A Landau pole - an infinity occurring in the running of coupling constants in QFT is a known phenomena. How does the Landau pole energy scale behave if we increase the order of our calculation, (more loops) especially in the case of Higgs quadrilinear coupling?

  • 2
    $\begingroup$ Thank you for the answers. I was particularly interested in 1. if there is a chance that this problem does go away if treated non-perturbatively and if that does not happen 2. is it possible that it gets worse - the energy scale at which our theory breaks down gets smaller while including more orders of calculation. But as for the second question, as Moshe wrote, now I think it should depend on a particular case. $\endgroup$ – AAA Nov 6 '11 at 22:48
  • 4
    $\begingroup$ Removed a link in the previous comment. This is not a place to advertise alternatives to mainstream physics. $\endgroup$ – user566 Nov 7 '11 at 15:45
  • 2
    $\begingroup$ This a specific technical question, confused and misguided philosophizing on one’s dissatisfaction with renormalization is irrelevant, and will not be allowed to dominate the discussion here. Comments directly related to this specific question are welcome, but any subsequent attempts to divert the conversation have been/will be deleted. $\endgroup$ – user566 Nov 7 '11 at 16:00
  • 5
    $\begingroup$ @Vladimir: This is a forum for professional physicists to discuss current mainstream publishable research. Most of us would not like to be repeatedly dragged into discussions we have no interest in, on things which do not fall into that category. My job as a moderator is to provide our user base with a pleasant and productive environment, and I take that job seriously. $\endgroup$ – user566 Nov 7 '11 at 16:24
  • 2
    $\begingroup$ Yes, I have. But, this is besides the point. None of your comments has anything to do with the question. You should not take any mention of the word “renormalization” as an invitation to start a discussion of your own issues with the subject. $\endgroup$ – user566 Nov 7 '11 at 16:33

The $\beta$ - function of a coupling determines its energy dependence. This in turn is a function of all the couplings in the theory, usually calculated in perturbation theory. So, things could be complicated for multi-dimensional coupling space.

For a single coupling, assume the one loop result is positive. This means that as long as the coupling is weak, it will grow with the energy scale. If you extrapolate that result way beyond its region of validity, you find that the coupling becomes infinite at some finite energy scale (but, long before that perturbation theory breaks down). This is such a fantastically high energy scale that this so-called Landua pole is an academic issue. Any QFT typically has energy range where it is useful as an effective field theory, and it is not typically valid or useful in such a huge range of energy scales. In any event, at these enormous energy scales quantum gravity is definitely relevant, and it is unlikely to be a quantum field theory at all. For these reasons the Landau pole is no longer a concern for most people, it was more of an issue when QFT was thought to be well-defined at all energy scales.

To your question, since the coupling becomes strong, pretty much anything can happen. It may be that the coupling does diverges at some energy scale (higher or lower than the initial estimate), though to make that statement with confidence you'd need to be able to calculate the $\beta$ - function at strong coupling. If this is the case, your QFT is an effective field theory defined only at sufficiently low energy scales.

It may also be that the $\beta$ - function gets some negative contributions and starts decreasing, whereas a zero becomes possible. When this happens the coupling constant increases initially, but stops running at some specific value. This is the scenario of UV fixed point, which makes the theory well-defined at all energy scales. In this case the problem, such as it is, indeed goes away.


Landau pole is not a mathematically consistent object. The reason relies on its derivation based on a few terms of a perturbative expansion. A typical case of this is provided by the scalar field. Just consider the following academic case

$$ L=\frac{1}{2}(\partial\phi)^2-\frac{\lambda}{4}\phi^4. $$

This field has the following behaviors:

$$ \beta(\lambda)=\frac{3^3\lambda^2}{4\pi^2}, \qquad \lambda\rightarrow 0 $$

and, as proved by several authors (e.g. see http://arxiv.org/abs/1102.3906 and http://arxiv.org/abs/1011.3643),

$$ \beta(\lambda)=4\lambda, \qquad \lambda\rightarrow\infty $$

This implies that, by a continuity argument, the Landau pole simply does not exist for the scalar field but this is anyhow trivial. The factor 4 in the infrared limit is indeed the space-time dimensionality.

  • $\begingroup$ So $\beta (\lambda) \approx \frac{4\lambda^2}{ \lambda + \frac{16 \pi^2}{27}}$ is a good approximation for all $\lambda$? You wanted to say "the ultraviolet limit" here? $\endgroup$ – Vladimir Kalitvianski Nov 8 '11 at 20:09
  • $\begingroup$ Yes, the first beta function is for the ultraviolet limit but I do not know the full beta function. We can only state the ones at limits. Yours is just a guess. $\endgroup$ – Jon Nov 8 '11 at 20:55
  • $\begingroup$ The factor 4 stands at $\lambda\to\infty$ and you call it "IR limit". But it is a strong coupling limit, isn't it? $\endgroup$ – Vladimir Kalitvianski Nov 8 '11 at 20:58
  • $\begingroup$ Infrared limit and strong coupling limit are the same thing as the ultraviolet limit corresponds to the weak coupling limit. The factor 4 is the space-time dimensionality. $\endgroup$ – Jon Nov 8 '11 at 22:24
  • $\begingroup$ You omitted the mass term in order to simulate a non-Abelian gauge field, I guess. Another question, if the beta-function is known exactly, does that mean you can fulfill the renormalization exactly and get rid of all bare stuff? Can the exactly renormalized theory be the desired physical theory to deal with from the very beginning? $\endgroup$ – Vladimir Kalitvianski Nov 9 '11 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.