# Performing Wick Rotation to get Euclidean action of a scalar field

I'm working with the signature $(+,-,-,-)$ and with a Minkowski space-time 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. The topic is something has been already asked (look here). The problem is that the answer to the question is not satisfactory for me and for this special case.

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 always the same negative euclidean interval but different results.

If I choose $t = i\tau$ I get $$S_M ~=~i\int_{+i\infty}^{-i\infty} d\tau d^3x \Psi^\dagger\left(i\frac{\partial}{\partial i\tau} + \frac{\nabla^2}{2m}\right)\Psi ~=~ -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi$$

If I choose $t = -i\tau$ I get $$S_M ~=~-i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(-i\frac{\partial}{\partial i\tau} + \frac{\nabla^2}{2m}\right)\Psi ~=~ -i\int_{-i\infty}^{+i\infty} d\tau d^3x \Psi^\dagger\left(-\frac{\partial}{\partial \tau} + \frac{\nabla^2}{2m}\right)\Psi$$

I think there is a slight difference that I don't understand

1. Is the Euclidean action defined by $S_M = i S_E$ or by $S_M = -iS_E$?

For the Euclidean path integral to be convergent, the Boltzmann factor should be an exponentially decaying function of the scalar field $$\phi$$. This in turn dictates that the direction of the Wick rotation. (It should perhaps be stressed that Wick rotation is not merely a renaming of time variables, but that it involves an actual deformation of the time-integration contour in the complex time plane.) Standard conventions for the Wick rotation are
\begin{align} -S_E~=~&iS_M, \cr t_E~=~&it_M, \cr {\cal L}_E~=~&-{\cal L}_M,\end{align} \tag{1}
where the subscripts $$E$$ and $$M$$ stand for Euclid and Minkowski, respectively. In OP's case \begin{align} {\cal L}_M~=~&i\phi^{\ast} \frac{d\phi}{dt_M}- {\cal V} ,\cr {\cal L}_E~=~&\phi^{\ast} \frac{d\phi}{dt_E}+ {\cal V} ,\cr {\cal V}~=~&\frac{1}{2m}\nabla\phi^{\ast}\cdot \nabla\phi .\end{align} \tag{2} See also this related Phys.SE post.
This is happening because your field depends on $t$, $\Psi = \Psi (x,t)$. Therefore, when you perform the Wick rotation $t = i\tau$, you also Wick rotate to your field, and obtain an action for $\Psi(x,i\tau)$. In the second case, you obtain an action for $\Psi(x,-i\tau)$. Those are not the same $\Psi$'s.
• Yes, it's true but the classical Euclidian action is the second I wrote with $\Psi = \Psi(x,t)$ not? Jan 24, 2016 at 14:29
Note that $\int^{\infty}_{-\infty}f(t)dt=\int^{\infty}_{-\infty}f(-t)dt$ and also $\int^{\infty}_{-\infty}\frac{df(t)}{dt}dt=\int^{\infty}_{-\infty}\frac{df(-t)}{-dt}dt$