It is well-known that QED and $\Phi_4^4$ quantum field theory have (in renormalized perturbation theory) a Landau pole and therefore are not asymptotically free. Is this specific to 4-dimensional QFT, or are there examples of theories with Landau poles in 2 and 3 dimensions?

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    $\begingroup$ The "therefore" in your first sentence should go the other-way. $\endgroup$ – Ron Maimon Apr 28 '12 at 1:30
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    $\begingroup$ I think it goes both ways. $\endgroup$ – Arnold Neumaier Apr 28 '12 at 7:53

One can make up a nonlocal theory to have a Landau pole in 3d, by using Levy-fields (fields with a wrong power-law propagator).

With a single scalar field $\phi$, consider the Euclidean action:

$$ S = \int |q|^{\alpha}|\phi(q)|^2 d^3q+ \int t\phi^2 + \lambda|\phi|^4 d^3x$$

This theory has a perturbation theory exactly like $\phi^4$ theory, with a propagator which has a different divergence structure:

$$ \langle\phi(q)\phi(q')\rangle_F = {1\over |q|^\alpha}$$

to make the divergence of the vertex renormalization logarithmic, you need the vertex renormalization integral to be

$$ \int {1\over |q|^{2\alpha}} $$

So that $\alpha=1.5$ is the log-diverent choice. This choice produces a log-running renormalizable theory in 3d. The renormalization running has the same exact form to one loop order as the usual $\phi^4$ theory, so it should have a Landau pole (otherwise, it could be asymptotically safe, but it is similar enough to $\phi^4_4$ that no way).

The lattice version of this theory provides a non-perturbative statistical model to define the long-distance theory, and it is an Ising model with a nonlocal coupling $J(x-y)$ which falls off as the appropriate powerlaw.

The scalar field correlations from a nonlocal J coupling are determined from the Fourier transform of J, so that to get the power-law ${1\over |q|^{1.5}}$ you want the correlation function:

$$ G(x-y) \approx {1\over|x-y|^{1.5}} $$

This comes from a J coupling which is the Fourier transform of ${|q|^{1.5}}$, which integral $\int |q|^{1.5} e^{iq\cdot x} d^3q$ falls off as ${1\over |x|^{4.5}}$ on dimensional grounds (with an unimportant compensating $\delta$ function at the origin to zero out the integral--- the Fourier transform vanishes at the origin). So the J function for the Ising-like model should fall off with a 4.5 exponent to produce the just-balanced correlation function at long distances.

This falloff in the coupling arranges this Ising model to be marginally mean-field in 3d, so that it runs to a weak-coupling long-distance limit. You can simulate this Ising model on a computer, and the long-distance fluctuations should be described by the nonlocal scalar theory above.

This means (with physicist standards of argument, this won't persuade a mathematician) that you get the Landau pole in the long-distance theory. The reason is that the Ising model is an infinite $\lambda$ starting point, and if it renormalizes to a weak coupling free field theory at long distances, the long-distance theory should reverse-renormalize to an infinite coupling $\lambda$ at distances corresponding to the lattice-spacing of the Ising model. This is heuristic of course, a rigorous version of renormalization is sorely lacking.

In 3d, if you don't do these nonlocal tricks, you don't usually get Landau poles, because the usual couplings are dimensional and have power-law running. This means that they get "stronger" at short distances as a particular power, which is not really getting stronger, because the whole theory acquires a different scale-invariant scaling from the free-field theory and you can't do small-coupling perturbation theory directly in 3d without tricks. The non-perturbative semi-strong-coupling fixed point is the Wilson Fisher theory, which can be described by $\epsilon$ expansion, which expands near 4 dimensions

The dimensional expansions of 1970s physics are best considered as streamlining an expansion in the power-law parameter $\alpha$ introduced in the nonlocal Lagrangian above, this is the point of view of "analytic renormalization" which is the neglected forerunner of both dimensional regularization and epsilon expansion. The guy who did analytic renormalization, Eugene Speer, is never properly credited, although he is largely responsible for this conceptual epiphany.

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    $\begingroup$ E.R. Speer, Analytic Renormalization, scitation.aip.org/content/aip/journal/jmp/9/9/10.1063/1.1664729 $\endgroup$ – Arnold Neumaier Apr 6 '15 at 13:51
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    $\begingroup$ E.R. Speer, Analytic renormalization using many space-time dimensions, projecteuclid.org/euclid.cmp/1103859813 $\endgroup$ – Arnold Neumaier Apr 6 '15 at 13:52
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    $\begingroup$ Nice answer! I think the correct power for $J$ is 4.5 instead of 1.5. One usually needs $J$ to be $L^1$ at infinity otherwise the spin system is rather sick. The good thing with $\alpha=1.5$ is that it is less than 2 so you should still get a unitary QFT. $\endgroup$ – Abdelmalek Abdesselam Apr 6 '15 at 21:36
  • $\begingroup$ @AbdelmalekAbdesselam: Whoops! The correlation function goes as $1/|x-y|^{1.5}$, not the J! The J powerlaw is fixed by demanding that the equation of motion gives this correlation function as a solution, I'll fix it now. For the $\alpha$, the range of allowed $\alpha$ which produce unitary field theories is precisely the ones for which the Schwinger representation is a sum over Levy flights with a sensible probability exponent, which is why I like to call these "Levy field theories". Generalizing traditional particle Brownian paths to Levy Flights was my path to these, not Speer. $\endgroup$ – Ron Maimon Apr 7 '15 at 4:19
  • $\begingroup$ @AbdelmalekAbdesselam: yes, you are right, it's 4.5 not 1.5, of course, I am sorry for the lapse. $\endgroup$ – Ron Maimon Apr 7 '15 at 4:28

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