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Let the action be $$S= \int \bigg\{ \frac{1}{2} \big(\frac{dX}{dt}\big)^2 - V(X) \bigg\} d\tau$$

and the corresponding Path-Integral

$$Z= \int DX(t) e^{iS}.$$

Since the convergence is not clear we Euclideanize the time coordinate $t$ by the Wick rotation

$$ t \rightarrow -i \tau$$

and get the Path-Integral

$$Z_E=\int DX(\tau) e^{-S_E},$$

with $$S_E= \int \bigg\{ \frac{1}{2} \big(\frac{dX}{d\tau}\big)^2 + V(X) \bigg\} d\tau.$$

And now my question - the Euclideanized path-Integral allegedly has a better convergence property, but i do not quite see why this is the case?

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2 Answers 2

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In a nutshell, the Euclidean Boltzmann factor is exponentially suppressed because the Euclidean action is bounded from below (assuming the potential $V$ is bounded from below), while the Minkowskian Boltzmann factor has modulus 1 and is oscillatory.

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  • $\begingroup$ I know this is another question.. but.. you're solving another integral. I'm trying to find a good article or description on what observables can or can't be calculated after a wick rotation and solving. Can I setup and solve (in theory) for observables in an experiment involving destructive interference (which clearly depends on the cancellation of factors in the normal e^iS integral)? If you get what I'm hinting at.. $\endgroup$
    – BjornW
    Jun 15, 2020 at 14:55
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You just have to look at the exponential. So assuming your Lagrangian is bounded from below when written in its euclidean version, you can see that the Wick rotation is taking an imaginary exponential which is oscillatory and turning it into a decaying exponential which can be approximated more easily, for example summing over its extremal points.

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