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Consider the divergent series $$S = 1 + 1 + 1 + \ldots$$ which may appear in some calculations involving the Casimir effect. There are two main ways to evaluate this series. One can perform analytic continuation (e.g. by the zeta function), yielding $$S = \zeta(0) = -1/2.$$ The other common way is to parametrize the divergence by regulating the series. For example, we can add an exponential regulator, which turns the series into a convergent geometric series: $$S = 1 + (1-r) + (1-r)^2 + \ldots = \frac{1}{r}.$$ Then, we can cancel pure divergences (i.e. terms of the form $1/r^n$) using local counterterms, for $$S = 0.$$ These two methods seem to disagree.


Both of these methods (which I'll call 'analytic continuation' and 'UV reg. + counterterms') are used heavily in physics.

  • In the usual Casimir effect, we have the series $1+2+3+\ldots$. Analytic continuation yields $\zeta(-1) = -1/12$, and parametrization gives $1/r^2 - 1/12$, so the two methods agree.
  • When computing loop integrals, dimensional regularization is the first method, and most other methods (Pauli-Villars, hard cutoff) use the second method.
  • Sometimes, both methods have to be used. Dimensional regularization only removes scaleless divergences $\int dk/k^n$, and other divergences manifest as poles in $\epsilon$. Then we cancel those with counterterms.

Given that these two renormalization methods are both very commonly (and interchangeably) used, I'm disturbed that they disagree on the series $S$ defined above. I've heard several possible resolutions to this problem, some in the (deleted) comments below.

  • In any physical situation, the regulator shouldn't matter. So in this case, $S$ would not be physically observable; it would have to be subtracted from some other divergent series, and the result would not depend on the method used.
  • Analytic continuation is generally more reliable. If the two methods disagree, analytic continuation is right.

Which of these, if any, is the right resolution to the paradox?


Here are some references to related questions asked recently:

  • This question asks if analytic continuation may yield a nonunique answer.
  • This question presents a third option for renormalization, which is to postulate that $$\sum_{n=0}^\infty a_n = a_0 + \sum_{n=1}^\infty a_n.$$ This method agrees with analytic continuation and UV reg. sometimes, and disagrees other times. It's unclear to me where this one should go.
  • This answer presents a fourth option, smoothed asymptotics, and shows that its result is not unique. It is also unclear to me how this method is related to the other three.
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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Jul 6 '16 at 17:37
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    $\begingroup$ You may also want to link to my old question which puts a slightly different spin in the question of whether analytic continuation in the context of dimensional regularization always yields a unique result. $\endgroup$ – tparker Jul 10 '16 at 21:08
  • $\begingroup$ Related: physics.stackexchange.com/q/26877 $\endgroup$ – tparker Jul 11 '16 at 6:37
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Your example has no counterterms to cancel the 1/r singularity! This explains the discrepancy.

In a correctly regularized expression, the counterterms are not introduced in an ad hoc way but by renormalizing some constants in the original problem that gives rise to the series. These modify all terms in the sum and make the divergent part vanish for particular choices, so that the singular limit can be taken in the result. Having to discard an infinity without having been able to cancel it properly is always a sign of having made a mistake.

You might be interested in reading my tutorial paper ''Renormalization without infinities'' - a tutorial that discusses renormalization on a much simpler level than quantum field theory.

In general, to know what the result should be one must start with a well-defined expression whose limit is sought and then transform the parameters in this expression in a way that after the transformation the limit exists. Then the result should be regulator independent.

But if one doesn't have a finite context to start with there is no way to know what one should do. Detached from physical content, a divergent series is just that - divergent and hence meaningless. The series $\sum_{i=0}^\infty (−1)^i$ can be given any integer value by appropriately reordering and taking partial sums, and all these values are different from the value you get by interpreting it as a limit of a geometric series.

Thus it is only the context that makes the value of a divergent series possibly well-defined.

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  • $\begingroup$ Thanks for the response! However, I attempted to regularize the vacuum energy contribution and got the wrong answer: in analogy with the series, it is $\int_0^\infty (1-r)^x dx = -1/\log(1-r) = 1/r-1/2 + O(r)$. The difference in energies is $-1/2$, rather than $+1/2$ with zeta regularization. Do you have an idea what went wrong, if there are extra terms I'm forgetting? $\endgroup$ – knzhou Jul 11 '16 at 20:27
  • $\begingroup$ @knzhou: just taking a series or an integral, renormalization is ill-defined. To know what the result should be one must start with a well-defined expression whose limit is soughtr and then transform the parameters in this expression in a way that after the transformation the limit exists. Then the result should be regulator independent. But if one doesn't have a finite context to start with there is no way to know what one should do. $\endgroup$ – Arnold Neumaier Jul 12 '16 at 9:00
  • $\begingroup$ So the conclusion from my contradictory result trying to subtract $1+1+...$ from $\int 1 dx$ is that such a series/counterterm would never arise in a physical problem? $\endgroup$ – knzhou Jul 15 '16 at 0:54
  • $\begingroup$ This doesn't give me much faith in the method though... how do I know when it will work, if I see a divergent series "in the wild", detached from physical context? $\endgroup$ – knzhou Jul 15 '16 at 0:58
  • $\begingroup$ Detached from physical content, a divergent series is just that - divergent and hence meaningless. The series $\sum_{i=0}^\infty (-1)^i$ can be given any integer value by appropriately reordering and taking partial sums, and all these are different from the value you get by interpreting it as a limit of a geometric series. It is only the context that makes the value of a divergent series possibly well-defined. $\endgroup$ – Arnold Neumaier Jul 15 '16 at 8:07
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In fact, analytical continuation in some sense is subtracting off the divergence using local counter terms. A very clear treatment has been made in Terry Tao's blog: https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/ (See section 2)

Note the result of the regularization does dependent on the regulator you choose, and this is essentially explicit in Zhengyan Shi's answer. (The $f(0)=1 $ condition) In your case, the analytical continuation can be thought of using a compactly supported function $\eta(x)$ on $(0,1)$,with $\eta(0)=1$.Thus one can consider the sum $$\sum_{n=0}^{\infty}n^s\eta(\frac{n}{N})$$

In Terry Tao's blog, he treats it with general complex $s$, with $Re(s)>-1$, and it turns out that with $\eta(0)=1$, one can actually analytically continue zeta function by subtracting of the divergent piece in the asymptotic expansion of $N$, i.e. $$\zeta(-s)=\lim_{N\to \infty}(\sum_{n=0}^{\infty}n^s\eta(\frac{n}{N})-N^{s+1}\int_0^{\infty}x^s\eta(x)dx)$$ and it does not depend on what $\eta(x)$ one chooses as long as $\eta(0)=1$. One can see in this sense analytical continuation is subtracting of the divergence using counter terms.As mentioned above,the result depends on what $\eta(0)$ is.

For example, if we let $s=0$,and consider the sum $\sum_{n=0}^{\infty}\eta(\frac{n}{N})$, which has asymptotic expansion $$N\int_0^{\infty}\eta(x)dx-\frac{1}{2}\eta(0)+\mathcal{O}(\frac{1}{N})$$

The first term is the divergent piece and thus is subtracted off by "counter terms".The second term is the regulated sum.

For zeta regularization $\eta(0)=1$ and thus the answer is $-\frac{1}{2}$. For your $(1-r)^n$ regularization, in this language (you can off course cut your function off beyond $x=1$), is $\eta(x)=(1-r)^x$ for given $r$, and $\eta(0)=1$ as well. So if you consider the finite part of $\lim_{N\to \infty}\sum_{n=0}^{\infty}(1-r)^{\frac{n}{N}}$, you would indeed get $-\frac{1}{2}$ in agreement with the zeta regularization.

But instead of looking at $N\to \infty $ limit with finite $r$, the second method in your question looks at $r\to 1$ limit with finite $N$. The function $x^y$ is not continuous at $(0,0)$ so you are bound to get different asymptotic behaviour depending on which limit you take first.

So as a summary, 1) Zeta function regularization can be thought of subtracting countqerterms 2) Depending on how you choose the regulator and take the limit, you can get different answers, and there is no right answer a priori. The "right" regularization must take into account of the actual physical context. (such as not breaking various symmetries the theory has.)

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I don't know much about QFT at all, but I have read some discussions of Casimir's effect in Matthew Schwartz's QFT textbook and I hope I can offer some valuable ideas about why the different regulators agree in some cases but not others.

Essentially, in Schwartz's treatment, there is a class of regulators that yield the same physical result, namely the $-\frac{1}{24r}$ term in the energy. More precisely, he derived the necessary constraints on the regulators.

Here are the details. Suppose the cutoff angular frequency in the hard-cutoff regulator is given by $\omega_{cutoff} = \pi \Lambda$ (the energy will be shown to be independent of the cutoff!). Then a general regulator $f$ turns the original sum:$$ E(r) = \sum_n \frac{\omega_n}{2} \quad \omega_n = \frac{n \pi}{r}$$

Into the sum: $$E(r) = \sum_n \frac{\omega_n}{2} f(\frac{\omega_n}{\omega_{cutoff}})$$

Using the general expression to calculate the total energy, we find two constraints that are necessary. First of all, if $$ \lim_{x ->\infty} xf(x) = 0$$ Then we end up with a nice and clean expression for the total energy (the detailed calculation is a straightforward application of Euler-MacLaurin Series): $$E_{total} = E(r) + E(L-r) = \frac{\pi}{2} \Lambda^2 L - \frac{\pi f(0)}{24r} + ...$$ If we then add the constraint that $f(0) = 1$, then the $r$ dependent energy term matches the experimental results. And the two constraints are not very strong at all. A wide variety of regulators satisfy these constraints. For example:

  1. The heat kernel regulator: $f(x) = e^{-x}$

  2. The Gaussian regulator: $f(x) = e^{-x^2}$

  3. The Hard cut-off: $f(x) = \theta(1-x)$

For your case of $S$, you used the regulator $(1-r)^n$. Maybe you can write down more general regulators and derive some constraints to ensure consistency? That's a just a thought.

P.S. In the Casimir case, if you just started with analytic continuation on the series: $$ E(r) = \sum_n \frac{\omega_n}{2}$$ The regulator is simply $f(x) = 1$. That does not satisfy the vanishing constraint mentioned above. But a direct application of analytic continuation still gives you the correct result... this is confusing for me too. Hopefully, people who actually know QFT can add much more insightful answers.

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