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The procedure of dimensional regularization for UV-divergent integrals is generally described as first evaluating the integral in dimensions low enough for it to converge, then "analytically continuing" this result in the number of dimensions $d$. I don't understand how this could possibly work conceptually, because a d-dimensional integral $I_d$ is only defined when $d$ is an integer greater than or equal to 1, so the domain of $I_d$ is discrete, and there's no way to analytically continue a function defined on a discrete set.

For example, in Srednicki's QFT book, the key equation from which all the dim reg results come is (pg. 101) "... the area $\Omega_d$ of the unit sphere in $d$ dimensions ... is $\Omega_d = \frac{2 \pi ^{d/2}}{\Gamma \left( \frac{d}{2} \right) };$ (14.23)". (Note: see edit below.) But this is highly misleading at best. The area of the unit sphere in $d$ dimensions is $\frac{2 \pi^{d/2}}{\left( \frac{d}{2} - 1 \right) !}$ if $d$ is even and $\geq 2$, it is $\frac{2^d \pi^\frac{d-1}{2} \left( \frac{d-1}{2} \right)! }{(d-1)!}$ if $d$ is odd and $\geq 1$, and it is nothing at all if $d$ is not a positive integer. These formulas agree with Srednicki's when $d$ is a positive integer, but they avoid giving the misleading impression that there is a natural value to assign to $\Omega_d$ when it isn't.

Beyond purely mathematical objections, there's a practical ambiguity in this framework - how do you interpolate the factorial function to the complex plane? Srednicki chooses to do so via the Euler gamma function without any explanation. But there are other possible interpolations which seem equally natural - for example, the Hadamard gamma function or Luschny's factorial function. (See http://www.luschny.de/math/factorial/hadamard/HadamardsGammaFunction.html for more examples.) Why not use those?

In fact, these two alternative functions are both analytic everywhere, so you can't use them to extract the integral's pole structure, which you need in order to cancel the UV infinities. To me, this suggests that the final results of dim reg might be highly dependent on your choice of interpolation scheme, therefore requiring a justification for using the Euler gamma function. Could we prove to a dim reg skeptic that all results for physical observables are independent of the interpolation scheme? (Note that this is a stronger requirement than showing they are independent of the fictitious mass parameter $\tilde{\mu}$.)

(I know that the Bohr-Mollerup theorem shows that the Euler gamma function uniquely has certain "nice" properties, but I don't see why those properties are helpful for doing dim reg.)

I'm not looking for a hyper-technical treatment of dim reg, just a conceptual picture of what it even means to analytically continue a function from the discrete set of positive integers.

Edit: It appears that the details of exactly which field-theory results do and do not depend on the choice of regularization scheme are not well-understood; see this paper for one discussion.

Edit: kaylimekay correctly points out below that the relevant $\Gamma(d)$ term is actually the one that comes from the radial integral, not the one that comes from the angular integral. But I don't think that this really helps solve the problem at all.

The issue basically boils down to ambiguous notation; two qualitatively different "exponentiation" functions use the same notation. To be explicit, I'll give them two different names.

The first is the function $\mathrm{expNat}: \mathbb{R} \times \mathbb{N} \to \mathbb{R}$ defined by the usual "repeated multiplication" that we learn in Algebra 1 (or whenever):

$$\mathrm{expNat}(x, n) := \underbrace{x \times x \times \dots \times x}_{n \text{ times}}.$$

The second is the more complicated function $\mathrm{expReal}: \mathbb{R^+} \times \mathbb{R} \to \mathbb{R^+}$ (where the exponent can be an arbitrary real number) that we learn in Calculus 1 (or whenever). There are several equivalent ways to define this function, but for concreteness we'll take

$$\mathrm{expReal}(x, y) := \exp(y \ln(x)),$$ with $\exp(z)$ defined as the unique solution to the initial value problem $\exp'(z) \equiv \exp(z),\ \exp(0) = 1$, and $\ln(x)$ defined as its inverse.

The parameter that kaylimekay calls $\alpha$ comes from using Feynman's formula to rewrite the product of $\alpha$ different terms, and it is always a natural number. I'll rename it $n$ to make this clear. Rewritten in more explicit notation, the relevant angular integral is

$$ \int_0^\infty dk\, \frac{\mathrm{expNat}(k, d-1)}{\mathrm{expNat}(k^2 + \Delta^2, n)}. $$

It's not clear to me why we can extend $\mathrm{expNat}$ to $\mathrm{expReal}$; as explained above, this isn't a legitimate analytic continuation because it starts from a discrete set. And there are other possible extensions as well; for example, it would be equally correct to write kaylimekay's expression as $$ \int_0^\infty dk\, \frac{k^{(d-1) \cos(2 \pi d)}}{(k^2 + \Delta^2)^n}, $$ but writing it this way would suggest a different extension to the real domain for $d$, which (as far as I can tell) could give you different final answers after analytic continuation.

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  • $\begingroup$ I've deleted some off-topic comments. $\endgroup$ – David Z May 5 '16 at 9:36
  • $\begingroup$ Have you heard of the Gamma function? It allows you to analytically continue factorials away from integers. $\endgroup$ – Prahar Mitra Nov 11 '16 at 6:09
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    $\begingroup$ @Prahar Uh, did you read the question? It's about the gamma function. $\endgroup$ – tparker Nov 11 '16 at 17:33
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    $\begingroup$ What I’m saying is maybe it’s only this pole structure that matters when you’re doing dimensional regularization, since we’re only really interested in values of the functions infinitesimally close to integral dimensions. $\endgroup$ – AfterShave Jun 4 '20 at 1:29
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    $\begingroup$ The functional relation fixes the order and the residue of the poles which I think is all that matters for renormalization. Any gamma function which satisfies the relation will have the same poles and thus give the same results. $\endgroup$ – AfterShave Jun 4 '20 at 1:37
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Every regularization scheme is somewhat arbitrary. There are three popular regularization schemes when it comes to path integrals and their associated perturbative divergent integrals: time slicing, mode regularization, and dimensional regularization.

  • Time slicing is the usual procedure used to derive the path integral, and it is the discretization of time into finite time intervals.

  • Mode regularization is essentially an UV cut off, i.e. the truncation of the high-energy modes in the Fourier expansion of the path.

  • Dimensional regularization is performed as you described exploiting (one) generalization of the factorial to complex numbers.

In any case, the regularization is a limiting procedure, a finite (or different from zero) parameter is introduced such that all the integrals become finite, then they're manipulated in a way that the result in the limit when the parameter goes to infinity (zero) remains finite. In principle, and in practice, the final result may be dependent on the scheme chosen. Therefore it is necessary to introduce counterterms such that all the results agree with each other. This is done somewhat ad hoc, but luckily the counterterms are fixed at a low (second) order in the perturbation expansion in many situations.

The procedures chosen are all in some sense arbitrary, for (at least for the moment) there is not a satisfactory and unambiguous mathematical definition of the involved path integrals/QFT perturbative expansions. The dimensional regularization is often preferred for essentially one reason (as far as I know), it is the easiest to deal with: the resulting counterterm in fact is relativistically covariant (and that is important in relativistic theories/in the presence of a curved background) and the additional vertices coming form the counterterm at higher loops are easy to compute.

Now my guess is that it could be possible to regularize also using one of the other "complex extensions" of the factorial you mentioned, but in all likelihood the resulting counterterms would be different and maybe not covariant.

For a more detailed discussion on regularization schemes I suggest to read this book of Bastianelli and van Nieuwnehuizen.

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    $\begingroup$ Gauge invariance is also an important property of dim reg. $\endgroup$ – Adam May 5 '16 at 7:12
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    $\begingroup$ Is it just me or do both answers basically ignore the question and tell the OP other stuff he probably already knows instead. The OP claims there is another complex function that coincides for integer d that has another Pole structure. He is not talking about another way to do the UV regularization. (continues) $\endgroup$ – Kvothe Jul 14 '17 at 8:27
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    $\begingroup$ The difference between continuations would not be like an arbitrary constant that needs to be fixed (like some measured mass). And this does not seem to be like using another scheme. This would really give another result. Is the OP incorrect? Or is yuggib saying that such a procedure would really be equivalent. In that case he should be more clear on why. Is it just that if we use an analytic continuation without poles we get the same finite scattering amplitudes without including the counterterms? $\endgroup$ – Kvothe Jul 14 '17 at 8:27
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In dimensional regularization, $d$ is a complex number, not a true dimension. The $d$-dimensional integrals of a rational function are defined for any complex $d$ with sufficiently negative real part (the threshold depending on the integrand), and therefore can be analytically continued to a (provably meromorphic) function for all $d$. For a concise, mathematically sound definition see the reference to Etingov in the wikipedia article on dimensional regularization.

For $d=4$ there is typically a pole in individual contributions from Feynman diagrams, but none in the sum defining the full contributions to the S-matrix elements at any fixed order.

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This is a very old question, but it doesn't seem that the existing answers took you seriously. I see that you are still active, so I'll try to reactivate this thread, although maybe you've answered this question for yourself in the intervening time. :)

There is one aspect of your post that I disagree with, so I'll address that first. You say that the key equation from which all of dim reg comes is the result for the area of the $d$-sphere, and then wonder what would happen if one used a different extension of the factorial to non-integer values.

However, this $d$-sphere area is not the important factor for the calculation of whatever loop diagram. This factor is anyway finite for any positive number of dimensions, including numbers of dimensions large enough that the loop integral diverges.

That said, I do agree that you are correct when you say that there is no canonical way to extend this function over the integers to a function over the complex numbers. As you say, there are other choices one could make, and without some addition constraining principle being introduced, there is no way to choose. But, it also doesn't matter. It doesn't matter because this factor is always finite, so as we take the limit of $d$ approaching whatever positive integer, it will always converge back to the correct value, regardless of which extension you used.

The actually important factor is the radial part of the loop integral. This is the part that diverges if $d$ is too large.

There's a funny coincidence here that I hadn't specifically thought about until seeing your question. It turns out that both the surface area factor and the radial factor of the loop integral can both be expressed in terms of the gamma function. So it might seem like they have a lot in common. But in fact the issues surrounding these two factors are very different.

Let's think about the radial part of the loop integral, which is the part that can be divergent. That has the form $$ \int_0^\infty dk \frac{k^{d-1}}{(k^2+\Delta^2)^\alpha} .$$ In the original formulation of the loop integral, there was a measure like $d^dk$ and this is just the radial part. And of course in that original formulation, non-integer $d$ makes no sense. But the radial part written as above is well defined for any $d$ such that $0<\mathrm{Re}~d<2\alpha$. (I'm assuming $\Delta\ne 0$, or else we have to talk about IR divergences too.) I can therefore use this form to calculate a value for any such value of $d$ and that provides a unique extension of the result away from the integers in the convergent region. If I used one of the many other possible extensions of the factorial, I would get a different value away from those integers. (The important distinction here, in my view, is that the original physical formulation of the problem automatically gave us a unique way to define the values away from the integers, so that is what we should use.)[Edit: this statement does not rigorously address the original question.]

Now, again, if we were only going to consider values of $d$ where the original integral converges, this would be an academic distinction, because we'd always take the limit and get back the same result regardless of which extension we used. But in fact we are precisely interested in continuing further, to values of $d$ where the original does not converge. To do that, we need to analytically continue the function defined in the convergent region. But a theorem from complex analysis tells us that an analytic continuation is unique if it is analytic on an open subset of the region that you want to cover. So in order to extend the value of the integral to the region of the complex $d$ plane where it diverges, we need to commit to some form that holds not just on the convergent integers, but on some extended region around them. (That is what the above integral form lets us do, and that form is uniquely inspired by the original, physical integral.)[Same edit]

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    $\begingroup$ Thanks for the thoughtful answer, and for catching that the relevant $\Gamma(d)$ term comes from the radial rather than the angular integral. But I don't think that this really addressed my main issue; in particular, I disagree with you that the form that you write above is "uniquely inspired by the original, physical integral". See my edit to my question. $\endgroup$ – tparker Jan 3 at 16:05
  • $\begingroup$ Also, it looks like the condition for convergence is not $\mathrm{Re}(d) < 2 \alpha + 1$, but instead $\mathrm{Re}(d) \in (0, 2 \alpha)$. $\endgroup$ – tparker Jan 3 at 18:47
  • $\begingroup$ @tparker Thanks, apparently I can't do arithmetic sometimes. I was also sloppy and didn't consider $d<0$ but obviously that is a valid possibility when analytically continuing, so I will fix it. $\endgroup$ – kaylimekay Jan 4 at 4:08
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    $\begingroup$ @tparker ok I read your edit, and I agree with you. The last sentence of my last two paragraphs is too strong, and I will add a disclaimer to my post. I think the rest of my last paragraph agrees with your question, that is, one first must commit to an extension away from integers in the convergent region, and then that fixes the behavior in the divergent region, which is what we are really after. I'll think about this a bit more. $\endgroup$ – kaylimekay Jan 4 at 4:27

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