11
$\begingroup$

There is an old question posted (Regularization) which did not get an answer, about the validation of analytic continuation as regularization. It did get some discussion in the comments, referring to Terence Tao's blog T.Tao, but questions seem to remain:

  1. Why is analytical continuation allowed to regulate a diverging sum?

  2. If Tao's article solves this, then where exactly in the text is this proof?

  3. If analytical continuation is allowed, how do we avoid ambiguity?

To start with the last point, it should be clear that we can do this in several ways, for instance with a Taylor series or with a Dirichlet series. Example:

$$ S = 1+2+3\ +... = 1 z^1 + 2 z^2 + 3 z^3 + ... \quad \mbox{for} \ z=1 $$ $$ \ \ \ \ \ \ \ \ \ \Rightarrow S = \lim_{z \uparrow 1} \sum_{n=0}^\infty\ n\ z^n\ = \lim_{z \uparrow 1} \frac{z}{(z-1)^2}\ = \infty, $$ $$ \mbox{or:}\ \ \ \ \ \ S = 1+2+3\ +... = \frac{1}{1^{\mbox{$s$}}}+\frac{1}{2^\mbox{$s$}}+\frac{1}{3^\mbox{$s$}}+... \quad \mbox{for} \ s=-1 $$ $$ \Rightarrow \ \mbox{(by analytical continuation:)} $$ $$ S = \sum_{n=1}^\infty\ \frac{1}{n^s} = \zeta(s)\ = \frac{-1}{12} \quad \mbox{for} \ s=-1, $$ so that does not help. Of course the analytical continuation of a function is unique, but here we see that the choice of the function is not unique. The results are not the same for the Taylor series and the Dirichlet series. We also see that in the second case the analytical continuation ($s\rightarrow-1$) goes far outside the region where the original sum would converge ($\mbox{Re}\ s>1$), whereas in the first case the function is evaluated just at the boundary of the convergence region.

As for the question of physical meaningfulness, in the following I have put together the Casimir effect derivation in two ways. To some extent the results are the same, but it is not clear why the analytical continuation procedure should be reliable.

A) With exponential regularization, we compute the vacuum energy between the plates with: $$\omega = c\ \sqrt{q^2+k_z^2}$$ $$ E = \sum_{k_z=n\pi/a} (2-\delta_{n0}) \int \frac{q\ dq}{2\pi} \ \frac{{\small \hbar}}{\small 2} \ \omega\ e^{-\omega/C} $$ with the $2-\delta_{n0}$ because for $n>0$ we have a TE and TM mode and only a TM mode for $n=0$. This would be the energy per unit area below the piston in this drawing:

Casimir force

provided the area of the piston is very large, so for the $xy$-plane we can use an integral instead of a sum. Putting $\hbar=c=1$ for convenience, let's see how this works out: (for computational procedure see also math problem) $$ E = \sum_{k_z=n\pi/a} (2-\delta_{n0}) \int \frac{q\ dq}{2\pi} \ \frac{{\small \omega}}{\small 2} \ e^{-\omega/C} $$ $$ = \sum_{k_z=n\pi/a} (2-\delta_{n0}) \ \frac1{4\pi} \ \left(2C^3+2C^2 k_z + C k_z^2\right) \ e^{-k_z/C} $$ $$ = \frac{C^3}{2\pi} \ \coth\frac{\pi}{2a C} + \frac C{8a^2} \left(2 a C +\pi \coth \frac{\pi}{2a C}\right) \ \mbox{csch}^2 (\frac{\pi}{2a C}) $$ $$ = \frac{3 a \ C^4}{\pi^2} - \frac{\pi^2}{720\ a^3} - \frac{\pi^4}{5040\ a^5 C^2} + O(\frac1{C^4}) $$ Clearly there's a vacuum energy term proportional to $a$, as expected, but also to $C^4$ so we cannot let the cutoff go to $\infty$. For the force on the bottom of the piston we find: $$ F(a)= \frac {-dE}{da} = \frac{-\pi^2}{16\ a^2} \cdot \frac{2+\cosh(\pi/(2a C))}{\sinh(\pi/(2a C))} $$ $$ = \frac{-3\ C^4}{\pi^2} + \frac{-\pi^2}{240\ a^4} + \frac{\pi^4}{1008\ a^6C^2} + O(\frac1{C^4}). $$ If we then look at the total force and include the force on the top side: $$ F_{\mbox{cas}} = F(a) - F(H-a) = \frac{-\pi^2}{240\ a^4} + \frac{\pi^2}{240\ (H-a)^4} + O(\frac1{C^2}) $$ So this two-sided force is finite even if we let $C$ go to $\infty$. For the energy this was not possible and also not for the one-sided force. Physically this all makes perfect sense.

B) Now if we do this with Zeta regularization, we replace the exponential regulator with $1/\omega^{\ \large s}$, giving: $$ E = \sum_{k_z=n\pi/a} (2-\delta_{n0}) \int \frac{q\ dq}{2\pi} \ \frac{{\small \omega}}{\small 2} \ \omega^{-s} = $$ $$ = \sum_{k_z=n\pi/a} (2-\delta_{n0}) \ \frac{k_z^{3-s}}{4\pi(s-3)} = $$ $$ = \frac{\pi^{2-s} \zeta(s-3)}{2(s-3)\ a^{3-s}} \ \ \rightarrow \ \ \frac{-\pi^2}{720 \ a^3}, \ \ \ \mbox{if} \ s \rightarrow 0. $$ Here we do not have to take any limit to remove the regulator, we can just insert $s=0$ because there is no singularity. And that is exactly the worrying thing! We do not see a very large vacuum energy, becoming even infinite if the regulator is removed. We do not have to go first to the force instead of the energy and then to the two-sided force to get a finite result. It seems to defy the underlying physical picture... Of course for completeness we can still look at the force before setting $s=0$: $$ F(a)= \frac {-dE}{da} = \frac{-\pi^{2-s}}{2\ a^{4-s}} \zeta(s-3) \ \ \rightarrow \ \ \frac{-\pi^2}{240 \ a^4}, \ \ \ \mbox{if} \ s = 0. $$ Again, we get immediately the Casimir force, but not the large force due to volume-dependent vacuum energy, and we do not have to go to the two-sided force to keep the answer finite. So this regularization works fine, but it seems to work too well! It is missing the complications of the other approach, but those were actually natural from a physical perspective...

$\endgroup$
2
  • $\begingroup$ For your second question, it's in equation (24) $\endgroup$
    – LPZ
    Mar 1 at 13:38
  • $\begingroup$ There is a particularly nice discssion of some of these points in Terry Tao's blog:terrytao.wordpress.com/2010/04/10/… $\endgroup$
    – mike stone
    Mar 3 at 14:15

1 Answer 1

3
$\begingroup$

For your first question, analytical continuation is one way out of many ways to regulate a sum. As you pointed out, there is always an ambiguity in attributing a sum to a diverging series. Analytical continuation is one way to do this. What is especially relevant in physics is that in some cases, it is consistent with regularisation approaches in the following sense.

A regularisation allows you to render the original divergent series to a convergent one. However the resulting sum now depends on the cutoff and the regulator. In some cases, you can check that in the asymptotic expansion of the cutoff scale, the constant term is independent of the regulator, and corresponds to the analytically continued value.

For your second question, it is essentially answered in equation (24) and the following paragraph. The case of Riemann series, you have the general asymptotic expansion for $\Re s <1$: $$ \sum_{n=1}^\infty n^{-s}\eta(n/N) = C_{\eta,-s}N^{1-s}+\zeta(s)+O(1/N) $$ with $\eta$ the regulator and $N$ the cutoff scale. As expected the regularised sum depends both on $N$ and $\eta$. This is the case for the first term, which is leading order. However, the constant term by definition does not depend on $N$, but surprisingly does not depend on $\eta$ either. This is why it is a natural candidate for the sum of the divergent series. It turns out to match the analytical continuation approach. Tao proposes two approaches to see this. One is to define $\zeta$ by the previous equation and check that it is analytic. A second approach is to derive the asymptotic expansion from the analytically continued $\zeta$.

For your third question I will treat your examples in more detail. Using the entire series (of the polylogarithm), you can also recover the zeta function. You just need to recognise that the Abel summation amounts to using the regulator: $$ \eta(x) = e^{-ax} $$ with $\Re a>0$. In other words, you can relate $z$ to the cutoff $z = e^{-a/N}$, and you need to do an asymptotic expansion in terms of $N = -\frac a{\ln(z)}$. In this case, you get: $$ \begin{align} \text{Li}_{-1}(z) &= \frac z{(1-z)^2} \\ &= \frac{e^{-an/N}}{(e^{-an/N}-1)^2} \\ &= \left(\frac Na\right)^2-\frac1{12}+O(1/N) \\ &= \frac1{(\ln z)^2}-\frac1{12}+O(\ln(z)) \end{align} $$ so you recover the cutoff and regulator independent value $-\frac1{12}$.

For the second method, you do see that energy is infinite since the original sum diverges at $s=0$. To define the finite energy at $s=0$, you need to invoke analytic continuation, which directly gives you the cutoff and regulator independent value.

I think that part of your uneasiness is that you don't see a leading infinite term in the analytical continuation method. It turns out that the Casimir force is a bit of an exception in this regard. In your typical QFT calculations, the value you re interested in is often a pole of the analytic continuation. You therefore still have an infinite result that you'll need to absorb into the renormalised parameters.

For example, you can see this kind of behaviour in simple theories like vertex corrections of $\phi^4$ in $D=3+1$ dimensions. Whether you use zeta or dimensional regularisation, you'll get a pole that needs to be compensated by renormalising the coupling constant $\lambda$. The "tameness" of the Casimir force comes from the fact that parameters of the theory don't need to be renormalised, which is rather exceptional in QFT.

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.