In CFT 2D, we often like to wick rotate the minkowski theory, to end up with an euclidean theory. Often, we then use complex coordinates to parametrize the plane, to make things easier.

For instance, the accelerated trajectory, $x^2-t^2=1$ is mapped to the circle through the Wick rotation, $t\rightarrow i \tau$, where we have $x^2+\tau^2=1$. If we consider $z=x+i\tau$ as a complex variable, we can write it as $z\bar{z}=1$. In this coordinate system, we can for instance easily see that by using a well-chosen Mobius transformation $z\rightarrow \frac{az+b}{cz+d}$ we can map the circle to the real line. This, in real space, corresponds to a conformal transformation that maps the accelerated trajectory to a static one.

However, let us know consider a slightly different trajectory/locus. Consider the equation

$$ (x+t)^3(x-t)=1 $$

This curves looks very similar to the accelerating one in real space. However, after the mapping it goes to :

$$ z^3 \bar{z}=1 $$

Solving this equation yields only two points: $-1$ and $1$.

This seems very strange, as the topology itself changes from a curve to a set of points. This is not unheard of, as it happens also in other cases (c.f. wick rotating the black hole spacetime). What I find surprising is that here, I could certainly find a change of coordinate in real space that maps my curve to the accelerated one.

But from the point of view of the complex coordinate, there is no way one could find such a transformation, since we would need to map two points to a whole curve.

It is clear here that the wick rotation "trick" breaks down. But why is that so? And how could I predict when I am allowed to use such tools? Usually, people gloss over this trick, and so I'm having trouble understanding where things went wrong.


1 Answer 1


There is probably a much more nuanced explanation with regards to what goes wrong with Wick rotations in such cases, but to me the issue from a practical standpoint which indicates that the mapping is problematic is that the quantity $(x+t)^{3} (x-t)$ under a Wick rotation becomes complex. Even after a Wick rotation, $x^{2} - t^{2} \rightarrow x^{2} + \tau ^{2}$ is still going to be strictly real. For a general curve such as yours that satisfies:

\begin{equation} \tag{1} f(x,t) = 1 \end{equation}

but does not remain real under a Wick rotation:

\begin{equation} \tag{2} [f(x,it)]^{*} \neq f(x,it) \end{equation}

The reformulation $f(x,it) = 1$ as you implicitly demand in your second equation implies:

\begin{equation} \tag{3} Im[f(x,it)] = 0 \end{equation}

which is in direct contradiction with equation (2). Of course, contradiction in this case means you have arbitrarily set a new constraint on your coordinates, hence understandably further reduces the number of accepted points on the plane that satisfy your equation.

  • $\begingroup$ Indeed, I agree with your comment. But that seems to be an extremely steep restriction. I would have thought that there may be some way of circumventing it... Maybe I am wrong $\endgroup$
    – Frotaur
    Jun 21, 2022 at 11:42
  • $\begingroup$ It might seem like a steep restriction, but this is only with regards to mappings like this, so it doesn't seem too unreasonable when it comes to the efficacy of Wick rotations. Also, there are tricks to circumvent this, even though the physical aptitude of the results are up for debate and might need some extra attention. Off the top of my head, express your initial trajectory as $\frac{1}{2} [f(x,t) + f^{*} (x,t)] = 1$. This retains an identical real parameter curve, but manages to make the Wick-rotated form strictly real. $\endgroup$
    – rhomaios
    Jun 21, 2022 at 12:17
  • $\begingroup$ So your are saying essentially that the Wick rotation is not uniquely defined in this case. The thing is that this "curve" is actually a physical interface in my problem, so I have trouble understanding the meaning of all this in the Euclidean theory. Probably that here the Euclidean treatment is ill-defined, it seems. $\endgroup$
    – Frotaur
    Jun 21, 2022 at 14:22
  • $\begingroup$ Perhaps a more rigorous (and more physically understandable) mapping would be to Wick-rotate half of the interface towards one "direction" $t =i\tau$, and the other half towards the other $t=-i\tau$. This again ensures the entire Wick-rotated interface is strictly real. It is or course tantamount to my earlier suggestion, but seems to have a better intuitive idea behind it. Without knowing what you want to study I can't know whether this helps of course, but it's a cute little idea to try perhaps. $\endgroup$
    – rhomaios
    Jun 21, 2022 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.