• 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
5,054 questions , 2,207 unanswered
5,345 answers , 22,721 comments
1,470 users with positive rep
818 active unimported users
More ...

  Wick rotation in field theory - rigorous justification?

+ 5 like - 0 dislike

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?

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user Little Brown One
asked Apr 28, 2014 in Theoretical Physics by Little Brown One (25 points) [ no revision ]
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

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user Qmechanic
@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.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user Little Brown One

2 Answers

+ 6 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user V. Moretti
answered Apr 28, 2014 by Valter Moretti (2,085 points) [ no revision ]
+ 4 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-05-04 11:30 (UCT), posted by SE-user user1504
answered Apr 28, 2014 by user1504 (1,110 points) [ no revision ]

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