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Zeta function regularisation

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data

    Hi I need to regularize [itex]\sum_{r \in Z+1/2} r [/itex]

    In my opinion there are two ways of going about it either re-express it as [itex] \sum_{r \in Z+1/2} r = \sum_{r =1} r - \frac{1}{2} \sum_{r =1} = \zeta (-1) - \zeta (0) = \frac{1}{6} [/itex]

    or

    [itex] \sum_{r \in Z+1/2} r = \frac{1}{2} \sum_{r =1} r - \sum_{r =1} r = - \frac{1}{2} \zeta (-1) = \frac{1}{24} [/itex]

    I know I need the second answer however I don't see any reason why the first answer is not valid. In fact I think it more so, since the first sum goes term for term with the second, whereas in the second method the r =2 term of the first sum is cancelled by the r=1 of the second thus having a staggered structure if the sum was finite. Any thoughts?
     
  2. jcsd
  3. Sep 15, 2014 #2

    nrqed

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    You are making a good point. I am assuming you are encountering this in a quantum field theory calculation?

    The unfortunate answer (which, I know, will feel unsatisfactory) is that both expressions are equally valid. This is because the sum is actually infinite, as you know, and therefore regularizing can give different finite results, depending on how one proceed. So the finite answer is pretty much arbitrary! But the key point is that after renormalization, all infinities cancel out (in a renormalizable theory). The key point is that one must regularize all divergent expressions in a consistent way. So if one uses the second expression to regularize an expression in one step of the calculation, one must use the same expression to regularize other divergent sums and then the final, renormalized result is finite and well-defined.

    I hope this helps a bit.
     
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