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What is the rigorous justification of Wick rotation in QFT? I'm aware that it is very useful when calculating loop integrals and one can very easily justify it there. However, I haven't seen a convincing proof that it can be done at the level of path integral.

How do we know for sure that Minkowski action and Euclidean action lead to the equivalent physical result? Is there an example where they differ by e.g. a contribution from a pole not taken into account while performing Wick rotation?

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Comment to the question (v2): What do you mean by rigorous? Cf. physics.stackexchange.com/q/6530/2451 , physics.stackexchange.com/q/27665/2451 and links therein. Also note that Wick rotation of spinor fields is non-trivial, cf. e.g. physics.stackexchange.com/q/21261/2451 –  Qmechanic Apr 28 at 15:22
    
@Qmechanic, by rigour I mean that there are some conditions that need to be fulfilled in order to perform the Wick rotation. I would like to see some analysis which establishes that it is ok to perform Wick rotation on path integrals, instead of just presuming it's ok to do it. What if the measure of the path integral has a pole somewhere? That could invalidate the naive Wick rotation. –  Little Brown One Apr 28 at 17:18

2 Answers 2

The path integral, mathematically speaking, does not exist as an integral: It is not associated with any positive or complex measure. Conversely, the Euclidean path integral does exist. The Wick rotation is a way to "construct" the Feynman integral as a limit case of the well-defined Euclidean one. If, instead, you are interested in an axiomatic approach connecting the Lorentzian n-point functions (verifying Wightman axioms) with corresponding Euclidean n-point functions (and vice versa), there is a well-known theory based on the so called Osterwalder-Schrader reconstruction theorem rigorously discussing the "Wick rotation" in a generalized fashion.

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1st comment:

It's worth thinking for a second about where Wick rotation comes from. You can do this in the context of the quantum mechanics of a free particle. In QFT, all of the details are more complicated, but the basic idea is the same.

In free particle, QM, we get the path integral by inserting sums over intermediate states at various times. The need for Wick rotation arises as soon as you do this just once.

$\langle q' | e^{- \frac{ iP^2 t}{2m\hbar}}|q\rangle = \int_{-\infty}^\infty \langle q'| p \rangle \langle p |e^{- \frac{ iP^2 t}{2m\hbar}}|q\rangle dp = \frac{1}{2\pi\hbar} \int_{-\infty}^\infty e^{\frac{-i t }{2m\hbar} p^2 + i\frac{q' - q}{\hbar} p} dp$.

This is an oscillatory integral. The integrand has norm 1 because the argument of the exponential is purely imaginary. Such integrals don't converge absolutely, so the right hand side of this equation is not a well-defined expression as it stands. To make it well defined we need to supply some additional information.

Wick rotation provides a way of doing this. You observe that the left hand side is analytic in $t$, and that the right hand side is well-defined if $Im(t) < 0$. Then you can define the integral for real $t$ by saying that it's analytic continued from complex $t$ with negative imaginary part.

2nd comment:

As V. Moretti pointed out, in QFT, it's in some sense backwards to think of analytically continuing from Minkowski signature to Euclidean signature. Rather, one finds something in Euclidean signature which has nice properties and then analytically continues from Euclidean to Minkowski. However, one can often begin this process by taking a Minkowski action and finding its Euclidean version, and then trying to build up a QFT from there. There's no guarantee that this will work though. Spinor fields may have reality conditions that depend on the signature of spacetime. Or the Euclidean action you derive may be badly behaved. This is famously the case for Einstein's gravity; the Euclidean action is not bounded below, so one does not get a sensible Euclidean theory.

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