• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,728 comments
1,470 users with positive rep
818 active unimported users
More ...

  What is the corresponding history in Minkowski spacetime of an instanton?

+ 4 like - 0 dislike

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
$$S_E=\int d\tau[\frac{1}{2}(\frac{dx}{d\tau})^2+V(x)]\tag{4}$$
we can see that the potential is upside-down now $-V(x)$.

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 know kink solutioin:
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:

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:
$$\lim_{\tau\rightarrow\infty}\langle R|e^{-H\tau}|L\rangle\sim e^{-S_E[\bar{x}(\tau)]}.\tag{8}$$

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 a rigorous proof is rather missed in most references. I would appreciate it very much if somebody can give a rigorous proof.

(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?

Of course, if the answer to (1) is yes, then the interpretation of the Euclidean transition amplitude as a tunneling is right even though there is no corresponding interpretation of the kink solution in Minkowski spacetime. But the profile of the kink is so clear such that one can not stop recognizing it as a true (quantum) process in Minkowski spacetime, that is
$$\bar{x}(\tau)=\tanh t,$$
where the substitution is $\tau=t$ rather $\tau=i t$ in the field configuration (viewed as one-dimensional field theory). If this is true, then there is something need to be understood for the Wick rotation.


asked Apr 4, 2017 in Theoretical Physics by Wein Eld (195 points) [ revision history ]
edited Apr 4, 2017 by Wein Eld

Wick rotation is dangerous - it causes vertigo.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights