Renormalisation Question

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fzero
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Is the way to see that 4D only gives logarithmic because we produced a [itex]\frac{1}{\epsilon}[/itex] pole by dim reg and since a [itex]\frac{1}{\epsilon}[/itex] in dim reg corresponds to a [itex]\log{\frac{\Lambda}{m}}[/itex] in UV cutoff we only have a log. Is that correct?
Dimensional regularization of any single divergent loop integral always gives the divergence in terms of a simple pole, because the singularities of the Gamma function are simple poles. It seems like logarithmic singularities occur when the divergence is of the form [tex]\Gamma(\epsilon)[/tex]. Higher-order singularities will appear as [tex]\Gamma(-n+\epsilon)[/tex], where [tex]n[/tex] is an integer. For example, this quadratic divergence appears as [tex]\Gamma(-1)[/tex], but it's still a simple pole. We can't use the power of [tex]\epsilon[/tex] to determine the degree of divergence.
 

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