Casimir effect regularization for every divergent sum or series

Question

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??

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.