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Using analytic continuation in dim-reg

+ 5 like - 0 dislike
803 views

In a book by Wise and Manohar, Heavy Quark Physics (pg 80), they discuss the limit 

\begin{equation}\lim _{\lambda\rightarrow \infty} \lambda^{\,z\,(\epsilon)}\end{equation}

where $z$ is some function of an infinitesimal parameter, $\epsilon$. Then they say "as long as $z$ depends on $\epsilon$ in a way that allows one to analytically continue $z$ to negative values" then this limit is zero. I'm not very familiar with analytic continuation (other then the qualitative idea of what it means) but this seems very strange to me. 

I understand why the paths of contour integrals can be morphed between one another (due to Residue theorem), but why should such arguments hold for limits as well?

asked Jun 2, 2014 in Mathematics by JeffDror (650 points) [ revision history ]

The limit of analytic functions is not analytic, so this statement is at best misleading, but more appropriately called "dead wrong". You should look to see what they apply this to do, the conclusion can be correct even if the intermediate steps are brain-damaged.

1 Answer

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The point is that an analytic function is determined by the values in an arbitrary line segment. If $z(\epsilon_0)<0$ then the function $f(\epsilon,\lambda):=\lambda^{z(\epsilon)} $ is analytic in a neighborhood of $(\epsilon_0,\infty)$, and $f(\epsilon,\infty)=0$ for all $\epsilon$ close to $\epsilon_0$.Thus it agrees with the zero function on a line segment, hence is the zero function.

answered Jun 3, 2014 by Arnold Neumaier (12,570 points) [ no revision ]

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