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What's the radius of convergence for 1/N expansion in QCD?

  1. Mar 14, 2010 #1
    I'm asking this question as someone who has not studied this topic technically. By radius of convergence, I mean exact results, not just approximations, can be obtained by summing sufficiently many terms. (does this ever happen?) I don't mind if you need 1000 terms, as long as the series is convergent mathematically. Is N=3 QCD within the radius of convergence? And how about N=2, N=1?
  2. jcsd
  3. Mar 15, 2010 #2

    Physics Monkey

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    Though I know of no precise statement to this effect, I think the answer is that the 1/N series does not converge. I am aware of statements that some perturbative results in the 't Hooft coupling can be convergent exactly at large N, but this is different expansion. There definitely can be non-perturbative in 1/N corrections; one has to worry about such things in AdS/CFT, for example.

    There are lots of subtle issues relating to large N. For example, the limits [tex] T \rightarrow 0 [/tex] and [tex] N \rightarrow \infty [/tex] may not commute. More generally, the limits of infinite volume and large N may not commute as in Eguchi-Kawai reduction. The limits [tex] \omega \rightarrow 0 [/tex] and [tex] N \rightarrow \infty [/tex] relevant for response functions do not commute. There is plenty more.

    Hope this helps.
  4. Mar 15, 2010 #3


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    Interesting question; I studied large-N 1+1 dim. QCD up to first loop order (= with mesonic fluctuations) w/o ever wondering about convergence ...
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