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Instantons are classical solutions of the Euclidean (i.e. imaginary time) classical equations of motion.

The standard example from single particle 1D QM is taking a potential of the form $V(x)=(x^2-a^2)^2$ and looking for solutions $\gamma$ for $$ \ddot{\gamma} = +V'(\gamma) = 4 (\gamma^2-a^2)\gamma \tag{CoM}$$ (note the plus sign on the potential--this is Euclidean time).

An approximate solution for this is for example given by $\gamma(t) = a \tanh(\pm\sqrt{2}a(t-t_0))$ for any $t_0\in\mathbb{R}$ (there's a family of solutions indexed by $t_0$).

Say we are looking for solutions which obey the boundary conditions $\gamma(\pm T) = \pm a$ over a time period $[-T,T]$ for some $T$. Then the solution with the $+$ variant, the so-called single instanton solution, will obey the boundary conditions only in the limit $T\to\infty$. Be that as it may, we press on and realize that we can also concatenate solutions with alternating $\pm$ variatns in order to still obey the boundary conditions, so we can get instanton-anti-instanton-instanton string which is called the 3-instanton solution. This is, again, only an approximate solution of CoM.

One verifies that if the action of one instanton is $S_0$ then the action of a concatenation of $n$ instantons is $n S_0$.

My question: One could just as well characterize solutions of CoM by saying they minimize the action. When stated like that, the multi-instanton solutions are not minimizers of the action since they are actually integer multiples of it. So how can they be called approximate classical solutions, as opposed to the single instanton solutions, which are approximate only in the sense that they obey the BC only in the $T\to\infty$ limit?

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There are two issues here. The first is about boundary conditions in the single instanton sector. This is easily remedied, there is an exact solution of the equations of motion that satisfies the boundary condition for finite $T$, see here.

The second issue is more complicated. The question is whether the two-instanton (instanton-anti-instanton), or three instanton (two-instanton-anti-instanton) etc, configurations are saddle points of the path integral. Indeed, they cannot be ordinary saddle points, because the instanton-anti-instanton configuration lives in the same topological sector as the perturbative vacuum, so the configuration is smoothly connected to the vacuum solution. These configurations exist as saddle points of the complexified path integral. The corresponding saddle point manifolds are known as Lefshetz thimbles, see, for example, here.

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  • $\begingroup$ Thanks for your answer. So just to double-check: my main misconception was that I conflated solutions of CoM as minimizers of the action, whereas they are extremum points of the action, which could just as well be saddle points, and the multi-instanton solutions are precisely saddle points? Furthermore, you seem to indicate that they are not "ordinary saddle points", but can you elaborate why true saddle points (how are they defined? as zeros of S'?) cannot be in the same topological sector of the vacuum? $\endgroup$ – PPR Dec 6 '19 at 15:54
  • $\begingroup$ @PPR Sorry for increasing confusion. The term "saddle" is just a generic term, because we are trying to find the path integral analog of the saddle point approximation to ordinary integrals. For a purely real integral of the form $e^{-S}$ we would seem to need true minima of the action (we can have exactly flat directions, leading to collective coordinate integrals). The instanton is indeed a true minimum in QM (and has exactly flat directions in YM theory). $\endgroup$ – Thomas Dec 6 '19 at 16:46
  • $\begingroup$ However, the instanton-anti-instanton configuration is certainly not a minimum (indeed, not even an ordinary saddle point), because at finite sepration it does not solve the equations of motion. It has to be interpreted in terms of Picard-Lefshetz theory, as explained in the linked paper. $\endgroup$ – Thomas Dec 6 '19 at 16:48
  • $\begingroup$ thanks for clarifying. I am indeed interested in the simplest, QM 1D double well example which in Euclidean time is a real path integral. However, let me insist: if we _do take the limit $T\to\infty$, then the instanton-anti-instanton-instanton string is still not a minimum, right? Because its action equals $3 S_0$. So why are we looking at these strings when performing the dilute instanton gas approximation? $\endgroup$ – PPR Dec 6 '19 at 17:19
  • $\begingroup$ @PPR The IAI has an action $\sim 3S_0$, but it is continuously connected to the I configuration, so the minimum action in this sector is $S_0$. If the IAI are well separated, then the dependence of the action on IA separation is exponentially small, so people have the intuition that it should be possible to treat this as a flat direction, which gives the standard instanton gas formula. This intuition is ultimately correct, but it takes some work to justify it more rigorously, which is what the linked paper attempts to do. $\endgroup$ – Thomas Dec 6 '19 at 19:48

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