Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  How can dimensional regularization "analytically continue" from a discrete set?

+ 11 like - 0 dislike
3212 views

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)". 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.

This post imported from StackExchange Physics at 2017-09-17 12:55 (UTC), posted by SE-user tparker
asked May 5, 2016 in Theoretical Physics by tparker (305 points) [ no revision ]
retagged Sep 17, 2017
Have you heard of the Gamma function? It allows you to analytically continue factorials away from integers.

This post imported from StackExchange Physics at 2017-09-17 12:55 (UTC), posted by SE-user Prahar
@Prahar Uh, did you read the question? It's about the gamma function.

This post imported from StackExchange Physics at 2017-09-17 12:55 (UTC), posted by SE-user tparker

I am not sure that the Gamma function (integral) is a unique way to represent $n!$ for "continuous" representation of $n!$ for "fractional" values:  $n\ne \text{integer}$.

This resembles me a joke about $1/r^{12}$ potential in a certain book of physisicts' jokes, especially when they ask you to consider the limit $12\to \infty$.

While there are many different continuous interpolants, the Gamma function is, I think, the unique analytic function interpolating the factorial.

@VladimirKalitvianski  Can you cite the the physicist's jokes' book? even if it's in Russian.

I remember my Nonlinear control theory teacher told us in some practical engineering application that for some really big number (greater than 1) they took 2... :-D

That's Engineering for you!

@MathematicalPhysicist: I am not in a mood to make a research in Internet, so I will limit my answer with mentioning one book published in the West with a title like "Physicists are joking". In the USSR this book has been translated and completed with some Russian physicist's jokes, under a title like "Physicists are continuing joking".

Also, a colleague of mine informed me that his boss considered the number 1/3 as already a sufficiently small parameter to develop the solution is a series. (I agree, nowadays we can use non linear summation of series and extend their region of "applicability", i.e., the precision that may be sufficient for our requirements.)

2 Answers

+ 7 like - 0 dislike

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.

This post imported from StackExchange Physics at 2017-09-17 12:55 (UTC), posted by SE-user yuggib
answered May 5, 2016 by yuggib (360 points) [ no revision ]
Gauge invariance is also an important property of dim reg.

This post imported from StackExchange Physics at 2017-09-17 12:55 (UTC), posted by SE-user Adam
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)

This post imported from StackExchange Physics at 2017-09-17 12:55 (UTC), posted by SE-user Kvothe
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?

This post imported from StackExchange Physics at 2017-09-17 12:55 (UTC), posted by SE-user Kvothe
+ 5 like - 0 dislike

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.

This post imported from StackExchange Physics at 2017-09-17 12:55 (UTC), posted by SE-user Arnold Neumaier
answered May 8, 2016 by Arnold Neumaier (15,787 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...