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I am referring to this question, and especially this answer.

In addition, QCD has - like all field theories - only an asymptotic perturbation series, which means that the series itself will also diverge if all terms are summed.

What does it mean? From what I know, if the sum over a series diverge, that means that the summation doesn't work, which means that the quantity you are trying to calculate, you can't get answer for that, for any quantity that comes back from your calculation must be of finite value.

But in QCD and QED things seem a lot more complicated, since some divergences are allowed:

This doesn't mean that QCD perturbation theory doesn't have ultraviolet divergences, it has those like any other unitary interacting field theory in 4d. These ultraviolet divergences though are not a sign of a problem with the theory, since the lattice definition works fine. This is in contrast to, say, QED, where the short lattice spacing limit requires the bare coupling to blow up, and it is likely that the theory blows up to infinite coupling at some small but finite distance. This is certainly what happens in the simplest interacting field theory, the quartically self-interacting scalar

My questions:

  1. How many kinds of divergence there are in QCD and QED?
  2. And how do we know what kind of divergence is acceptable ( in the sense that we can still extract values out for prediction after some renormalization process)?
  3. If the sum diverge, then we won't be able to calculate the series's sum. Isn't that is defeating the purpose of the series? For any series, if the sum diverge after summing all the terms, then we know that the formula must be wrong or the series have no physical meaning. But why is it that for QCD series, the formula is still correct ( because it is used to extract coupling constants) and has physical meaning ( QCD series must correspond to something in reality)?
  4. The fact that QCD has non-convergent series means that it cannot be the fundamental theory of nature, right?
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Hi Graviton, it strikes me quite odd that all of your recent questions end with "some theory can not be a fundamental theory of naure, right?". This gives the impression that you dont want "some theory" to be a fundamental theory of nature and gives the questions a negatively opinionated touch ... –  Dilaton Jul 9 '13 at 10:14
    
@Dilaton, what is so odd about me asking? And furthermore, even if it was odd, the question that whether if "X has divergent series rules out X as the fundamental theory of nature" is important, and an interesting one, right? –  Graviton Jul 9 '13 at 12:06
    
Yes, the physics issues themself are interesting and I like the very nice answers that are coming in ... –  Dilaton Jul 9 '13 at 12:11
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Regarding question 4: It's only a problem if the theory is defined by its perturbation series. QCD is not defined by its perturbation series. The failure of the perturbation series to converge just means that it's not a very good approximation. –  user1504 Jul 9 '13 at 12:12

1 Answer 1

up vote 22 down vote accepted

In QED there are 4 kinds of divergences:

  1. Ultraviolet divergences. Naive calculations depend on the cut-off in such a way that they go to infinity as the cut-off do. However, QED is a perturbatively renormalizable theory so that non-naive, well-done computations (see regularization and renormalization) give sensible results.

  2. Landau pole. The coupling constant $\alpha={e^2\over \hbar \, c}$, which is the expansion parameter in the perturbative series, grows with energy and goes to infinity for a finite value of the energy. It turns out that this finite value of energy is larger than the electroweak scale, where QED merges with the weak interaction and QED is not a good theory of nature anymore. Therefore, it isn't a real (phenomenological) problem.

  3. Infrared divergences. These are due to the fact that photons are massless. They however cancel out once one takes into account all the effects that contribute to a measurable observable.

  4. Non-convergent series. The $n$-th term of the perturbative expansion is of the form $\left({\alpha\over 2\pi}\right)^{n}\, (2n-1)!!$, so that the series is not convergent but asymptotic because the factor $(2n-1)!!$ grows very fast for large values of $n$. This means that we cannot give a non-perturbative definition of QFT by summing up all the terms of the series. However, the first terms are meaningful and actually give predictions that accurately agree with observations. The 'first terms' are approximately $n\sim {\pi\over \alpha}\sim 430$. And for this value of $n$, $\left({\alpha\over 2\pi}\right)^{n}\, (2n-1)!!\sim 10^{-187}$. Therefore, as long as we are not interested in a precision of one part in $10^{187}$ , this is not a real problem either. Note that QED is the theory of nature that has been confirmed with greatest precision — one part in $10^{9}$ in electron's anomalous magnetic dipole, for which $n=4$.

For QCD points 1, 3, and 4 are more o less the same. However, point 2 doesn't apply since in QCD the coupling constant $\alpha_s$ gets lower with the increasing of energy, and in fact it goes to zero as energy goes to infinity. See asymptotic freedom.

To summarize, infrared divergences are due to not taking into account effects that contribute to the observable magnitude. The asymptotic nature of QFT perturbative expansions prevents a non-perturvative (exact) definition of the theory (through its series), but doesn't entail a practical problem when comparing predictions with measurements. The lack of perturbative divergences and Landau-like poles are a necessary condition for a theory to be well-defined at arbitrarily high energies. However, theories that contain these divergences (ultraviolet or Landau-like poles) can still be very useful at energies above some scale. On the other hand, theories without these divergencies (ultraviolet or Landau-like poles), such as QCD, don't have to be valid to all energies as theories of nature.

As M. Brown points out in the comments, there is a relation between instantons and renormalons and the asymptotic nature of series. Please, see these notes snd the questions Instantons and Non Perturbative Amplitudes in Gravity and Asymptoticity of Pertubative Expansion of QFT

Reply to Graviton's comment: In my opinion, a fundamental theory of nature (whatever it means) should have a non-perturbative definition. If the perturbative expansion is not convergent, it cannot provide this non-perturbative definition. However, in principle, this doesn't necessarily mean that theory cannot have a non-perturbative definition or an exact solution, but this must be given by other means.

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+1 good answer. Regarding point 4: there's a lot that could be said here about instantons and renormalons (by somebody more knowledgeable than me!). My very vague understanding is that there are some deep connections between the (exponentially small) ambiguity of the asymptotic series and the (also exponentially small) instanton contributions, but how it all works and how general it is is still fuzzy to me. –  Michael Brown Jul 9 '13 at 8:04
    
If I understand it correctly, the fact that QCD has non-convergent series means that it cannot be the fundamental theory of nature, right? –  Graviton Jul 9 '13 at 8:17
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@Graviton: In my opinion, a fundamental theory of nature (whatever it means) should have a non-perturbative definition. If the perturbative expansion is not convergent, it cannot provide this non-perturbative definition. However, in principle, this doesn't necessarily mean that theory cannot have a non-perturbative definition or an exact solution, but this must be given by other means. –  drake Jul 9 '13 at 8:37
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@Neuneck: It might be better to say that 'QCD is a continuum limit of lattice QCD' is a non-perturbative definition. Just writing the Lagrangian down obscures the additional choices you have to make when defining the lattice path integral. –  user1504 Jul 9 '13 at 12:05
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To amplify drake's answer to Graviton's question from the opposite angle: an asymptotic series can well be the perturbative expansion of a non-pertrubatively defined fundamental theory of nature. Having the asymptotic perturbation series is not yet having the "fundamental theory", but it can be a step in that direction. –  Urs Schreiber Jul 10 '13 at 10:51

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