY

The discussion centers on the necessity of counterterms in the Lagrangian to address logarithmic divergences as \(\lambda \rightarrow \infty\). Specifically, it examines cases involving logarithmic forms such as log(a+\(\lambda^{n}\)), log(\(\lambda\)), and log\(^k\)(\(\lambda\)). Utilizing dimensional regularization, the participants conclude that the number of counterterms required depends on the specific logarithmic divergences present. Additionally, operator-regularization is proposed as an alternative method that may eliminate the need for extra terms in the Lagrangian.

PREREQUISITES
  • Understanding of logarithmic divergences in quantum field theory
  • Familiarity with dimensional regularization techniques
  • Knowledge of Lagrangian formulations in physics
  • Concept of operator-regularization and zeta-function regularization
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  • Research the application of dimensional regularization in quantum field theories
  • Study the role of counterterms in renormalization processes
  • Explore operator-regularization and its advantages over traditional methods
  • Investigate specific examples of logarithmic divergences in theoretical physics
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The discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory, as well as researchers focused on renormalization techniques and regularization methods.

zetafunction
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If i have ONLY logarithmic divergences as \lambda \rightarrow \infty of the form

log(a+\lambda ^{n}) or log (\lambda ) or log^{k}(\lambda) for some real numbers a,n and k HOW many counterterms should i put into de Lagrangian in order to make it FINITE ?? , the idea is let us suppose we use DIMENSIONAL REGULARIZATION so we only had logarithmic divergent integrals (and assuming that power law divergences can be reduced by dimensional regularization or other method to only logarithmic divergences), how many counterterms should i add to the original lagrangian to obtain finite results ??
 
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One way around is to use operator-regularization (a generalization of the zeta-function) that will take care of everything, and preclude the need to add extra terms to the original Lagrangian (should they be needed).
 

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