We consider that if the classical vacuua of a theory are degenerate then each of them can be non-invariant under one or more of the symmetries of the Lagrangian. We can choose one of the vacuua and construct our perturbative QFT around that vacuum. Via choosing a vacuum, we spontaneously break the symmetry and our perturbative QFT constructed around the chosen vacuum would have this (spontaneously) broken symmetry and not the original symmetry of the theory we started out with.

However, in principle, there would be instanton corrections to the classically obtained degenerate vacuua and the degeneracy would break and we would, in fact, have a true vacuum which would have the symmetry of the full theory. So, it seems that if we add the non-perturbative corrections, the spontaneously broken symmetry should be unbroken. However, this seems strange because this seems to suggest that non-perturbative corrections won't just change the results of perturbative calculations by tiny amounts but, rather, would qualitatively change predictions such as if the gauge boson should have mass or not. Or am I mixing non-perturbative and perturbative results in a naive incoherent way? If so, exactly how do we reconcile the two views? Or do we even need to?

  • $\begingroup$ That's why instantons are called non-perturbative corrections. You cannot treat them as small and do perturbation theory. $\endgroup$
    – Kosm
    Commented Jun 13, 2019 at 16:37
  • $\begingroup$ @Kosm So, if I include instantons then my vacuum would be non-degenerate and the symmetry would remain unbroken? What I am confused about is the fact that since instanton corrections always remain smaller than perturbative corrections, I can't see how such small non-perturbative corrections would bring back the massive gauge boson of perturbative QFT to being massless in the non-perturbative theory. $\endgroup$
    – user87745
    Commented Jun 13, 2019 at 16:51
  • $\begingroup$ that depends. Say, if you have axionic continuous shift symmetry and add instanton term for the axion, the continuous symmetry will be broken to a discrete one $\endgroup$
    – Kosm
    Commented Jun 13, 2019 at 17:03
  • $\begingroup$ ... But the vacuum will still break that shift symmetry $\endgroup$
    – Kosm
    Commented Jun 13, 2019 at 17:09
  • $\begingroup$ I see you are talking about gauged symmetry. Then instanton terms ($\sim aF\tilde{F}$, $a$ - axion) are not allowed by gauge symmetry unless it is anomalous. But if it is anomalous, you must add the instanton term to cancel the (cubic) anomaly by Green-Schwarz mechanism. Either way the massive vector boson will remain massive. Also keep in mind that gauge symmetry strictly speaking cannot be spontaneously broken since you can fix the gauge by hand, thus removing the symmetry. $\endgroup$
    – Kosm
    Commented Jun 14, 2019 at 3:59


Your Answer

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