quantum mechanics - How to understand "analytical continuation" in the context of instantons?

Since this is a subtle and interesting question to me. I will give a rather detailed description. I hope you can keep reading it and find it interesting too.

For simplicity, in the following I will only discuss the one-dimensional instanton, that is, quantum mechanics. But the question applies to more general instantons such as BPST instanton in $SU(2)$ Yang-Mills theory. Let us begin with a simple quantum mechanical problem $S_M=\int dt\, L_M=\int dt\,[\frac{1}{2}(\frac{dx}{dt})^2-V(x)]$ with the potential being the double-well, shown as following:

Let us denote the state when the particle is located at the left and right minima as $|L\rangle$ and $|R\rangle$, respectively. Let us consider the Euclidean transition amplitude: $$\lim_{\tau\rightarrow\infty}\langle R|e^{-H\tau}|L\rangle .\tag{1}$$ In the path integral formalism, it is $$\lim_{\tau\rightarrow\infty}\langle R|e^{-H\tau}|L\rangle=\int\mathcal{D}x\,e^{-\int d\tau[\frac{1}{2}(\frac{dx}{d\tau})^2+V(x)]},\tag{2}$$ with all the paths being fixed at left minimum at initial and right minimum at final Euclidean time $\tau$. Before we talk about how to evaluate it. Let us first compare it to the Minkowski path integral $$\lim_{t\rightarrow\infty}\langle R|e^{-iHt}|L\rangle=\int\mathcal{D}x\,e^{i\int d t[\frac{1}{2}(\frac{dx}{d t})^2-V(x)]}.\tag{3}$$ Eq.(2) can be obtained by a formal substitution $t=-i\tau$ in Eq.(3). Note that from the Euclidean action $$S_E=\int d\tau[\frac{1}{2}(\frac{dx}{d\tau})^2+V(x)]\tag{4}$$ Now in the action $S_E$, we can see that the potential is $-V(x)$, i.e., it is upside-down compared to the original in the $S_M$.

The path integral (2) can be evaluated through the method of steepest descent: expanding around the minima of the Euclidean action $S_E$. One of the minimum of Euclidean action gives a solution in the Euclidean spacetime (since we are discussing quantum mechanical situation, the Euclidean spacetime is one-dimensional): $$\frac{\delta S_E}{\delta x}=0,\tag{5}$$ which has the well-known kink solutioin: $$\bar{x}(\tau)=\tanh(\tau-\tau_0),\tag{6}$$ where $\tau_0$ is an arbitrary constant, originating from the $\tau-$translation symmetry of $S_E$. In the path integral we need to integrate $\tau_0$ to sum over all translated paths of expression (6). For simplicity, let us take $\tau_0=0$ to look into the profile of the solution. It is shown as following:

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Note that there is no classical solution in Minkowski spacetime to $$\frac{\delta S_M}{\delta x}=0 \tag{7}$$ with the same initial and final conditions because any path will break the energy conservation law. Now we can actually proceed further upon the method of steepest decent and obtain at the leading order: $$\lim_{\tau\rightarrow\infty}\langle R|e^{-H\tau}|L\rangle\sim e^{-S_E[\bar{x}(\tau)]}.\tag{8}$$

Up to here, everything is fine. But somehow people interpret the above result as the tunnelling rate from the left minimum to the right minimum. I know there are more things in the whole story. For instance, one may recognize the result Eq.(8) as the exponential suppression that one will obtain in a WKB calculation of solving the stationary Schrödinger equation and hence justify the interpretation.

But what I encounter more frequently is that, the reason we interpret Eq.(8) as the tunnelling rate because we can continue it back to the Minkowski result. Perhaps the logic is the following (Note, I infer this logic from sentences and there may be something deeper they mean. This is what I want to discuss in this post). Since the instanton action is ocassionally independent of $\tau$, when we continue back, formally substitute $\tau=it$, we still have such an exponential suppression.

The problem is that, as explained by Qmechanic's answer, we can never stop at the stationary point, which is only one piece of the complete Euclidean path integral, do the calculation and say, look, let us now continue the result back into the Minkowski spacetime. Actually, the stationary point itself will change simutaneously when we rotate the time. When we rotate back into the Minkowski spacetime, we do not have any stationary point at all if we still believe the positions are real numbers.

In particular, in the context of the instantons in gauge theory (with interactions with fermions). One may calculate the following transition amplitude in a gauge field instanton background $$\langle 0|0\rangle_{1-inst}$$ Due to fermioinc zero modes, one obtain a vanishing result. In order to obtain a non-vanishing result, we must insert the t' Hooft operators like $\bar\psi_R\psi_L$, that is $$\langle 0|\bar\psi_R\psi_L|0\rangle.$$

At this point, now it is really very frequent to see people say, this represents the chiral violation in real Minkowski spacetime through analytical continuation. But how? How can we continue a result near a Euclidean stationary point back into the Minkowski spacetime? When we do the continuation, should we simutaneousely continue the stationary point itself, i.e., continue the Euclidean instanton to the Minkowski spacetime?

In conclusion, my question is, how exactly understand the hand-wave word "analytical continuation" in these specific cases? I believe this has an answers in the Picard–Lefschetz theory.

The following is the original question which now is of relatively less relevance.

My question is about the tunneling interpretation of the kink solution and the Euclidean transition amplitude. People always say that a kink solution describes that a tunneling process happens from the left minimum at far past to the right minimum at far future. This picture to me is a bit vague. The questions is

(1) Is $\lim_{t\rightarrow\infty}\langle R|e^{-iHt}|L\rangle=\lim_{\tau\rightarrow\infty}\langle R|e^{-H\tau}|L\rangle$? Well, it seems that this is true usually by rough argument of Wick rotation (or I was mistaken). But even it is true, It is not justified at all to have $$\lim_{\tau\rightarrow\infty}\langle R|e^{-H\tau}|L\rangle_{\rm inst}={\rm leading\ order\ of} \lim_{t\rightarrow\infty}\langle R|e^{-iHt}|L\rangle$$

(2) Though the Minkowski action does not contain a classical solution. There should be quantum paths that can break the energy conservation law due to the uncertainty principle. What is the (dominant) quantum paths in the tunneling process. Our first guess may be $\bar{x}(t)=\tanh(it)$ with formal substitution $\tau=it$ back to the kink solution $\bar{x}(\tau)=\tanh(\tau)$. But $\tanh(it)$ is imaginary therefore unphysical as the position $x$. Is there any interpretation of the kink solution in Minkowski spacetime?

Answer

TL;DR: OP's title question (v7) about instantons in the Minkowski signature is physically meaningless. It is an irrelevant mathematical detour run amok. The connection to physics/Nature is established via a Wick rotation of the full Euclidean path integral, not bits and pieces thereof. Within the Euclidean path integral, it is possible to consistently expand over Euclidean instantons, but it meaningless to Wick rotation the instanton picture to Minkowski signature.

In more details, let there be given a double-well potential

$$V(x)~=~\frac{1}{2}(x^2-a^2)^2. \tag{A}$$

The Minkowskian and Euclidean formulations are connected via a Wick rotation

$$ t^E e^{i\epsilon}~=~e^{i\frac{\pi}{2}} t^M e^{-i\epsilon}.\tag{B} $$

We have included Feynman's $i\epsilon$-prescription in order to help convergence and avoid branch cuts and singularities. See also this related Phys.SE post.

I) On one hand, the Euclidean partition function/path integral is

$$ Z^E~=~Z(\Delta t^E e^{i\epsilon})~=~ \langle x_f | \exp\left[-\frac{H \Delta t^E e^{i\epsilon}}{\hbar}\right] | x_i \rangle ~=~N \int [dx] \exp\left[-\frac{S^E[x]}{\hbar} \right],\tag{C} $$

with Euclidean action

$$ S^E[x]~=~\int_{t^E_i}^{t^E_f} \! dt^E \left[ \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^E}\right)^2+e^{i\epsilon}V(x)\right] $$ $$~=~ \int_{t^E_i}^{t^E_f} \! dt^E \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^E}\mp e^{i\epsilon}\sqrt{2V(x)}\right)^2 \pm \int_{x_i}^{x_f} \! dx ~\sqrt{2V(x)}\tag{D}.$$

and real regular kink/anti-kink solution$^1$

$$ \frac{dx}{dt^E}\mp e^{i\epsilon}\sqrt{2V(x)}~\approx~0 \qquad\Leftrightarrow \qquad x(t^E)~\approx~\pm a\tanh(e^{i\epsilon}\Delta t^E). \tag{E}$$

Note that a priori space $x$ and time $t^E$ are real coordinates in the path integral (C). To evaluate the Euclidean path integral (C) via the method of steepest descent, we need not complexify space nor time. We are already integrating in the direction of steepest descent!

II) On the other hand, the corresponding Minkowskian partition function/path integral is $$ Z^M~=~Z(i \Delta t^M e^{-i\epsilon}) ~=~ \langle x_f | \exp\left[-\frac{iH \Delta t^M e^{-i\epsilon}}{\hbar} \right] | x_i \rangle ~=~N \int [dx] \exp\left[\frac{iS^M[x]}{\hbar} \right],\tag{F} $$

with Minkowskian action

$$ S^M[x]~=~\int_{t^M_i}^{t^M_f} \! dt^M \left[ \frac{e^{i\epsilon}}{2} \left(\frac{dx}{dt^M}\right)^2-e^{-i\epsilon}V(x)\right]$$ $$~=~ \int_{t^M_i}^{t^M_f} \! dt^M \frac{e^{-i\epsilon}}{2} \left(\frac{dx}{dt^M}\mp i e^{i\epsilon}\sqrt{2V(x)}\right)^2 \pm i \int_{x_i}^{x_f} \! dx ~\sqrt{2V(x)},\tag{G}$$

and imaginary singular kink/anti-kink solution

$$ \frac{dx}{dt^M}\mp i e^{i\epsilon}\sqrt{2V(x)}~\approx~0 \qquad\Leftrightarrow \qquad x(t^M)~\approx~\pm i a\tan(e^{-i\epsilon}\Delta t^M)~=~\pm a\tanh(i e^{-i\epsilon}\Delta t^M). \tag{H}$$

It is reassuring that the $i\epsilon$ regularization ensures that the particle starts and ends at the potential minima:

$$ \lim_{\Delta t^M\to \pm^{\prime}\infty} x(t^M)~=~(\pm a) (\pm^{\prime} 1). \tag{I}$$

Unfortunately, that seems to be just about the only nice thing about the solution (H). Note that a priori space $x$ and time $t^M$ are real coordinates in the path integral (F). We cannot directly apply the method of steepest descent to evaluate the Minkowski path integral. We need to deform the integration contour and/or complexify time and space in a consistent way. This is governed by Picard-Lefschetz theory & the Lefschetz thimble. In particular, the role of the imaginary singular kink/anti-kink solution (H) looses its importance, because we cannot expand perturbatively around it in any meaningful way.

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$^1$ The explicit (hyperbolic) tangent solution (E) is an over-simplified toy solution. It obscures the dependence of finite initial (and final) time $t^E_i$ (and $t^E_f$), moduli parameters, and multi-instantons. We refer to the literature for details.