We know that Euclidean QFT originates from path integral formalism of $$\langle\phi_f|e^{-\beta\hat{H}}|\phi_x\rangle.\tag{1}$$ We can understand that for $\beta\rightarrow\infty$, we can obtain the ground state via: $$\langle\phi_f|e^{-\beta\hat{H}}|\phi_x\rangle=\sum_n\langle\phi_f|e^{-\beta\hat{H}}|n\rangle\langle n|\phi_x\rangle\stackrel{\beta\rightarrow\infty}{=}e^{-E_0\cdot\infty}\langle\phi_f|n\rangle\langle n|\phi_x\rangle.$$

This is easy to understand. When we take $\beta\rightarrow\infty$, equivalently, we are taking the temperature to zero ($\beta$ is the inverse temperature in the thermal partition functional), therefore, all states are frozen to the ground state. All this stuff is rigorous enough to my taste. But in many cases, people are saying that Euclidean QFT also describes the tunneling in real Minkowski spacetime such as in instanton contex. However, I never saw a rigour proof of such statement. Minkowski QFT is related to Euclidean QFT neither by coordinate transformation nor by analytical continuation:

  1. If we view $t=-i\beta$ as a coordinate transformation, then $t$ and $\beta$ can not be simultaneously real
  2. if we view Euclidean QFT as the analytical continuation of Minkowski QFT, then $\phi(t,\vec{x})$ and $\phi(\beta,\vec{x})$ can not be simultaneously real.

What I think is the rigour way to treat Euclidean QFT is that, we shall view it simply a path integral formalism of Eq.(1) and is derived independently of Minkowski QFT which is derived from $$\langle\phi_f|e^{-iHt}|\phi_i\rangle.$$

Then, how can we justify the statement that the Euclidean QFT describes the tunneling in Minkowski spacetime?

  • $\begingroup$ Instantons is what you are looking for. These localized solutions to the field equations describe tunneling processes in a qft. $\endgroup$
    – Moe
    Dec 28, 2016 at 10:50
  • $\begingroup$ @Moe, I understand instantons. I can accept comfortably that instantons contribute new vacua as described in the post. In particular, I think the introduction of $\theta$ vacua in QCD is quite rigour. But what I can not accept is that people state directly that a Euclidean process represents the tunneling in Minkowski spacetime. $\endgroup$
    – Wein Eld
    Dec 28, 2016 at 10:57
  • $\begingroup$ Well, I'm with you there I don't like this statement either. I'm on mobile atm and therefore to lazy to give a proper answer but I was confused with this for a long time too. There were 2 papers which helped me in getting a better grasp on Instantons. I hope I don't mix them up now but I think it was a paper from Atiyah and Manton on Skyrmions which helped me with the tunneling aspect. $\endgroup$
    – Moe
    Dec 28, 2016 at 11:01
  • $\begingroup$ Related: physics.stackexchange.com/q/127879/50583 $\endgroup$
    – ACuriousMind
    Dec 28, 2016 at 13:25
  • $\begingroup$ Also, why do you say that Euclidean QFT is not related to Minkowski QFT by analytic continuation? That's precisely what the Osterwalder-Schrader reconstruction theorem is about! $\endgroup$
    – ACuriousMind
    Dec 28, 2016 at 13:27

1 Answer 1


The relation between tunneling and classical paths in imaginary time can already be seen in single particle quantum mechanics. In the path integral formulation, each path $x(t)$ contributes with amplitude $e^{iS(x)}$. The action is \begin{equation} S(x)=\int dt\,\left(\frac{m\ddot{x}}{2}-V(x)\right). \end{equation} In the case that is interesting for us, $V(x)$ has a barrier between points $x_0$ and $x_1$, where tunneling is to happen. Suppose that we change variables to an imaginary time $\tau\in i\mathbb{R}$ such that $t=-i\tau$, so that \begin{equation} S(x)=i\int d\tau\,\left(\frac{m x''}{2}+V(x)\right), \end{equation} where primes denote derivatives with respect to $\tau$. Now we are dealing with an action $S_E=iS$ whose potential between $x_0$ and $x_1$ is just a well. The amplitude associated with a path $x$ can be written as $e^{-S_E(x)}$. The minimum for $S_E$ is just the action for the classical path $x_{cl}$, so the dominant contribution to the probabilitity for the tunneling is $e^{-S_E(x_{cl})}$. The conclusion is:

The dominant contribution to the probability of tunneling is given by the classical path for the action $S_E=iS$ in imaginary time $\tau=it$.

Now we can proceed to the quantum field theory case. The reasoning is very similar. The lagrangian can be split in kinetic terms $\mathcal{L}_{kin}$ (quadratic in the first derivatives of the field, for bosons) and the rest $V$. When changing variables to $\tau=it\in i\mathbb{R}$, the action is \begin{equation} S(\phi)=i\int d^4 x \left(\mathcal{L}^E_{kin}(\phi)+V(x)\right) \end{equation}

where the signs in the kinetic terms $\mathcal{L}^E_{kin}$ are now those corresponding to a euclidean metric, so $S_E(\phi)=iS(\phi)$ is known as the euclidean action. Thus, as before: for transitions between field configurations $\phi_0$ and $\phi_1$ at constant time with minimal energy, the leading contribution will be given by the classical field $\phi$ for the euclidean action.

Note 1: I would say that there's no problem with the variable $\tau$ being imaginary. In QFT, for example, we can think of our "euclidean fields" as defined over the space time $(i\mathbb{R})\times\mathbb{R}^3$ which is diffeomorphic to $\mathbb{R}^4$ and on which we have defined the corresponding metric, that is just the euclidean one. So we are working on euclidean $\mathbb{R}^4$.

Note 2: This answer is specifically about the relation between tunneling in Minkowski space and classical paths in euclidean space. It doesn't address the rigorous formulation of Wick rotation, which is given by the Osterwalder-Schrader theorem, as has been stated in the comments.

  • $\begingroup$ In both Minkowski QFT and Euclidean QFT, we are working with fields on $R^4$. And in your answer, you simply gave the statement and not the (rigorous) explanation to which I am really looking forward. $\endgroup$
    – Wein Eld
    Dec 28, 2016 at 19:13
  • $\begingroup$ Of course, in both cases we are working on $\mathbb{R}^4$, did I say we were not? Do you mean that my answer is not an explanation? Or is it that it doesn't have enough rigour? Maybe I can help you in any of those cases. If it is the first, what kind of explanation are you expecting? In the second, I can try to improve the part that you feel lacks rigour $\endgroup$
    – coconut
    Dec 28, 2016 at 19:32
  • $\begingroup$ For instance, you said that `` The minimum for $S_E$ is just the action for the classical path $x_{cl}$, so the dominant contribution to the probabilitity for the tunneling is $e^{−S_E(xcl)}$", How can we prove the latter? $\endgroup$
    – Wein Eld
    Dec 28, 2016 at 19:44
  • $\begingroup$ Well, each contribution is proportional to $e^{-S_E}$, so the one that maximizes it is the dominant one. Maximizing $e^{-S_E}$ is equivalent to minimizing $S_E$, because $e^{-x}$ is a decreasing function. $\endgroup$
    – coconut
    Dec 28, 2016 at 20:45
  • $\begingroup$ Yes, I know minimizing $S_E$ is equivalent to minimizing $e^{-S_E}$. What I can not accept is that $e^{-S_E}$ represents the tunneling probability which you gave as a statement. $\endgroup$
    – Wein Eld
    Dec 28, 2016 at 20:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.