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?


1 Answer 1


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.

  • $\begingroup$ 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
    Commented Jul 21, 2018 at 17:42
  • $\begingroup$ @NameYYY I added a few remarks about three or more flavors. $\endgroup$
    – Thomas
    Commented Jul 22, 2018 at 1:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.