can we use the tools of renormalization of casimir effect to get finite results for any divergent series in QFT?

for example let be the divergent series $ \sum_{n=1}^{\infty}n^{l} $ for positive 'l' then instead of the sum we interpretate this as the difference

$$ \sum_{n=1}^{\infty}n^{l}e^{-n\epsilon}-\int_{0}^{\infty}dtt^{l}e^{-t\epsilon}=finite$$

this is made to evaluate the infinite sums of the vaccuum energy in casimir effect but could it be done for any divergent series??


1 Answer 1


You're asking a mathematical question. Given a function $f$, let $a_n = f(n)$, then does the limit $\epsilon \downarrow 0$ exist for

$$ \sum a_n e^{-n \epsilon} - \int f(t) e^{-t\epsilon}?$$ The answer is no: the integral will generically only get ride of the 'hardest' divergence. The sum $\sum a_n e^{-n \epsilon}$ typically (if $\{a_n\}$ behaves reasonably well) has an expansion of the form $$ \alpha \, \epsilon^{-\nu} + \beta \, \epsilon^{-\nu + 1/2} + \ldots$$ and similarly for the integral, but you can only show that the first terms of both small-$\epsilon$ expansions agree. An example if $f(x) = x^{3/2}$, $a_n = n^{3/2}$. The integral has a single pole in $\epsilon$ but the sum has 3 divergent terms.

There is also an ambiguity, i.e. what should the lower integration boundary be? $t=0$ is just a choice and changing to, say, $t=1/2$ changes the answer/divergent pieces. Look into Eucler-Maclaurin theory for a deeper understanding of this.

There is also a physical misunderstanding. You are "heat kernel" regulating the series $\sum a_n$, but it doesn't always make sense to tag on a factor of $\exp(-n \epsilon)$. Otherwise, why not pick $\exp(-\sqrt{n} \epsilon)$ or $\exp(-n^{2013} \epsilon)$? In practice you often need to pick something covariant, say an energy or eigenvalue $\lambda_n$ of an operator. It just happens that for bosonic strings, the spectrum consists of integers $n$, so that's why textbooks don't mention this.

The right thing to do then is to look for counterterms that can cancel the negative powers of $\epsilon$ in the small-epsilon expansion shown above.


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