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Recently I read the path integral of double well tunnelling. I am puzzled about the Wick rotation calculation. For example, I choose potential like $V(x)=(x^2-1)^2$ and Lagragian $L= \frac{1}{2} \dot x^2-V(x)$

My questions:

  1. Every book will choose the boundary condition $x(-\infty)=-1$ and $x(+\infty)=1$ and says there is no classical Newtonian trajectory satisfying such boundary condition. Why there is no Newtonian trajectory? Because when your initial velocity is larger than $\sqrt{2}$, then you can climb over the hill and oscillate between the double well. And at some specific time plus the $n\times$ period, you can always go to $x=+1$. Ok, you may say we need a solution that particle almost always stay around at $x=+1$ at $t$ is very large. I admit that we don't have solution satisfying such requirement. So my question is why the tunnelling problem is related to boundary condition $x(-\infty)=-1$ and $x(+\infty)=1$ rather than the oscillating solution?

  2. The most magic thing that puzzles me is the Wick rotation. My current understanding is that when there is no classical trajectory, we cannot use stationary phase approximation. In principle, we need to sum over all paths. However before and after Wick rotation the values are same. When doing the Wick rotation, we can find there is a leading contribution. Ok, no problem. But what should I do when there is a classical trajectory $1$ before the Wick rotation and there will be an another different classical trajectory $2$ after the Wick rotation which all satisfy the same boundary condition? For example, in $2$ dim a potential $V(x,y)$ has some hills and wells. Should I only compute the classical trajectory before Wick rotation or I only compute the classical trajectory after Wick rotation or I sum over these two ?

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marked as duplicate by Qmechanic quantum-mechanics Jul 4 '18 at 8:54

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