Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Landau pole

  1. Apr 21, 2013 #1
    I read on Wikipedia (and verified myself) that if the beta function is proportional to the coupling raised to a power greater than 1, then there is a Landau pole, i.e., the coupling blows up at high energy.

    Here is what I got, for β=gα, where 1∠α, where g is the coupling and μ is scale:

    [tex]g_2^{\alpha-1}=\frac{g_1^{\alpha-1}}{1-(\alpha-1)g_1^{\alpha-1}\log\frac{\mu_2}{\mu_1}} [/tex]

    The denominator will blow up for large μ2 if [tex]g_1^{\alpha-1} [/tex] is positive.

    But doesn't QCD, QED, scalar theory, all have α>1 ??? So is there a Landau pole in all of them?
     
  2. jcsd
  3. Apr 21, 2013 #2
    QED and phi^4 theory have a Landau pole (if you trust perturbation theory). QCD, though, has a negative beta function. As a result the coupling goes to zero at high energies, and the "pole" is at low energies, around 200 MeV. At this energy scale, QCD perturbation theory breaks down and interesting nonperturbative phenomena like confinement and chiral symmetry breaking appear. The theory remains perfectly consistent, as we can verify by performing lattice gauge theory simulations.

    Lattice simulations of phi-fourth theory, though, suggest that there really is a Landau pole in that theory. It is found numerically that if you increase the energy scale while adjusting the coupling at that scale to maintain the same low-energy physics, you have to increase the coupling to infinity at a finite energy. This is the Landau pole.
     
  4. Apr 21, 2013 #3

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For QCD, ##\beta = - b g^3##, with ##b>0##, so the coupling blows up at low energy. This is the famous asymptotic freedom: the coupling actually goes to zero at high energies, in the reverse behavior of QED and scalar field theories.

    So there is no Landau pole in QCD. There is probably no real Landau pole in QED. The scale at which it would occur is over a couple hundred orders of magnitude beyond the Planck scale. It is obvious that we can't just extrapolate a Standard Model result so far into energies where we cannot neglect quantum gravity.

    The Higgs sector may or may not have a Landau pole. The most recent published LHC results did not pin the potential down accurately enough to conclude that the Landau pole occurs before we reach the Planck scale. If future results indicate that there is a Landau pole at some scale below the Planck scale, it could still be the case that new physics enters at some intermediate scale, making the Landau pole "safe."
     
  5. Apr 21, 2013 #4
    oops...I calculated β=gα for α>1 and not β=Cgα. If you include the C it becomes:

    [tex]g_2^{\alpha-1}=\frac{g_1^{\alpha-1}}{1-C(\alpha-1)g_1^{\alpha-1}\log\frac{\mu_2}{\mu_1}} [/tex]

    so now I see that since C is negative in QCD, you get:

    [tex]g_2^{\alpha-1}=\frac{g_1^{\alpha-1}}{1+|C|(\alpha-1)g_1^{\alpha-1}\log\frac{\mu_2}{\mu_1}} [/tex]

    and there is no blowup at high μ2, but there is blowup at low μ2 when the log becomes negative. Is this low point μ2 also called a Landau pole? Or is the term Landau pole only for blowups at high energy?

    I have a question about how you calculate the scale at which the beta function for QCD blows up. Presumably you have to make a measurement of g1 at some scale μ1. But since quarks aren't free, how does one do this? You can't collide quarks. You have to work with hadrons. Is the coupling constant of hadrons related somehow to g1?

    So if I'm understanding this right, these infinities of the coupling only mean the theory is no good at that scale, but can still be used for scales where the coupling is not infinity (we just say that new physics enters at the troublesome Landau scale)?

    Sorry. Last question :smile:, but from the above expressions it looks like theories that have positive coupling constants have a high energy Landau pole and are asymoptotically free or trivial at low energies, and those with negative coupling constants have a low energy Landau pole and are trivial or asymptotically free at high energies. Is this generally true?

    But what's the big deal about the Landau pole? For QCD we say the pole leads to new physics like confinement. So why not for scalar theory we say that the Landau pole at high energy just leads to new physics?
     
  6. Apr 21, 2013 #5
    I think "Landau pole" is reserved for the high-energy pole. In particular we know that the fact the QCD's coupling blows up at low energies does *not* indicate a problem with the theory, while it is believed the the QED/phi-fourth Landau poles indicate that the theory is sick at some level.

    You essentially can collide quarks and gluons, if you do it at high energies. In high-energy collisions, hadrons act like bags of loosely associated partons and their collisions can be analyzed in terms of the collisions between the individual partons, namely the quarks and gluons.

    I think one way to measure g at high energies is to look at the ratio of three-jet events to two-jet events. Jets are what result after the individual quarks or gluons involved in a collision hadronizes. If you have two quarks that scatter elastically off each other, you get two jets, one for each quark. But if one of those quarks also radiates a hard gluon, that gluon can show up as a third jet. The amplitude to radiate such a gluon is determined by g, so you can look at the frequency of three-jet events to measure g.

    If you can you might check out Weinberg Vol II section 18.3, "Varieties of Asymptotic Behavior," which enumerates the various possibilities for the limiting behavior of coupling constants.

    If you're talking about QED/phi-fourth theory, yes, that's right. But QCD is perfectly good at arbitrarily low energies; it's only perturbation theory that breaks down.

    Well, as I mentioned above, nonperturbative numerical simulations of QCD indicate that it is well-defined at low energies. Confinement is an effect that emerges *within* QCD. But nonperturbative numerical simulations of phi-fourth theory indicate that it is *not* well-defined at high energies: no new effects emerge to save the theory. If you want to save the theory you have to add something in by hand. So, for instance, presumably new physics (QG or something else) at some high energy scale is needed in order to save the Higgs sector of the SM from inconsistency. I'm not sure what the status is for QED.
     
    Last edited: Apr 21, 2013
  7. Apr 21, 2013 #6

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For QCD, the "new physics" of confinement has consequences that can be compared with observation. Things like hadrons being apparently color neutral and strongly bound are exactly what would be expected if we have color confinement.

    On the other hand, to say that there is new physics at high energy scales (beyond the reach of experiment) is not always so satisfying. We indeed expect quantum effects of gravity to modify Standard Model results and even perhaps expect some GUT theory to emerge a few orders of magnitude below the Planck scale. Either of these could protect us from any putative Landau pole in the Higgs self-coupling. However, it is a substantially weak argument (as compared to say QCD) to say that the answer is some (miraculous) new physics that we can't precisely pin down the nature of, or hope to measure soon.

    This is a conservative, perhaps pessimistic, view that is commonly held. Perhaps it errs on the side of being too critical, since history has many examples where scary looking divergences led to the discovery of new physics. But this is a main difference between QCD and scalar QFT.
     
  8. Apr 21, 2013 #7
    Thank you both!

    QED can't be entirely sick since it works very accurately for present energies. So I think new physics has to save it once we explore higher energies.

    As for scalar Higgs and QED, I'm a little confused because at low energies don't the couplings become zero, i.e. triviality? Is that observed right now? Does the charge of the electron go to zero at low energies (those are what the above formulas suggest)? But that seems silly as two electrons that are not moving (zero energy) exert a force on each other?
     
  9. Apr 21, 2013 #8

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    QED ceases to be a good description of reality beyond momenta transfer of 10's of GeV. Then you need the full electroweak theory. So the Landau pole is somewhat academic.
     
  10. Apr 21, 2013 #9

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Can you be a bit clearer what you mean? In either case there is a natural IR cutoff for the 1-loop beta function set by the mass of the lightest particle that can appear in the loop. Furthermore, the bare couplings are not expected to be well-defined on their own, but the combination of bare coupling and UV cutoff appearing the integrated equation is set to the experimentally observed value, which at least for the fine-structure constant, is perfectly non-zero in the IR.
     
  11. Apr 21, 2013 #10
    Triviality is the following problem:

    Suppose you have a cutoff ##\Lambda## for your theory. You specify some coupling ##g(\Lambda)## at the cutoff. Then you can use the renormalization group to run this coupling down to some fixed low energy scale, call it ##M##, giving ##g(M)##. Now, we actually do experiments at the low energy scale ##M##. These experiments fix ##g(M)##. So you'll need to adjust ##g(\Lambda)## to give the right ##g(M)## when you run the coupling down to ##M##.

    Now you decide you want to take the cutoff ##\Lambda## to infinity. As you increase ##\Lambda##, you need to increase ##g(\Lambda)##. It may be that you can take this limit without a problem. But you might encounter the following difficulty.

    You might find that in order to keep ##g(M)## unchanged while you increase ##\Lambda##, you have to let ##g(\Lambda)## go to infinity at a finite value of ##\Lambda##. That is, an infinite value for ##g(\Lambda)## will run down to a finite value for ##g(M)##. Well, fine, we can let ##g(\Lambda) = \infty##. But when we try to increase ##\Lambda## further, we can no longer keep ##g(M)## fixed, because we can no longer increase ##g(\Lambda)## to compensate for the renormalization group running. So now as we increase ##\Lambda## to infinity, ##g(M)## inexorably decreases to zero.

    As you point out, for QED and for the SM Higgs sector we measure nonzero values for ##g(M)##. What this tells you is that these theories must have a finite cutoff ##\Lambda##: they must be effective theories that are only valid below some energy scale, above which they must be supplemented by something new.
     
  12. Apr 21, 2013 #11

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There's still (at least naively) a Landau pole for the ##U(1)_Y## coupling. I assume that at least some of the literature addresses the difference between this and the QED RG.
     
  13. Apr 21, 2013 #12

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    The problem with worrying about just hypercharge is a) there are no particles that have hypercharge and no isospin, so you don't have a hypercharge-only theory, and b) even if you did, loops would bring in isospin anyway (and the Landau pole is a feature of loops).
     
  14. Apr 21, 2013 #13

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    Exactly.

    It is way outside the known region of applicability of QED and well before that is reached the electroweak theory comes into play. That's assuming you can trust perturbation theory for values that blow up to infinity - which I doubt.

    But just out of curiosity, because I have never come across any literature about it, does anyone know if the electroweak theory has similar issues ie its own Landau pole? I suspect it does but have not seen anything on it.

    Thanks
    Bill
     
    Last edited: Apr 21, 2013
  15. Apr 21, 2013 #14

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    I suspect you're right, because the beta functions are all positive for EWK, but I don't know of anyone who went through the trouble of working it all out.
     
  16. Apr 22, 2013 #15

    DrDu

    User Avatar
    Science Advisor

    I read some time ago that proving that electroweak coupling is a valid mathematical theory was chosen as one of the millenium problems. The author (which I don't remember) mentioned that electroweak theory whas probably chosen instead of QED because there are no Landau poles.
     
  17. Apr 22, 2013 #16

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    There is something in the back of my mind that the issue with the Electroweak and what the Millennium prize is for is the general issue of Yang-Mills theory namely showing they actually exist or something like that. But I certainly haven't seen anything about Electroweak having issues like the Landau pole so you may be correct about that. However I think its primary reason for existence is it unifies two separate theories like EM unified electricity and magnetism. Now if only the others can be unified - string theory perhaps? However despite its great promise and dazzling math it seems to caught up in issues of its own rather than making progress towards that goal. Not that I think there is really any other viable candidate. My suspicion is we need a few more String Theory revolutions before its full promise is realized - but nature I suspect has something rather unexpected in store - if only we knew what it was.

    Thanks
    Bill
     
    Last edited: Apr 22, 2013
  18. Apr 22, 2013 #17
    Sorry, but I think the reason I'm not clear is because my grasp of the subject is a bit weak. The β function for QED is [tex]\beta=\frac{e^3}{12 \pi^2} [/tex] so the equation:

    [tex]e_2^{2}=\frac{e_1^{2}}{1-\frac{1}{6 \pi^2}e_1^{2}\log\frac{\mu_2}{\mu_1}}[/tex]

    suggests that if you run μ2 to low energies, e2 goes to zero since the denominator blows up. The vertex function for QED is dimensionally regularized as:

    [tex]V^\mu(p,p')=e\gamma^\mu+\frac{e^3}{8 \pi^2} \left[ \left(\frac{1}{\epsilon}-1-\frac{1}{2}\int dF_3 \ln(D/\mu^2)\right) \gamma^\mu+...\right] [/tex]

    and MS- bar just chops off the 1/ε pole using a counterterm. So the expression in parenthesis becomes finite. Now if μ2 goes to zero, doesn't the vertex vanish? The combination of the coupling e3 (=0) multiplied by the finite [tex]\left[ \left(-1-\frac{1}{2}\int dF_3 \ln(D/\mu^2)\right) \gamma^\mu+...\right] [/tex] should be zero? And also the tree term: [tex]e\gamma^\mu=0[/tex]

    I can sort of see that since μ is going to zero in the denominator of the loop term, that 0*∞=finite. However, the μ occurs in a log, and 0^3*Log[0]=0.
     
    Last edited: Apr 22, 2013
  19. Apr 22, 2013 #18

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    OK, this is the argument I thought you were using. The problem is that you are regulating an integral over the momentum in a loop. Since everything that is electrically charged is massive, there is a minimum amount of momentum that you can have, equal to the mass of the lightest particle. So the integral is cut off in the IR by the electron mass.
     
  20. Apr 22, 2013 #19
    What you seem to be saying seems different than what others have been saying.

    So what I took from your comments is that at the Landau pole of QED, the coupling is ∞. If you go beyond the Landau pole, it's still ∞. When you run the coupling from these two points downwards to a small scale M, they should give the same result, but since you had to run the latter case over a larger energy interval to get to M, it should be smaller. The only way to reconcile this is to say that the coupling has a fixed point of zero in the IR? Doesn't this imply that any theory with a high energy Landau pole is pretty much trivial unless new physics can save it at the Landau pole?

    I think what f-zero is saying is that the coupling μ can't be made arbitrarily small to zero, because an experiment always has some energy in it, namely the rest mass of the smallest particle involved in the interaction (although you could envision an interaction between three heavy particles, and another interaction between a heavy particle and a light particle: then you could have a process where you had external light particles, with an interaction involving heavy particles: then the beta function of just the heavy particles might require that you insert the light particle mass for μ).
     
  21. Apr 24, 2013 #20
    I don't think so?

    The way to reconcile this is to say that you're not allowed to push the cutoff ##\Lambda## above point where ##g(\Lambda)## blows up, if you want to keep ##g(M)## fixed.

    Yes; if you want to take ##\Lambda \to \infty## you have to accept that ##g(M) = 0##, which gives a trivial noninteracting theory. Alternatively you could accept that the theory is really an effective theory with some finite cutoff ##\Lambda##; that is, unless new physics comes in at some high energy scale.

    I believe what fzero is saying is that while the beta function of QED is usually quoted as ##\beta(e) = \frac{e^3}{12 \pi^2}##, this is only an approximation valid for energies ##\mu >> m_e##. For ##\mu << m_e## the beta function goes to zero. The electron charge essentially stops running below ##m_e##, and goes to a constant (the observed charge), not zero. This is because if you evaluate the vacuum polarization diagrams that contribute to the beta function, they are independent of the external momenta for external momenta << ##m_e##. Intuitively, electron loops are highly suppressed below ##m_e## because there is so little energy available that the electrons must be highly virtual.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Landau pole
  1. Landau levels (Replies: 1)

Loading...