The question arises when I'm reading over the section "3.3.1 Minkowski Space" in page 16-17 in the following link: https://www-thphys.physics.ox.ac.uk/people/JohnCardy/qft/qftcomplete.pdf

It is discussing the technique of using Wick Rotation to calculate the generating function in Minkowski space.

It mentioned that simply inserting τ=it into the results of the generating function in Euclidean space (i.e. imaginary time) provides the generating function in Minkowski space.

However, on top of page 17, it mentioned that I also have to let p0 --> ip0 as well. Why do I have to do this as well? How is that related to defining τ=it and making a Wick Rotation?


Cardy discusses how to pass from euclidean space to Minkowski space.

The Wick rotation can be thought of as a coordinate transformation $x' \rightarrow x$, where the $x' \equiv (\tau, \vec{x}')$ are the euclidean ones and the $x \equiv (t,\vec{x})$ are the ones for Minkowski space (see below for a caveat). As stated in the question $\tau = \mathrm{i} t$.

A covector transforms according to

$$ \omega_\mu = \frac{ \partial x'^\nu }{ \partial x^\mu } \omega'_\nu. $$ Using this transformation law for the vector $p_\mu$ we get for $p_0$ $$ p_0 = \frac{ \partial x'^\mu }{ \partial x^0 } p'_\mu = \frac{ \partial x'^0 }{ \partial x^0 } p'_0 = \frac{ \partial \tau }{ \partial t } p'_0 = \mathrm{i} p'_0 . $$

Now for the caveat. Viewing the Wick transformation as a coordinate transormation gives the following metric

$$ g_{\mu\nu} = \frac{ \partial x'^\alpha }{ \partial x^\mu } \frac{ \partial x'^\beta }{ \partial x^\nu } g'_{\alpha\beta} = \mathrm{diag}(-1,1,1,1)_{\mu\nu} $$ which in Cardy's conventions is the negative of the Minkowski metric.

  • $\begingroup$ Interesting... I'm wondering why is p a covariant vector? Before the transformation p is just defined (in the middle of page 14 of the document in the link) to be the Fourier Transformation pair of x, but Cardy never mentioned that it is a covariant vector. $\endgroup$ – user148792 Aug 30 '18 at 18:28
  • $\begingroup$ @TaylorTiger $p$ is not naturlly a covector. However, the component of the corresponding covector is what is needed when computing the inner product $x\cdot p = x^\mu p_\mu \equiv x^0 p_0 + \vec{x} \cdot \vec{p}$. The thing is, that when you are in euclidean space the components of the covectors and vectors are the same, so index position does not matter. However, after the Wick rotation you are in Minkowski space and there the position of the index plays an important role. $\endgroup$ – yalda Aug 31 '18 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy