• 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.


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

202 submissions , 160 unreviewed
4,981 questions , 2,140 unanswered
5,339 answers , 22,624 comments
1,470 users with positive rep
813 active unimported users
More ...

  Performing Wick Rotation to get Euclidean action of scalar field

+ 2 like - 0 dislike

I'm working with the signature $(+,-,-,-)$ and with a Minkowski space-stime Lagrangian

\mathcal{L}_M = \Psi^\dagger\left(i\partial_0 + \frac{\nabla^2}{2m}\right)\Psi
The Minkowski action is
S_M = \int dt d^3x \mathcal{L}_M
I should obtain the Euclidean action by Wick rotation.

My question is about the way with that I should perform the Wick rotation. 

Since the spacetime interval is defined by $ds^2 = dt^2 - d\vec{x}^2$, If I perform a Wick rotation (just rotating the time axis) I get a negative Euclidean interval. 

1. What's the sense of that? What's the connection between physical actions calculated in two different signature?

2. I can perform the rotation with different signs $t =\pm i\tau$. I know that, if there exist any poles, I must choose the correct sign in order to not cross them. But in this case, apparently I can choose both, and I get

If I choose $ t = i\tau $ I get

\[i\int_{+i\infty}^{-i\infty} d\tau d^3x \Psi^\dagger\left(i\frac{\partial}{\partial i\tau} + \frac{\nabla^2}{2m}\right)\Psi =\\ -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger(x,i\tau)\left(\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi(x,i\tau) \]

That's different from the standard euclidean action which is with a minus between $\partial_t$ and $\nabla^2$.

asked Jan 24, 2016 in Theoretical Physics by apt45 (50 points) [ revision history ]
edited Jan 25, 2016 by apt45

Your action is not Minkowski - the time derivative has first order only but should have second order!

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).
To avoid this verification in future, please log in or register.

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights