In Modern Supersymmetry, John Terning, on pgs 151, and 153 performs a so called 'reliable' instanton calculation when dealing with the ADS superpotential 'since the gauge group is completely broken'. Alternatively, see his slides #16, and 19 here.

My question has 2 parts, depending on the answer to the first part:

(1) What does he mean by 'reliable'? Is it a question of rigor?

Depending on the answer to (1):

(2) What is an example of an 'unreliable' instanton calculation? Is it back-of-the-envelope?


2 Answers 2


Instanton calculations involve integrals over collective coordinates. One of these is the instanton size $\rho$. Reliable instanton calculations are those for which the integral over instanton sizes is dominated by small sizes so that (for asymptotically free theories) the coupling constant is small and higher order corrections in the semi-classical expansion are small. This can be achieved in Higgs phases (large instantons are suppressed as $\exp(-\rho^2 v^2)$, where $v$ is the Higgs vev), at high temperature, or at high baryon density.


I don't know what Terning has in mind, but one problem does spring to mind: If the gauge symmetry isn't completely broken, we might be dealing with a nonabelian gauge theory in a confining phase. When the coupling is strong, the instanton approximation (in which we do perturbative corrections around each instanton, then sum over instantons) isn't trustworthy.

  • $\begingroup$ is this then just the same as the usual problem of strong coupling in no-abelian gauge theories? $\endgroup$
    – DJBunk
    Commented Jan 6, 2013 at 1:45
  • $\begingroup$ Yes, that's what I was speculating. But, judging by Thomas's answer, I was looking in the right direction, but missing a key part of the problem. $\endgroup$
    – user1504
    Commented Jan 6, 2013 at 2:22

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