11
$\begingroup$

The question is about why analytical continuation as a regularization scheme works at all, and whether there are some physical justifications. However, as this is a relatively general question, I shall use the following examples to make the question more concrete.

Let us consider the Casimir energy first, which is usually the first place in QFT where the notion of regularization shows up.In the calculation of Casimir energy, one encounters the sum $\sum_{n=1}^{\infty}n$ and to regularize it, one way of doing it is to use zeta function regularization, i.e. by writing $\lim_{s\to 1}\sum_{n=1}^{\infty}n^{-s}$, and then use Riemann zeta function, as an analytical continuation. Different from other regularization schemes(such as hard cutoff, attaching a regulator, whose pole term could be canceled by adding counter terms in the Lagrangian), it seems that this does not have a physical picture behind it. Moreover, different from the other two methods, there is also ambiguity in the answer: One can simply change it to $\lim_{s\to 0}\sum_{n=1}^{\infty}\frac{n}{(n+\alpha)^s} $, where $\alpha$ is a free parameter. Thus one can analytical continue it to $\zeta(-1,\alpha+1)-\alpha\zeta(0,\alpha+1)=\frac{1}{2}\alpha^2-\frac{1}{12}$, which gives $-\frac{1}{12}$ when $\alpha=0$, but in general a polynomial of $\alpha$. Note this term remains in the Casimir force because $\alpha$ has no $r$ dependence a priori.(1-d as an example, $E(r)=\frac{\pi}{4r}(\alpha^2-\frac{1}{6})$). Thus it seems with zeta function regularization that we can only give some lower bound of the magnitude of the force instead of the force it self.

Casimir energy is not an observable. So maybe one might think the problem is not so serious. However, let us consider another example- in AdS/CFT correspondence, one has, roughly speaking, $Z_{CFT}=Z_{AdS}|_{boundary}$ so that $F_{CFT}=F_{AdS}|_{boundary}$,where $F$ is free energy.The claim does not make sense at the first sight as naively speaking one can always shift energy by a constant. However, both theories in AdS/CFT are supersymmetric theory so one could no longer arbitrarily shift the Lagrangian by a constant. In this context $F$ is rigid. However, calculation of $F$ again involves regularization(such as when computing one loop determinant, summing over Klauz Klein levels, or something similar such as section 3.5 of this paper), and if one uses hurwitz zeta to do analytical continuation, the above arbitrariness is still there.

(Maybe one would expect that some regulators are compatible with super-symmetry and some are not, and in this way the answer becomes unambigious. But I haven't seen any discussions like that. )

$\endgroup$
  • $\begingroup$ Forgive my naive question... but what, exactly, does the theory predict for the spectrum of the force? Is it a 1/f law? How is that captured by the theory and the different regularization schemes? $\endgroup$ – CuriousOne Feb 3 '16 at 7:25
  • $\begingroup$ Minor comment to the post (v3): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1401.0825 $\endgroup$ – Qmechanic Feb 3 '16 at 13:57
  • 2
    $\begingroup$ Terry Tao has an excellent blog post about the relations between smoothly regularized divergent sums, their finite parts and analytic continuations. $\endgroup$ – ACuriousMind Feb 3 '16 at 15:10
  • $\begingroup$ @ACuriousMind Thanks for the excellent suggestion! It solves the problem. $\endgroup$ – user110373 Feb 26 '16 at 5:44
  • $\begingroup$ @user110373 If you don't mind, can you explain how it solved the problem of non-uniqueness? $\endgroup$ – user76284 Jul 21 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.