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Dimensional Regularization Problem

  1. Jul 15, 2009 #1
    Hi guys, this is my first post.

    I recently realized that there is something odd going on with dimensional regularization so I figured I could ask here.

    So let's take equation (A.44) in Peskin's book. Now if we set n=1 and d=3-e, this integral is obviously ultraviolet diverging(in fact quadratically). On the other hand if we believe what dimensional regularization is telling us, this integral behaves like [tex]\Gamma(-1/2+e/2)[/tex]. The gamma function only has poles for negative integer and zero values of it's argument, so in this case the integral is perfectly finite.

    I have been looking up old papers trying to figure out why this is happening. I believe that there must be some kind of a problem with analytically continuing the function but I'm not quite sure what this problem is.

    Anyone have any ideas?

    Also if we consider a massless [tex]\phi^4[/tex] theory, the tadpole diagram goes like:
    [tex]\int \frac{d^d q}{(2 \pi)^d)} \frac{1}{k^2}[/tex] In d=4 this is infinite again but dimensional analysis gives us zero. I have been doing some reading on this and it seems that if we employ a more sophisticated dimensional regularization procedure, we can prove this to be zero. We need to introduce a function that will allow us to exchange the integrals in the Schwinger trick.
     
    Last edited: Jul 15, 2009
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  3. Jul 20, 2009 #2

    blechman

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    welcome to the forum!

    Acutally, it is linearly divergent in 3 dimensions, not quadratic. DimReg is insensitive to such divergences. It only feels the logarithmic divergence, and there is none here (the term proportional to [itex]\log\Lambda[/itex] vanishes if you were doing this in momentum cutoff). This is why DimReg is so useful (since only log divergences tell us something interesting, namely, beta functions!).

    You can also go ahead and carefully do the integral in 3-e dimensions (rather than just blindly using the tables) using the tricks described in, for example, Pierre Ramond's textbook. And you would find that the integral is finite. Cute, huh?!


    Another awesome trick (and a dangerous one!) is that scaleless integrals always vanish in DimReg. This is obvious just by dimensional analysis - the integral you wrote down has mass dimension 2-e, but there is nothing with mass in the problem so it must vanish! (BTW: I assume by "q" you mean "k"!!)

    To see it more explicitly, look at the "radial" part of the integral, and you get ([itex]x\equiv k^2[/itex]):

    [tex]I=\int_0^\infty\frac{dx}{x^{1+e/2}}[/tex]

    This integral is BOTH UV and IR divergent, so split the integral at some random point y and do both integrals separately, where e<0 for the IR-divergent term and e>0 for the UV-divergent term:

    [tex]I=(\int_0^y+\int_y^\infty)\frac{dx}{x^{1+e/2}}=\frac{2}{e_{IR}}(y^{e_{IR}/2}-0)+\frac{2}{e_{UV}}(0-y^{e_{UV}/2})[/tex]

    Now that we have a finite answer, we analytically continue [itex]e_{IR}=e_{UV}[/itex] and the integral vanishes.

    This is very dangerous! It's useful if you're computing a cross section, but if you are interested in the beta function, for example, the UV and IR divergences must be isolated. This is usually done by replacing the denominator by [itex]k^2-m^2[itex] and THEN doing DimReg. Now you will get an answer proportional to m^2 and the 1/e pole is pure UV and therefore related to the beta function. If you didn't do this, you would conclude (WRONGLY!!) that the beta function vanishes in this theory.

    Moral: DimReg is very, VERY useful, but it has several pitfalls. Beware! And have fun! :wink:
     
    Last edited by a moderator: Nov 21, 2013
  4. Jul 20, 2009 #3

    blechman

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    actually, let me clarify the last point I made: if the mass is zero, then the "beta" function for the mass vanishes as well by dimensional analysis. What I was thinking about was the coupling constant loops (related to scalar-scalar scattering, for instance) which have an integral (for massless scalars) of the form

    [tex]\int \frac{d^dk}{(2\pi)^d}\frac{1}{k^4}[/tex]

    This integral vanishes by the same argument I gave above, but the beta function obviously doesn't vanish. If you regulate the IR with a mass, you get a 1/e pole (beta function) and a log(m) (IR divergence). If you didn't do it this way, it would vanish and you'd be confused.

    *********************
    In fact, I was being a little dumb (doing this from memory, sorry!), and the integral SHOULD have been

    [tex]\int\frac{dx}{x^{e/2}}[/tex]

    Now do the same thing I did above, appropriately changed, and you will get the same result (after analytically continuing in e).
     
    Last edited: Jul 20, 2009
  5. Jul 20, 2009 #4
    would not be better 'Zeta regularization' ?? i found this paper

    http://www.wbabin.net/science/moreta23.pdf

    he uses zeta function in order to get rid of divergences, valid for any UV except the logarithmic one :( , the advantage of zeta regularization is that provides similar results but you do not need an non-integer dimension.
     
  6. Jul 21, 2009 #5

    blechman

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    But the log divergence is the only one I care about!! It tells you about the running.

    Well, look, use whatever regulator you feel like, but in DimReg, this is how it works.
     
  7. Jul 21, 2009 #6
    Thanks a lot guys. I have done some reading and some serious thinking on these issues. I'm still pretty freaked out by the fact that I start with an integral that is clearly diverging, I employ a correct STRICTLY MATHEMATICAL procedure and I end up with a finite answer. This seems to be the case though.

    In an odd dimensional theory, we would need a counterterm with an odd number of derivatives to cancel our "infinity" coming from the first integral I wrote. This can not be the case since this term violates Lorentz invariance and DimReg has Lorentz invariance "built in" to it.

    Also I have seen a paper(Leibrandt, Rev. Mod. Phys., Vol 47,No 4, Page 849, 1975) that explains the massless business more concisely.

    Thanks a lot.
     
  8. Jul 21, 2009 #7

    blechman

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    there's nothing wrong with it! it takes some getting used to, but it really is totally fine.

    This is wrong! You only need n-derivative operators if you have a divergence that is proportional to [itex]p^n[/itex], the EXTERNAL momentum - it has nothing whatsoever to do with the power-law divergence of the counterterm. Lorentz invariance requires n to be even, so you have nothing to worry about.

    Have fun!
     
  9. Jul 21, 2009 #8
    the question is that even the integral [tex] \int_{0}^{\infty}dx (x+a)^{-1} [/tex] is divergent the series

    [tex] \sum _{n=0}^{\infty} (x+a)^{-1} [/tex] is SUMMABLE in the Ramanujan sense and gives the value

    [tex] \frac{ \Gamma' (a)}{\Gamma (a)} [/tex] which is finite.
     
  10. Aug 13, 2009 #9
    I had the opportunity to study dimensional regularization in these two years, but still it charms me and leave me a little wordless. Anyway I would like to make you notice one thing:
    try to calculate in dim reg the two integrals
    1) [tex]\int { d^n k \over k^2( k^2 + m^2) }[/tex]
    2) [tex]\int {d^n k \over k^2 + m^2} [/tex] which is the tadpole

    by simple partial fractioning you can see that

    [tex]\int { d^n k \over k^2( k^2 + m^2) } = {1 \over m^2} \left[ \int {d^n k \over k^2} - \int{d^n k \over k^2 + m^2} \right] [/tex]

    but calculating explicitly the two integrals you find that

    [tex]\int { d^n k \over k^2( k^2 + m^2) } = - {1 \over m^2} \int {d^n k \over k^2 + m^2} [/tex]

    this necessarely requires [tex]\int {d^n k \over k^2} = 0[/tex]for internal consistency...
    in this way you can show that a massless tadpole has to be null to give a consistent theory.
     
    Last edited: Aug 13, 2009
  11. Aug 13, 2009 #10
    there is a missing [tex] m^2 [/tex] on the numerator of your third equation :D

    an even most misterious formula [tex] \int_{0}^{\infty}x^{k}dx =0 [/tex] according to Hadamard finite part integral and dimensional regularization.
     
  12. Aug 13, 2009 #11
    Yes I'm sorry, it's dimensionally wrong, but you can put m = 1 and everything is ok ;)
    i wrote it too fast, now I edit it...

    p.s.
    by the way, the misterious integral you have written up is nothing but the massless tadpole itself

    [tex] \int {d^n k \over k^\alpha} = \Omega(n) \int_0^\infty k^{n-1-\alpha} dk = 0[/tex]

    where obviously [tex] \Omega(n) [/tex] is the solid angle in n dimensions
     
    Last edited: Aug 13, 2009
  13. May 25, 2011 #12
    Could you explain in more detail why:

    1. DR only feels the logarithmic divergence.

    2. only log divergences tell us something interesting.

    Thank you.
     
  14. May 26, 2011 #13
    I don't think it's right saying that dimensional regularization feels only log divergences.

    There is something magical happening. Take the tadpole in n dimensions
    $$\int {d^n k \over k^2 + m^2}$$
    It is obvious that this integral is divergent in n >= 2,
    in particular for n = 2 log div, for n = 3 linear div, for n = 4 quadratically div, etc...

    Now if you evaluate it in terms of beta function you would find something like
    $$\int {d^n k \over k^2 + m^2 } = m^{n-2} {\Omega(n) \over 2 } B( n/2, (2-n)/2 )$$

    which is now divergent in n =2,4,6,8,10 etc...
    but not in n = 3,5,7,9...

    and so you would have the at least magical result
    $$\int {d^3 k \over k^2 + m^2 } = - 2 \pi^2 m$$
    SO it's not true that it is sensible only to log div, but it seems to be sensible to divergences with these strange jumps of 1 dimension :)
     
  15. May 26, 2011 #14

    vanhees71

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    I also don't understand what you mean that dim. reg. is only "sensitive to logarithmic divergences". The idea behind dim. reg. is to define the class of integrals that appear when evaluating Feynman diagrams not only as a function of external momenta and masses but also the space-time dimension.

    Let's discuss the real-world case of d=4. You start with the one-loop integrals. If they are divergent in d=4 they have a pole there, i.e., there are terms prop. to 1/(d-4) in the (Laurent) expansion of the integral wrd. d around d=4. In this way you have nicely separated the divergent piece and can interpret it, depending on the model in question. If you have a Dyson-renormalizable theory like the Standard Model you can absorb these terms in wave-function normalization factors, masses, and couplings that are already present in the original Lagrangian. Then expressing everything in observable parameters rather than bare ones you have nice finite results, and the observable parameters can be fitted to data and used for predictions for other observables. This works for all kinds of divergences, not only the logarithmic ones.

    If you have more than 1 loop, the issue is a bit more complicated. Then you have to renormalize the sub-divergences first. Then, for renormalizable theories, due to BPHZ, the remaining overall divergence can again be renormalized with help of counterterms of the same form as in the original Lagrangian, i.e., it can be lumped into the wave-function normalization factors, masses, and couplings. This scheme you use iteratively, order by order in perturbation theory.

    Dim. reg. is particularly convenient for that systematic evaluation of divergent radiatiative corrections. The only exception are chiral models with problems, how to treat specifically four-dimensional objects like gamma5 or the Levi-Civita tensor. Here, one has to apply additional model dependent constraints. E.g., in the Standard Model one has to find prescriptions of these quantities in dimensions other than 4 such that the anomalies refer to the axial current, not the vector one. Otherwise you'd destroy gauge invariance, and this means desaster. This has been worked out by 't Hooft and Veltman.

    You find a quite thorough treatment of dim. reg. in my QFT manuscript:

    http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf
     
    Last edited by a moderator: Apr 25, 2017
  16. Jun 6, 2011 #15

    blechman

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    People seem very offended by my "dim reg only sees log divergences" statement. I should defend myself....

    What I mean to say is that if you have a power divergence in an integral that goes like [itex]\Lambda^n[/itex] for [itex]n\ge 0[/itex] (where n=0 is "log divergence") where [itex]\Lambda[/itex] is a cutoff, then when you regulate the same integral in DimReg, you will pick up those divergences as poles in d, located at d=4-n. So for instance, a quadratic divergence appears as a pole at d=2, a log divergence at d=4, etc.

    Since in the end of the day we set d=4, we have that DimReg only reproduces the log divergent terms. The quadratically divergent terms, for example, are automatically made finite in DimReg without any need to subtract them.

    This is a GOOD thing: only the log divergences have "meaning" - they represent violations in scaling and lead to renormalization group evolution. Power divergences do not do anything - they are just "swept under the rug" in the usual renormalization algorithm. But the log divergences are more interesting since they represent something real (namely, resumming large logs with Callan-Symanzik equation). All divergences in DimReg therefore have "meaning", rather than in a cutoff method.
     
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