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I've heard that instantons in QCD generate quark bilinear condensate $\langle \bar{q}_{L}q_{R}\rangle$ which is responsible for spontaneous symmetry breaking. Is there any clear and simple way to explain this?

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Short answer: Instantons explain chiral symmetry breaking qualitatively (and semi-quantitatively) in strongly coupled gauge theories such as QCD, and quantitatively in certain cousins of QCD.

Instantons have fermion zero modes (as a consequence of the index theorem), and 't Hooft explained that the effect of zero modes on correlation functions can be summarized in terms of an effective vertex $$ {\cal L } \sim \det_f (\bar\psi_L^f\psi_R^g) + {\it h.c.} $$ where $f,g$ are flavor indices. In $N_f=1$ QCD this directly generates $\langle\bar\psi_L\psi_R\rangle$, but the overall coefficient cannot be reliably determined.

In theories with $N_f>1$ the 't Hooft vertex directly generates condensates of the form $\langle(\bar\psi_L\psi_R)^{N_f}\rangle$. In some cases, the overall coefficient can be determined, see, for example this work.

The 't Hooft vertex in theories with $N_f>1$ can be understood as generating an effective Nambu-Jona-Lasinio model. Take $N_f=2$. Then $$ {\cal L} \sim G (\bar\psi_L\psi_R)^2 + \ldots $$ which is known to spontaneously break chiral symmetry beyond some critical coupling $G$ (as explained in many papers, this is easily seen using the mean field approximation or Dyson-Schwinger equations). We cannot reliably compute $\langle\bar\psi_L\psi_R\rangle$, beacuse $G$ is sensitive to large instantons (and strong coupling), and the NJL model is non-renormalizable (it requires a cutoff). A more microscopic picture of how instantons break chiral symmetry follows from the Casher-Banks relation, and the existence of fermion zero modes, see here.

For $N_f>2$ the effective 't Hooft vertex is not of the standard (4-fermion) NJL-model form, but in the mean field approximation 6 (and higher) fermion langrangians can also lead to spontaneous symmetry breaking. There are reasons to believe that there is a critical $N_f$ (smaller than the critical $N_f$ where asymptotic freedom disapears) beyond which chiral symmetry is unbroken.

QCD with realistic quark masses is intermediate between $N_f=2$ and $N_f=3$. The effective lagrangian is of the form $$ {\cal L} \sim G_s m_s (\bar u_L u_R)(\bar d_L d_R) + G(\bar\psi_L\psi_R)^3 + \ldots $$ and it is not apriori clear which of the two terms is more important.

There are some QCD-like model, where chiral symmetry breaking and instantons can be studied reliably, see this work.

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  • $\begingroup$ Thank you for the answer! Could you please explain why you take $N_{f} = 2$ for qualitative description of the relation between instantons and quark VEV? Is this related with the fact that $u,d$ flavors are the lightest? And what is changed if we choose $N_{f} = 3$? This will not be Nambu-Jona-Lasinio model, but will anything be changed? $\endgroup$
    – Name YYY
    Jul 21, 2018 at 17:42
  • $\begingroup$ @NameYYY I added a few remarks about three or more flavors. $\endgroup$
    – Thomas
    Jul 22, 2018 at 1:20

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