# Dimensional Regularization and epsilon

earth2
Hi!

Quick question: Does it make a difference if i choose my dim reg. to be D=4-2epsilon or D=4+2epsilon (i suppose in both cases epsilon >0).

I mean opinion i should not matter but standard qft books normally don't touch this question very deeply...

Cheers,
earth2

Homework Helper
Gold Member
Lowering the spacetime dimension has a net effect of weakening divergences. However, it's likely that the effect of changing the sign of $$\epsilon$$ could be made up by changing some other signs, e.g. in the counterterms.

earth2
Thanks fzero!

So, just to get this straight... if i put aside the aspect of renormalization for a sec on focus only on regularization: I calculate the integrals in d-dimension and then after integration set d to d=4 +/- 2epsilon. But which sign i use and if episilon is > 0 or <0 should not matter on this very level, right? The effect only comes into play if I consider renormalization and construct counterterms... Have I understood you correctly? :)

Homework Helper
Gold Member
If you are trying to isolate the divergent part of an integral, the sign probably doesn't matter. I hesitate to claim that the sign never matters, since it's clear that choosing $$d=4+2\epsilon$$ makes the divergences infinitesimally worse.

actually, it DOES matter: You go to 4-2e dimensions to regulate UV divergences, and 4+2e dimensions to regulate IR divergences (where e>0 in all cases). This is an important distinction, since IR divergences are NOT canceled by counterterms, while UV divergences are. So, for example, when trying to compute RG behavior, you really need to make the distinction.

You also have to be careful in general, since the sign of e can potentially do things like introduce imaginary parts to Feynman amplitudes, which are very strongly constrained by things like the Optical Theorem. This might not matter at 1 loop, but it can be a serious concern at higher orders in perturbation theory. If you start haphazardly changing the sign of e, you risk violating unitarity!

I am afraid that in order to see where the sign of epsilon may matter one has to go through all the steps of the calculation. But: epsilon is an unphysical "thing" and therefore its sign must not matter - or let's say, if one finds out that it matters then this is only an indication that the "modified trick using negative epsilon" to regularize an amplitiude fails. That would mean that the choice of tricks in regularization is constrained somehow; I think we could live with that.

earth2
Hey folks,

thanks for the replies so far. Let's look at a concrete example.

$$\int \frac{d^D l }{l^2(l-p)^2}$$ with $$p^2=0$$. The integral is in principal trivial and gives approximately

$$\text{integral}=\Gamma(\epsilon)(1/p^2)^{2-d/2}$$. So in the case $$d=4-2\epsilon$$ setting epsilon to zero I'll get $$\Gamma(0) * \infty \approx \infty (?)$$ while for $$d=4+2\epsilon$$ it will be $$\Gamma(0)*0 \approx 0 (?)$$. This is where my confusions comes from. Shouldn't the result be equal?

Cheers, earth2

Last edited:
earth2
Ok, found my mistake...both expressions are indeterminate... :) Thanks guys!

I am afraid that in order to see where the sign of epsilon may matter one has to go through all the steps of the calculation. But: epsilon is an unphysical "thing" and therefore its sign must not matter - or let's say, if one finds out that it matters then this is only an indication that the "modified trick using negative epsilon" to regularize an amplitiude fails. That would mean that the choice of tricks in regularization is constrained somehow; I think we could live with that.

It can matter, but in subtle ways. Let me give you an easier example: the Feynman propagator is:

$$D_F(x-y)=\int\frac{d^dk}{(2\pi)^4}\frac{ie^{ik\cdot (x-y)}}{k^2-m^2+i\varepsilon}$$

Now what about that $\varepsilon$? That is not a "physical" thing - it just tells us how to deform the contour to make sense of this integral. On the other hand, if you chose to make $\varepsilon<0$, you would violate causality (effectively switching "retarded" and "advanced" Green's functions)! So you see, it DOES matter how you deform the contour in the complex plane, even though the actual parameter $\varepsilon$ isn't really "physical".

The same idea applies to the dimensionality of spacetime in DimReg, where you analytically continue d to the complex plane and solve the integral in a domain where it converges. But by going above or below d=4, you are in a different domain of convergence, and that DOES have physical consequences.

That being said: just as the Feynman Prop. example, if all you are interested in is a brute-force regulating and renormalizing at a single subtraction point at 1 loop, then it shouldn't matter. But if that's all you are interested it, I would claim that you are a student doing a hw problem and nothing else!! Once you go to higher loops, consider finite-T effects, RG flow, IR safety, etc., then you have to be much more careful.

Last edited:
Convincing, I agree

actually, it DOES matter: You go to 4-2e dimensions to regulate UV divergences, and 4+2e dimensions to regulate IR divergences (where e>0 in all cases). This is an important distinction, since IR divergences are NOT canceled by counterterms, while UV divergences are. So, for example, when trying to compute RG behavior, you really need to make the distinction.

But QED has both UV and IR divergences at the same time. No wonder the IR is ill-understood....

But QED has both UV and IR divergences at the same time. No wonder the IR is ill-understood....

huh??

It can matter, but in subtle ways. Let me give you an easier example: the Feynman propagator is:

$$D_F(x-y)=\int\frac{d^dk}{(2\pi)^4}\frac{ie^{ik\cdot (x-y)}}{k^2-m^2+i\varepsilon}$$

But this is a _very_ different epsilon!

The same idea applies to the dimensionality of spacetime in DimReg, where you analytically continue d to the complex plane and solve the integral in a domain where it converges. But by going above or below d=4, you are in a different domain of convergence, and that DOES have physical consequences.

Really? Please give an example where this happens. (In textbook calculations, you usually get poles at integral values of d, so the analytic continuation near d=4 is single-sheeted,
and the problems you conjure are absent.)

huh??

In the presence of both UV and IR divergences, how do you choose e? positive or negative?

But this is a _very_ different epsilon!

Of course. I was only trying to give an example of how what you might call an "unphysical" parameter's sign can have physical implications. That is all I meant. I am not suggesting that the DimReg epsilon is related to the Feynman epsilon - sorry if I gave that impression.

Really? Please give an example where this happens. (In textbook calculations, you usually get poles at integral values of d, so the analytic continuation near d=4 is single-sheeted,
and the problems you conjure are absent.)

The real problem I am worried about (and has plagued the SCET community for years) is the question of IR safety - positive/negative epsilon regulates different divergences and these must be handled very differently. See Sterman's textbook (or one of his many TASI lectures) for examples.

Funny, I was going to suggest the same thing to you!!

In the presence of both UV and IR divergences, how do you choose e? positive or negative?

You first do one, and then the other: First you do d=4-2e, which gives you UV divergences explicitly regulated (although still IR divergent). Then you subtract the UV divergences with a counterterm so that the only divergences left are IR. Then you analytically continue to d=4+2e dimensions to regulate these IR divergences. These are then canceled the usual way by constructing IR safe observables. This is also explained in Sterman's text/lecture notes.

The point is that in this procedure you must be VERY careful which divergences are UV and which are IR. If you treat them all the same way (ignore the sign of epsilon), you might (incorrectly) subtract an IR divergence with a UV counterterm, which could lead to trouble when trying to construct IR safe observables. See, for example, the textbook by Manohar and Wise.

You first do one, and then the other: First you do d=4-2e, which gives you UV divergences explicitly regulated (although still IR divergent). Then you subtract the UV divergences with a counterterm so that the only divergences left are IR. Then you analytically continue to d=4+2e dimensions to regulate these IR divergences. These are then canceled the usual way by constructing IR safe observables. This is also explained in Sterman's text/lecture notes.

The point is that in this procedure you must be VERY careful which divergences are UV and which are IR. If you treat them all the same way (ignore the sign of epsilon), you might (incorrectly) subtract an IR divergence with a UV counterterm, which could lead to trouble when trying to construct IR safe observables. See, for example, the textbook by Manohar and Wise.

Thanks. I am not an expert on dimensional regularization though I read a fair bit about it, though not in the last few years. So I don't know what IR safety is. Is
http://xxx.lanl.gov/pdf/0807.5118
good enough to address the things you said? Otherwise, please give online sources if possible.

Sterman can be a hard read.

"IR Safety" is basically just the statement that there are no IR divergences. For example, in QED the (pure) Bhabha scattering cross section is NOT IR safe, but when you include the soft bremsstrahlung cross section it IS.

For an example in QCD (where it becomes more important): the quark and gluon pdf for the proton is NOT IR safe, yet the deep-inelastic scattering cross section and Drell-Yan cross sections are. That is why there are so many different "pdf"s out there - because it depends on how you subtract the IR divergences. But in the IR safe DIS cross section, there is no such ambiguity, since those IR divergences always cancel.

Gold Member
There is also a major conceptual difference between UV and IR divergences. UV divergences arise from the theory being poorly defined. Removing the UV divergences through renormalization is part of correctly constructing the theory.

IR divergences on the other hand occur in well-defined models, even rigorously constructed models (e.g. $$QED_{2}$$). This is because IR divergences come about because of massless particles. In this case the asymptotic in/out spaces are not Fock spaces, if you act like they are you get IR divergences. So basically in $$QED_{2}$$ for example $$e^{-} + e^{-} \rightarrow e^{-} + e^{-}$$ doesn't exist as process, the states will always contain soft (not virtual) photon clouds. So the asymptotic states must contain soft-photon clouds. The divergence only occurs from trying to calculate fictitious processes.

One must always calculate real processes and therefore you need to include soft-photon clouds. Any process containing the required soft photon clouds (or one where they genuinely are not produced) is called infra-red safe.

"IR Safety" is basically just the statement that there are no IR divergences.

Thanks. I now have sort of an overview of what is in the paper, and understand what is meant by IR safety.

Could you now please go back to your original statement that the limits eps to 0 for dimension 4-eps and 4+eps can give different results, and give an example where this happens? It must involve computations where the dependence on eps is not meromorphic at eps=0. But how do these arise?

There is also a major conceptual difference between UV and IR divergences. UV divergences arise from the theory being poorly defined. Removing the UV divergences through renormalization is part of correctly constructing the theory.

IR divergences on the other hand occur in well-defined models, even rigorously constructed models (e.g. $$QED_{2}$$). This is because IR divergences come about because of massless particles.

Since you mention rigorous methods in the context of a thread on dimensional regularization: Is there a version of the latter that can be used for a rigorous analysis for some nontrivial <4D model? (No matter whether for UV or IR regularization.)

Gold Member
Since you mention rigorous methods in the context of a thread on dimensional regularization: Is there a version of the latter that can be used for a rigorous analysis for some nontrivial <4D model? (No matter whether for UV or IR regularization.)
Good question. Regrettably, no there is not. There have been several attempts to obtain a generalisation of dimensional regularisation at the rigorous constructive level, but the success rate has been zero. You'll see this point mentioned several times in Rivasseau's monograph "From perturbative to constructive Renormalization", as it really provides a stumbling block with Yang-Mills. It also prevents a good analysis of anomalies, where at present all we can say is the general mechanism behind anomalies in the Algebraic approach, but nothing as detailed as one would like.

Good question. Regrettably, no there is not. There have been several attempts to obtain a generalisation of dimensional regularisation at the rigorous constructive level, but the success rate has been zero. You'll see this point mentioned several times in Rivasseau's monograph "From perturbative to constructive Renormalization", as it really provides a stumbling block with Yang-Mills. It also prevents a good analysis of anomalies, where at present all we can say is the general mechanism behind anomalies in the Algebraic approach, but nothing as detailed as one would like.

Isn't the latter covered in Scharf's ghost story?
https://www.amazon.com/dp/0471414808/?tag=pfamazon01-20

Last edited by a moderator:
Gold Member
Isn't the latter covered in Scharf's ghost story?
https://www.amazon.com/dp/0471414808/?tag=pfamazon01-20
Sorry, I assumed you meant a rigorous non-perturbative definition/generalisation of dimensional regularization.
There is a rigorous version of perturbative dimensional regularization covered in Scharf's book, also books by the mathematician Gerald Folland. In this case it is essentially an application of the Hahn-Banach theorem to a given distribution. A beautiful example is given in some papers of Folland and his book "Quantum Field Theory: A tourist guide for mathematicians".

Last edited by a moderator:
Sorry, I assumed you meant a rigorous non-perturbative definition/generalisation of dimensional regularization.
Well, I am fishing in the dark, knowing roughly but not precisely what exists in the literature -- hoping to learn from you.
There is a rigorous version of perturbative dimensional regularization covered in Scharf's book, also books by the mathematician Gerald Folland. In this case it is essentially an application of the Hahn-Banach theorem to a given distribution. A beautiful example is given in some papers of Folland and his book "Quantum Field Theory: A tourist guide for mathematicians".
Thanks for the references. -- Which papers?

Could you now please go back to your original statement that the limits eps to 0 for dimension 4-eps and 4+eps can give different results, and give an example where this happens? It must involve computations where the dependence on eps is not meromorphic at eps=0. But how do these arise?

I'm afraid I cannot think of a good counterexample, so I'm going to have to retract my claim. As you say, DimReg typically only introduces poles in epsilon, so that shouldn't be a problem. I was just being silly, although if anyone knows of a counterexample I can't think of, I'm happy to hear about it....

The big issue about the sign of epsilon is whether the divergence is UV or IR, and as I and others have already said, this is a very important point, as UV and IR divergences are fundamentally different things and are not treated the same way, and they must be kept separate.

I'm afraid I cannot think of a good counterexample, so I'm going to have to retract my claim. As you say, DimReg typically only introduces poles in epsilon, so that shouldn't be a problem. I was just being silly, although if anyone knows of a counterexample I can't think of, I'm happy to hear about it....

The big issue about the sign of epsilon is whether the divergence is UV or IR, and as I and others have already said, this is a very important point, as UV and IR divergences are fundamentally different things and are not treated the same way, and they must be kept separate.

But why do you then still connect the UV/IR issue with the sign of epsilon?

Since you seem to be sure that a very different treatment of UV and IR in dimensional regularization is necessary, but you cannot think of an example where the sign of eps makes a difference, this different treatment must be related to something else - perhaps to the + i eps you once introduced?

But why do you then still connect the UV/IR issue with the sign of epsilon?

Since you seem to be sure that a very different treatment of UV and IR in dimensional regularization is necessary, but you cannot think of an example where the sign of eps makes a difference, this different treatment must be related to something else - perhaps to the + i eps you once introduced?

Here's an example of an integral that has a UV divergence:

$$\int_0^\infty\frac{dx}{(x+m^2)^{1+\epsilon}}$$

This is a typical result of a loop integral after DimReg (with $x=k^2$). If $\epsilon\le 0$, the integral remains divergent, but if $\epsilon>0$ the integral is convergent, although it will give you a pole in $\epsilon$.

Here's an example of an integral with an IR divergence:

$$\int_0^1 dz~z^{1-2\epsilon}}$$

This is a typical Feynman parameter integral in $d=4-2\epsilon$. Notice that this integral is also ill-defined at the lower limit of integration unless $\epsilon<0$, precisely the reverse of the first (UV divergent case).

So in summary: UV divergent integrals are made finite for $\epsilon>0$, while IR divergent integrals are made finite for $\epsilon<0$, where in all cases, $d=4-2\epsilon$.

This is what I've been trying to say.

Here's an example of an integral that has a UV divergence:

$$\int_0^\infty\frac{dx}{(x+m^2)^{1+\epsilon}}$$

This is a typical result of a loop integral after DimReg (with $x=k^2$). If $\epsilon\le 0$, the integral remains divergent, but if $\epsilon>0$ the integral is convergent, although it will give you a pole in $\epsilon$.

Here's an example of an integral with an IR divergence:

$$\int_0^1 dz~z^{1-2\epsilon}}$$

This is a typical Feynman parameter integral in $d=4-2\epsilon$. Notice that this integral is also ill-defined at the lower limit of integration unless $\epsilon<0$, precisely the reverse of the first (UV divergent case).

So in summary: UV divergent integrals are made finite for $\epsilon>0$, while IR divergent integrals are made finite for $\epsilon<0$, where in all cases, $d=4-2\epsilon$.

This is what I've been trying to say.

OK, now I understand. Thanks.