# infinity of running couplings

+ 5 like - 0 dislike
93 views

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
retagged Mar 7, 2014
I don't think there is a universal answer: you calculate the relevant beta function to higher order in the theory you are interested in, and see what happens. The relevant calculations are standard QFT ones, described in many textbooks.

This post has been migrated from (A51.SE)
Fine, Moshe, but there's still a well-defined question (implicitly contained above) whether the Landau pole of $\lambda\phi^4$ in $d=4$ may go away if we calculate it more accurately, e.g. to all orders or exactly non-perturbatively. It almost certainly doesn't go away but not so much due to explicit calculations but because there would have to be a UV fixed point that flows to the interacting scalar theory. It doesn't seem to exist - as far as we know the candidate theories - so the theory should disappear at some scale, the Landau pole.

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

This post has been migrated from (A51.SE)
I don't have time for an answer now, but quick summary is that anything is possible. The complete running is encoded in the beta function, which has the answer to all your questions. One loop results only give you the first Taylor coefficient of that function at small coupling. This small piece of information is consistent with lots of different scenarios, including those you mention.

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

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.

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

This post has been migrated from (A51.SE)

+ 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.

This post has been migrated from (A51.SE)
answered Nov 8, 2011 by (2,395 points)
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.

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

This post has been migrated from (A51.SE)
answered Nov 8, 2011 by (345 points)
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?

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

This post has been migrated from (A51.SE)
The factor 4 stands at $\lambda\to\infty$ and you call it "IR limit". But it is a strong coupling limit, isn't it?

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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOverf$\varnothing$owThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.