For starters, let me say that although the Casimir effect is standard textbook stuff, the only QFT textbook I have in reach is Weinberg and he doesn't discuss it. So the only source I currently have on the subject is Wikipedia. Nevertheless I suspect this question is appropriate since I don't remember it being addressed in other textbooks

Naively, computation of the Casimir pressure leads to infinite sums and therefore requires regularization. Several regulators can be used that yield the same answer: zeta-function, heat kernel, Gaussian, probably other too. The question is:

What is the mathematical reason all regulators yield the same answer?

In physical terms it means the effect is insensitive to the detailed physics of the UV cutoff, which in realistic situation is related to the properties of the conductors used. The Wikipedia mentions that for some more complicated geometries the effect is sensitive to the cutoff, so why for the classic parallel planes example it isn't?

EDIT: Aaron provided a wonderful Terry Tao ref relevant to this issue. From this text is clear that the divergent sum for vacuum energy can be decomposed into a finite and an infinite part, and that the finite part doesn't depend on the choice of regulator. However, the infinite part does depend on the choice of regulator (see eq 15 in Tao's text). Now, we have another parameter in the problem: the separation between the conductor planes L. What we need to show is that the infinite part doesn't depend on L. This still seams like a miracle since it should happen for all regulators. Moreover, unless I'm confused it doesn't work for the toy example of a massless scalar in 2D. For this example, all terms in the vacuum energy sum are proportional to 1/L hence the infinite part of the sum asymptotics is also proportional to 1/L. So we have a "miracle" that happens only for specific geometries and dimensions

  • $\begingroup$ I think the answer to the question is that for renormalizable theories the answer must be independent of the regularization (once you send the cut-off/regularization away). $\endgroup$
    – lcv
    Dec 9, 2018 at 19:25

5 Answers 5


Check out Terry Tao's brilliant post on zeta function regularization here (I never understood the subject until I read this post). The short answer is that they're all computing the same thing, the (suitably defined) asymptotics of the divergent sum.

  • 2
    $\begingroup$ Thx Aaron, a great ref! I think something is still unclear though. I'm going to edit the question accordingly $\endgroup$
    – Squark
    Dec 30, 2011 at 15:29
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    $\begingroup$ Tao is great but this is simply not a valid answer to the original question. Tao's text is pretty much purely mathematical and makes no physics (QFT, renormalization) analysis of the situation whatsoever - namely the coupling constants, counterterms or regulators, symmetries constraining the laws of physics if any, and the dependence or independence of physical observables on these things. So -1. $\endgroup$ Jan 2, 2012 at 20:01

Although, I do not know if a general proof exists, I think that the Casimir effect of a renormalizable quantum field theory should be completely understood by means of a theory of renormalization on manifolds with boundary. The key feature is that one cannot, in general, neglect the renormalization of the coupling constants in the boundary terms. Using this strategy, Bondag and Vassilevich observed that the renormalization of the surface tension term (The full action including the surface terms is given in equation 44) provides a counterterm which cancels a divergent term in the Casimir energy of a dielectric ball which cannot be canceled by the zero point energy subtraction (as was pointed out by the same authers together with Kirsten in a a previous work). In the later work, the authors verified that the model including the surface terms is renormalizable at one loop.

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    $\begingroup$ Good physics answer, +1, as opposed to just cheap production of URLs to worshiped Fields medal winners who, despite their greatness, provide pretty much zero contribution to the answer to the original question, namely what is adjustable about the parameters describing these physical situations (plates vs other geometries). $\endgroup$ Jan 2, 2012 at 19:58
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    $\begingroup$ One has to understand that any boundary condition (like E=0) is a solution to the coupled equations of charges and fields. So the Casimir effect is an interaction of charges in QFT, not just a "vacuum field effect". $\endgroup$ Jan 3, 2012 at 9:22

(This argument is for a one-dimensional system, but similar arguments can be given in higher dimensions. We work in units with $c = \hbar = e = 1$).

Suppose we have some regulator procedure parameterized by a momentum cutoff $\Lambda$. Then, for distance $L$ between two parallel plates, we can expand the regularized energy sum in powers of the cutoff as $$E = a_L \Lambda^2 + b_L \Lambda + c_L + O\left(\frac{1}{\Lambda}\right).$$ We can deduce the dependence of $a_L$ and $b_L$ by dimensional analysis. The first two terms come from the contribution of very high energies to the sum. Hence, they cannot depend on things like, say, the electron mass, which is negligible at high energies; the only available energy scale is 1/L. Thus, the only dependence allowed by dimensional analysis is $a_L = \alpha L, b_L = \beta$ for some dimensionless constants $\alpha$ and $\beta$. So we obtain $$E = \alpha L \Lambda^2 + \beta \Lambda + c_L + O\left(\frac{1}{\Lambda}\right).$$ It's clear that the second term will drop out when we calculate the force $dE/dL$. What about the first term? Observe that there is always a vacuum on both sides of any plate. If we move a plate by $dx$, then the first term gives an increase the energy of the vacuum on the left by $\alpha \Lambda^2 dx$, but it also gives a decrease of the energy of the vacuum on the right by the same amount. Hence, so long as we calculate the derivative of the total vacuum energy, the first term also drops out.

This shows that (in the limit of large cutoff, so that the $O(1/\Lambda)$ terms disappear) the force comes solely from $c_L$, which is in fact independent of the cutoff! One would hope that the term is actually independent of all the details of the regulator; the post by Terence Tao (referenced already in Aaron's answer) proves this in a special case, i.e. the free massless boson.


You might well have asked the same question of any calculation in quantum mechanics: why do different regulators for formally divergent quantities give the same result? The answer is that assuming those regulators don't break important symmetries that affect our measurements (like for instance Lorentz invariance) then regulators only mess with physics in the UV, and leave IR physics the same. If we're interested in low energy physics, then different regulators will all leave the quantities we're interested in untouched, and modify only irrelevant high energy details. This isn't a miracle! See also the discussion in: Which renormalisation techniques are available for 3+1 QED?

In Casimir's original paper, he presented a derivation that shows any "sensible" choice of regulator will give the same result. This is essentially what Terry Tao runs through in his blog post, but his post is quite detailed and maybe a more digestible presentation could be useful! It goes like this:

In a one-dimensional box, the frequencies are quantised to those of plane waves, $\omega_n=\tfrac{\pi}{\ell}n$ where $\ell$ is the size of the box. Let us then define \begin{align} E(\ell)=\sum_n\frac{1}{2}\omega_n \end{align} the sum of the zero point energies of these modes. Let's now modify this sum with a (mostly) arbitrary regulator, \begin{align} E(\ell)=\frac{\pi}{2}\sum_n\frac{n}{\ell}\ f\ \left(\frac{n}{\ell\Lambda}\right) \end{align} where we just multiplied each term in the sum by $f(\frac{n}{\ell\Lambda})$. We assume only that this function $f(x)$ dies off faster than $x^{-1}$, and that $f(0)=1$. The first requirement is simply that the regulator only shuts off modes at high energies - the UV frequencies go right through the box. The second requirement is just that the regulator shouldn't affect the spectrum in the IR.

Let's now envisage we have a pair of places separated at a distance $a$, inside a larger box of size $L$, shown below$^1$:

enter image description here

If we leave $L$ fixed, and vary $a$, the force on the walls of our plates at distance $a$ is $-dE/da$. The energy of the $L-a$ side of the box is \begin{align} E(L-a)=\frac{\pi}{2}\sum_n\frac{n}{(L-a)}\ f\ \left(\frac{n}{(L-a)\Lambda}\right) \end{align}

and if we take the continuum limit $L\rightarrow\infty$, we get \begin{align} E(L-a)\rightarrow\frac{\pi}{2}(L-a)\Lambda^2\int xf(x) dx \end{align} So now adding the discrete sum for between the plates to the continuum sum for the outside of the plates, we get \begin{align} E_{total}=E(a)+E(L-a)=\frac{\pi}{2}L\Lambda^2\int x\ f(x) dx+\frac{\pi}{2a}\left(\sum_n\ n\ f\left(\frac{n}{a\Lambda}\right)-\int \ n\ f\left(\frac{n}{a\Lambda}\right) dn\right) \end{align} The difference between the sum of a function and its integral is given by the Euler-Maclaurin formula: \begin{align} \sum_{n=1}^Nf(n)-\int_0^Nf(n)dn=\frac{f(0)+f(N)}{2}+\frac{f'(0)-f(0)}{12}+...B_j\frac{f^{(j-1)}(0)+f^{(j-1)}(N)}{j!} \end{align} where $B_j$ are the Bernoulli numbers.

Okay, now we only care about the pieces of this expression that depend on $a$, since we're going to differentiate with respect to $a$. Isolating these terms, \begin{align} E_{total}=-\frac{\pi f(0)}{24a}-\frac{B_4}{4!}\frac{3\pi}{2a^3\Lambda^2}f^{''}(0)+... \end{align} where the $...$ terms are suppressed by increasing powers of $\Lambda$. If $\Lambda$ is taken to be large, we throw away everything except the first term; recalling we required $f(0)=1$ we then obtain the Casimir force \begin{align} F=-\frac{dE}{da}=-\frac{\pi}{24a^2} \end{align}

Now, it's often mysterious to people why for e.g. esoteric manipulations like zeta regularisation give the right answer. The above argument demystifies why good regulators are good regulators: you can expand the difference between the actual sum summed up to a cut off and the zeta sum (or whatever sum - heat kernel sum perhaps) using Euler-Maclaurin and show that there's a universal piece which isn't touched by the choice of cut-off. Bad things can only happen if the regulators break symmetries.

You're worried that different geometries or dimensions this doesn't work - but it does! So long as the regulator doesn't break the symmetry of the geometry, you're fine. It's not clear what your $2d$ scalar example you have in mind is, but if it's a Casimir type problem, then the infinite part certainly will not depend on the plate separation, so we're fine. An earlier and interesting answer points out that sometimes if the plates have a strange boundary, the regulator needs to do more than just "subtract the zero point energy" - the bottom line though is you need to mess with the UV physics only in a way that respects the symmetries of the boundary.

$^1$Image (and derivation) stolen shamelessly from Matt Schwartz's QFT textbook.

  • $\begingroup$ pair of places -> pair of plates $\endgroup$
    – Qmechanic
    Jan 6, 2019 at 13:16
  • $\begingroup$ How does this explain zeta regularization?$ \left|ω_{0}\right|^{-s}$ does not equal 1! $\endgroup$
    – Matrix001
    Oct 10, 2022 at 7:58

Quantities we can measure in the lab are finite. We have mathematical models which, when used naively, yield infinite values for quantities that we can measure. Consider these infinite values artifacts of the specific mathematical model/tool/computation method. If one can identify ways in which a particular type of infinity consistently arises in a mathematical model/method, then one can keep track of it and "subtract if off" (i.e. renormalize it away) in a consistent manner.

If, for a particular way of calculating, infinities arises in a haphazard way ... then, there is no way to remove it in a mathematically consistent manner and you don't hear about that particular way.

  • $\begingroup$ every infinite can be regularized mathematically :) another question is if it is admitted by physicist.. $\endgroup$ Dec 27, 2012 at 23:02

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