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  infinity of running couplings

+ 5 like - 0 dislike

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?

This post has been migrated from (A51.SE)
asked Nov 6, 2011 in Theoretical Physics by AAB (35 points) [ no revision ]
retagged Mar 7, 2014 by dimension10
Most voted comments show all comments
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.

This post has been migrated from (A51.SE)
@Moshe: I invite you to discuss (what is confused and misguided in my article) in my chat http://chat.stackexchange.com/rooms/1606/renormalizations-pros-and-contras

This post has been migrated from (A51.SE)
@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.

This post has been migrated from (A51.SE)
@Moshe: I respect your job and have nothing against it but you deliberately include my research results into "non professional" category.

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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.

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AAB, no doubt about it, the higher orders at least modify the speed with which the Landau pole is approaching. It can slow it down or speed it up. The scalar theory probably has to break down at some point but gauge theories may sometimes continue, via S-dualities, and one gets interesting "cascades of Seiberg dualities" where one may switch from a divergent coupling to an equivalent tiny one many times as the energy is being raised.

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Any relationship (exact or approximate) between real and bare charges is meaningless since there are no bare charges.

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2 Answers

+ 5 like - 0 dislike

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.

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answered Nov 8, 2011 by Moshe (2,405 points) [ no revision ]
One can see M. Gell-Mann's interview about this in episode 53, although it is worth to watch episodes 50-55. http://www.webofstories.com/play/10607?o=MS

This post has been migrated from (A51.SE)
Interesting. Thanks for that.

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+ 1 like - 0 dislike

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.

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answered Nov 8, 2011 by JonLester (345 points) [ no revision ]
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?

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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.

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The factor 4 stands at $\lambda\to\infty$ and you call it "IR limit". But it is a strong coupling limit, isn't it?

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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.

This post has been migrated from (A51.SE)
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?

This post has been migrated from (A51.SE)
@Vladimir, we all know where your leading questions are leading. Your contributions are appreciated (for example, I think your previous comment here is correct), but I think you will have a more valuable and pleasant experience here if you stop trying to hijack threads and lead them to your "reformulation" issue. You will not be able to reach that particular destination here, there are not going to be any discussions of alternatives to established physics here.

This post has been migrated from (A51.SE)

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