Why do we have a non-zero quark vacuum condensate even though the QCD coupling goes to zero in the deep infrared?

It is well-known that QCD has a Landau pole at $\Lambda_{\rm QCD}\sim 200$ MeV, which means that the perturbative QCD coupling becomes strong at this scale. Conventionally, this is claimed to be the reason why quarks condense at this scale and why we obtain a quark vacuum condensate with such an energy density.

However, for energies below $\Lambda_{\rm QCD}$, QCD again becomes weakly coupled, as it was shown, for example, in beyond-perturbation-theory lattice computations. That puzzles me: why are quarks with energies below $\Lambda_{\rm QCD}$ still condensed, even though their coupling goes to zero for zero momentum?

And a more conceptual question: how can we at all talk about a vacuum condensate, if we say that this condensate is connected to some non-zero energy scale?

Edit: Thanks for your answer, Cosmas Zachos. I understand now that chiral symmetry breaking occurs due to non-perturbative effects, so that a priori the condensation and hadronization are unrelated to the perturbative coupling. However, does this mean than non-perturbative effects have to be strong for all energies below $\Lambda_{\rm QCD}$?

And I still don't have a physical intuition why the quarks condense and hadronize at 200 MeV and not at lower (zero) energies, if the condensate apparently is a vacuum effect. I know that this is "the point of SSB", but I still lack the deeper physical understanding of this scale of the vacuum condensate.

• At energies below Λ, so distances longer than a fermi, you are outside the hadron's radius and α_s is not involved directly anymore. Whether nuclear physicists choose to perturb in the large pion-nucleon constant or not, it is certainly a risky operation. – Cosmas Zachos Oct 18 '16 at 14:53

The leading "fact" in the first paragraph is wrong: Scales of chiral symmetry breaking demonstrates, nonperturbatively, that chiral condensation occurs near but not at the confinement scale, and can be probed by different color representations of fermions, resulting in a variety of scales.

So you may think of the hadron as, 1), a central region of light current quarks and gluons, with small, perturbative, coupling; 2) a shell of chiral condensation, strong coupling; 2') past which an effective theory obtains, of constituent quarks one hundred times heavier than their current ancestors, by interacting through pions (χSB goldstons); finally, 3), only then, further out yet the ultra-stong coupling region confinement radius beyond which color is not manifest directly.

Outside that radius you will not see gluons but only colorless hadrons, also strongly interacting, but not through a weaker αs ; you are almost certainly misreading the paper you cite.

There are theorems, like the Vafa-Witten theorem compelling χSB through confinement, and there has been ample speculation of one without the other, but the basic picture above is the mainstream view, rarely controverted.

The QCD chiral condensate is a phase property of the QCD vacuum, that dictates nonvanishing v.e.v.s for chirally non-invariant quark bilinears,
$$\langle \bar{q}^a_R q^b_L \rangle = v \delta^{ab} ~,$$ formed through nonperturbative action of "low energy" (sub-GeV) QCD gluons, with v ≈ −(250 MeV)3.

This is the scale of χSB, and if it were vanishing, as you insinuate, the SBB phenomenon under discussion would simply not occur.

There is another type of QCD vacuum condensation, Gluon condensation omitted here to spare confusion, as it is not obviously related to χSB.

Edit on exhibiting the χSB factor of 100 mentioned: In MeVs, the current up quark of mass 2-8 bloats up to the constituent up quark of mass 336. The down quark, 5-15 ⟶ 340.

Edit in response to comment: The magic of dynamical mass generation out of infrared nonlinearities in a given medium is a prerequisite for the Goldstone theorem, as indicated above: without it, we simply wouldn't have SSB. The modality and precise scale of this outcome, if a theory supports it, as QCD evidently does, is inevitably technical, and no simplistic narrative is available to anyone's satisfaction. One simply has to do the math. Nambu, of course, got the 2008 Nobel prize precisely for illustrating it cogently with a simple fermion model half a century ago (Nambu & Jona-Lasinio (1961) PhysRev 122 345). The only "poetic" answer to your question, beyond cold math, is that the QCD vacuum is supercharged with energy available for the condensation, and that scale is generated in connection to the strong coupling involved, so at scales of the order of magnitude of, but not equal to, Λ.

• This was exactly my conceptual question: why does the condensation already take place at 200 MeV and not at lower (zero) energies, if it is a vacuum effect? So why can a quark bilinear have a certain energy density even though it has zero momentum? And do I understand you correctly that we assume chiral symmetry breaking to occur due to non-perturbative effects, so that a priori the condensation and hadronization is unrelated to the perturbative coupling? If that's the case, is the strong coupling for energies below MeV a perturbative or a non-perturbative coupling? – Thomas Oct 17 '16 at 1:54
• That's the point of all SSB phenomena: they are a property of the medium vacuum, like EW breaking, the vacuum emits and absorbs massless goldstons freely, but it generates a scale--if it did not, you wouldn't have SSB!! Correct, the perturbative coupling is a canard in such phenomena. Sub MeV implies furiously nonperturbative, as per the links provided. – Cosmas Zachos Oct 17 '16 at 2:27